# Euler's sum of powers conjecture

Euler's sum of powers conjecture
You are encouraged to solve this task according to the task description, using any language you may know.

There is a conjecture in mathematics that held for over two hundred years before it was disproved by the finding of a counterexample in 1966 by Lander and Parkin.

This conjecture is called Euler's sum of powers conjecture and can be stated as such:

At least k positive kth powers are required to sum to a kth power, except for the trivial case of one kth power: yk = yk.

In 1966, Leon J. Lander and Thomas R. Parkin used a brute-force search on a CDC 6600 computer restricting numbers to those less than 250.

The task consists in writing a program to search for an integer solution of $x_0^5 + x_1^5 + x_2^5 + x_3^5 = y^5$ where all $x_i$ and $y$ are distinct integers between 0 and 250 (exclusive). Show an answer here.

## 11l

Translation of: Python
```F eulers_sum_of_powers()
V max_n = 250
V pow_5 = (0 .< max_n).map(n -> Int64(n) ^ 5)
V pow5_to_n = Dict(0 .< max_n, n -> (Int64(n) ^ 5, n))

L(x0) 1 .< max_n
L(x1) 1 .< x0
L(x2) 1 .< x1
L(x3) 1 .< x2
V pow_5_sum = pow_5[x0] + pow_5[x1] + pow_5[x2] + pow_5[x3]
I pow_5_sum C pow5_to_n
V y = pow5_to_n[pow_5_sum]
R (x0, x1, x2, x3, y)

V r = eulers_sum_of_powers()
print(‘#.^5 + #.^5 + #.^5 + #.^5 = #.^5’.format(r, r, r, r, r))```
Output:
`133^5 + 110^5 + 84^5 + 27^5 = 144^5`

## 360 Assembly

In the program we do not user System/360 integers (31 bits) unable to handle the problem, but System/360 packed decimal (15 digits). 250^5 needs 12 digits.
This program could have been run in 1964. Here, for maximum compatibility, we use only the basic 360 instruction set. Macro instruction XPRNT can be replaced by a WTO.

``` EULERCO  CSECT
USING  EULERCO,R13
B      80(R15)
DC     17F'0'
DC     CL8'EULERCO'
STM    R14,R12,12(R13)
ST     R13,4(R15)
ST     R15,8(R13)
LR     R13,R15
ZAP    X1,=P'1'
LOOPX1   ZAP    PT,MAXN            do x1=1 to maxn-4
SP     PT,=P'4'
CP     X1,PT
BH     ELOOPX1
ZAP    PT,X1
AP     PT,=P'1'
ZAP    X2,PT
LOOPX2   ZAP    PT,MAXN            do x2=x1+1 to maxn-3
SP     PT,=P'3'
CP     X2,PT
BH     ELOOPX2
ZAP    PT,X2
AP     PT,=P'1'
ZAP    X3,PT
LOOPX3   ZAP    PT,MAXN            do x3=x2+1 to maxn-2
SP     PT,=P'2'
CP     X3,PT
BH     ELOOPX3
ZAP    PT,X3
AP     PT,=P'1'
ZAP    X4,PT
LOOPX4   ZAP    PT,MAXN            do x4=x3+1 to maxn-1
SP     PT,=P'1'
CP     X4,PT
BH     ELOOPX4
ZAP    PT,X4
AP     PT,=P'1'
ZAP    X5,PT              x5=x4+1
ZAP    SUMX,=P'0'         sumx=0
ZAP    PT,X1              x1
BAL    R14,POWER5
AP     SUMX,PT
ZAP    PT,X2              x2
BAL    R14,POWER5
AP     SUMX,PT
ZAP    PT,X3              x3
BAL    R14,POWER5
AP     SUMX,PT
ZAP    PT,X4              x4
BAL    R14,POWER5
AP     SUMX,PT            sumx=x1**5+x2**5+x3**5+x4**5
ZAP    PT,X5              x5
BAL    R14,POWER5
ZAP    VALX,PT            valx=x5**5
LOOPX5   CP     X5,MAXN            while x5<=maxn & valx<=sumx
BH     ELOOPX5
CP     VALX,SUMX
BH     ELOOPX5
CP     VALX,SUMX          if valx=sumx
BNE    NOTEQUAL
MVI    BUF,C' '
MVC    BUF+1(79),BUF      clear buffer
ED     WC,X1              x1
MVC    BUF+0(8),WC+8
ED     WC,X2              x2
MVC    BUF+8(8),WC+8
ED     WC,X3              x3
MVC    BUF+16(8),WC+8
ED     WC,X4              x4
MVC    BUF+24(8),WC+8
ED     WC,X5              x5
MVC    BUF+32(8),WC+8
XPRNT  BUF,80             output x1,x2,x3,x4,x5
B      ELOOPX1
NOTEQUAL ZAP    PT,X5
AP     PT,=P'1'
ZAP    X5,PT              x5=x5+1
ZAP    PT,X5
BAL    R14,POWER5
ZAP    VALX,PT            valx=x5**5
B      LOOPX5
ELOOPX5  AP     X4,=P'1'
B      LOOPX4
ELOOPX4  AP     X3,=P'1'
B      LOOPX3
ELOOPX3  AP     X2,=P'1'
B      LOOPX2
ELOOPX2  AP     X1,=P'1'
B      LOOPX1
ELOOPX1  L      R13,4(0,R13)
LM     R14,R12,12(R13)
XR     R15,R15
BR     R14
POWER5   ZAP    PQ,PT              ^1
MP     PQ,PT              ^2
MP     PQ,PT              ^3
MP     PQ,PT              ^4
MP     PQ,PT              ^5
ZAP    PT,PQ
BR     R14
MAXN     DC     PL8'249'
X1       DS     PL8
X2       DS     PL8
X3       DS     PL8
X4       DS     PL8
X5       DS     PL8
SUMX     DS     PL8
VALX     DS     PL8
PT       DS     PL8
PQ       DS     PL8
WC       DS     CL17
BUF      DS     CL80
YREGS
END```
Output:
```      27      84     110     133     144
```

## 6502 Assembly

This is long enough as it is, so the code here is just the main task body. Details like the BASIC loader (for a C-64, which is what I ran this on) and the output routines (including the very tedious conversion of a 40-bit integer to decimal) were moved into external include files. Also in an external file is the prebuilt table of the first 250 fifth powers ... actually, just for ease of referencing, it's the first 256, 0...255, in five tables each holding one byte of the value.

```; Prove Euler's sum of powers conjecture false by finding
; positive a,b,c,d,e such that a⁵+b⁵+c⁵+d⁵=e⁵.

; we're only looking for the first counterexample, which occurs with all
; integers less than this value
max_value      = \$fa  ; decimal 250

; this header turns our code into a LOADable and RUNnable BASIC program

; this contains the first 256 integers to the power of 5 broken up into
; 5 tables of one-byte values (power5byte0 with the LSBs through
; power5byte4 with the MSBs)
.include "power5table.s"

; this defines subroutines and macros for printing messages to
; the console, including `puts` for printing out a NUL-terminated string,
; puthex to display a one-byte value in hexadecimal, and putdec through
; putdec5 to display an N-byte value in decimal
.include "output.s"

; label strings for the result output
.feature string_escapes
success:   .asciiz "\r\rFOUND EXAMPLE:\r\r"
between:   .asciiz "^5 + "
penult:    .asciiz "^5 = "
eqend:     .asciiz "^5\r\r(SUM IS "

; the BASIC loader program prints the elapsed time at the end, so we include a
; label for that, too
tilabel:   .asciiz ")\r\rTIME:"

; ZP locations to store the integers to try
x0 = \$f7
x1 = x0 + 1
x2 = x0 + 2
x3 = x0 + 3

; we use binary search to find integer roots; current bounds go here
low = x0 + 4
hi  = x0 + 5

; sum of powers of current candidate integers
sum: .res 5

; when we find a sum with an integer 5th root, we put it here
x4: .res 1

main: ; loop for x0 from 1 to max_value
ldx #01
stx x0

loop0: ; loop for x1 from x0+1 to max_value
ldx x0
inx
stx x1

loop1: ; loop for x2 from x1+1 to max_value
ldx x1
inx
stx x2

loop2: ; loop for x3 from x2+1 to max_value
ldx x2
inx
stx x3

loop3: ; add up the fifth powers of the four numbers
; initialize to 0
lda #00
sta sum
sta sum+1
sta sum+2
sta sum+3
sta sum+4

; we use indexed addressing, taking advantage of the fact that the xn's
; are consecutive, so x0,1 = x1, etc.
ldy #0
ldx x0,y
lda sum
clc
sta sum
lda sum+1
sta sum+1
lda sum+2
sta sum+2
lda sum+3
sta sum+3
lda sum+4
sta sum+4
iny
cpy #4
; now sum := x₀⁵+x₁⁵+x₂⁵+x₃⁵
; set initial bounds for binary search
ldx x3
inx
stx low
ldx #max_value
dex
stx hi

binsearch:
; compute midpoint
lda low
cmp hi
beq notdone
bcs done_search
notdone:
ldx #0
clc

; now a + carry bit = low+hi; rotating right will get the midpoint
ror
; compare square of midpoint to sum
tax
lda sum+4
cmp power5byte4,x
bne notyet
lda sum+3
cmp power5byte3,x
bne notyet
lda sum+2
cmp power5byte2,x
bne notyet
lda sum+1
cmp power5byte1,x
bne notyet
lda sum
cmp power5byte0,x
beq found
notyet:
bcc sum_lt_guess
inx
stx low
bne endbin
beq endbin
sum_lt_guess:
dex
stx hi
endbin:
bne binsearch
beq binsearch

done_search:
inc x3
lda x3
cmp #max_value
bcs end_loop3
jmp loop3

end_loop3:
inc x2
lda x2
cmp #max_value
bcs end_loop2
jmp loop2

end_loop2:
inc x1
lda x1
cmp #max_value
bcs end_loop1
jmp loop1

end_loop1:
inc x0
lda x0
cmp #max_value
bcs end_loop0
jmp loop0

end_loop0:
; should never get here, means we didn't find an example.
brk

found: stx x4
puts success
putdec x0
ldy #1

ploop: puts between
putdec {x0,y}
iny
cpy #4
bcc ploop
puts penult
putdec {x0,y}
puts eqend
putdec5 sum
puts tilabel
rts```
Output:
```    **** COMMODORE 64 BASIC V2 ****

64K RAM SYSTEM  38911 BASIC BYTES FREE

SEARCHING FOR ESOP
RUN:

FOUND EXAMPLE:

27^5 + 84^5 + 110^5 + 133^5 = 144^5

(SUM IS 61917364224)

TIME:095617

... almost ten hours, but it did find it!

```with Ada.Text_IO;

procedure Sum_Of_Powers is

type Base is range 0 .. 250; -- A, B, C, D and Y are in that range
type Num is range 0 .. 4*(250**5); -- (A**5 + ... + D**5) is in that range
subtype Fit is Num range 0 .. 250**5; -- Y**5 is in that range

Modulus: constant Num := 254;
type Modular is mod Modulus;

type Result_Type is array(1..5) of Base; -- this will hold A,B,C,D and Y

type Y_Type is array(Modular) of Base;
type Y_Sum_Type is array(Modular) of Fit;

Y_Sum: Y_Sum_Type := (others => 0);
Y: Y_Type := (others => 0);
-- for I in 0 .. 250, we set Y_Sum(I**5 mod Modulus) := I**5
--                       and Y(I**5 mod Modulus) := I
-- Modulus has been chosen to avoid collisions on (I**5 mod Modulus)
-- later we will compute Sum_ABCD := A**5 + B**5 + C**5 + D**5
-- and check if Y_Sum(Sum_ABCD mod modulus) = Sum_ABCD

function Compute_Coefficients return Result_Type is

Sum_A: Fit;
Sum_AB, Sum_ABC, Sum_ABCD: Num;
Short: Modular;

begin
for A in Base(0) .. 246 loop
Sum_A := Num(A) ** 5;
for B in A .. 247 loop
Sum_AB := Sum_A + (Num(B) ** 5);
for C in Base'Max(B,1) .. 248 loop -- if A=B=0 then skip C=0
Sum_ABC := Sum_AB + (Num(C) ** 5);
for D in C .. 249 loop
Sum_ABCD := Sum_ABC + (Num(D) ** 5);
Short    := Modular(Sum_ABCD mod Modulus);
if Y_Sum(Short) = Sum_ABCD then
return A & B & C & D & Y(Short);
end if;
end loop;
end loop;
end loop;
end loop;
return 0 & 0 & 0 & 0 & 0;
end Compute_Coefficients;

Tmp: Fit;
ABCD_Y: Result_Type;

begin -- main program

-- initialize Y_Sum and Y
for I in Base(0) .. 250 loop
Tmp := Num(I)**5;
if Y_Sum(Modular(Tmp mod Modulus)) /= 0 then
raise Program_Error with "Collision: Change Modulus and recompile!";
else
Y_Sum(Modular(Tmp mod Modulus)) := Tmp;
Y(Modular(Tmp mod Modulus)) := I;
end if;
end loop;

-- search for a solution (A, B, C, D, Y)
ABCD_Y := Compute_Coefficients;

-- output result
for Number of ABCD_Y loop
end loop;

end Sum_Of_Powers;
```
Output:
``` 27 84 110 133 144
```

## ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.8.win32
```# max number will be the highest integer we will consider                    #
INT max number = 250;

# Construct a table of the fifth powers of 1 : max number                    #
[ max number ]LONG INT fifth;
FOR i TO max number DO
LONG INT i2 =  i * i;
fifth[ i ] := i2 * i2 * i
OD;

# find the first a, b, c, d, e such that a^5 + b^5 + c^5 + d^5 = e^5         #
# as the fifth powers are in order, we can use a binary search to determine  #
# whether the value is in the table                                          #
BOOL found := FALSE;
LONG INT sum   = fifth[a] + fifth[b] + fifth[c] + fifth[d];
INT      low  := d;
INT      high := max number;
WHILE low < high
DO
INT e := ( low + high ) OVER 2;
IF fifth[ e ] = sum
THEN
# the value at e is a fifth power                    #
found := TRUE;
print( ( ( whole( a, 0 ) + "^5 + " + whole( b, 0 ) + "^5 + "
+ whole( c, 0 ) + "^5 + " + whole( d, 0 ) + "^5 = "
+ whole( e, 0 ) + "^5"
)
, newline
)
)
ELIF sum < fifth[ e ]
THEN high := e - 1
ELSE low  := e + 1
FI
OD
OD
OD
OD
OD```

Output:

```27^5 + 84^5 + 110^5 + 133^5 = 144^5
```

## ALGOL W

As suggested by the REXX solution, we find a solution to a^5 + b^5 + c^5 = e^5 - d^5 which results in a significant reduction in run time.
Algol W integers are 32-bit only, so we simulate the necessary 12 digit arithmetic with pairs of integers.

```begin
% find a, b, c, d, e such that a^5 + b^5 + c^5 + d^5 = e^5                              %
%                        where 1 <= a <= b <= c <= d <= e <= 250                        %
% we solve this using the equivalent equation a^5 + b^5 + c^5 = e^5 - d^5               %
% 250^5 is 976 562 500 000 - too large for a 32 bit number so we will use pairs of      %
% integers and constrain their values to be in 0..1 000 000                             %
% Note only positive numbers are needed                                                 %
integer MAX_NUMBER, MAX_V;
MAX_NUMBER := 250;
MAX_V      := 1000000;
begin
% quick sorts the fifth power differences table                                     %
procedure quickSort5 ( integer value lb, ub ) ;
if ub > lb then begin
% more than one element, so must sort                                       %
integer left, right, pivot, pivotLo, pivotHi;
left    := lb;
right   := ub;
% choosing the middle element of the array as the pivot %
pivot   := left + ( ( ( right + 1 ) - left ) div 2 );
pivotLo := loD( pivot );
pivotHi := hiD( pivot );
while begin
while left  <= ub
and begin integer cmp;
cmp := hiD( left ) - pivotHi;
if cmp = 0 then cmp := loD( left ) - pivotLo;
cmp < 0
end
do left := left + 1;
while right >= lb
and begin integer cmp;
cmp := hiD( right ) - pivotHi;
if cmp = 0 then cmp := loD( right ) - pivotLo;
cmp > 0
end
do right := right - 1;
left <= right
end do begin
integer swapLo, swapHi, swapD, swapE;
swapLo       := loD( left  );
swapHi       := hiD( left  );
swapD        := Dd(  left  );
swapE        := De(  left  );
loD( left  ) := loD( right );
hiD( left  ) := hiD( right );
Dd(  left  ) := Dd(  right );
De(  left  ) := De(  right );
loD( right ) := swapLo;
hiD( right ) := swapHi;
Dd(  right ) := swapD;
De(  right ) := swapE;
left         := left  + 1;
right        := right - 1
end while_left_le_right ;
quickSort5( lb,   right );
quickSort5( left, ub    )
end quickSort5 ;
% table of fifth powers                                                             %
integer array lo5, hi5         ( 1 :: MAX_NUMBER );
% table if differences between fifth powers                                         %
integer array loD, hiD, De, Dd ( 1 :: MAX_NUMBER * MAX_NUMBER );
integer dUsed, dPos;
% compute fifth powers                                                              %
for i := 1 until MAX_NUMBER do begin
lo5( i ) := i * i; hi5( i ) := 0;
for p := 3 until 5 do begin
integer carry;
lo5( i ) := lo5( i ) * i;
carry    := lo5( i ) div MAX_V;
lo5( i ) := lo5( i ) rem MAX_V;
hi5( i ) := hi5( i ) * i;
hi5( i ) := hi5( i ) + carry
end for_p
end for_i ;
% compute the differences between fifth powers e^5 - d^5, 1 <= d < e <= MAX_NUMBER  %
dUsed := 0;
for e := 2 until MAX_NUMBER do begin
for d := 1 until e - 1  do begin
dUsed := dUsed + 1;
De(  dUsed ) := e;
Dd(  dUsed ) := d;
loD( dUsed ) := lo5( e ) - lo5( d );
hiD( dUsed ) := hi5( e ) - hi5( d );
if loD( dUsed ) < 0 then begin
loD( dUsed ) := loD( dUsed ) + MAX_V;
hiD( dUsed ) := hiD( dUsed ) - 1
end if_need_to_borrow
end for_d
end for_e;
% sort the fifth power differences                                                  %
quickSort5( 1, dUsed );
% attempt to find a^5 + b^5 + c^5 = e^5 - d^5                                       %
for a := 1 until MAX_NUMBER do begin
integer loA, hiA;
loA := lo5( a ); hiA := hi5( a );
for b := a until MAX_NUMBER do begin
integer loB, hiB;
loB := lo5( b ); hiB := hi5( b );
for c := b until MAX_NUMBER do begin
integer low, high, loSum, hiSum;
loSum :=                       loA + loB + lo5( c );
hiSum := ( loSum div MAX_V ) + hiA + hiB + hi5( c );
loSum :=   loSum rem MAX_V;
% look for hiSum,loSum in hiD,loD                                       %
low   := 1;
high  := dUsed;
while low < high do begin
integer mid, cmp;
mid := ( low + high ) div 2;
cmp := hiD( mid ) - hiSum;
if cmp = 0 then cmp := loD( mid ) - loSum;
if cmp = 0 then begin
% the value at mid is the difference of two fifth powers        %
write( i_w := 1, s_w := 0
, a, "^5 + ", b, "^5 + ", c, "^5 + "
, Dd( mid ), "^5 = ", De( mid ), "^5"
);
go to found
end
else if cmp > 0 then high := mid - 1
else                 low  := mid + 1
end while_low_lt_high
end for_c
end for_b
end for_a ;
found :
end
end.```
Output:
```27^5 + 84^5 + 110^5 + 133^5 = 144^5
```

## Arturo

Translation of: Nim
```eulerSumOfPowers: function [top][
p5: map 0..top => [& ^ 5]

loop 4..top 'a [
loop 3..a-1 'b [
loop 2..b-1 'c [
loop 1..c-1 'd [
s: (get p5 a) + (get p5 b) + (get p5 c) + (get p5 d)
if integer? index p5 s ->
return ~"|a|^5 + |b|^5 + |c|^5 + |d|^5 = |index p5 s|^5"
]
]
]
]

]

print eulerSumOfPowers 249
```
Output:
`133^5 + 110^5 + 84^5 + 27^5 = 144^5`

## AWK

```# syntax: GAWK -f EULERS_SUM_OF_POWERS_CONJECTURE.AWK
BEGIN {
start_int = systime()
main()
printf("%d seconds\n",systime()-start_int)
exit(0)
}
function main(  sum,s1,x0,x1,x2,x3) {
for (x0=1; x0<=250; x0++) {
for (x1=1; x1<=x0; x1++) {
for (x2=1; x2<=x1; x2++) {
for (x3=1; x3<=x2; x3++) {
sum = (x0^5) + (x1^5) + (x2^5) + (x3^5)
s1 = int(sum ^ 0.2)
if (sum == s1^5) {
printf("%d^5 + %d^5 + %d^5 + %d^5 = %d^5\n",x0,x1,x2,x3,s1)
return
}
}
}
}
}
}
```
Output:
```133^5 + 110^5 + 84^5 + 27^5 = 144^5
15 seconds
```

## BCPL

Translation of: Go
```GET "libhdr"

LET solve() BE {
LET pow5 = VEC 249
LET sum = ?

FOR i = 1 TO 249
pow5!i := i * i * i * i * i

FOR w = 4 TO 249
FOR x = 3 TO w - 1
FOR y = 2 TO x - 1
FOR z = 1 TO y - 1 {
sum := pow5!w + pow5!x + pow5!y + pow5!z
FOR a = w + 1 TO 249
IF pow5!a = sum {
writef("solution found: %d  %d  %d  %d  %d *n", w, x, y, z, a)
RETURN
}
}
writef("Sorry, no solution found.*n")
}

LET start() = VALOF {
solve()
RESULTIS 0
}```
Output:
```solution found: 133 110 84 27 144
```

## Bracmat

```  0:?x0
&   whl
' ( 1+!x0:<250:?x0
& out\$(x0 !x0)
& 0:?x1
&   whl
' ( 1+!x1:~>!x0:?x1
& out\$(x0 !x0 x1 !x1)
& 0:?x2
&   whl
' ( 1+!x2:~>!x1:?x2
& 0:?x3
&   whl
' ( 1+!x3:~>!x2:?x3
&   (!x0^5+!x1^5+!x2^5+!x3^5)^1/5
: (   #?y
& out\$(x0 !x0 x1 !x1 x2 !x2 x3 !x3 y !y)
& 250:?x0:?x1:?x2:?x3
| ?
)
)
)
)
)```

Output

`x0 133 x1 110 x2 84 x3 27 y 144`

## C

The trick to speed up was the observation that for any x we have x^5=x modulo 2, 3, and 5, according to the Fermat's little theorem. Thus, based on the Chinese Remainder Theorem we have x^5==x modulo 30 for any x. Therefore, when we have computed the left sum s=a^5+b^5+c^5+d^5, then we know that the right side e^5 must be such that s==e modulo 30. Thus, we do not have to consider all values of e, but only values in the form e=e0+30k, for some starting value e0, and any k. Also, we follow the constraints 1<=a<b<c<d<e<N in the main loop.

```// Alexander Maximov, July 2nd, 2015
#include <stdio.h>
#include <time.h>
typedef long long mylong;

void compute(int N, char find_only_one_solution)
{	const int M = 30;   /* x^5 == x modulo M=2*3*5 */
int a, b, c, d, e;
mylong s, t, max, *p5 = (mylong*)malloc(sizeof(mylong)*(N+M));

for(s=0; s < N; ++s)
p5[s] = s * s, p5[s] *= p5[s] * s;
for(max = p5[N - 1]; s < (N + M); p5[s++] = max + 1);

for(a = 1; a < N; ++a)
for(b = a + 1; b < N; ++b)
for(c = b + 1; c < N; ++c)
for(d = c + 1, e = d + ((t = p5[a] + p5[b] + p5[c]) % M); ((s = t + p5[d]) <= max); ++d, ++e)
{	for(e -= M; p5[e + M] <= s; e += M); /* jump over M=30 values for e>d */
if(p5[e] == s)
{	printf("%d %d %d %d %d\r\n", a, b, c, d, e);
if(find_only_one_solution) goto onexit;
}
}
onexit:
free(p5);
}

int main(void)
{
int tm = clock();
compute(250, 0);
printf("time=%d milliseconds\r\n", (int)((clock() - tm) * 1000 / CLOCKS_PER_SEC));
return 0;
}
```
Output:

The fair way to measure the speed of the code above is to measure it's run time to find all possible solutions to the problem, given N (and not just a single solution, since then the time may depend on the way and the order we organize for-loops).

```27 84 110 133 144
time=235 milliseconds
```

Another test with N=1000 produces the following results:

```27 84 110 133 144
54 168 220 266 288
81 252 330 399 432
108 336 440 532 576
135 420 550 665 720
162 504 660 798 864
time=65743 milliseconds
```

PS: The solution for C++ provided below is actually quite good in its design idea behind. However, with all proposed tricks to speed up, the measurements for C++ solution for N=1000 showed the execution time 81447ms (+23%) on the same environment as above for C solution (same machine, same compiler, 64-bit platform). The reason that C++ solution is a bit slower is, perhaps, the fact that the inner loops over rs have complexity ~N/2 steps in average, while with the modulo 30 trick that complexity can be reduced down to ~N/60 steps, although one "expensive" extra %-operation is still needed.

## C#

### Loops

Translation of: Java
```using System;

namespace EulerSumOfPowers {
class Program {
const int MAX_NUMBER = 250;

static void Main(string[] args) {
bool found = false;
long[] fifth = new long[MAX_NUMBER];

for (int i = 1; i <= MAX_NUMBER; i++) {
long i2 = i * i;
fifth[i - 1] = i2 * i2 * i;
}

for (int a = 0; a < MAX_NUMBER && !found; a++) {
for (int b = a; b < MAX_NUMBER && !found; b++) {
for (int c = b; c < MAX_NUMBER && !found; c++) {
for (int d = c; d < MAX_NUMBER && !found; d++) {
long sum = fifth[a] + fifth[b] + fifth[c] + fifth[d];
int e = Array.BinarySearch(fifth, sum);
found = e >= 0;
if (found) {
Console.WriteLine("{0}^5 + {1}^5 + {2}^5 + {3}^5 = {4}^5", a + 1, b + 1, c + 1, d + 1, e + 1);
}
}
}
}
}
}
}
}
```

### Paired Powers, Mod 30, etc...

Translation of: vbnet
```using System;
using System.Collections.Generic;
using System.Linq;

namespace Euler_cs
{
class Program
{
struct Pair
{
public int a, b;
public Pair(int x, int y)
{
a = x; b = y;
}
}

static int min = 1, max = 250;
static ulong[] p5;
static SortedDictionary<ulong, Pair>[] sum2 =
new SortedDictionary<ulong, Pair>;

static string Fmt(Pair p)
{
return string.Format("{0}^5 + {1}^5", p.a, p.b);
}

public static void InitM()
{
for (int i = 0; i <= 29; i++)
sum2[i] = new SortedDictionary<ulong, Pair>();
p5 = new ulong[max + 1];
p5[min] = Convert.ToUInt64(min) * Convert.ToUInt64(min);
p5[min] *= p5[min] * Convert.ToUInt64(min);
for (int i = min; i <= max - 1; i++)
{
for (int j = i + 1; j <= max; j++)
{
p5[j] = Convert.ToUInt64(j) * Convert.ToUInt64(j);
p5[j] *= p5[j] * Convert.ToUInt64(j);
if (j == max) continue;
ulong x = p5[i] + p5[j];
sum2[x % 30].Add(x, new Pair(i, j));
}
}
}

static List<string> CalcM(int m)
{
List<string> res = new List<string>();
for (int i = max; i >= min; i--)
{
ulong p = p5[i]; int pm = i % 30, mp = (pm - m + 30) % 30;
foreach (var s in sum2[m].Keys)
{
if (p <= s) break;
ulong t = p - s;
if (sum2[mp].Keys.Contains(t) && sum2[mp][t].a > sum2[m][s].b)
res.Add(string.Format("  {1} + {2} = {0}^5",
i, Fmt(sum2[m][s]), Fmt(sum2[mp][t])));
}
}
return res;
}

static int Snip(string s)
{
int p = s.IndexOf("=") + 1;
return Convert.ToInt32(s.Substring(p, s.IndexOf("^", p) - p));
}

static int CompareRes(string x, string y)
{
int res = Snip(x).CompareTo(Snip(y));
if (res == 0) res = x.CompareTo(y);
return res;
}

static int Validify(int def, string s)
{
int res = def, t = 0; int.TryParse(s, out t);
if (t >= 1 && t < Math.Pow((double)(ulong.MaxValue >> 1), 0.2))
res = t;
return res;
}

static void Switch(ref int a, ref int b)
{
int t = a; a = b; b = t;
}

static void Main(string[] args)
{
if (args.Count() > 1)
{
min = Validify(min, args);
max = Validify(max, args);
if (max < min) Switch(ref max, ref min);
}
else if (args.Count() == 1)
max = Validify(max, args);
Console.WriteLine("Mod 30 shortcut with threading, checking from {0} to {1}...", min, max);
List<string> res = new List<string>();
DateTime st = DateTime.Now;
InitM();
for (int j = 0; j <= 29; j++)
{
var jj = j;
}
foreach (var item in taskList.Select(t => t.Result))
res.Sort(CompareRes);
foreach (var item in res)
Console.WriteLine(item);
Console.WriteLine("  Computation time to check entire space was {0} seconds",
(DateTime.Now - st).TotalSeconds);
if (System.Diagnostics.Debugger.IsAttached)
}
}
}
```
Output:
(no command line arguments)
```Mod 30 shortcut with threading, checking from 1 to 250...
27^5 + 84^5 + 110^5 + 133^5 = 144^5

Computation time to check entire space was 0.0838058 seconds```
(command line argument = "1000")
```Mod 30 shortcut with threading, checking from 1 to 1000...
27^5 + 84^5 + 110^5 + 133^5 = 144^5
54^5 + 168^5 + 220^5 + 266^5 = 288^5
81^5 + 252^5 + 330^5 + 399^5 = 432^5
108^5 + 336^5 + 440^5 + 532^5 = 576^5
135^5 + 420^5 + 550^5 + 665^5 = 720^5
162^5 + 504^5 + 660^5 + 798^5 = 864^5
Computation time to check entire space was 5.4109744 seconds```

## C++

### First version

The simplest brute-force find is already reasonably quick:

```#include <algorithm>
#include <iostream>
#include <cmath>
#include <set>
#include <vector>

using namespace std;

bool find()
{
const auto MAX = 250;
vector<double> pow5(MAX);
for (auto i = 1; i < MAX; i++)
pow5[i] = (double)i * i * i * i * i;
for (auto x0 = 1; x0 < MAX; x0++) {
for (auto x1 = 1; x1 < x0; x1++) {
for (auto x2 = 1; x2 < x1; x2++) {
for (auto x3 = 1; x3 < x2; x3++) {
auto sum = pow5[x0] + pow5[x1] + pow5[x2] + pow5[x3];
if (binary_search(pow5.begin(), pow5.end(), sum))
{
cout << x0 << " " << x1 << " " << x2 << " " << x3 << " " << pow(sum, 1.0 / 5.0) << endl;
return true;
}
}
}
}
}
return false;
}

int main(void)
{
int tm = clock();
if (!find())
cout << "Nothing found!\n";
cout << "time=" << (int)((clock() - tm) * 1000 / CLOCKS_PER_SEC) << " milliseconds\r\n";
return 0;
}
```
Output:
```133 110 84 27 144
time=234 milliseconds
```

We can accelerate this further by creating a parallel std::set<double> p5s containing the elements of the std::vector pow5, and using it to replace the call to std::binary_search:

```	set<double> pow5s;
for (auto i = 1; i < MAX; i++)
{
pow5[i] = (double)i * i * i * i * i;
pow5s.insert(pow5[i]);
}
//...
if (pow5s.find(sum) != pow5s.end())
```

This reduces the timing to 125 ms on the same hardware.

A different, more effective optimization is to note that each successive sum is close to the previous one, and use a bidirectional linear search with memory. We also note that inside the innermost loop, we only need to search upward, so we hoist the downward linear search to the loop over x2.

```bool find()
{
const auto MAX = 250;
vector<double> pow5(MAX);
for (auto i = 1; i < MAX; i++)
pow5[i] = (double)i * i * i * i * i;
auto rs = 5;
for (auto x0 = 1; x0 < MAX; x0++) {
for (auto x1 = 1; x1 < x0; x1++) {
for (auto x2 = 1; x2 < x1; x2++) {
auto s2 = pow5[x0] + pow5[x1] + pow5[x2];
while (rs > 0 && pow5[rs] > s2) --rs;
for (auto x3 = 1; x3 < x2; x3++) {
auto sum = s2 + pow5[x3];
while (rs < MAX - 1 && pow5[rs] < sum) ++rs;
if (pow5[rs] == sum)
{
cout << x0 << " " << x1 << " " << x2 << " " << x3 << " " << pow(sum, 1.0 / 5.0) << endl;
return true;
}
}
}
}
}
return false;
}
```

This reduces the timing to around 25 ms. We could equally well replace rs with an iterator inside pow5; the timing is unaffected.

For comparison with the C code, we also check the timing of an exhaustive search up to MAX=1000. (Don't try this in Python.) This takes 87.2 seconds on the same hardware, comparable to the results found by the C code authors, and supports their conclusion that the mod-30 trick used in the C solution leads to better scalability than the iterator optimizations.

Fortunately, we can incorporate the same trick into the above code, by inserting a forward jump to a feasible solution mod 30 into the loop over x3:

```				for (auto x3 = 1; x3 < x2; x3++)
{
// go straight to the first appropriate x3, mod 30
if (int err30 = (x0 + x1 + x2 + x3 - rs) % 30)
x3 += 30 - err30;
if (x3 >= x2)
break;
auto sum = s2 + pow5[x3];
```

With this refinement, the exhaustive search up to MAX=1000 takes 16.9 seconds.

Thanks, C guys!

### Second version

We can create a more efficient method by using the idea (taken from the EchoLisp listing below) of precomputing difference between pairs of fifth powers. If we combine this with the above idea of using linear search with memory, this still requires asymptotically O(N^4) time (because of the linear search within diffs), but is at least twice as fast as the solution above using the mod-30 trick. Exhaustive search up to MAX=1000 took 6.2 seconds for me (64-bit on 3.4GHz i7). It is not clear how it can be combined with the mod-30 trick.

The asymptotic behavior can be improved to O(N^3 ln N) by replacing the linear search with an increasing-increment "hunt" (and the outer linear search, which is also O(N^4), with a call to std::upper_bound). With this replacement, the first solution was found in 0.05 seconds; exhaustive search up to MAX=1000 took 2.80 seconds; and the second nontrivial solution (discarding multiples of the first solution), at y==2615, was found in 94.6 seconds. Note: there is no solution 2615, because 645^5 + 1523^5 + 1722^5 +2506^5 = 122 280 854 808 884 376, but 2615^5=122 280 854 808 884 375. This is an error due to limitation in mantissa of double type (52 bits). 128 bit type is required for the next solution 85359.

```template<class C_, class LT_> C_ Unique(const C_& src, const LT_& less)
{
C_ retval(src);
std::sort(retval.begin(), retval.end(), less);
retval.erase(unique(retval.begin(), retval.end()), retval.end());
return retval;
}

template<class I_, class P_> I_ HuntFwd(const I_& hint, const I_& end, const P_& less)	// if less(x) is false, then less(x+1) must also be false
{
I_ retval(hint);
int step = 1;
// expanding phase
while (end - retval > step)
{
I_ test = retval + step;
if (!less(test))
break;
retval = test;
step <<= 1;
}
// contracting phase
while (step > 1)
{
step >>= 1;
if (end - retval <= step)
continue;
I_ test = retval + step;
if (less(test))
retval = test;
}
if (retval != end && less(retval))
++retval;
return retval;
}

bool DPFind(int how_many)
{
const int MAX = 1000;
vector<double> pow5(MAX);
for (int i = 1; i < MAX; i++)
pow5[i] = (double)i * i * i * i * i;
vector<pair<double, int>> diffs;
for (int i = 2; i < MAX; ++i)
{
for (int j = 1; j < i; ++j)
diffs.emplace_back(pow5[i] - pow5[j], j);
}
auto firstLess = [](const pair<double, int>& lhs, const pair<double, int>& rhs) { return lhs.first < rhs.first; };
diffs = Unique(diffs, firstLess);

for (int x4 = 4; x4 < MAX - 1; ++x4)
{
for (int x3 = 3; x3 < x4; ++x3)
{
// if (133 * x3 == 110 * x4) continue;	// skip duplicates of first solution
const auto s2 = pow5[x4] + pow5[x3];
auto pd = upper_bound(diffs.begin() + 1, diffs.end(), make_pair(s2, 0), firstLess) - 1;
for (int x2 = 2; x2 < x3; ++x2)
{
const auto sum = s2 + pow5[x2];
pd = HuntFwd(pd, diffs.end(), [&](decltype(pd) it){ return it->first < sum; });
if (pd != diffs.end() && pd->first == sum && pd->second < x3)	// find each solution only once
{
const double y = pow(pd->first + pow5[pd->second], 0.2);
cout << x4 << " " << x3 << " " << x2 << " " << pd->second << " -> " << static_cast<int>(y + 0.5) << "\n";
if (--how_many <= 0)
return true;
}
}
}
}
return false;
}
```

Thanks, EchoLisp guys!

## Clojure

```(ns test-p.core
(:require [clojure.math.numeric-tower :as math])
(:require [clojure.data.int-map :as i]))

(defn solve-power-sum [max-value max-sols]
" Finds solutions by using method approach of EchoLisp
Large difference is we store a dictionary of all combinations
of y^5 - x^5 with the x, y value so we can simply lookup rather than have to search "
(let [pow5 (mapv #(math/expt % 5) (range 0 (* 4 max-value)))                  ; Pow5 = Generate Lookup table for x^5
y5-x3 (into (i/int-map) (for [x (range 1 max-value)                     ; For x0^5 + x1^5 + x2^5 + x3^5  = y^5
y (range (+ 1 x) (* 4 max-value))]        ; compute y5-x3 = set of all possible differnences
[(- (get pow5 y) (get pow5 x)) [x y]])) ; note: (get pow5 y) is closure for: pow5[y]
solutions-found (atom 0)]

(for [x0 (range 1 max-value)                                    ; Search over x0, x1, x2 for sums equal y5-x3
x1 (range 1 x0)
x2 (range 1 x1)
:when (< @solutions-found max-sols)
:let [sum (apply + (map pow5 [x0 x1 x2]))]         ; compute sum of items to the 5th power
:when (contains? y5-x3 sum)]                       ; check if sum is in set of differences
(do
(swap! solutions-found inc)                          ; increment counter for solutions found
(concat [x0 x1 x2] (get y5-x3 sum))))))              ; create result (since in set of differences)

; Output results with numbers in ascending order placing results into a set (i.e. #{}) so duplicates are discarded
; CPU i7 920 Quad Core @2.67 GHz clock Windows 10
(println (into #{} (map sort (solve-power-sum 250 1))))   ; MAX = 250, find only 1 value: Duration was 0.26 seconds
(println (into #{} (map sort (solve-power-sum 1000 1000))));MAX = 1000, high max-value so all solutions found: Time = 4.8 seconds
```

Output

```1st Solution with MAX = 250 (Solution Time: 260 ms CPU i7 920 Quad Core)
#{(27 84 110 133 144))

All Solutions with MAX = 1000 (Solution Time: 4.8 seconds CPU i7 920 Quad Core)
#{(27 84 110 133 144)
(162 504 660 798 864)
(135 420 550 665 720)
(108 336 440 532 576)
(189 588 770 931 1008)
(54 168 220 266 288)
(81 252 330 399 432)}
```

## COBOL

```       IDENTIFICATION DIVISION.
PROGRAM-ID. EULER.
DATA DIVISION.
FILE SECTION.
WORKING-STORAGE SECTION.
1   TABLE-LENGTH CONSTANT 250.
1   SEARCHING-FLAG     PIC 9.
88  FINISHED-SEARCHING VALUE IS 1
WHEN SET TO FALSE IS 0.
1  CALC.
3  A               PIC 999 USAGE COMPUTATIONAL-5.
3  B               PIC 999 USAGE COMPUTATIONAL-5.
3  C               PIC 999 USAGE COMPUTATIONAL-5.
3  D               PIC 999 USAGE COMPUTATIONAL-5.
3  ABCD            PIC 9(18) USAGE COMPUTATIONAL-5.
3  FIFTH-ROOT-OFFS PIC 999 USAGE COMPUTATIONAL-5.
3  POWER-COUNTER   PIC 999 USAGE COMPUTATIONAL-5.
88 POWER-MAX       VALUE TABLE-LENGTH.

1   PRETTY.
3  A               PIC ZZ9.
3  FILLER          VALUE "^5 + ".
3  B               PIC ZZ9.
3  FILLER          VALUE "^5 + ".
3  C               PIC ZZ9.
3  FILLER          VALUE "^5 + ".
3  D               PIC ZZ9.
3  FILLER          VALUE "^5 = ".
3  FIFTH-ROOT-OFFS PIC ZZ9.
3  FILLER          VALUE "^5.".

1   FIFTH-POWER-TABLE   OCCURS TABLE-LENGTH TIMES
ASCENDING KEY IS FIFTH-POWER
INDEXED BY POWER-INDEX.
3  FIFTH-POWER PIC 9(18) USAGE COMPUTATIONAL-5.

PROCEDURE DIVISION.
MAIN-PARAGRAPH.
SET FINISHED-SEARCHING TO FALSE.
PERFORM POWERS-OF-FIVE-TABLE-INIT.
PERFORM VARYING
A IN CALC
FROM 1 BY 1 UNTIL A IN CALC = TABLE-LENGTH

AFTER B IN CALC
FROM 1 BY 1 UNTIL B IN CALC = A IN CALC

AFTER C IN CALC
FROM 1 BY 1 UNTIL C IN CALC = B IN CALC

AFTER D IN CALC
FROM 1 BY 1 UNTIL D IN CALC = C IN CALC

IF FINISHED-SEARCHING
STOP RUN
END-IF

PERFORM POWER-COMPUTATIONS

END-PERFORM.

POWER-COMPUTATIONS.

MOVE ZERO TO ABCD IN CALC.

FIFTH-POWER(B IN CALC)
FIFTH-POWER(C IN CALC)
FIFTH-POWER(D IN CALC)
TO ABCD IN CALC.

SET POWER-INDEX TO 1.

SEARCH ALL FIFTH-POWER-TABLE
WHEN FIFTH-POWER(POWER-INDEX) = ABCD IN CALC
MOVE POWER-INDEX TO FIFTH-ROOT-OFFS IN CALC
MOVE CORRESPONDING CALC TO PRETTY
DISPLAY PRETTY END-DISPLAY
SET FINISHED-SEARCHING TO TRUE
END-SEARCH

EXIT PARAGRAPH.

POWERS-OF-FIVE-TABLE-INIT.
PERFORM VARYING POWER-COUNTER FROM 1 BY 1 UNTIL POWER-MAX
COMPUTE FIFTH-POWER(POWER-COUNTER) =
POWER-COUNTER *
POWER-COUNTER *
POWER-COUNTER *
POWER-COUNTER *
POWER-COUNTER
END-COMPUTE
END-PERFORM.
EXIT PARAGRAPH.

END PROGRAM EULER.
```

Output

```133^5 + 110^5 +  84^5 +  27^5 = 144^5.
```

## Common Lisp

```(ql:quickload :alexandria)
(let ((fifth-powers (mapcar #'(lambda (x) (expt x 5))
(alexandria:iota 250))))
(loop named outer for x0 from 1 to (length fifth-powers) do
(loop for x1 from 1 below x0 do
(loop for x2 from 1 below x1 do
(loop for x3 from 1 below x2 do
(let ((x-sum (+ (nth x0 fifth-powers)
(nth x1 fifth-powers)
(nth x2 fifth-powers)
(nth x3 fifth-powers))))
(if (member x-sum fifth-powers)
(return-from outer (list x0 x1 x2 x3 (round (expt x-sum 0.2)))))))))))
```
Output:
`(133 110 84 27 144)`

## D

### First version

Translation of: Rust
```import std.stdio, std.range, std.algorithm, std.typecons;

enum maxN = 250;
auto pow5 = iota(size_t(maxN)).map!(i => ulong(i) ^^ 5).array.assumeSorted;

foreach (immutable x0; 1 .. maxN)
foreach (immutable x1; 1 .. x0)
foreach (immutable x2; 1 .. x1)
foreach (immutable x3; 1 .. x2) {
immutable powSum = pow5[x0] + pow5[x1] + pow5[x2] + pow5[x3];
if (pow5.contains(powSum))
return tuple(x0, x1, x2, x3, pow5.countUntil(powSum));
}
assert(false);
}

void main() {
writefln("%d^5 + %d^5 + %d^5 + %d^5 == %d^5", eulersSumOfPowers[]);
}
```
Output:
`133^5 + 110^5 + 84^5 + 27^5 == 144^5`

Run-time about 0.64 seconds. A Range-based Haskell-like solution is missing because of Issue 14833.

### Second version

Translation of: Python
```void main() {
import std.stdio, std.range, std.algorithm, std.typecons;

enum uint MAX = 250;
uint[ulong] p5;
Tuple!(uint, uint)[ulong] sum2;

foreach (immutable i; 1 .. MAX) {
p5[ulong(i) ^^ 5] = i;
foreach (immutable j; i .. MAX)
sum2[ulong(i) ^^ 5 + ulong(j) ^^ 5] = tuple(i, j);
}

const sk = sum2.keys.sort().release;
foreach (p; p5.keys.sort())
foreach (immutable s; sk) {
if (p <= s)
break;
if (p - s in sum2) {
writeln(p5[p], " ", tuple(sum2[s][], sum2[p - s][]));
return; // Finds first only.
}
}
}
```
Output:
`144 Tuple!(uint, uint, uint, uint)(27, 84, 110, 133)`

### Third version

Translation of: C++

This solution is a brutal translation of the iterator-based C++ version, and it should be improved to use more idiomatic D Ranges.

```import core.stdc.stdio, std.typecons, std.math, std.algorithm, std.range;

alias Pair = Tuple!(double, int);
alias PairPtr = Pair*;

// If less(x) is false, then less(x + 1) must also be false.
PairPtr huntForward(Pred)(PairPtr hint, const PairPtr end, const Pred less) pure nothrow @nogc {
PairPtr result = hint;
int step = 1;

// Expanding phase.
while (end - result > step) {
PairPtr test = result + step;
if (!less(test))
break;
result = test;
step <<= 1;
}

// Contracting phase.
while (step > 1) {
step >>= 1;
if (end - result <= step)
continue;
PairPtr test = result + step;
if (less(test))
result = test;
}
if (result != end && less(result))
++result;
return result;
}

bool dPFind(int how_many) nothrow {
enum MAX = 1_000;

double[MAX] pow5;
foreach (immutable i; 1 .. MAX)
pow5[i] = double(i) ^^ 5;

Pair[] diffs0; // Will contain (MAX-1) * (MAX-2) / 2 pairs.
foreach (immutable i; 2 .. MAX)
foreach (immutable j; 1 .. i)
diffs0 ~= Pair(pow5[i] - pow5[j], j);

// Remove pairs with duplicate first items.
diffs0.length -= diffs0.sort!q{ a < b }.uniq.copy(diffs0).length;
auto diffs = diffs0.assumeSorted!q{ a < b };

foreach (immutable x4; 4 .. MAX - 1) {
foreach (immutable x3; 3 .. x4) {
immutable s2 = pow5[x4] + pow5[x3];
auto pd0 = diffs[1 .. \$].upperBound(Pair(s2, 0));
PairPtr pd = &pd0 - 1;
foreach (immutable x2; 2 .. x3) {
immutable sum = s2 + pow5[x2];
const PairPtr endPtr = &diffs[\$ - 1] + 1;
// This lambda heap-allocates.
pd = huntForward(pd, endPtr, (in PairPtr p) pure => (*p) < sum);
if (pd != endPtr && (*pd) == sum && (*pd) < x3) { // Find each solution only once.
immutable y = ((*pd) + pow5[(*pd)]) ^^ 0.2;
printf("%d %d %d %d : %d\n", x4, x3, x2, (*pd), cast(int)(y + 0.5));
if (--how_many <= 0)
return true;
}
}
}
}

return false;
}

void main() nothrow {
if (!dPFind(100))
printf("Search finished.\n");
}
```
Output:
```133 110 27 84 : 144
133 110 84 27 : 144
266 220 54 168 : 288
266 220 168 54 : 288
399 330 81 252 : 432
399 330 252 81 : 432
532 440 108 336 : 576
532 440 336 108 : 576
665 550 135 420 : 720
665 550 420 135 : 720
798 660 162 504 : 864
798 660 504 162 : 864
Search finished.```

See Pascal.

## EasyLang

```n = 250
len p5[] n
len h5[] 65537
for i = 0 to n - 1
p5[i + 1] = i * i * i * i * i
h5[p5[i + 1] mod 65537 + 1] = 1
.
func search a s . y .
y = -1
b = n
while a + 1 < b
i = (a + b) div 2
if p5[i + 1] > s
b = i
elif p5[i + 1] < s
a = i
else
a = b
y = i
.
.
.
for x0 = 0 to n - 1
for x1 = 0 to x0
sum1 = p5[x0 + 1] + p5[x1 + 1]
for x2 = 0 to x1
sum2 = p5[x2 + 1] + sum1
for x3 = 0 to x2
sum = p5[x3 + 1] + sum2
if h5[sum mod 65537 + 1] = 1
call search x0 sum y
if y >= 0
print x0 & " " & x1 & " " & x2 & " " & x3 & " " & y
break 4
.
.
.
.
.
.
```

## EchoLisp

To speed up things, we search for x0, x1, x2 such as x0^5 + x1^5 + x2^5 = a difference of 5-th powers.

```(define dim 250)

;; speed up n^5
(define (p5 n) (* n n n n n))
(remember 'p5) ;; memoize

;; build vector of all  y^5 - x^5 diffs - length 30877
(define all-y^5-x^5
(for*/vector
[(x (in-range 1 dim))  (y (in-range (1+ x) dim))]
(- (p5 y) (p5 x))))

;; sort to use vector-search
(begin (vector-sort! <  all-y^5-x^5) 'sorted)

;; find couple (x y) from y^5 - x^5
(define (x-y y^5-x^5)
(for*/fold (x-y null)
[(x (in-range 1 dim)) (y (in-range (1+ x ) dim))]
(when
(= (- (p5 y) (p5 x)) y^5-x^5)
(set! x-y (list x y))
(break #t)))) ; stop on first

;; search
(for*/fold  (sol null)
[(x0 (in-range 1 dim)) (x1 (in-range (1+ x0) dim)) (x2 (in-range (1+ x1) dim))]
(set! sol (+ (p5 x0) (p5 x1) (p5 x2)))
(when
(vector-search sol all-y^5-x^5)  ;; x0^5 + x1^5 + x2^5 = y^5 - x3^5 ???
(set! sol (append (list x0 x1 x2) (x-y  sol))) ;; found
(break #t))) ;; stop on first

→   (27 84 110 133 144) ;; time 2.8 sec
```

## Elixir

Translation of: Ruby
```defmodule Euler do
def sum_of_power(max \\ 250) do
{p5, sum2} = setup(max)
sk = Enum.sort(Map.keys(sum2))
Enum.reduce(Enum.sort(Map.keys(p5)), Map.new, fn p,map ->
sum(sk, p5, sum2, p, map)
end)
end

defp setup(max) do
Enum.reduce(1..max, {%{}, %{}}, fn i,{p5,sum2} ->
i5 = i*i*i*i*i
add = for j <- i..max, into: sum2, do: {i5 + j*j*j*j*j, [i,j]}
end)
end

defp sum([], _, _, _, map), do: map
defp sum([s|_], _, _, p, map) when p<=s, do: map
defp sum([s|t], p5, sum2, p, map) do
if sum2[p - s],
do:   sum(t, p5, sum2, p, Map.put(map, Enum.sort(sum2[s] ++ sum2[p-s]), p5[p])),
else: sum(t, p5, sum2, p, map)
end
end

Enum.each(Euler.sum_of_power, fn {k,v} ->
IO.puts Enum.map_join(k, " + ", fn i -> "#{i}**5" end) <> " = #{v}**5"
end)
```
Output:
```27**5 + 84**5 + 110**5 + 133**5 = 144**5
```

## ERRE

```PROGRAM EULERO

CONST MAX=250

!\$DOUBLE

FUNCTION POW5(X)
POW5=X*X*X*X*X
END FUNCTION

!\$INCLUDE="PC.LIB"

BEGIN
CLS
FOR X0=1 TO MAX DO
FOR X1=1 TO X0 DO
FOR X2=1 TO X1 DO
FOR X3=1 TO X2 DO
LOCATE(3,1) PRINT(X0;X1;X2;X3)
SUM=POW5(X0)+POW5(X1)+POW5(X2)+POW5(X3)
S1=INT(SUM^0.2#+0.5#)
IF SUM=POW5(S1) THEN PRINT(X0,X1,X2,X3,S1) END IF
END FOR
END FOR
END FOR
END FOR
END PROGRAM```
Output:
```133 110 84 27 144
```

## F#

```//Find 4 integers whose 5th powers sum to the fifth power of an integer (Quickly!) - Nigel Galloway: April 23rd., 2015
let G =
let GN = Array.init<float> 250 (fun n -> (float n)**5.0)
let rec gng (n, i, g, e) =
match (n, i, g, e) with
| (250,_,_,_) -> "No Solution Found"
| (_,250,_,_) -> gng (n+1, n+1, n+1, n+1)
| (_,_,250,_) -> gng (n, i+1, i+1, i+1)
| (_,_,_,250) -> gng (n, i, g+1, g+1)
| _ -> let l = GN.[n] + GN.[i] + GN.[g] + GN.[e]
match l with
| _ when l > GN.           -> gng(n,i,g+1,g+1)
| _ when l = round(l**0.2)**5.0 -> sprintf "%d**5 + %d**5 + %d**5 + %d**5 = %d**5" n i g e (int (l**0.2))
| _                             -> gng(n,i,g,e+1)
gng (1, 1, 1, 1)
```
Output:
```"27**5 + 84**5 + 110**5 + 133**5 = 144**5"
```

## Factor

This solution uses Factor's `backtrack` vocabulary (based on continuations) to simplify the reduction of the search space. Each time `xn` is called, a new summand is introduced which can only take on a value as high as the previous summand - 1. This also creates a checkpoint for the backtracker. `fail` causes the backtracking to occur.

```USING: arrays backtrack kernel literals math.functions
math.ranges prettyprint sequences ;

CONSTANT: pow5 \$[ 0 250 [a,b) [ 5 ^ ] map ]

: xn ( n1 -- n2 n2 ) [1,b) amb-lazy dup ;

250 xn xn xn xn drop 4array dup pow5 nths sum dup pow5
member? [ pow5 index suffix . ] [ 2drop fail ] if
```
Output:
```{ 133 110 84 27 144 }
```

## Forth

Translation of: Go
```: sq  dup * ;
: 5^  dup sq sq * ;

create pow5 250 cells allot
:noname
250 0 DO  i 5^  pow5 i cells + !  LOOP ; execute

: @5^  cells pow5 + @ ;

: solution? ( n -- n )
pow5 250 cells bounds DO
dup i @ = IF  drop i pow5 - cell / unloop EXIT  THEN
cell +LOOP drop 0 ;

\ GFORTH only provides 2 index variables: i, j
\ so the code creates locals for two outer loop vars, k & l

: euler  ( -- )
250 4 DO i { l }
l 3 DO i { k }
k 2 DO
i 1 DO
i @5^ j @5^ + k @5^ + l @5^ + solution?
dup IF
l . k . j . i . . cr
unloop unloop unloop unloop EXIT
ELSE
drop
THEN
LOOP
LOOP
LOOP
LOOP ;

euler
bye
```
Output:
```\$ gforth-fast ./euler.fs
133 110 84 27 144
```

## Fortran

### FORTRAN IV

To solve this problem, we must handle integers up 250**5 ~= 9.8*10**11 . So we need integers with at less 41 bits. In 1966 all Fortrans were not equal. On IBM360, INTEGER was a 32-bit integer; on CDC6600, INTEGER was a 60-bit integer. And Leon J. Lander and Thomas R. Parkin used the CDC6600.

```C EULER SUM OF POWERS CONJECTURE - FORTRAN IV
C FIND I1,I2,I3,I4,I5 : I1**5+I2**5+I3**5+I4**5=I5**5
INTEGER I,P5(250),SUMX
MAXN=250
DO 1 I=1,MAXN
1  P5(I)=I**5
DO 6 I1=1,MAXN
DO 6 I2=1,MAXN
DO 6 I3=1,MAXN
DO 6 I4=1,MAXN
SUMX=P5(I1)+P5(I2)+P5(I3)+P5(I4)
I5=1
2  IF(I5-MAXN) 3,3,6
3  IF(P5(I5)-SUMX) 5,4,6
4  WRITE(*,300) I1,I2,I3,I4,I5
STOP
5  I5=I5+1
GOTO 2
6  CONTINUE
300  FORMAT(5(1X,I3))
END
```
Output:
```  27  84 110 133 144
```

### Fortran 95

Works with: Fortran version 95 and later
```program sum_of_powers
implicit none

integer, parameter :: maxn = 249
integer, parameter :: dprec = selected_real_kind(15)
integer :: i, x0, x1, x2, x3, y
real(dprec) :: n(maxn), sumx

n = (/ (real(i, dprec)**5, i = 1, maxn) /)

outer: do x0 = 1, maxn
do x1 = 1, maxn
do x2 = 1, maxn
do x3 = 1, maxn
sumx = n(x0)+ n(x1)+ n(x2)+ n(x3)
y = 1
do while(y <= maxn .and. n(y) <= sumx)
if(n(y) == sumx) then
write(*,*) x0, x1, x2, x3, y
exit outer
end if
y = y + 1
end do
end do
end do
end do
end do outer

end program
```
Output:
`          27          84         110         133         144`

## FreeBASIC

```' version 14-09-2015
' compile with: fbc -s console

' some constants calculated when the program is compiled

Const As UInteger max = 250
Const As ULongInt pow5_max = CULngInt(max) * max * max * max * max
' limit x1, x2, x3
Const As UInteger limit_x1 = (pow5_max / 4) ^ 0.2
Const As UInteger limit_x2 = (pow5_max / 3) ^ 0.2
Const As UInteger limit_x3 = (pow5_max / 2) ^ 0.2

' ------=< MAIN >=------

Dim As ULongInt pow5(max), ans1, ans2, ans3
Dim As UInteger x1, x2, x3, x4, x5 , m1, m2

Cls : Print

For x1 = 1 To max
pow5(x1) = CULngInt(x1) * x1 * x1 * x1 * x1
Next x1

For x1 = 1 To limit_x1
For x2 = x1 +1 To limit_x2
m1 = x1 + x2
ans1 = pow5(x1) + pow5(x2)
If ans1 > pow5_max Then Exit For
For x3 = x2 +1 To limit_x3
ans2 = ans1 + pow5(x3)
If ans2 > pow5_max Then Exit For
m2 = (m1 + x3) Mod 30
If m2 = 0 Then m2 = 30
For x4 = x3 +1 To max -1
ans3 = ans2 + pow5(x4)
If ans3 > pow5_max Then Exit For
For x5 = x4 + m2 To max Step 30
If ans3 < pow5(x5) Then Exit For
If ans3 = pow5(x5) Then
Print x1; "^5 + "; x2; "^5 + "; x3; "^5 + "; _
x4; "^5 = "; x5; "^5"
Exit For, For
EndIf
Next x5
Next x4
Next x3
Next x2
Next x1

Print
Print "done"

' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
```
Output:
`27^5 + 84^5 + 110^5 + 133^5 = 144^5`

## Go

Translation of: Python
```package main

import (
"fmt"
"log"
)

func main() {
fmt.Println(eulerSum())
}

func eulerSum() (x0, x1, x2, x3, y int) {
var pow5 int
for i := range pow5 {
pow5[i] = i * i * i * i * i
}
for x0 = 4; x0 < len(pow5); x0++ {
for x1 = 3; x1 < x0; x1++ {
for x2 = 2; x2 < x1; x2++ {
for x3 = 1; x3 < x2; x3++ {
sum := pow5[x0] +
pow5[x1] +
pow5[x2] +
pow5[x3]
for y = x0 + 1; y < len(pow5); y++ {
if sum == pow5[y] {
return
}
}
}
}
}
}
log.Fatal("no solution")
return
}
```
Output:
```133 110 84 27 144
```

## Groovy

Translation of: Java
```class EulerSumOfPowers {
static final int MAX_NUMBER = 250

static void main(String[] args) {
boolean found = false
long[] fifth = new long[MAX_NUMBER]

for (int i = 1; i <= MAX_NUMBER; i++) {
long i2 = i * i
fifth[i - 1] = i2 * i2 * i
}

for (int a = 0; a < MAX_NUMBER && !found; a++) {
for (int b = a; b < MAX_NUMBER && !found; b++) {
for (int c = b; c < MAX_NUMBER && !found; c++) {
for (int d = c; d < MAX_NUMBER && !found; d++) {
long sum = fifth[a] + fifth[b] + fifth[c] + fifth[d]
int e = Arrays.binarySearch(fifth, sum)
found = (e >= 0)
if (found) {
println("\${a + 1}^5 + \${b + 1}^5 + \${c + 1}^5 + \${d + 1}^5 + \${e + 1}^5")
}
}
}
}
}
}
}
```
Output:
`27^5 + 84^5 + 110^5 + 133^5 + 144^5`

```import Data.List
import Data.List.Ordered

main :: IO ()
main = print \$ head [(x0,x1,x2,x3,x4) |
-- choose x0, x1, x2, x3
-- so that 250 < x3 < x2 < x1 < x0
x3 <- [1..250-1],
x2 <- [1..x3-1],
x1 <- [1..x2-1],
x0 <- [1..x1-1],

let p5Sum = x0^5 + x1^5 + x2^5 + x3^5,

-- lazy evaluation of powers of 5
let p5List = [i^5|i <- [1..]],

-- is sum a power of 5 ?
member p5Sum p5List,

-- which power of 5 is sum ?
let Just x4 = elemIndex p5Sum p5List ]
```
Output:
```(27,84,110,133,144)
```

Or, using dictionaries of powers and sums, and thus rather faster:

Translation of: Python
```import qualified Data.Map.Strict as M
import Data.List (find, intercalate)
import Data.Maybe (maybe)

------------- EULER'S SUM OF POWERS CONJECTURE -----------

counterExample :: (M.Map Int (Int, Int), M.Map Int Int) -> Maybe (Int, Int)
counterExample (sumMap, powerMap) =
find
(\(p, s) -> M.member (p - s) sumMap)
(M.keys powerMap >>=
(((>>=) . flip takeWhile (M.keys sumMap) . (>)) <*> \ p s -> [(p, s)]))

sumMapForRange :: [Int] -> M.Map Int (Int, Int)
sumMapForRange xs =
M.fromList
[ ((x ^ 5) + (y ^ 5), (x, y))
| x <- xs
, y <- tail xs
, x > y ]

powerMapForRange :: [Int] -> M.Map Int Int
powerMapForRange = M.fromList . (zip =<< fmap (^ 5))

--------------------------- TEST -------------------------
main :: IO ()
main =
putStrLn \$
"Euler's sum of powers conjecture – " <>
maybe
("no counter-example found in the range " <> rangeString xs)
(showExample sumsAndPowers xs)
(counterExample sumsAndPowers)
where
xs = [1 .. 249]
sumsAndPowers = ((,) . sumMapForRange <*> powerMapForRange) xs

showExample :: (M.Map Int (Int, Int), M.Map Int Int) -> [Int] -> (Int, Int) -> String
showExample (sumMap, powerMap) xs (p, s) =
"a counter-example in range " <> rangeString xs <> ":\n\n" <>
intercalate "^5 + " (show <\$> [a, b, c, d]) <>
"^5 = " <>
show (powerMap M.! p) <>
"^5"
where
(a, b) = sumMap M.! (p - s)
(c, d) = sumMap M.! s

rangeString :: [Int] -> String
rangeString [] = "[]"
rangeString (x:xs) = '[' : show x <> " .. " <> show (last xs) <> "]"
```
Output:
```Euler's sum of powers conjecture – a counter-example in range [1 .. 249]:

133^5 + 110^5 + 84^5 + 27^5 = 144^5```

## J

```   require 'stats'
(#~ (= <.)@((+/"1)&.:(^&5)))1+4 comb 248
27 84 110 133
```

Explanation:

```1+4 comb 248
```
finds all the possibilities for our four arguments. Then,
```(#~ (= <.)@((+/"1)&.:(^&5)))
```
discards the cases we are not interested in. (It only keeps the case(s) where the fifth root of the sum of the fifth powers is an integer.)

Only one possibility remains.

Here's a significantly faster approach (about 100 times faster), based on the echolisp implementation:

```find5=:3 :0
y=. 250
n=. i.y
p=. n^5
a=. (#~ 0&<),-/~p
s=. /:~a
l=. (i.*:y)(#~ 0&<),-/~p
c=. 3 comb <.5%:(y^5)%4
t=. +/"1 c{p
x=. (t e. s)#t
|.,&<&~./|:(y,y)#:l#~a e. x
)
```

Use:

```   find5''
┌─────────────┬───┐
│27 84 110 133│144│
└─────────────┴───┘
```

Note that this particular implementation is a bit hackish, since it relies on the solution being unique for the range of numbers being considered. If there were more solutions it would take a little extra code (though not much time) to untangle them.

## Java

Translation of: ALGOL 68

Tested with Java 6.

```public class eulerSopConjecture
{

static final int    MAX_NUMBER = 250;

public static void main( String[] args )
{
boolean found = false;
long[]  fifth = new long[ MAX_NUMBER ];

for( int i = 1; i <= MAX_NUMBER; i ++ )
{
long i2 =  i * i;
fifth[ i - 1 ] = i2 * i2 * i;
} // for i

for( int a = 0; a < MAX_NUMBER && ! found ; a ++ )
{
for( int b = a; b < MAX_NUMBER && ! found ; b ++ )
{
for( int c = b; c < MAX_NUMBER && ! found ; c ++ )
{
for( int d = c; d < MAX_NUMBER && ! found ; d ++ )
{
long sum  = fifth[a] + fifth[b] + fifth[c] + fifth[d];
int  e = java.util.Arrays.binarySearch( fifth, sum );
found  = ( e >= 0 );
if( found )
{
// the value at e is a fifth power
System.out.print( (a+1) + "^5 + "
+ (b+1) + "^5 + "
+ (c+1) + "^5 + "
+ (d+1) + "^5 = "
+ (e+1) + "^5"
);
} // if found;;
} // for d
} // for c
} // for b
} // for a
} // main

} // eulerSopConjecture
```

Output:

```27^5 + 84^5 + 110^5 + 133^5 = 144^5
```

## JavaScript

### ES5

```var eulers_sum_of_powers = function (iMaxN) {

var aPow5 = [];
var oPow5ToN = {};

for (var iP = 0; iP <= iMaxN; iP++) {
var iPow5 = Math.pow(iP, 5);
aPow5.push(iPow5);
oPow5ToN[iPow5] = iP;
}

for (var i0 = 1; i0 <= iMaxN; i0++) {
for (var i1 = 1; i1 <= i0; i1++) {
for (var i2 = 1; i2 <= i1; i2++) {
for (var i3 = 1; i3 <= i2; i3++) {
var iPow5Sum = aPow5[i0] + aPow5[i1] + aPow5[i2] + aPow5[i3];
if (typeof oPow5ToN[iPow5Sum] != 'undefined') {
return {
i0: i0,
i1: i1,l
i2: i2,
i3: i3,
iSum: oPow5ToN[iPow5Sum]
};
}
}
}
}
}

};

var oResult = eulers_sum_of_powers(250);

console.log(oResult.i0 + '^5 + ' + oResult.i1 + '^5 + ' + oResult.i2 +
'^5 + ' + oResult.i3 + '^5 = ' + oResult.iSum + '^5');
```
Output:
```133^5 + 110^5 + 84^5 + 27^5 = 144^5
```
This
Translation of: D
that verify: a^5 + b^5 + c^5 + d^5 = x^5
```var N=1000, first=false
var ns={}, npv=[]
for (var n=0; n<=N; n++) {
var np=Math.pow(n,5); ns[np]=n; npv.push(np)
}
loop:
for (var a=1;   a<=N; a+=1)
for (var b=a+1; b<=N; b+=1)
for (var c=b+1; c<=N; c+=1)
for (var d=c+1; d<=N; d+=1) {
var x = ns[ npv[a]+npv[b]+npv[c]+npv[d] ]
if (!x) continue
print( [a, b, c, d, x] )
if (first) break loop
}
function print(c) {
var e='<sup>5</sup>', ep=e+' + '
document.write(c, ep, c, ep, c, ep, c, e, ' = ', c, e, '<br>')
}
```
Or this
Translation of: C
that verify: a^5 + b^5 + c^5 + d^5 = x^5
```var N=1000, first=false
var npv=[], M=30 // x^5 == x modulo M (=2*3*5)
for (var n=0; n<=N; n+=1) npv[n]=Math.pow(n, 5)
var mx=1+npv[N]; while(n<=N+M) npv[n++]=mx

loop:
for (var a=1;   a<=N; a+=1)
for (var b=a+1; b<=N; b+=1)
for (var c=b+1; c<=N; c+=1)
for (var t=npv[a]+npv[b]+npv[c], d=c+1, x=t%M+d; (n=t+npv[d])<mx; d+=1, x+=1) {
while (npv[x]<=n) x+=M; x-=M // jump over M=30 values for x>d
if (npv[x] != n) continue
print( [a, b, c, d, x] )
if (first) break loop;
}
function print(c) {
var e='<sup>5</sup>', ep=e+' + '
document.write(c, ep, c, ep, c, ep, c, e, ' = ', c, e, '<br>')
}
```
Or this
Translation of: EchoLisp
that verify: a^5 + b^5 + c^5 = x^5 - d^5
```var N=1000, first=false
var dxs={}, pow=Math.pow
for (var d=1; d<=N; d+=1)
for (var dp=pow(d,5), x=d+1; x<=N; x+=1)
dxs[pow(x,5)-dp]=[d,x]
loop:
for (var a=1; a<N; a+=1)
for (var ap=pow(a,5), b=a+1; b<N; b+=1)
for (var abp=ap+pow(b,5), c=b+1; c<N; c+=1) {
var dx = dxs[ abp+pow(c,5) ]
if (!dx || c >= dx) continue
print( [a, b, c].concat( dx ) )
if (first) break loop
}
function print(c) {
var e='<sup>5</sup>', ep=e+' + '
document.write(c, ep, c, ep, c, ep, c, e, ' = ', c, e, '<br>')
}
```
Or this
Translation of: Python
that verify: a^5 + b^5 = x^5 - (c^5 + d^5)
```var N=1000, first=false
var is={}, ipv=[], ijs={}, ijpv=[], pow=Math.pow
for (var i=1; i<=N; i+=1) {
var ip=pow(i,5); is[ip]=i; ipv.push(ip)
for (var j=i+1; j<=N; j+=1) {
var ijp=ip+pow(j,5); ijs[ijp]=[i,j]; ijpv.push(ijp)
}
}
ijpv.sort( function (a,b) {return a - b } )
loop:
for (var i=0, ei=ipv.length; i<ei; i+=1)
for (var xp=ipv[i], j=0, je=ijpv.length; j<je; j+=1) {
var cdp = ijpv[j]
if (cdp >= xp) break
var cd = ijs[xp-cdp]
if (!cd) continue
var ab = ijs[cdp]
if (ab >= cd) continue
print( [].concat(ab, cd, is[xp]) )
if (first) break loop
}
function print(c) {
var e='<sup>5</sup>', ep=e+' + '
document.write(c, ep, c, ep, c, ep, c, e, ' = ', c, e, '<br>')
}
```
Output:
``` 275 + 845 + 1105 + 1335 = 1445
545 + 1685 + 2205 + 2665 = 2885
815 + 2525 + 3305 + 3995 = 4325
1085 + 3365 + 4405 + 5325 = 5765
1355 + 4205 + 5505 + 6655 = 7205
1625 + 5045 + 6605 + 7985 = 8645
```

### ES6

#### Procedural

```(() => {
'use strict';

const eulersSumOfPowers = intMax => {
const
pow = Math.pow,
xs = range(0, intMax)
.map(x => pow(x, 5)),
dct = xs.reduce((a, x, i) =>
(a[x] = i,
a
), {});

for (let a = 1; a <= intMax; a++) {
for (let b = 2; b <= a; b++) {
for (let c = 3; c <= b; c++) {
for (let d = 4; d <= c; d++) {
const sumOfPower = dct[xs[a] + xs[b] + xs[c] + xs[d]];
if (sumOfPower !== undefined) {
return [a, b, c, d, sumOfPower];
}
}
}
}
}
return undefined;
};

// range :: Int -> Int -> [Int]
const range = (m, n) =>
Array.from({
length: Math.floor(n - m) + 1
}, (_, i) => m + i);

// TEST
return soln ? soln.slice(0, 4)
.map(x => `\${x}^5`)
.join(' + ') + ` = \${soln}^5` : 'No solution found.'

})();
```
Output:
`133^5 + 110^5 + 84^5 + 27^5 = 144^5`

#### Functional

Using dictionaries of powers and sums, and a little faster than the procedural version above:

Translation of: Python
```(() => {
'use strict';

const main = () => {

const
iFrom = 1,
iTo = 249,
xs = enumFromTo(1, 249),
p5 = x => Math.pow(x, 5);

const
// powerMap :: Dict Int Int
powerMap = mapFromList(
zip(map(p5, xs), xs)
),
// sumMap :: Dict Int (Int, Int)
sumMap = mapFromList(
bind(
xs,
x => bind(
tail(xs),
y => Tuple(
p5(x) + p5(y),
Tuple(x, y)
)
)
)
);

// mbExample :: Maybe (Int, Int)
const mbExample = find(
tpl => member(fst(tpl) - snd(tpl), sumMap),
bind(
map(x => parseInt(x, 10),
keys(powerMap)
),
p => bind(
takeWhile(
x => x < p,
map(x => parseInt(x, 10),
keys(sumMap)
)
),
s => [Tuple(p, s)]
)
)
);

// showExample :: (Int, Int) -> String
const showExample = tpl => {
const [p, s] = Array.from(tpl);
const [a, b] = Array.from(sumMap[p - s]);
const [c, d] = Array.from(sumMap[s]);
return 'Counter-example found:\n' + intercalate(
'^5 + ',
map(str, [a, b, c, d])
) + '^5 = ' + str(powerMap[p]) + '^5';
};

return maybe(
'No counter-example found',
showExample,
mbExample
);
};

// GENERIC FUNCTIONS ----------------------------------

// Just :: a -> Maybe a
const Just = x => ({
type: 'Maybe',
Nothing: false,
Just: x
});

// Nothing :: Maybe a
const Nothing = () => ({
type: 'Maybe',
Nothing: true,
});

// Tuple (,) :: a -> b -> (a, b)
const Tuple = (a, b) => ({
type: 'Tuple',
'0': a,
'1': b,
length: 2
});

// bind (>>=) :: [a] -> (a -> [b]) -> [b]
const bind = (xs, mf) => [].concat.apply([], xs.map(mf));

// concat :: [[a]] -> [a]
// concat :: [String] -> String
const concat = xs =>
0 < xs.length ? (() => {
const unit = 'string' !== typeof xs ? (
[]
) : '';
return unit.concat.apply(unit, xs);
})() : [];

// enumFromTo :: (Int, Int) -> [Int]
const enumFromTo = (m, n) =>
Array.from({
length: 1 + n - m
}, (_, i) => m + i);

// find :: (a -> Bool) -> [a] -> Maybe a
const find = (p, xs) => {
for (let i = 0, lng = xs.length; i < lng; i++) {
if (p(xs[i])) return Just(xs[i]);
}
return Nothing();
};

// fst :: (a, b) -> a
const fst = tpl => tpl;

// intercalate :: [a] -> [[a]] -> [a]
// intercalate :: String -> [String] -> String
const intercalate = (sep, xs) =>
0 < xs.length && 'string' === typeof sep &&
'string' === typeof xs ? (
xs.join(sep)
) : concat(intersperse(sep, xs));

// intersperse(0, [1,2,3]) -> [1, 0, 2, 0, 3]

// intersperse :: a -> [a] -> [a]
// intersperse :: Char -> String -> String
const intersperse = (sep, xs) => {
const bln = 'string' === typeof xs;
return xs.length > 1 ? (
(bln ? concat : x => x)(
(bln ? (
xs.split('')
) : xs)
.slice(1)
.reduce((a, x) => a.concat([sep, x]), [xs])
)) : xs;
};

// keys :: Dict -> [String]
const keys = Object.keys;

// Returns Infinity over objects without finite length.
// This enables zip and zipWith to choose the shorter
// argument when one is non-finite, like cycle, repeat etc

// length :: [a] -> Int
const length = xs =>
(Array.isArray(xs) || 'string' === typeof xs) ? (
xs.length
) : Infinity;

// map :: (a -> b) -> [a] -> [b]
const map = (f, xs) =>
(Array.isArray(xs) ? (
xs
) : xs.split('')).map(f);

// mapFromList :: [(k, v)] -> Dict
const mapFromList = kvs =>
kvs.reduce(
(a, kv) => {
const k = kv;
return Object.assign(a, {
[
(('string' === typeof k) && k) || JSON.stringify(k)
]: kv
});
}, {}
);

// Default value (v) if m.Nothing, or f(m.Just)

// maybe :: b -> (a -> b) -> Maybe a -> b
const maybe = (v, f, m) =>
m.Nothing ? v : f(m.Just);

// member :: Key -> Dict -> Bool
const member = (k, dct) => k in dct;

// snd :: (a, b) -> b
const snd = tpl => tpl;

// str :: a -> String
const str = x => x.toString();

// tail :: [a] -> [a]
const tail = xs => 0 < xs.length ? xs.slice(1) : [];

// take :: Int -> [a] -> [a]
// take :: Int -> String -> String
const take = (n, xs) =>
'GeneratorFunction' !== xs.constructor.constructor.name ? (
xs.slice(0, n)
) : [].concat.apply([], Array.from({
length: n
}, () => {
const x = xs.next();
return x.done ? [] : [x.value];
}));

// takeWhile :: (a -> Bool) -> [a] -> [a]
// takeWhile :: (Char -> Bool) -> String -> String
const takeWhile = (p, xs) =>
xs.constructor.constructor.name !==
'GeneratorFunction' ? (() => {
const lng = xs.length;
return 0 < lng ? xs.slice(
0,
until(
i => lng === i || !p(xs[i]),
i => 1 + i,
0
)
) : [];
})() : takeWhileGen(p, xs);

// until :: (a -> Bool) -> (a -> a) -> a -> a
const until = (p, f, x) => {
let v = x;
while (!p(v)) v = f(v);
return v;
};

// Use of `take` and `length` here allows for zipping with non-finite
// lists - i.e. generators like cycle, repeat, iterate.

// zip :: [a] -> [b] -> [(a, b)]
const zip = (xs, ys) => {
const lng = Math.min(length(xs), length(ys));
return Infinity !== lng ? (() => {
const bs = take(lng, ys);
return take(lng, xs).map((x, i) => Tuple(x, bs[i]));
})() : zipGen(xs, ys);
};

// MAIN ---
return main();
})();
```
Output:
```Counter-example found:
133^5 + 110^5 + 84^5 + 27^5 = 144^5```

## jq

Works with: jq version 1.4

This version finds all non-decreasing solutions within the specified bounds, using a brute-force but not entirely blind approach.

```# Search for y in 1 .. maxn (inclusive) for a solution to SIGMA (xi ^ 5) = y^5
# and for each solution with x0<=x1<=...<x3, print [x0, x1, x3, x3, y]
#
def sum_of_powers_conjecture(maxn):
def p5: . as \$in | (.*.) | ((.*.) * \$in);
def fifth: log / 5 | exp;

# return the fifth root if . is a power of 5
def integral_fifth_root: fifth | if . == floor then . else false end;

(maxn | p5) as \$uber
| range(1; maxn) as \$x0
| (\$x0 | p5) as \$s0
| if \$s0 < \$uber then range(\$x0; (\$uber - \$s0 | fifth) + 1) as \$x1
| (\$s0 + (\$x1 | p5)) as \$s1
| if \$s1 < \$uber then range(\$x1; (\$uber - \$s1 | fifth) + 1) as \$x2
| (\$s1 + (\$x2 | p5)) as \$s2
| if \$s2 < \$uber then range(\$x2; (\$uber - \$s2 | fifth) + 1) as \$x3
| (\$s2 + (\$x3 | p5)) as \$sumx
| (\$sumx | integral_fifth_root)
| if . then [\$x0,\$x1,\$x2,\$x3,.] else empty end
else empty
end
else empty
end
else empty
end ;```

`sum_of_powers_conjecture(249)`
Output:
```\$ jq -c -n -f Euler_sum_of_powers_conjecture_fifth_root.jq
[27,84,110,133,144]
```

## Julia

```const lim = 250
const pwr = 5
const p = [i^pwr for i in 1:lim]

x = zeros(Int, pwr-1)
y = 0

for a in combinations(1:lim, pwr-1)
b = searchsorted(p, sum(p[a]))
0 < length(b) || continue
x = a
y = b
break
end

if y == 0
println("No solution found for power = ", pwr, " and limit = ", lim, ".")
else
s = [@sprintf("%d^%d", i, pwr) for i in x]
s = join(s, " + ")
println("A solution is ", s, " = ", @sprintf("%d^%d", y, pwr), ".")
end
```
Output:
```A solution is 27^5 + 84^5 + 110^5 + 133^5 = 144^5.
```

## Kotlin

```fun main(args: Array<String>) {
val p5 = LongArray(250){ it.toLong() * it * it * it * it }
var sum: Long
var y: Int
var found = false
loop@ for (x0 in 0 .. 249)
for (x1 in 0 .. x0 - 1)
for (x2 in 0 .. x1 - 1)
for (x3 in 0 .. x2 - 1) {
sum = p5[x0] + p5[x1] + p5[x2] + p5[x3]
y = p5.binarySearch(sum)
if (y >= 0) {
println("\$x0^5 + \$x1^5 + \$x2^5 + \$x3^5 = \$y^5")
found = true
break@loop
}
}
if (!found) println("No solution was found")
}
```
Output:
```133^5 + 110^5 + 84^5 + 27^5 = 144^5
```

## Lua

Brute force but still takes under two seconds with LuaJIT.

```-- Fast table search (only works if table values are in order)
function binarySearch (t, n)
local start, stop, mid = 1, #t
while start < stop do
mid = math.floor((start + stop) / 2)
if n == t[mid] then
return mid
elseif n < t[mid] then
stop = mid - 1
else
start = mid + 1
end
end
return nil
end

-- Test Euler's sum of powers conjecture
function euler (limit)
local pow5, sum = {}
for i = 1, limit do pow5[i] = i^5 end
for x0 = 1, limit do
for x1 = 1, x0 do
for x2 = 1, x1 do
for x3 = 1, x2 do
sum = pow5[x0] + pow5[x1] + pow5[x2] + pow5[x3]
if binarySearch(pow5, sum) then
print(x0 .. "^5 + " .. x1 .. "^5 + " .. x2 .. "^5 + " .. x3 .. "^5 = " .. sum^(1/5) .. "^5")
return true
end
end
end
end
end
return false
end

-- Main procedure
if euler(249) then
print("Time taken: " .. os.clock() .. " seconds")
else
print("Looks like he was right after all...")
end
```
Output:
```133^5 + 110^5 + 84^5 + 27^5 = 144^5
Time taken: 1.247 seconds```

## Mathematica/Wolfram Language

```Sort[FindInstance[
x0^5 + x1^5 + x2^5 + x3^5 == y^5 && x0 > 0 && x1 > 0 && x2 > 0 &&
x3 > 0, {x0, x1, x2, x3, y}, Integers][[1, All, -1]]]
```
Output:
`{27,84,110,133,144}`

## Microsoft Small Basic

```' Euler sum of powers conjecture - 03/07/2015
'find: x1^5+x2^5+x3^5+x4^5=x5^5
'-> x1=27 x2=84 x3=110 x4=133 x5=144
maxn=250
For i=1 to maxn
p5[i]=Math.Power(i,5)
EndFor
For x1=1 to maxn-4
For x2=x1+1 to maxn-3
'TextWindow.WriteLine("x1="+x1+", x2="+x2)
For x3=x2+1 to maxn-2
'TextWindow.WriteLine("x1="+x1+", x2="+x2+", x3="+x3)
For x4=x3+1 to maxn-1
'TextWindow.WriteLine("x1="+x1+", x2="+x2+", x3="+x3+", x4="+x4)
x5=x4+1
valx=p5[x5]
sumx=p5[x1]+p5[x2]+p5[x3]+p5[x4]
While x5<=maxn and valx<=sumx
If valx=sumx Then
TextWindow.WriteLine("Found!")
TextWindow.WriteLine("-> "+x1+"^5+"+x2+"^5+"+x3+"^5+"+x4+"^5="+x5+"^5")
TextWindow.WriteLine("x5^5="+sumx)
Goto EndPgrm
EndIf
x5=x5+1
valx=p5[x5]
EndWhile 'x5
EndFor 'x4
EndFor 'x3
EndFor 'x2
EndFor 'x1
EndPgrm:```
Output:
```Found!
-> 27^5+84^5+110^5+133^5=144^5
x5^5=61917364224 ```

## Modula-2

 This example does not show the output mentioned in the task description on this page (or a page linked to from here). Please ensure that it meets all task requirements and remove this message. Note that phrases in task descriptions such as "print and display" and "print and show" for example, indicate that (reasonable length) output be a part of a language's solution.

```MODULE EulerConjecture;
FROM FormatString IMPORT FormatString;

PROCEDURE Pow5(a : LONGINT) : LONGINT;
BEGIN
RETURN a * a * a * a * a
END Pow5;

VAR
buf : ARRAY[0..63] OF CHAR;
a,b,c,d,e,sum,curr : LONGINT;
BEGIN
FOR a:=0 TO 250 DO
FOR b:=a TO 250 DO
IF b=a THEN CONTINUE END;
FOR c:=b TO 250 DO
IF (c=a) OR (c=b) THEN CONTINUE END;
FOR d:=c TO 250 DO
IF (d=a) OR (d=b) OR (d=c) THEN CONTINUE END;
sum := Pow5(a) + Pow5(b) + Pow5(c) + Pow5(d);
FOR e:=d TO 250 DO
IF (e=a) OR (e=b) OR (e=c) OR (e=d) THEN CONTINUE END;
curr := Pow5(e);
IF (sum#0) AND (sum=curr) THEN
FormatString("%l^5 + %l^5 + %l^5 + %l^5 = %l^5\n", buf, a, b, c, d, e);
WriteString(buf)
ELSIF curr > sum THEN
BREAK
END
END;
END;
END;
END;
END;

WriteString("Done");
WriteLn;
END EulerConjecture.
```

## Nim

Translation of: PureBasic
```# Brute force approach

import times

# assumes an array of non-decreasing positive integers
proc binarySearch(a : openArray[int], target : int) : int =
var left, right, mid : int
left = 0
right = len(a) - 1
while true :
if left > right : return 0  # no match found
mid = (left + right) div 2
if a[mid] < target :
left = mid + 1
elif a[mid] > target :
right = mid - 1
else :
return mid  # match found

var
p5 : array[250, int]
sum = 0
y, t1 : int

let t0 = cpuTime()

for i in 1 .. 249 :
p5[i] = i * i * i * i * i

for x0 in 1 .. 249 :
for x1 in 1 .. x0 - 1 :
for x2 in 1 .. x1 - 1 :
for x3 in 1 .. x2 - 1 :
sum = p5[x0] + p5[x1] + p5[x2] + p5[x3]
y = binarySearch(p5, sum)
if y > 0 :
t1 = int((cputime() - t0) * 1000.0)
echo "Time : ", t1, " milliseconds"
echo  \$x0 & "^5 + " & \$x1 & "^5 + " & \$x2 & "^5 + " & \$x3 & "^5 = " & \$y & "^5"
quit()

if y == 0 :
echo "No solution was found"
```
Output:
```Time : 156 milliseconds
133^5 + 110^5 + 84^5 + 27^5 = 144^5
```

## Oforth

```: eulerSum
| i j k l ip jp kp |
250 loop: i [
i 5 pow ->ip
i 1 + 250 for: j [
j 5 pow ip + ->jp
j 1 + 250 for: k [
k 5 pow jp + ->kp
k 1 + 250 for: l [
kp l 5 pow + 0.2 powf dup asInteger == ifTrue: [ [ i, j, k, l ] println ]
]
]
]
] ;```
Output:
```>eulerSum
[27, 84, 110, 133]
```

## PARI/GP

Naive script:

`forvec(v=vector(4,i,[0,250]), if(ispower(v^5+v^5+v^5+v^5,5,&n), print(n" "v)), 2)`
Output:
`144 [27, 84, 110, 133]`

Naive + caching (`setbinop`):

```{
v2=setbinop((x,y)->[min(x,y),max(x,y),x^5+y^5],[0..250]); \\ sums of two fifth powers
for(i=2,#v2,
for(j=1,i-1,
if(v2[i]<v2[j] && ispower(v2[i]+v2[j],5,&n) && #(v=Set([v2[i],v2[i],v2[j],v2[j]]))==4,
print(n" "v)
)
)
)
}```
Output:
`144 [27, 84, 110, 133]`

## Pascal

Works with: Free Pascal

slightly improved.Reducing calculation time by temporary sum and early break.

```program Pot5Test;
{\$IFDEF FPC} {\$MODE DELPHI}{\$ELSE]{\$APPTYPE CONSOLE}{\$ENDIF}
type
tTest = double;//UInt64;{ On linux 32Bit double is faster than  Uint64 }
var
Pot5 : array[0..255] of tTest;
res,tmpSum : tTest;
x0,x1,x2,x3, y : NativeUint;//= Uint32 or 64 depending on OS xx-Bit
i : byte;
BEGIN
For i := 1 to 255 do
Pot5[i] := (i*i*i*i)*Uint64(i);

For x0 := 1 to 250-3 do
For x1 := x0+1 to 250-2 do
For x2 := x1+1 to 250-1 do
Begin
//set y here only, because pot5 is strong monoton growing,
//therefor the sum is strong monoton growing too.
y := x2+2;// aka x3+1
tmpSum := Pot5[x0]+Pot5[x1]+Pot5[x2];
For x3 := x2+1 to 250 do
Begin
res := tmpSum+Pot5[x3];
while (y< 250) AND (res > Pot5[y]) do
inc(y);
IF y > 250 then BREAK;
if res = Pot5[y] then
writeln(x0,'^5+',x1,'^5+',x2,'^5+',x3,'^5 = ',y,'^5');
end;
end;
END.
```
output
```27^5+84^5+110^5+133^5 = 144^5
real  0m1.091s {Uint64; Linux 32}real  0m0.761s {double; Linux 32}real  0m0.511s{Uint64; Linux 64}
```

## Perl

Brute Force:

```use constant MAX => 250;
my @p5 = (0,map { \$_**5 } 1 .. MAX-1);
my \$s = 0;
my %p5 = map { \$_ => \$s++ } @p5;
for my \$x0 (1..MAX-1) {
for my \$x1 (1..\$x0-1) {
for my \$x2 (1..\$x1-1) {
for my \$x3 (1..\$x2-1) {
my \$sum = \$p5[\$x0] + \$p5[\$x1] + \$p5[\$x2] + \$p5[\$x3];
die "\$x3 \$x2 \$x1 \$x0 \$p5{\$sum}\n" if exists \$p5{\$sum};
}
}
}
}
```
Output:
`27 84 110 133 144`

Adding some optimizations makes it 5x faster with similar output, but obfuscates things.

Translation of: C++
```use constant MAX => 250;
my @p5 = (0,map { \$_**5 } 1 .. MAX-1);
my \$rs = 5;
for my \$x0 (1..MAX-1) {
for my \$x1 (1..\$x0-1) {
for my \$x2 (1..\$x1-1) {
my \$s2 = \$p5[\$x0] + \$p5[\$x1] + \$p5[\$x2];
\$rs-- while \$rs > 0 && \$p5[\$rs] > \$s2;
for (my \$x3 = 1;  \$x3 < \$x2;  \$x3++) {
my \$e30 = (\$x0 + \$x1 + \$x2 + \$x3 - \$rs) % 30;
\$x3 += (30-\$e30) if \$e30;
last if \$x3 >= \$x2;
my \$sum = \$s2 + \$p5[\$x3];
\$rs++ while \$rs < MAX-1 && \$p5[\$rs] < \$sum;
die "\$x3 \$x2 \$x1 \$x0 \$rs\n" if \$p5[\$rs] == \$sum;
}
}
}
}
```

## Phix

Translation of: Python

Around four seconds, not spectacularly fast. My naive brute force was over a minute. This is not where Phix shines.
Quitting when the first is found drops the main loop to 0.7s, so 1.1s in all, vs 4.3s for the full search.
Without the return 0, you just get six permutes (of ordered pairs) for 144.

```with javascript_semantics
constant MAX = 250

constant p5 = new_dict(),
sum2 = new_dict()

atom t0 = time()
for i=1 to MAX do
atom i5 = power(i,5)
setd(i5,i,p5)
for j=1 to i-1 do
atom j5 = power(j,5)
setd(j5+i5,{j,i},sum2)
end for
end for

?time()-t0

function forsum2(object s, object data, object p)
if p<=s then return 0 end if
integer k = getd_index(p-s,sum2)
if k!=NULL then
?getd(p,p5)&data&getd_by_index(k,sum2)
return 0 -- (show one solution per p)
end if
return 1
end function

function forp5(object key, object /*data*/, object /*user_data*/)
traverse_dict(forsum2,key,sum2)
return 1
end function

traverse_dict(forp5,0,p5)

?time()-t0
```
Output:
```0.421
{144,27,84,110,133}
4.312
```

## PHP

Translation of: Python
```<?php

function eulers_sum_of_powers () {
\$max_n = 250;
\$pow_5 = array();
\$pow_5_to_n = array();
for (\$p = 1; \$p <= \$max_n; \$p ++) {
\$pow5 = pow(\$p, 5);
\$pow_5 [\$p] = \$pow5;
\$pow_5_to_n[\$pow5] = \$p;
}
foreach (\$pow_5 as \$n_0 => \$p_0) {
foreach (\$pow_5 as \$n_1 => \$p_1) {
if (\$n_0 < \$n_1) continue;
foreach (\$pow_5 as \$n_2 => \$p_2) {
if (\$n_1 < \$n_2) continue;
foreach (\$pow_5 as \$n_3 => \$p_3) {
if (\$n_2 < \$n_3) continue;
\$pow_5_sum = \$p_0 + \$p_1 + \$p_2 + \$p_3;
if (isset(\$pow_5_to_n[\$pow_5_sum])) {
return array(\$n_0, \$n_1, \$n_2, \$n_3, \$pow_5_to_n[\$pow_5_sum]);
}
}
}
}
}
}

list(\$n_0, \$n_1, \$n_2, \$n_3, \$y) = eulers_sum_of_powers();

echo "\$n_0^5 + \$n_1^5 + \$n_2^5 + \$n_3^5 = \$y^5";

?>
```
Output:
`133^5 + 110^5 + 84^5 + 27^5 = 144^5`

## Picat

```import sat.
main =>
X = new_list(5), X :: 1..150, decreasing_strict(X),
X**5 #= sum([X[I]**5 : I in 2..5]),
solve(X), printf("%d**5 = %d**5 + %d**5 + %d**5 + %d**5", X, X, X, X, X).```
Output:
`144**5 = 133**5 + 110**5 + 84**5 + 27**5`
`CPU time 6.626 seconds`

## PicoLisp

```(off P)
(off S)

(for I 250
(idx
'P
(list (setq @@ (** I 5)) I)
T )
(for (J I (>= 250 J) (inc J))
(idx
'S
(list (+ @@ (** J 5)) (list I J))
T ) ) )
(println
(catch 'found
(for A (idx 'P)
(for B (idx 'S)
(T (<= (car A) (car B)))
(and
(lup S (- (car A) (car B)))
(throw 'found
(conc
(cadr (lup S (- (car A) (car B))))
(cdr (lup P (car A))) ) ) ) ) ) ) )```
Output:
`(27 84 110 133 144)`

## PowerShell

Brute Force Search
This is a slow algorithm, so attempts have been made to speed it up, including pre-computing the powers, using an ArrayList for them, and using [int] to cast the 5th root rather than use truncate.

```# EULER.PS1
\$max = 250

\$powers =  New-Object System.Collections.ArrayList
for (\$i = 0; \$i -lt \$max; \$i++) {
}

for (\$x0 = 1; \$x0 -lt \$max; \$x0++) {
for (\$x1 = 1; \$x1 -lt \$x0; \$x1++) {
for (\$x2 = 1; \$x2 -lt \$x1; \$x2++) {
for (\$x3 = 1; \$x3 -lt \$x2; \$x3++) {
\$sum = \$powers[\$x0] + \$powers[\$x1] + \$powers[\$x2] + \$powers[\$x3]
\$S1 = [int][Math]::pow(\$sum,0.2)

if (\$sum -eq \$powers[\$S1]) {
Write-host "\$x0^5 + \$x1^5 + \$x2^5 + \$x3^5 = \$S1^5"
return
}
}
}
}
}
```
Output:
```PS > measure-command { .\euler.ps1 | out-default }
133^5 + 110^5 + 84^5 + 27^5 = 144^5

Days              : 0
Hours             : 0
Minutes           : 0
Seconds           : 31
Milliseconds      : 608
Ticks             : 316082251
TotalDays         : 0.000365835938657407```

## Prolog

```makepowers :-
retractall(pow5(_, _)),
between(1, 249, X),
Y is X * X * X * X * X,
assert(pow5(X, Y)),
fail.
makepowers.

within(A, Bx, N) :-  % like between but with an exclusive upper bound
succ(B, Bx),
between(A, B, N).

solution(X0, X1, X2, X3, Y) :-
makepowers,
within(4, 250, X0), pow5(X0, X0_5th),
within(3, X0,  X1), pow5(X1, X1_5th),
within(2, X1,  X2), pow5(X2, X2_5th),
within(1, X2,  X3), pow5(X3, X3_5th),
Y_5th is X0_5th + X1_5th + X2_5th + X3_5th,
pow5(Y, Y_5th).
```
Output:
```?- solution(X0,X1,X2,X3,Y).
X0 = 133,
X1 = 110,
X2 = 84,
X3 = 27,
Y = 144 .
```

## PureBasic

```EnableExplicit

; assumes an array of non-decreasing positive integers
Procedure.q BinarySearch(Array a.q(1), Target.q)
Protected l = 0, r = ArraySize(a()), m
Repeat
If l > r : ProcedureReturn 0 : EndIf; no match found
m = (l + r) / 2
If a(m) < target
l = m + 1
ElseIf a(m) > target
r = m - 1
Else
ProcedureReturn m ; match found
EndIf
ForEver
EndProcedure

Define i, x0, x1, x2, x3, y
Define.q sum
Define Dim p5.q(249)

For i = 1 To 249
p5(i) = i * i * i * i * i
Next

If OpenConsole()
For x0 = 1 To 249
For x1 = 1 To x0 - 1
For x2 = 1 To x1 - 1
For x3 = 1 To x2 - 1
sum = p5(x0) + p5(x1) + p5(x2) + p5(x3)
y = BinarySearch(p5(), sum)
If y > 0
PrintN(Str(x0) + "^5 + " + Str(x1) + "^5 + " + Str(x2) + "^5 + " + Str(x3) + "^5 = " + Str(y) + "^5")
Goto finish
EndIf
Next x3
Next x2
Next x1
Next x0

PrintN("No solution was found")
finish:
PrintN("")
PrintN("Press any key to close the console")
Repeat: Delay(10) : Until Inkey() <> ""
CloseConsole()
EndIf
```
Output:
```133^5 + 110^5 + 84^5 + 27^5 = 144^5
```

## Python

### Procedural

```def eulers_sum_of_powers():
max_n = 250
pow_5 = [n**5 for n in range(max_n)]
pow5_to_n = {n**5: n for n in range(max_n)}
for x0 in range(1, max_n):
for x1 in range(1, x0):
for x2 in range(1, x1):
for x3 in range(1, x2):
pow_5_sum = sum(pow_5[i] for i in (x0, x1, x2, x3))
if pow_5_sum in pow5_to_n:
y = pow5_to_n[pow_5_sum]
return (x0, x1, x2, x3, y)

print("%i**5 + %i**5 + %i**5 + %i**5 == %i**5" % eulers_sum_of_powers())
```
Output:
`133**5 + 110**5 + 84**5 + 27**5 == 144**5`

The above can be written as:

Works with: Python version 2.6+
```from itertools import combinations

def eulers_sum_of_powers():
max_n = 250
pow_5 = [n**5 for n in range(max_n)]
pow5_to_n = {n**5: n for n in range(max_n)}
for x0, x1, x2, x3 in combinations(range(1, max_n), 4):
pow_5_sum = sum(pow_5[i] for i in (x0, x1, x2, x3))
if pow_5_sum in pow5_to_n:
y = pow5_to_n[pow_5_sum]
return (x0, x1, x2, x3, y)

print("%i**5 + %i**5 + %i**5 + %i**5 == %i**5" % eulers_sum_of_powers())
```
Output:
`27**5 + 84**5 + 110**5 + 133**5 == 144**5`

It's much faster to cache and look up sums of two fifth powers, due to the small allowed range:

```MAX = 250
p5, sum2 = {}, {}

for i in range(1, MAX):
p5[i**5] = i
for j in range(i, MAX):
sum2[i**5 + j**5] = (i, j)

sk = sorted(sum2.keys())
for p in sorted(p5.keys()):
for s in sk:
if p <= s: break
if p - s in sum2:
print(p5[p], sum2[s] + sum2[p-s])
exit()
```
Output:
`144 (27, 84, 110, 133)`

### Composition of pure functions

Works with: Python version 3.7
```'''Euler's sum of powers conjecture'''

from itertools import (chain, takewhile)

# main :: IO ()
def main():
'''Search for counter-example'''

xs = enumFromTo(1)(249)

powerMap = {x**5: x for x in xs}
sumMap = {
x**5 + y**5: (x, y)
for x in xs[1:]
for y in xs if x > y
}

# isExample :: (Int, Int) -> Bool
def isExample(ps):
p, s = ps
return p - s in sumMap

# display :: (Int, Int) -> String
def display(ps):
p, s = ps
a, b = sumMap[p - s]
c, d = sumMap[s]
return '^5 + '.join([str(n) for n in [a, b, c, d]]) + (
'^5 = ' + str(powerMap[p]) + '^5'
)

print(__doc__ + ' – counter-example:\n')
print(
maybe('No counter-example found.')(display)(
find(isExample)(
bind(powerMap.keys())(
lambda p: bind(
takewhile(
lambda x: p > x,
sumMap.keys()
)
)(lambda s: [(p, s)])
)
)
)
)

# ----------------------- GENERIC ------------------------

# Just :: a -> Maybe a
def Just(x):
'''Constructor for an inhabited Maybe (option type) value.
Wrapper containing the result of a computation.
'''
return {'type': 'Maybe', 'Nothing': False, 'Just': x}

# Nothing :: () -> Maybe a
def Nothing():
'''Constructor for an empty Maybe (option type) value.
Empty wrapper returned where a computation is not possible.
'''
return {'type': 'Maybe', 'Nothing': True}

# bind (>>=) :: [a] -> (a -> [b]) -> [b]
def bind(xs):
Two computations sequentially composed,
with any value produced by the first
passed as an argument to the second.
'''
def go(f):
return chain.from_iterable(map(f, xs))
return go

# enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
'''Integer enumeration from m to n.'''
return lambda n: range(m, 1 + n)

# find :: (a -> Bool) -> [a] -> Maybe a
def find(p):
'''Just the first element in the list that matches p,
or Nothing if no elements match.
'''
def go(xs):
try:
return Just(next(x for x in xs if p(x)))
except StopIteration:
return Nothing()
return go

# maybe :: b -> (a -> b) -> Maybe a -> b
def maybe(v):
'''Either the default value v, if m is Nothing,
or the application of f to x,
where m is Just(x).
'''
return lambda f: lambda m: v if (
None is m or m.get('Nothing')
) else f(m.get('Just'))

# MAIN ---
if __name__ == '__main__':
main()
```
Output:
```Euler's sum of powers conjecture – counter-example:

133^5 + 110^5 + 84^5 + 27^5 = 144^5```

## QL SuperBASIC

This program enhances a modular brute-force search, posted on fidonet in the 1980s, via number theoretic enhancements as used by the program for ZX Spectrum Basic, but without the early decrement of the RHS in the control framework due to being backward compatible with the lack of floating point in Sinclair ZX80 BASIC (whereby the latter is truly 'zeroeth' generation). To emulate running on a ZX80 (needing a 16K RAM pack & MODifications, some given below) that completes the task sooner than the program for the OEM Spectrum, it relies entirely on integer functions, their upper limit being 2^15- 1 in ZX80 BASIC as well. Thus, the "slide rule" calculation of each percentage on the Spectrum is replaced by that of ones' digits across "abaci" of relatively prime bases Pi. Given that each Pi is to be <= 2^7 for said limit's sake, it takes six prime numbers or powers thereof to serve as bases of such a mixed-base number system, since it is necessary that ΠPi > 249^5 for unambiguous representation (as character strings). On ZX80s there are at most 64 consecutive printable characters (in the inverse video block: t%=48 thus becomes T=128). Just seven bases Pi <= 2^6 will be needed when the difference between 64 & a base is expressible as a four-bit offset, by which one must 'multiply' (since Z80s lack MUL) in the reduction step of the optimal assembly algorithm for MOD Pi. Such bases are: 49, 53, 55, 57, 59, 61, 64. In disproving Euler's conjecture, the program demonstrates that using 60 bits of integer precision in 1966 was 2-fold overkill, or even more so in terms of overhead cost vis-a-vis contemporaneous computers less sophisticated than a CDC 6600.

```
1 CLS
2 DIM i%(255,6) : DIM a%(6) : DIM c%(6)
3 DIM v%(255,6) : DIM b%(6) : DIM d%(29)
4 RESTORE 137
6      FOR m=0 TO 6
8    FOR j=1 TO 255
11 LET i%(j,m)=j MOD t%
12 LET v%(j,m)=(i%(j,m) * i%(j,m))MOD t%
14 LET v%(j,m)=(v%(j,m) * v%(j,m))MOD t%
15 LET v%(j,m)=(v%(j,m) * i%(j,m))MOD t%
17  END FOR j : END FOR m
21 FOR j=10 TO 29: d%(j)=210+ j
24 FOR j=0 TO 9: d%(j)=240+ j
25 LET t%=48
30      FOR w=6 TO 246 STEP 3
33     LET o=w
42     FOR x=4 TO 248 STEP 2
44    IF o<x THEN o=x
46    FOR m=1 TO 6: a%(m)=i%((v%(w,m)+v%(x,m)),m)
50    FOR y=10 TO 245 STEP 5
54   IF o<y THEN o=y
56   FOR m=1 TO 6: b%(m)=i%((a%(m)+v%(y,m)),m)
57   FOR z=14 TO 245 STEP 7
59  IF o<z THEN o=z
60  FOR m=1 TO 6: c%(m)=i%((b%(m)+v%(z,m)),m)
65  LET s\$="" : FOR m=1 TO 6: s\$=s\$&CHR\$(c%(m)+t%)
70  LET o=o+1 : j=d%(i%((i%(w,0)+i%(x,0)+i%(y,0)+i%(z,0)),0))
75  IF j<o THEN NEXT z
80  FOR k=j TO o STEP -30
85 LET e\$="" : FOR m=1 TO 6: e\$=e\$&CHR\$(v%(k,m)+t%)
90 IF s\$=e\$ THEN PRINT w,x,y,z,k,s\$,e\$: STOP
95  END FOR k : END FOR z : END FOR y : END FOR x : END FOR w
137 DATA 30,97,113,121,125,127,128
```
Output:
```

27      84     110     133     144   bT`íα0   bT`íα0

```

## Racket

Translation of: C++
```#lang racket
(define MAX 250)
(define pow5 (make-vector MAX))
(for ([i (in-range 1 MAX)])
(vector-set! pow5 i (expt i 5)))
(define pow5s (list->set (vector->list pow5)))
(let/ec break
(for* ([x0 (in-range 1 MAX)]
[x1 (in-range 1 x0)]
[x2 (in-range 1 x1)]
[x3 (in-range 1 x2)])
(define sum (+ (vector-ref pow5 x0)
(vector-ref pow5 x1)
(vector-ref pow5 x2)
(vector-ref pow5 x3)))
(when (set-member? pow5s sum)
(displayln (list x0 x1 x2 x3 (inexact->exact (round (expt sum 1/5)))))
(break))))
```
Output:
```(133 110 84 27 144)
```

## Raku

(formerly Perl 6)

Translation of: Python
```constant MAX = 250;

my  %po5{Int};
my %sum2{Int};

for 1..MAX -> \i {
%po5{i⁵} = i;
for 1..MAX -> \j {
%sum2{i⁵ + j⁵} = i, j;
}
}

%po5.keys.sort.race.map: -> \p {
for %sum2.keys.sort -> \s {
if p > s and %sum2{p - s} {
say ((sort |%sum2{s},|%sum2{p-s}) X~ '⁵').join(' + '), " = %po5{p}", "⁵" and exit
}
}
}
```
Output:
`27⁵ + 84⁵ + 110⁵ + 133⁵ = 144⁵`

## REXX

### fast computation

Programming note:   the 3rd argument can be specified which causes an attempt to find   N   solutions.
The starting and ending (low and high) values can also be specified   (to limit or expand the search range).
If any of the arguments are omitted, they default to the Rosetta Code task's specifications.

The method used is:

•   precompute all powers of five   (within the confines of allowed integers)
•   precompute all (positive) differences between two applicable 5th powers
•   see if any of the sums of any three   5th   powers are equal to any of those (above) differences
•           {thanks to the real nifty idea   (↑↑↑)   from user ID   G. Brougnard}
•   see if the sum of any four   5th   powers is equal to   any   5th power
•           (this is needed as the fourth number   d   isn't known yet).
•   {all of the above utilizes REXX's   sparse stemmed array hashing   which eliminates the need for sorting.}

By implementing (user ID)   G. Brougnard's   idea of   differences of two 5th powers,
the time used for computation was reduced by over a factor of   seventy.

In essence, the new formula being solved is:       a5   +   b5   +   c5     ==     x5   ─   d5

which lends itself to algorithm optimization by (only) having to:

•   [the right side of the above equation]   pre-compute all possible differences between any two applicable
integer powers of five   (there are   30,135   unique differences)
•   [the   left side of the above equation]   sum any applicable three integer powers of five
•   [the   ==   part of the above equation]   see if any of the above left─side sums match any of the   ≈30k   right─side differences
```/*REXX program finds unique positive integers for ────────── aⁿ+bⁿ+cⁿ+dⁿ==xⁿ  where n=5 */
parse arg L H N .                                /*get optional  LOW, HIGH,  #solutions.*/
if L=='' | L==","  then L=   0  + 1              /*Not specified?  Then use the default.*/
if H=='' | H==","  then H= 250  - 1              /* "      "         "   "   "     "    */
if N=='' | N==","  then N=   1                   /* "      "         "   "   "     "    */
w= length(H)                                     /*W:  used for display aligned numbers.*/
say center(' 'subword(sourceLine(1), 9, 3)" ", 70 +5*w, '─')  /*show title from 1st line*/
numeric digits 1000                              /*be able to handle the next expression*/
numeric digits max(9, length(3*H**5) )           /* "   "   "    "   3* [H to 5th power]*/
bH= H - 2;                 cH= H - 1             /*calculate the upper  DO  loop limits.*/
!.= 0                                            /* [↓]  define values of  5th  powers. */
do pow=1  for H;    @.pow= pow**5;     _= @.pow;        !._= 1;          \$._= pow
end   /*pow*/
?.= !.
do    j=4   for H-3                       /*use the range of:   four  to   cH.   */
do k=j+1  to H;  _= @.k - @.j;  ?._= 1 /*compute the   xⁿ - dⁿ    differences.*/
end   /*k*/                            /* [↑]  diff. is always positive as k>j*/
end      /*j*/                            /*define [↑]    5th  power differences.*/
#= 0                                             /*#:  is the number of solutions found.*/   /* [↓]  for N=∞ solutions.*/
do       a=L    to H-3                       /*traipse through possible  A  values. */   /*◄──done       246 times.*/
do     b=a+1  to bH;      s1= @.a + @.b    /*   "       "        "     B    "     */   /*◄──done    30,381 times.*/
do   c=b+1  to cH;      s2= s1  + @.c    /*   "       "        "     C    "     */   /*◄──done 2,511,496 times.*/
if ?.s2  then do d=c+1  to H;  s= s2+@.d /*find the appropriate solution.       */
if !.s  then call show     /*Is it a solution?   Then display it. */
end   /*d*/                /* [↑]    !.S  is a boolean.           */
end                 /*c*/
end                   /*b*/
end                     /*a*/

if #==0  then say "Didn't find a solution.";           exit 0
/*──────────────────────────────────────────────────────────────────────────────────────*/
show: _= left('', 5);     #= # + 1               /*_:  used as a spacer; bump # counter.*/
say _  'solution'   right(#, length(N))":"  _  'a='right(a, w)   _  "b="right(b, w),
_  'c='right(c, w)     _    "d="right(d, w)     _    'x='right(\$.s, w+1)
if #<N  then return                        /*return, keep searching for more sols.*/
exit #                                     /*stick a fork in it,  we're all done. */
```
output   when using the default input:
```──────────────────────────── aⁿ+bⁿ+cⁿ+dⁿ==xⁿ  where ⁿ=5 ─────────────────────────────
solution 1:       a= 27       b= 84       c=110       d=133       x= 144
```
output   when using the input of:     1   4000   999
```─────────────────────────────── aⁿ+bⁿ+cⁿ+dⁿ==xⁿ  where ⁿ=5 ───────────────────────────────
solution   1:       a=  27       b=  84       c= 110       d= 133       x=  144
solution   2:       a=  54       b= 168       c= 220       d= 266       x=  288
solution   3:       a=  81       b= 252       c= 330       d= 399       x=  432
solution   4:       a= 108       b= 336       c= 440       d= 532       x=  576
solution   5:       a= 135       b= 420       c= 550       d= 665       x=  720
solution   6:       a= 162       b= 504       c= 660       d= 798       x=  864
solution   7:       a= 189       b= 588       c= 770       d= 931       x= 1008
solution   8:       a= 216       b= 672       c= 880       d=1064       x= 1152
solution   9:       a= 243       b= 756       c= 990       d=1197       x= 1296
solution  10:       a= 270       b= 840       c=1100       d=1330       x= 1440
solution  11:       a= 297       b= 924       c=1210       d=1463       x= 1584
solution  12:       a= 324       b=1008       c=1320       d=1596       x= 1728
solution  13:       a= 351       b=1092       c=1430       d=1729       x= 1872
solution  14:       a= 378       b=1176       c=1540       d=1862       x= 2016
solution  15:       a= 405       b=1260       c=1650       d=1995       x= 2160
solution  16:       a= 432       b=1344       c=1760       d=2128       x= 2304
solution  17:       a= 459       b=1428       c=1870       d=2261       x= 2448
solution  18:       a= 486       b=1512       c=1980       d=2394       x= 2592
solution  19:       a= 513       b=1596       c=2090       d=2527       x= 2736
solution  20:       a= 540       b=1680       c=2200       d=2660       x= 2880
solution  21:       a= 567       b=1764       c=2310       d=2793       x= 3024
solution  22:       a= 594       b=1848       c=2420       d=2926       x= 3168
solution  23:       a= 621       b=1932       c=2530       d=3059       x= 3312
solution  24:       a= 648       b=2016       c=2640       d=3192       x= 3456
solution  25:       a= 675       b=2100       c=2750       d=3325       x= 3600
solution  26:       a= 702       b=2184       c=2860       d=3458       x= 3744
solution  27:       a= 729       b=2268       c=2970       d=3591       x= 3888
```

### lightning-fast computation

Programming note:   it can be observed from the 1st REXX program's output   (2nd example)   that all of the index solutions are just multiples of the 1st known set:

```     for A,   a multiple of   27
for B,   a multiple of   84
for C,   a multiple of  110
for D,   a multiple of  133
for X,   a multiple of  144
```

where "a multiple" is some positive integer.

Intrepid and resourceful Rosetta Code userid   Pat Garrett   has found that in a research paper that   Jim Frye   found another solution:

555   +   31835   +   289695   +   852825     =     853595

The paper can be seen at:   A variety of Euler's conjecture.

So this 2nd known set was added to the program below.

So, index solutions are also multiples of the 2nd known set:

```     for A,   a multiple of      55
for B,   a multiple of   3,183
for C,   a multiple of  28,969
for D,   a multiple of  85,282
for X,   a multiple of  85,359
```

Execution time for computing/displaying/writing   200   numbers on a slow PC is about   1/16   second.

Execution time on a slow PC for computing (alone)   for   6,000   numbers is less than   one   second.

Execution time on a fast PC for computing (alone)   for   23,686   numbers is exactly   1.00   seconds.

```/*REXX program shows unique positive integers for ────────── aⁿ+bⁿ+cⁿ+dⁿ==xⁿ  where n=5 */
numeric digits 1000                              /*ensure enough decimal digs for powers*/
parse arg N oFID .                               /*obtain optional arguments from the CL*/
if N=='' | N==","  then N= 200                   /*Not specified?  Then use the default.*/
if oFID==''|oFID==","  then oFID= 'EULERSUM.OUT' /* "      "         "   "   "     "    */
tell= N>=0                                       /*if N is ≥ 0, show output to terminal.*/
N= abs(N)                                        /*use the absolute value of  N.        */
a.1=  27  ;   a.2=    55   /*the   A   values for the two sets.   */
b.1=  84  ;   b.2=  3183   /* "    B      "    "   "   "    "     */
c.1= 110  ;   c.2= 28969   /* "    C      "    "   "   "    "     */
d.1= 133  ;   d.2= 85282   /* "    D      "    "   "   "    "     */
x.1= 144  ;   x.2= 85359   /* "    X      "    "   "   "    "     */
w= length( commas(N * x.2) )                     /*W:  used to align displayed numbers. */
\$= center(' 'subword( sourceLine(1), 9, 3)" ", 70 +5*w, '─')           /*create a title.*/
call show                                        /*show a title  (from 1st line of pgm).*/
oo= 1;   tt= 1                                   /*a counter for the  A.1  &  A.2  sets.*/
#= 0                                             /*count of number of solutions so far. */
do j=1  until #>N                         /*step through the possible solutions. */
one= a.1 * oo                             /*calculate the 1st set's  A.1  value. */
two= a.2 * tt                             /*    "      "  2nd   "    A.2    "    */
use= min(one, two)                        /*pick which "set" that is to be used. */
#= # + 1                                  /*bump counter for number of solutions.*/
if one==use  then do;      mult=oo;      oo= oo + 1;      which= 1;      end
if two==use  then do;      mult=tt;      tt= tt + 1;      which= 2;      end
\$= pad  'solution'  right(#,length(N))":  "  'a='right( commas(a.which * mult), w),
pad     'b='right( commas(b.which * mult), w),
pad     'c='right( commas(c.which * mult), w),
pad     'd='right( commas(d.which * mult), w),
pad     'x='right( commas(x.which * mult), w)
call show                                 /*write; maybe show output to terminal.*/
res= (x.which * mult) **5                 /*compute the sum of the  right  side. */
sum= (a.which * mult) **5   +   ,         /*   "     "   "   "  "    left    "   */
(b.which * mult) **5   +   ,
(c.which * mult) **5   +   ,
(d.which * mult) **5
if sum==res  then iterate                 /*All is kosher?   Then keep truckin'. */
\$= "***error*** the left side sum   doesn't   equal the right side result (X**5)."
tell=1;  call show;  exit 13              /*force telling of error to terminal.  */
end   /*j*/
tell=1;                                                                          call show
\$= pad ' Showed '   commas(N)   " solutions,  output written to file: " oFID;    call show
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg _;  do jc=length(_)-3  to 1  by -3; _=insert(',', _, jc); end;  return _
show:   if tell  then say \$;  call lineout oFID, \$;  \$=;  return  /*show and/or write it*/
```
output   when using the default input of:     200

(Shown at three-quarter size.)

```────────────────────────────────────────────── aⁿ+bⁿ+cⁿ+dⁿ==xⁿ  where n=5 ──────────────────────────────────────────────
solution   1:   a=        27       b=        84       c=       110       d=       133       x=       144
solution   2:   a=        54       b=       168       c=       220       d=       266       x=       288
solution   3:   a=        55       b=     3,183       c=    28,969       d=    85,282       x=    85,359
solution   4:   a=        81       b=       252       c=       330       d=       399       x=       432
solution   5:   a=       108       b=       336       c=       440       d=       532       x=       576
solution   6:   a=       110       b=     6,366       c=    57,938       d=   170,564       x=   170,718
solution   7:   a=       135       b=       420       c=       550       d=       665       x=       720
solution   8:   a=       162       b=       504       c=       660       d=       798       x=       864
solution   9:   a=       165       b=     9,549       c=    86,907       d=   255,846       x=   256,077
solution  10:   a=       189       b=       588       c=       770       d=       931       x=     1,008
solution  11:   a=       216       b=       672       c=       880       d=     1,064       x=     1,152
solution  12:   a=       220       b=    12,732       c=   115,876       d=   341,128       x=   341,436
solution  13:   a=       243       b=       756       c=       990       d=     1,197       x=     1,296
solution  14:   a=       270       b=       840       c=     1,100       d=     1,330       x=     1,440
solution  15:   a=       275       b=    15,915       c=   144,845       d=   426,410       x=   426,795
solution  16:   a=       297       b=       924       c=     1,210       d=     1,463       x=     1,584
solution  17:   a=       324       b=     1,008       c=     1,320       d=     1,596       x=     1,728
solution  18:   a=       330       b=    19,098       c=   173,814       d=   511,692       x=   512,154
solution  19:   a=       351       b=     1,092       c=     1,430       d=     1,729       x=     1,872
solution  20:   a=       378       b=     1,176       c=     1,540       d=     1,862       x=     2,016
solution  21:   a=       385       b=    22,281       c=   202,783       d=   596,974       x=   597,513
solution  22:   a=       405       b=     1,260       c=     1,650       d=     1,995       x=     2,160
solution  23:   a=       432       b=     1,344       c=     1,760       d=     2,128       x=     2,304
solution  24:   a=       440       b=    25,464       c=   231,752       d=   682,256       x=   682,872
solution  25:   a=       459       b=     1,428       c=     1,870       d=     2,261       x=     2,448
solution  26:   a=       486       b=     1,512       c=     1,980       d=     2,394       x=     2,592
solution  27:   a=       495       b=    28,647       c=   260,721       d=   767,538       x=   768,231
solution  28:   a=       513       b=     1,596       c=     2,090       d=     2,527       x=     2,736
solution  29:   a=       540       b=     1,680       c=     2,200       d=     2,660       x=     2,880
solution  30:   a=       550       b=    31,830       c=   289,690       d=   852,820       x=   853,590
solution  31:   a=       567       b=     1,764       c=     2,310       d=     2,793       x=     3,024
solution  32:   a=       594       b=     1,848       c=     2,420       d=     2,926       x=     3,168
solution  33:   a=       605       b=    35,013       c=   318,659       d=   938,102       x=   938,949
solution  34:   a=       621       b=     1,932       c=     2,530       d=     3,059       x=     3,312
solution  35:   a=       648       b=     2,016       c=     2,640       d=     3,192       x=     3,456
solution  36:   a=       660       b=    38,196       c=   347,628       d= 1,023,384       x= 1,024,308
solution  37:   a=       675       b=     2,100       c=     2,750       d=     3,325       x=     3,600
solution  38:   a=       702       b=     2,184       c=     2,860       d=     3,458       x=     3,744
solution  39:   a=       715       b=    41,379       c=   376,597       d= 1,108,666       x= 1,109,667
solution  40:   a=       729       b=     2,268       c=     2,970       d=     3,591       x=     3,888
solution  41:   a=       756       b=     2,352       c=     3,080       d=     3,724       x=     4,032
solution  42:   a=       770       b=    44,562       c=   405,566       d= 1,193,948       x= 1,195,026
solution  43:   a=       783       b=     2,436       c=     3,190       d=     3,857       x=     4,176
solution  44:   a=       810       b=     2,520       c=     3,300       d=     3,990       x=     4,320
solution  45:   a=       825       b=    47,745       c=   434,535       d= 1,279,230       x= 1,280,385
solution  46:   a=       837       b=     2,604       c=     3,410       d=     4,123       x=     4,464
solution  47:   a=       864       b=     2,688       c=     3,520       d=     4,256       x=     4,608
solution  48:   a=       880       b=    50,928       c=   463,504       d= 1,364,512       x= 1,365,744
solution  49:   a=       891       b=     2,772       c=     3,630       d=     4,389       x=     4,752
solution  50:   a=       918       b=     2,856       c=     3,740       d=     4,522       x=     4,896
solution  51:   a=       935       b=    54,111       c=   492,473       d= 1,449,794       x= 1,451,103
solution  52:   a=       945       b=     2,940       c=     3,850       d=     4,655       x=     5,040
solution  53:   a=       972       b=     3,024       c=     3,960       d=     4,788       x=     5,184
solution  54:   a=       990       b=    57,294       c=   521,442       d= 1,535,076       x= 1,536,462
solution  55:   a=       999       b=     3,108       c=     4,070       d=     4,921       x=     5,328
solution  56:   a=     1,026       b=     3,192       c=     4,180       d=     5,054       x=     5,472
solution  57:   a=     1,045       b=    60,477       c=   550,411       d= 1,620,358       x= 1,621,821
solution  58:   a=     1,053       b=     3,276       c=     4,290       d=     5,187       x=     5,616
solution  59:   a=     1,080       b=     3,360       c=     4,400       d=     5,320       x=     5,760
solution  60:   a=     1,100       b=    63,660       c=   579,380       d= 1,705,640       x= 1,707,180
solution  61:   a=     1,107       b=     3,444       c=     4,510       d=     5,453       x=     5,904
solution  62:   a=     1,134       b=     3,528       c=     4,620       d=     5,586       x=     6,048
solution  63:   a=     1,155       b=    66,843       c=   608,349       d= 1,790,922       x= 1,792,539
solution  64:   a=     1,161       b=     3,612       c=     4,730       d=     5,719       x=     6,192
solution  65:   a=     1,188       b=     3,696       c=     4,840       d=     5,852       x=     6,336
solution  66:   a=     1,210       b=    70,026       c=   637,318       d= 1,876,204       x= 1,877,898
solution  67:   a=     1,215       b=     3,780       c=     4,950       d=     5,985       x=     6,480
solution  68:   a=     1,242       b=     3,864       c=     5,060       d=     6,118       x=     6,624
solution  69:   a=     1,265       b=    73,209       c=   666,287       d= 1,961,486       x= 1,963,257
solution  70:   a=     1,269       b=     3,948       c=     5,170       d=     6,251       x=     6,768
solution  71:   a=     1,296       b=     4,032       c=     5,280       d=     6,384       x=     6,912
solution  72:   a=     1,320       b=    76,392       c=   695,256       d= 2,046,768       x= 2,048,616
solution  73:   a=     1,323       b=     4,116       c=     5,390       d=     6,517       x=     7,056
solution  74:   a=     1,350       b=     4,200       c=     5,500       d=     6,650       x=     7,200
solution  75:   a=     1,375       b=    79,575       c=   724,225       d= 2,132,050       x= 2,133,975
solution  76:   a=     1,377       b=     4,284       c=     5,610       d=     6,783       x=     7,344
solution  77:   a=     1,404       b=     4,368       c=     5,720       d=     6,916       x=     7,488
solution  78:   a=     1,430       b=    82,758       c=   753,194       d= 2,217,332       x= 2,219,334
solution  79:   a=     1,431       b=     4,452       c=     5,830       d=     7,049       x=     7,632
solution  80:   a=     1,458       b=     4,536       c=     5,940       d=     7,182       x=     7,776
solution  81:   a=     1,485       b=    85,941       c=   782,163       d= 2,302,614       x= 2,304,693
solution  82:   a=     1,512       b=     4,704       c=     6,160       d=     7,448       x=     8,064
solution  83:   a=     1,539       b=     4,788       c=     6,270       d=     7,581       x=     8,208
solution  84:   a=     1,540       b=    89,124       c=   811,132       d= 2,387,896       x= 2,390,052
solution  85:   a=     1,566       b=     4,872       c=     6,380       d=     7,714       x=     8,352
solution  86:   a=     1,593       b=     4,956       c=     6,490       d=     7,847       x=     8,496
solution  87:   a=     1,595       b=    92,307       c=   840,101       d= 2,473,178       x= 2,475,411
solution  88:   a=     1,620       b=     5,040       c=     6,600       d=     7,980       x=     8,640
solution  89:   a=     1,647       b=     5,124       c=     6,710       d=     8,113       x=     8,784
solution  90:   a=     1,650       b=    95,490       c=   869,070       d= 2,558,460       x= 2,560,770
solution  91:   a=     1,674       b=     5,208       c=     6,820       d=     8,246       x=     8,928
solution  92:   a=     1,701       b=     5,292       c=     6,930       d=     8,379       x=     9,072
solution  93:   a=     1,705       b=    98,673       c=   898,039       d= 2,643,742       x= 2,646,129
solution  94:   a=     1,728       b=     5,376       c=     7,040       d=     8,512       x=     9,216
solution  95:   a=     1,755       b=     5,460       c=     7,150       d=     8,645       x=     9,360
solution  96:   a=     1,760       b=   101,856       c=   927,008       d= 2,729,024       x= 2,731,488
solution  97:   a=     1,782       b=     5,544       c=     7,260       d=     8,778       x=     9,504
solution  98:   a=     1,809       b=     5,628       c=     7,370       d=     8,911       x=     9,648
solution  99:   a=     1,815       b=   105,039       c=   955,977       d= 2,814,306       x= 2,816,847
solution 100:   a=     1,836       b=     5,712       c=     7,480       d=     9,044       x=     9,792
solution 101:   a=     1,863       b=     5,796       c=     7,590       d=     9,177       x=     9,936
solution 102:   a=     1,870       b=   108,222       c=   984,946       d= 2,899,588       x= 2,902,206
solution 103:   a=     1,890       b=     5,880       c=     7,700       d=     9,310       x=    10,080
solution 104:   a=     1,917       b=     5,964       c=     7,810       d=     9,443       x=    10,224
solution 105:   a=     1,925       b=   111,405       c= 1,013,915       d= 2,984,870       x= 2,987,565
solution 106:   a=     1,944       b=     6,048       c=     7,920       d=     9,576       x=    10,368
solution 107:   a=     1,971       b=     6,132       c=     8,030       d=     9,709       x=    10,512
solution 108:   a=     1,980       b=   114,588       c= 1,042,884       d= 3,070,152       x= 3,072,924
solution 109:   a=     1,998       b=     6,216       c=     8,140       d=     9,842       x=    10,656
solution 110:   a=     2,025       b=     6,300       c=     8,250       d=     9,975       x=    10,800
solution 111:   a=     2,035       b=   117,771       c= 1,071,853       d= 3,155,434       x= 3,158,283
solution 112:   a=     2,052       b=     6,384       c=     8,360       d=    10,108       x=    10,944
solution 113:   a=     2,079       b=     6,468       c=     8,470       d=    10,241       x=    11,088
solution 114:   a=     2,090       b=   120,954       c= 1,100,822       d= 3,240,716       x= 3,243,642
solution 115:   a=     2,106       b=     6,552       c=     8,580       d=    10,374       x=    11,232
solution 116:   a=     2,133       b=     6,636       c=     8,690       d=    10,507       x=    11,376
solution 117:   a=     2,145       b=   124,137       c= 1,129,791       d= 3,325,998       x= 3,329,001
solution 118:   a=     2,160       b=     6,720       c=     8,800       d=    10,640       x=    11,520
solution 119:   a=     2,187       b=     6,804       c=     8,910       d=    10,773       x=    11,664
solution 120:   a=     2,200       b=   127,320       c= 1,158,760       d= 3,411,280       x= 3,414,360
solution 121:   a=     2,214       b=     6,888       c=     9,020       d=    10,906       x=    11,808
solution 122:   a=     2,241       b=     6,972       c=     9,130       d=    11,039       x=    11,952
solution 123:   a=     2,255       b=   130,503       c= 1,187,729       d= 3,496,562       x= 3,499,719
solution 124:   a=     2,268       b=     7,056       c=     9,240       d=    11,172       x=    12,096
solution 125:   a=     2,295       b=     7,140       c=     9,350       d=    11,305       x=    12,240
solution 126:   a=     2,310       b=   133,686       c= 1,216,698       d= 3,581,844       x= 3,585,078
solution 127:   a=     2,322       b=     7,224       c=     9,460       d=    11,438       x=    12,384
solution 128:   a=     2,349       b=     7,308       c=     9,570       d=    11,571       x=    12,528
solution 129:   a=     2,365       b=   136,869       c= 1,245,667       d= 3,667,126       x= 3,670,437
solution 130:   a=     2,376       b=     7,392       c=     9,680       d=    11,704       x=    12,672
solution 131:   a=     2,403       b=     7,476       c=     9,790       d=    11,837       x=    12,816
solution 132:   a=     2,420       b=   140,052       c= 1,274,636       d= 3,752,408       x= 3,755,796
solution 133:   a=     2,430       b=     7,560       c=     9,900       d=    11,970       x=    12,960
solution 134:   a=     2,457       b=     7,644       c=    10,010       d=    12,103       x=    13,104
solution 135:   a=     2,475       b=   143,235       c= 1,303,605       d= 3,837,690       x= 3,841,155
solution 136:   a=     2,484       b=     7,728       c=    10,120       d=    12,236       x=    13,248
solution 137:   a=     2,511       b=     7,812       c=    10,230       d=    12,369       x=    13,392
solution 138:   a=     2,530       b=   146,418       c= 1,332,574       d= 3,922,972       x= 3,926,514
solution 139:   a=     2,538       b=     7,896       c=    10,340       d=    12,502       x=    13,536
solution 140:   a=     2,565       b=     7,980       c=    10,450       d=    12,635       x=    13,680
solution 141:   a=     2,585       b=   149,601       c= 1,361,543       d= 4,008,254       x= 4,011,873
solution 142:   a=     2,592       b=     8,064       c=    10,560       d=    12,768       x=    13,824
solution 143:   a=     2,619       b=     8,148       c=    10,670       d=    12,901       x=    13,968
solution 144:   a=     2,640       b=   152,784       c= 1,390,512       d= 4,093,536       x= 4,097,232
solution 145:   a=     2,646       b=     8,232       c=    10,780       d=    13,034       x=    14,112
solution 146:   a=     2,673       b=     8,316       c=    10,890       d=    13,167       x=    14,256
solution 147:   a=     2,695       b=   155,967       c= 1,419,481       d= 4,178,818       x= 4,182,591
solution 148:   a=     2,700       b=     8,400       c=    11,000       d=    13,300       x=    14,400
solution 149:   a=     2,727       b=     8,484       c=    11,110       d=    13,433       x=    14,544
solution 150:   a=     2,750       b=   159,150       c= 1,448,450       d= 4,264,100       x= 4,267,950
solution 151:   a=     2,754       b=     8,568       c=    11,220       d=    13,566       x=    14,688
solution 152:   a=     2,781       b=     8,652       c=    11,330       d=    13,699       x=    14,832
solution 153:   a=     2,805       b=   162,333       c= 1,477,419       d= 4,349,382       x= 4,353,309
solution 154:   a=     2,808       b=     8,736       c=    11,440       d=    13,832       x=    14,976
solution 155:   a=     2,835       b=     8,820       c=    11,550       d=    13,965       x=    15,120
solution 156:   a=     2,860       b=   165,516       c= 1,506,388       d= 4,434,664       x= 4,438,668
solution 157:   a=     2,862       b=     8,904       c=    11,660       d=    14,098       x=    15,264
solution 158:   a=     2,889       b=     8,988       c=    11,770       d=    14,231       x=    15,408
solution 159:   a=     2,915       b=   168,699       c= 1,535,357       d= 4,519,946       x= 4,524,027
solution 160:   a=     2,916       b=     9,072       c=    11,880       d=    14,364       x=    15,552
solution 161:   a=     2,943       b=     9,156       c=    11,990       d=    14,497       x=    15,696
solution 162:   a=     2,970       b=   171,882       c= 1,564,326       d= 4,605,228       x= 4,609,386
solution 163:   a=     2,997       b=     9,324       c=    12,210       d=    14,763       x=    15,984
solution 164:   a=     3,024       b=     9,408       c=    12,320       d=    14,896       x=    16,128
solution 165:   a=     3,025       b=   175,065       c= 1,593,295       d= 4,690,510       x= 4,694,745
solution 166:   a=     3,051       b=     9,492       c=    12,430       d=    15,029       x=    16,272
solution 167:   a=     3,078       b=     9,576       c=    12,540       d=    15,162       x=    16,416
solution 168:   a=     3,080       b=   178,248       c= 1,622,264       d= 4,775,792       x= 4,780,104
solution 169:   a=     3,105       b=     9,660       c=    12,650       d=    15,295       x=    16,560
solution 170:   a=     3,132       b=     9,744       c=    12,760       d=    15,428       x=    16,704
solution 171:   a=     3,135       b=   181,431       c= 1,651,233       d= 4,861,074       x= 4,865,463
solution 172:   a=     3,159       b=     9,828       c=    12,870       d=    15,561       x=    16,848
solution 173:   a=     3,186       b=     9,912       c=    12,980       d=    15,694       x=    16,992
solution 174:   a=     3,190       b=   184,614       c= 1,680,202       d= 4,946,356       x= 4,950,822
solution 175:   a=     3,213       b=     9,996       c=    13,090       d=    15,827       x=    17,136
solution 176:   a=     3,240       b=    10,080       c=    13,200       d=    15,960       x=    17,280
solution 177:   a=     3,245       b=   187,797       c= 1,709,171       d= 5,031,638       x= 5,036,181
solution 178:   a=     3,267       b=    10,164       c=    13,310       d=    16,093       x=    17,424
solution 179:   a=     3,294       b=    10,248       c=    13,420       d=    16,226       x=    17,568
solution 180:   a=     3,300       b=   190,980       c= 1,738,140       d= 5,116,920       x= 5,121,540
solution 181:   a=     3,321       b=    10,332       c=    13,530       d=    16,359       x=    17,712
solution 182:   a=     3,348       b=    10,416       c=    13,640       d=    16,492       x=    17,856
solution 183:   a=     3,355       b=   194,163       c= 1,767,109       d= 5,202,202       x= 5,206,899
solution 184:   a=     3,375       b=    10,500       c=    13,750       d=    16,625       x=    18,000
solution 185:   a=     3,402       b=    10,584       c=    13,860       d=    16,758       x=    18,144
solution 186:   a=     3,410       b=   197,346       c= 1,796,078       d= 5,287,484       x= 5,292,258
solution 187:   a=     3,429       b=    10,668       c=    13,970       d=    16,891       x=    18,288
solution 188:   a=     3,456       b=    10,752       c=    14,080       d=    17,024       x=    18,432
solution 189:   a=     3,465       b=   200,529       c= 1,825,047       d= 5,372,766       x= 5,377,617
solution 190:   a=     3,483       b=    10,836       c=    14,190       d=    17,157       x=    18,576
solution 191:   a=     3,510       b=    10,920       c=    14,300       d=    17,290       x=    18,720
solution 192:   a=     3,520       b=   203,712       c= 1,854,016       d= 5,458,048       x= 5,462,976
solution 193:   a=     3,537       b=    11,004       c=    14,410       d=    17,423       x=    18,864
solution 194:   a=     3,564       b=    11,088       c=    14,520       d=    17,556       x=    19,008
solution 195:   a=     3,575       b=   206,895       c= 1,882,985       d= 5,543,330       x= 5,548,335
solution 196:   a=     3,591       b=    11,172       c=    14,630       d=    17,689       x=    19,152
solution 197:   a=     3,618       b=    11,256       c=    14,740       d=    17,822       x=    19,296
solution 198:   a=     3,630       b=   210,078       c= 1,911,954       d= 5,628,612       x= 5,633,694
solution 199:   a=     3,645       b=    11,340       c=    14,850       d=    17,955       x=    19,440
solution 200:   a=     3,672       b=    11,424       c=    14,960       d=    18,088       x=    19,584

Showed  200  solutions,  output written to file:  EULERSUM.OUT
```

## Ring

```# Project : Euler's sum of powers conjecture

max=250
for w = 1 to max
for x = 1 to w
for y = 1 to x
for z = 1 to y
sum = pow(w,5) + pow(x,5) + pow(y,5) + pow(z,5)
s1  = floor(pow(sum,0.2))
if sum = pow(s1,5)
see "" + w + "^5 + " + x + "^5 + " + y + "^5 + " + z + "^5 = " + s1 + "^5"
ok
next
next
next
next```

Output:

```133^5 + 110^5 + 84^5 + 27^5 = 144^5
```

## Ruby

Brute force:

```power5 = (1..250).each_with_object({}){|i,h| h[i**5]=i}
result = power5.keys.repeated_combination(4).select{|a| power5[a.inject(:+)]}
puts result.map{|a| a.map{|i| "#{power5[i]}**5"}.join(' + ') + " = #{power5[a.inject(:+)]}**5"}
```
Output:
```27**5 + 84**5 + 110**5 + 133**5 = 144**5
```

Faster version:

Translation of: Python
```p5, sum2, max = {}, {}, 250
(1..max).each do |i|
p5[i**5] = i
(i..max).each{|j| sum2[i**5 + j**5] = [i,j]}
end

result = {}
sk = sum2.keys.sort
p5.keys.sort.each do |p|
sk.each do |s|
break if p <= s
result[(sum2[s] + sum2[p-s]).sort] = p5[p]  if sum2[p - s]
end
end
result.each{|k,v| puts k.map{|i| "#{i}**5"}.join(' + ') + " = #{v}**5"}
```

The output is the same above.

## Run BASIC

```max=250
FOR w = 1 TO max
FOR x = 1 TO w
FOR y = 1 TO x
FOR z = 1 TO y
sum = w^5 + x^5 + y^5 + z^5
s1  = INT(sum^0.2)
IF sum=s1^5 THEN
PRINT w;"^5 + ";x;"^5 + ";y;"^5 + ";z;"^5 = ";s1;"^5"
end
end if
NEXT z
NEXT y
NEXT x
NEXT w```
```133^5 + 110^5 + 84^5 + 27^5 = 144^5
```

## Rust

```const MAX_N : u64 = 250;

fn eulers_sum_of_powers() -> (usize, usize, usize, usize, usize) {
let pow5: Vec<u64> = (0..MAX_N).map(|i| i.pow(5)).collect();
let pow5_to_n = |pow| pow5.binary_search(&pow);

for x0 in 1..MAX_N as usize {
for x1 in 1..x0 {
for x2 in 1..x1 {
for x3 in 1..x2 {
let pow_sum = pow5[x0] + pow5[x1] + pow5[x2] + pow5[x3];
if let Ok(n) = pow5_to_n(pow_sum) {
return (x0, x1, x2, x3, n)
}
}
}
}
}

panic!();
}

fn main() {
let (x0, x1, x2, x3, y) = eulers_sum_of_powers();
println!("{}^5 + {}^5 + {}^5 + {}^5 == {}^5", x0, x1, x2, x3, y)
}
```
Output:
`133^5 + 110^5 + 84^5 + 27^5 == 144^5`

## Scala

### Functional programming

```import scala.collection.Searching.{Found, search}

object EulerSopConjecture extends App {

val (maxNumber, fifth) = (250, (1 to 250).map { i => math.pow(i, 5).toLong })

def binSearch(fact: Int*) = fifth.search(fact.map(f => fifth(f)).sum)

def sop = (0 until maxNumber)
.flatMap(a => (a until maxNumber)
.flatMap(b => (b until maxNumber)
.flatMap(c => (c until maxNumber)
.map { case x\$1@d => (binSearch(a, b, c, d), x\$1) }
.withFilter { case (f, _) => f.isInstanceOf[Found] }
.map { case (f, d) => (a + 1, b + 1, c + 1, d + 1, f.insertionPoint + 1) }))).take(1)
.map { case (a, b, c, d, f) => s"\$a⁵ + \$b⁵ + \$c⁵ + \$d⁵ = \$f⁵" }

println(sop)

}
```
Output:
Vector(27⁵ + 84⁵ + 110⁵ + 133⁵ = 144⁵)

## Seed7

```\$ include "seed7_05.s7i";

const func integer: binarySearch (in array integer: arr, in integer: aKey) is func
result
var integer: index is 0;
local
var integer: low is 1;
var integer: high is 0;
var integer: middle is 0;
begin
high := length(arr);
while index = 0 and low <= high do
middle := (low + high) div 2;
if aKey < arr[middle] then
high := pred(middle);
elsif aKey > arr[middle] then
low := succ(middle);
else
index := middle;
end if;
end while;
end func;

const proc: main is func
local
var array integer: p5 is 249 times 0;
var integer: i is 0;
var integer: x0 is 0;
var integer: x1 is 0;
var integer: x2 is 0;
var integer: x3 is 0;
var integer: sum is 0;
var integer: y is 0;
var boolean: found is FALSE;
begin
for i range 1 to 249 do
p5[i] := i ** 5;
end for;
for x0 range 1 to 249 until found do
for x1 range 1 to pred(x0) until found do
for x2 range 1 to pred(x1) until found do
for x3 range 1 to pred(x2) until found do
sum := p5[x0] + p5[x1] + p5[x2] + p5[x3];
y := binarySearch(p5, sum);
if y > 0 then
writeln(x0 <& "**5 + " <& x1 <& "**5 + " <& x2 <& "**5 + " <& x3 <& "**5 = " <& y <& "**5");
found := TRUE;
end if;
end for;
end for;
end for;
end for;
writeln("No solution was found");
end if;
end func;```
Output:
```133**5 + 110**5 + 84**5 + 27**5 = 144**5
```

## SenseTalk

```findEulerSumOfPowers
to findEulerSumOfPowers
set MAX_NUMBER to 250
set possibleValues to 1..MAX_NUMBER
set possible5thPowers to each item of possibleValues to the power of 5
repeat for x0 in 1..250
repeat for x1 in 1..x0
repeat for x2 in 1..x1
repeat for x3 in 1..x2
set possibleSum to item x0 of possible5thPowers \
plus item x1 of possible5thPowers \
plus item x2 of possible5thPowers \
plus item x3 of possible5thPowers
if possibleSum is in possible5thPowers
put x0 & "^5 + " & x1 & "^5 + " & x2 & "^5 + " & x3 & "^5 = " & the item number of possibleSum within possible5thPowers & "^5"
return
end if
end repeat
end repeat
end repeat
end repeat
end findEulerSumOfPowers```

## Sidef

Translation of: Raku
```define range = (1 ..^ 250)

var p5 = Hash()
var sum2 = Hash()

for i in (range) {
p5{i**5} = i
for j in (range) {
sum2{i**5 + j**5} = [i, j]
}
}

var sk = sum2.keys.map{ Num(_) }.sort

for p in (p5.keys.map{ Num(_) }.sort) {

var s = sk.first {|s|
p > s && sum2.exists(p-s)
} \\ next

var t = (sum2{s} + sum2{p-s} -> map{|n| "#{n}⁵" }.join(' + '))
say "#{t} = #{p5{p}}⁵"
break
}
```
Output:
```84⁵ + 27⁵ + 133⁵ + 110⁵ =  144⁵
```

## Swift

Translation of: Rust
```extension BinaryInteger {
@inlinable
public func power(_ n: Self) -> Self {
return stride(from: 0, to: n, by: 1).lazy.map({_ in self }).reduce(1, *)
}
}

func sumOfPowers(maxN: Int = 250) -> (Int, Int, Int, Int, Int) {
let pow5 = (0..<maxN).map({ \$0.power(5) })
let pow5ToN = {n in pow5.firstIndex(of: n)}

for x0 in 1..<maxN {
for x1 in 1..<x0 {
for x2 in 1..<x1 {
for x3 in 1..<x2 {
let powSum = pow5[x0] + pow5[x1] + pow5[x2] + pow5[x3]

if let idx = pow5ToN(powSum) {
return (x0, x1, x2, x3, idx)
}
}
}
}
}

fatalError("Did not find solution")
}

let (x0, x1, x2, x3, y) = sumOfPowers()

print("\(x0)^5 + \(x1)^5 + \(x2)^5 \(x3)^5 = \(y)^5")
```
Output:
`133^5 + 110^5 + 84^5 + 27^5 = 144^5`

## Tcl

```proc doit {{badx 250} {complete 0}} {
## NB: \$badx is exclusive upper limit, and also limits y!
for {set y 1} {\$y < \$badx} {incr y} {
set s [expr {\$y ** 5}]
set r5(\$s) \$y           ;# fifth roots of valid sums
}
for {set a 1} {\$a < \$badx} {incr a} {
set suma [expr {\$a ** 5}]
for {set b 1} {\$b <= \$a} {incr b} {
set sumb [expr {\$suma + (\$b ** 5)}]
for {set c 1} {\$c <= \$b} {incr c} {
set sumc [expr {\$sumb + (\$c ** 5)}]
for {set d 1} {\$d <= \$c} {incr d} {
set sumd [expr {\$sumc + (\$d ** 5)}]
if {[info exists r5(\$sumd)]} {
set e \$r5(\$sumd)
puts "\$e^5 = \$a^5 + \$b^5 + \$c^5 + \$d^5"
if {!\$complete} {
return
}
}
}
}
}
}
puts "search complete (x < \$badx)"
}
doit
```
Output:
```144^5 = 133^5 + 110^5 + 84^5 + 27^5
```

real 0m2.387s

## UNIX Shell

Works with: Bourne Again SHell
Works with: Korn Shell
Works with: Zsh

Shell is not the go-to language for number-crunching, but if you're going to use a shell, it looks like ksh is the fastest option, at about 8x faster than bash and 2x faster than zsh.

```MAX=250
pow5=()
for (( i=1; i<MAX; ++i )); do
pow5[i]=\$(( i*i*i*i*i ))
done
for (( a=1; a<MAX; ++a )); do
for (( b=a+1; b<MAX; ++b )); do
for (( c=b+1; c<MAX; ++c )); do
for (( d=c+1; d<MAX; ++d )); do
(( sum=pow5[a]+pow5[b]+pow5[c]+pow5[d] ))
(( low=d+3 ))
(( high=MAX ))
while (( low <= high )); do
(( guess=(low+high)/2 ))
if (( pow5[guess]  == sum )); then
printf 'Found example: %d⁵+%d⁵+%d⁵+%d⁵=%d⁵\n' "\$a" "\$b" "\$c" "\$d" "\$guess"
exit 0
elif (( pow5[guess] < sum )); then
(( low=guess+1 ))
else
(( high=guess-1 ))
fi
done
done
done
done
done
printf 'No examples found.\n'
exit 1
```
Output:
```\$ time bash esop.sh
Found example: 27⁵+84⁵+110⁵+133⁵=144⁵
bash esop.sh  6953.75s user 37.53s system 99% cpu 1:57:02.41 total
\$ time ksh esop.sh
Found example: 27⁵+84⁵+110⁵+133⁵=144⁵
ksh esop.sh  855.66s user 5.30s system 99% cpu 14:26.78 total
\$ time zsh esop.sh
Found example: 27⁵+84⁵+110⁵+133⁵=144⁵
zsh esop.sh  1969.48s user 250.82s system 99% cpu 37:11.62 total```

## VBA

Translation of: AWK
```Public Declare Function GetTickCount Lib "kernel32.dll" () As Long
Public Sub begin()
start_int = GetTickCount()
main
Debug.Print (GetTickCount() - start_int) / 1000 & " seconds"
End Sub
Private Function pow(x, y) As Variant
pow = CDec(Application.WorksheetFunction.Power(x, y))
End Function
Private Sub main()
For x0 = 1 To 250
For x1 = 1 To x0
For x2 = 1 To x1
For x3 = 1 To x2
sum = CDec(pow(x0, 5) + pow(x1, 5) + pow(x2, 5) + pow(x3, 5))
s1 = Int(pow(sum, 0.2))
If sum = pow(s1, 5) Then
Debug.Print x0 & "^5 + " & x1 & "^5 + " & x2 & "^5 + " & x3 & "^5 = " & s1
Exit Sub
End If
Next x3
Next x2
Next x1
Next x0
End Sub
```
Output:
```33^5 + 110^5 + 84^5 + 27^5 = 144
160,187 seconds```

## VBScript

Translation of: ERRE
```Max=250

For X0=1 To Max
For X1=1 To X0
For X2=1 To X1
For X3=1 To X2
Sum=fnP5(X0)+fnP5(X1)+fnP5(X2)+fnP5(X3)
S1=Int(Sum^0.2)
If Sum=fnP5(S1) Then
WScript.StdOut.Write X0 & " " & X1 & " " & X2 & " " & X3 & " " & S1
WScript.Quit
End If
Next
Next
Next
Next

Function fnP5(n)
fnP5 = n ^ 5
End Function
```
Output:
`133 110 84 27 144`

## Visual Basic .NET

### Paired Powers Algorithm

Translation of: Python
```Module Module1

Structure Pair
Dim a, b As Integer
Sub New(x as integer, y as integer)
a = x : b = y
End Sub
End Structure

Dim max As Integer = 250
Dim p5() As Long,
sum2 As SortedDictionary(Of Long, Pair) = New SortedDictionary(Of Long, Pair)

Function Fmt(p As Pair) As String
Return String.Format("{0}^5 + {1}^5", p.a, p.b)
End Function

Sub Init()
p5(0) = 0 : p5(1) = 1 : For i As Integer = 1 To max - 1
For j As Integer = i + 1 To max
p5(j) = CLng(j) * j : p5(j) *= p5(j) * j
sum2.Add(p5(i) + p5(j), New Pair(i, j))
Next
Next
End Sub

Sub Calc(Optional findLowest As Boolean = True)
For i As Integer = 1 To max : Dim p As Long = p5(i)
For Each s In sum2.Keys
Dim t As Long = p - s : If t <= 0 Then Exit For
If sum2.Keys.Contains(t) AndAlso sum2.Item(t).a > sum2.Item(s).b Then
Console.WriteLine("  {1} + {2} = {0}^5", i, Fmt(sum2.Item(s)), Fmt(sum2.Item(t)))
If findLowest Then Exit Sub
End If
Next : Next
End Sub

Sub Main(args As String())
If args.Count > 0 Then
Dim t As Integer = 0 : Integer.TryParse(args(0), t)
If t > 0 AndAlso t < 5405 Then max = t
End If
Console.WriteLine("Checking from 1 to {0}...", max)
For i As Integer = 0 To 1
ReDim p5(max) : sum2.Clear()
Dim st As DateTime = DateTime.Now
Init() : Calc(i = 0)
Console.WriteLine("{0}  Computation time to {2} was {1} seconds{0}", vbLf,
(DateTime.Now - st).TotalSeconds, If(i = 0, "find lowest one", "check entire space"))
Next
End Sub
End Module
```
Output:
(No command line arguments)
```Checking from 1 to 250...
27^5 + 84^5 + 110^5 + 133^5 = 144^5

Computation time to find lowest one was 0.0807819 seconds

27^5 + 84^5 + 110^5 + 133^5 = 144^5

Computation time to check entire space was 0.3830103 seconds```
Command line argument = "1000"
```Checking from 1 to 1000...
27^5 + 84^5 + 110^5 + 133^5 = 144^5

Computation time to find lowest one was 0.3112007 seconds

27^5 + 84^5 + 110^5 + 133^5 = 144^5
54^5 + 168^5 + 220^5 + 266^5 = 288^5
81^5 + 252^5 + 330^5 + 399^5 = 432^5
108^5 + 336^5 + 440^5 + 532^5 = 576^5
135^5 + 420^5 + 550^5 + 665^5 = 720^5
162^5 + 504^5 + 660^5 + 798^5 = 864^5

Computation time to check entire space was 28.8847393 seconds```

### Paired Powers w/ Mod 30 Shortcut and Threading

If one divides the searched array of powers (sum2m()) into 30 pieces, the search time can be reduced by only searching the appropriate one (determined by the Mod 30 value of the value being sought). Once broken down by this, it is now easier to use threading to further reduce the computation time.
The following compares the plain paired powers algorithm to the plain powers plus the mod 30 shortcut algorithm, without and with threading.

```Module Module1

Structure Pair
Dim a, b As Integer
Sub New(x As Integer, y As Integer)
a = x : b = y
End Sub
End Structure

Dim min As Integer = 1, max As Integer = 250
Dim p5() As Long,
sum2 As SortedDictionary(Of Long, Pair) = New SortedDictionary(Of Long, Pair),
sum2m(29) As SortedDictionary(Of Long, Pair)

Function Fmt(p As Pair) As String
Return String.Format("{0}^5 + {1}^5", p.a, p.b)
End Function

Sub Init()
p5(0) = 0 : p5(min) = CLng(min) * min : p5(min) *= p5(min) * min
For i As Integer = min To max - 1
For j As Integer = i + 1 To max
p5(j) = CLng(j) * j : p5(j) *= p5(j) * j
If j = max Then Continue For
sum2.Add(p5(i) + p5(j), New Pair(i, j))
Next
Next
End Sub

Sub InitM()
For i As Integer = 0 To 29 : sum2m(i) = New SortedDictionary(Of Long, Pair) : Next
p5(0) = 0 : p5(min) = CLng(min) * min : p5(min) *= p5(min) * min
For i As Integer = min To max - 1
For j As Integer = i + 1 To max
p5(j) = CLng(j) * j : p5(j) *= p5(j) * j
If j = max Then Continue For
Dim x As Long = p5(i) + p5(j)
sum2m(x Mod 30).Add(x, New Pair(i, j))
Next
Next
End Sub

Sub Calc(Optional findLowest As Boolean = True)
For i As Integer = min To max : Dim p As Long = p5(i)
For Each s In sum2.Keys
Dim t As Long = p - s : If t <= 0 Then Exit For
If sum2.Keys.Contains(t) AndAlso sum2.Item(t).a > sum2.Item(s).b Then
Console.WriteLine("  {1} + {2} = {0}^5", i,
Fmt(sum2.Item(s)), Fmt(sum2.Item(t)))
If findLowest Then Exit Sub
End If
Next : Next
End Sub

Function CalcM(m As Integer) As List(Of String)
Dim res As New List(Of String)
For i As Integer = min To max
Dim pm As Integer = i Mod 30,
mp As Integer = (pm - m + 30) Mod 30
For Each s In sum2m(m).Keys
Dim t As Long = p5(i) - s : If t <= 0 Then Exit For
If sum2m(mp).Keys.Contains(t) AndAlso
sum2m(mp).Item(t).a > sum2m(m).Item(s).b Then
res.Add(String.Format("  {1} + {2} = {0}^5",
i, Fmt(sum2m(m).Item(s)), Fmt(sum2m(mp).Item(t))))
End If
Next : Next
Return res
End Function

Function Snip(s As String) As Integer
Dim p As Integer = s.IndexOf("=") + 1
Return s.Substring(p, s.IndexOf("^", p) - p)
End Function

Function CompareRes(ByVal x As String, ByVal y As String) As Integer
CompareRes = Snip(x).CompareTo(Snip(y))
If CompareRes = 0 Then CompareRes = x.CompareTo(y)
End Function

Function Validify(def As Integer, s As String) As Integer
Validify = def : Dim t As Integer = 0 : Integer.TryParse(s, t)
If t >= 1 AndAlso Math.Pow(t, 5) < (Long.MaxValue >> 1) Then Validify = t
End Function

Sub Switch(ByRef a As Integer, ByRef b As Integer)
Dim t As Integer = a : a = b : b = t
End Sub

Sub Main(args As String())
Select Case args.Count
Case 1 : max = Validify(max, args(0))
Case > 1
min = Validify(min, args(0))
max = Validify(max, args(1))
If max < min Then Switch(max, min)
End Select
Console.WriteLine("Paired powers, checking from {0} to {1}...", min, max)
For i As Integer = 0 To 1
ReDim p5(max) : sum2.Clear()
Dim st As DateTime = DateTime.Now
Init() : Calc(i = 0)
Console.WriteLine("{0}  Computation time to {2} was {1} seconds{0}", vbLf,
(DateTime.Now - st).TotalSeconds, If(i = 0, "find lowest one", "check entire space"))
Next
For i As Integer = 0 To 1
Console.WriteLine("Paired powers with Mod 30 shortcut (entire space) {2}, checking from {0} to {1}...",
min, max, If(i = 0, "sequential", "parallel"))
ReDim p5(max)
Dim res As List(Of String) = New List(Of String)
Dim st As DateTime = DateTime.Now
InitM()
Select Case i
Case 0
For j As Integer = 0 To 29
Next
Case 1
For j As Integer = 0 To 29 : Dim jj = j
Next
For Each item In taskList.Select(Function(t) t.Result)
End Select
For Each item In res
Console.WriteLine(item) : Next
Console.WriteLine("{0}  Computation time was {1} seconds{0}", vbLf, (DateTime.Now - st).TotalSeconds)
Next
End Sub
End Module
```
Output:
(No command line arguments)
```Paired powers, checking from 1 to 250...
27^5 + 84^5 + 110^5 + 133^5 = 144^5
Computation time to find lowest one was 0.0781252 seconds
27^5 + 84^5 + 110^5 + 133^5 = 144^5
Computation time to check entire space was 0.3280574 seconds
Paired powers with Mod 30 shortcut (entire space) sequential, checking from 1 to 250...
27^5 + 84^5 + 110^5 + 133^5 = 144^5
Computation time was 0.2655529 seconds
Paired powers with Mod 30 shortcut (entire space) parallel, checking from 1 to 250...
27^5 + 84^5 + 110^5 + 133^5 = 144^5

Computation time was 0.0624651 seconds```
(command line argument = "1000")
```Paired powers, checking from 1 to 1000...
27^5 + 84^5 + 110^5 + 133^5 = 144^5

Computation time to find lowest one was 0.2499343 seconds

27^5 + 84^5 + 110^5 + 133^5 = 144^5
54^5 + 168^5 + 220^5 + 266^5 = 288^5
81^5 + 252^5 + 330^5 + 399^5 = 432^5
108^5 + 336^5 + 440^5 + 532^5 = 576^5
135^5 + 420^5 + 550^5 + 665^5 = 720^5
162^5 + 504^5 + 660^5 + 798^5 = 864^5

Computation time to check entire space was 27.805961 seconds

Paired powers with Mod 30 shortcut (entire space) sequential, checking from 1 to 1000...
27^5 + 84^5 + 110^5 + 133^5 = 144^5
54^5 + 168^5 + 220^5 + 266^5 = 288^5
81^5 + 252^5 + 330^5 + 399^5 = 432^5
108^5 + 336^5 + 440^5 + 532^5 = 576^5
135^5 + 420^5 + 550^5 + 665^5 = 720^5
162^5 + 504^5 + 660^5 + 798^5 = 864^5

Computation time was 23.8068928 seconds

Paired powers with Mod 30 shortcut (entire space) parallel, checking from 1 to 1000...
27^5 + 84^5 + 110^5 + 133^5 = 144^5
54^5 + 168^5 + 220^5 + 266^5 = 288^5
81^5 + 252^5 + 330^5 + 399^5 = 432^5
108^5 + 336^5 + 440^5 + 532^5 = 576^5
135^5 + 420^5 + 550^5 + 665^5 = 720^5
162^5 + 504^5 + 660^5 + 798^5 = 864^5

Computation time was 5.4205943 seconds```
(command line arguments = "27 864")
```Paired powers, checking from 27 to 864...
27^5 + 84^5 + 110^5 + 133^5 = 144^5

Computation time to find lowest one was 0.1562309 seconds

27^5 + 84^5 + 110^5 + 133^5 = 144^5
54^5 + 168^5 + 220^5 + 266^5 = 288^5
81^5 + 252^5 + 330^5 + 399^5 = 432^5
108^5 + 336^5 + 440^5 + 532^5 = 576^5
135^5 + 420^5 + 550^5 + 665^5 = 720^5
162^5 + 504^5 + 660^5 + 798^5 = 864^5

Computation time to check entire space was 15.8243802 seconds

Paired powers with Mod 30 shortcut (entire space) sequential, checking from 27 to 864...
27^5 + 84^5 + 110^5 + 133^5 = 144^5
54^5 + 168^5 + 220^5 + 266^5 = 288^5
81^5 + 252^5 + 330^5 + 399^5 = 432^5
108^5 + 336^5 + 440^5 + 532^5 = 576^5
135^5 + 420^5 + 550^5 + 665^5 = 720^5
162^5 + 504^5 + 660^5 + 798^5 = 864^5

Computation time was 13.0438215 seconds

Paired powers with Mod 30 shortcut (entire space) parallel, checking from 27 to 864...
27^5 + 84^5 + 110^5 + 133^5 = 144^5
54^5 + 168^5 + 220^5 + 266^5 = 288^5
81^5 + 252^5 + 330^5 + 399^5 = 432^5
108^5 + 336^5 + 440^5 + 532^5 = 576^5
135^5 + 420^5 + 550^5 + 665^5 = 720^5
162^5 + 504^5 + 660^5 + 798^5 = 864^5

Computation time was 3.0305365 seconds```
(command line arguments = "189 1008")
```Paired powers, checking from 189 to 1008...
189^5 + 588^5 + 770^5 + 931^5 = 1008^5

Computation time to find lowest one was 14.6840411 seconds

189^5 + 588^5 + 770^5 + 931^5 = 1008^5

Computation time to check entire space was 14.7777685 seconds

Paired powers with Mod 30 shortcut (entire space) sequential, checking from 189 to 1008...
189^5 + 588^5 + 770^5 + 931^5 = 1008^5

Computation time was 12.4814705 seconds

Paired powers with Mod 30 shortcut (entire space) parallel, checking from 189 to 1008...
189^5 + 588^5 + 770^5 + 931^5 = 1008^5

Computation time was 2.7180777 seconds```

## Wren

Translation of: C
```var start = System.clock
var n = 250
var m = 30

var p5 = List.filled(n+m+1, 0)
var s = 0
while (s < n) {
var sq = s * s
p5[s] = sq * sq * s
s = s + 1
}
var max = p5[n-1]
while (s < p5.count) {
p5[s] = max + 1
s = s + 1
}
for (a in 1...n-3) {
for (b in a + 1...n-2) {
for (c in b + 1...n-1) {
var d = c + 1
var t = p5[a] + p5[b] + p5[c]
var e = d + (t % m)
s = t + p5[d]
while (s <= max) {
e = e - m
while (p5[e+m] <= s) e = e + m
if (p5[e] == s) {
System.print("%(a)⁵ + %(b)⁵ + %(c)⁵ + %(d)⁵ = %(e)⁵")
System.print("Took %(System.clock - start) seconds")
return
}
d = d + 1
e = e + 1
s = t + p5[d]
}
}
}
}
```
Output:

Timing is for an Intel Core i7-8565U machine running Wren 0.2.0 on Ubuntu 18.04.

```27⁵ + 84⁵ + 110⁵ + 133⁵ = 144⁵
Took 6.836934 seconds
```

## XPL0

Runs in 50.3 seconds on Pi4.

```func real Pow5(N);
int  N;
real X, P;
[X:= float(N);
P:= X*X;
P:= P*P;
return P*X;
];

int  X0, X1, X2, X3, Y;
real SP;
[for X0:= 1 to 250 do
for X1:= 1 to X0-1 do
for X2:= 1 to X1-1 do
for X3:= 1 to X2-1 do
[SP:= Pow5(X0) + Pow5(X1) + Pow5(X2) + Pow5(X3);
for Y:= X0+1 to 250 do
if Pow5(Y) = SP then
[IntOut(0, X0);  Text(0, "^^5 + ");
IntOut(0, X1);  Text(0, "^^5 + ");
IntOut(0, X2);  Text(0, "^^5 + ");
IntOut(0, X3);  Text(0, "^^5 = ");
IntOut(0, Y);   Text(0, "^^5^m^j");
exit;
];
];
]```
Output:
```133^5 + 110^5 + 84^5 + 27^5 = 144^5
```

## Zig

Translation of: Go
```const std = @import("std");
const stdout = std.io.getStdOut().outStream();

pub fn main() !void {
var pow5: i64 = undefined;
for (pow5) |*e, i| {
const n = @intCast(i64, i);
e.* = n * n * n * n * n;
}
var x0: u16 = 4;
while (x0 < pow5.len) : (x0 += 1) {
var x1: u16 = 3;
while (x1 < x0) : (x1 += 1) {
var x2: u16 = 2;
while (x2 < x1) : (x2 += 1) {
var x3: u16 = 1;
while (x3 < x2) : (x3 += 1) {
const sum = pow5[x0] + pow5[x1] + pow5[x2] + pow5[x3];
var y: u16 = x0 + 1;
while (y < pow5.len) : (y += 1) if (sum == pow5[y]) {
try stdout.print("{}⁵ + {}⁵ + {}⁵ + {}⁵ = {}⁵\n", .{ x0, x1, x2, x3, y });
return;
};
}
}
}
}

try stdout.print("Sorry, no solution found.\n", .{});
}
```
Output:
```133⁵ + 110⁵ + 84⁵ + 27⁵ = 144⁵
```

## zkl

Uses two look up tables for efficiency. Counts from 0 for ease of coding.

```pow5s:=[1..249].apply("pow",5); // (1^5, 2^5, 3^5 .. 249^5)
pow5r:=pow5s.enumerate().apply("reverse").toDictionary(); // [144^5:144, ...]
foreach x0,x1,x2,x3 in (249,x0,x1,x2){
sum:=pow5s[x0] + pow5s[x1] + pow5s[x2] + pow5s[x3];
if(pow5r.holds(sum))
println("%d^5 + %d^5 + %d^5 + %d^5 = %d^5"
.fmt(x3+1,x2+1,x1+1,x0+1,pow5r[sum]+1));
break(4);  // the foreach is actually four loops
}```
Output:
`27^5 + 84^5 + 110^5 + 133^5 = 144^5`

Using the Python technique of caching double sums results in a 5x speed up [to the first/only solution]; actually the speed up is close to 25x but creating the caches dominates the runtime to the first solution.

Translation of: Python
```p5,sum2:=Dictionary(),Dictionary();
foreach i in ([1..249]){
p5[i.pow(5)]=i;
foreach j in ([i..249]){ sum2[i.pow(5) + j.pow(5)]=T(i,j) } // 31,125 keys
}

sk:=sum2.keys.apply("toInt").copy().sort(); // RW list sorts faster than a RO one
foreach p,s in (p5.keys.apply("toInt"),sk){
if(p<=s) break;
if(sum2.holds(p - s)){
println("%d^5 + %d^5 + %d^5 + %d^5 = %d^5"
.fmt(sum2[s].xplode(),sum2[p - s].xplode(),p5[p]));
break(2);  // or get permutations
}
}```

Note: dictionary keys are always strings and copying a read only list creates a read write list.

Output:
`27^5 + 84^5 + 110^5 + 133^5 = 144^5`

## ZX Spectrum Basic

This "abacus revision" reverts back to an earlier one, i.e. the "slide rule" one calculating the logarithmic 'percentage' for each of 4 summands. It also calculates their ones' digits as if on a base m abacus, that complements the other base m digits embedded in their percentages via a check sum of the ones' digits when the percentages add up to 1. Because any other argument is less than m, there is sufficient additional precision so as to not find false solutions while seeking the first primitive solution. 1 is excluded a priori for (e.g.) w, since it is well-known that the only integral points on 1=m^5-x^5-y^5-z^5 are the obvious ones (that aren't distinct\all >0). Nonetheless, 1 (as the zeroeth 'prime' Po) can be the first of 3 factors (one of the others being a power of the first 4 primes) for specifying a LHS argument. Given 4 LHS arguments each raised to a 5th power as is the RHS for which m<=ΣΠPi<2^(3+5), i=0 to 4, the LHS solution (if one exists) will consist of 4 elements of a factorial domain that are each a triplet (a prime\Po * a prime^1st\2nd power * a prime^1st\4th power) multiple having a distinct pair of the first 4 primes present & absent. Such potential solutions are generated by aiming for simplicity rather than efficiency (e.g. not generating potential solutions where each multiple is even). Two theorems' consequences intended for an even earlier revision are also applied: Fermat's Little Theorem (as proven later by Euler) whereby Xi^5 mod P = Xi mod P for the first 3 primes; and the Chinese Remainder Theorem in reverse whereby m= k- i*2*3*5, such that k is chosen to be the highest allowed m congruent mod 30 to the sum of the LHS arguments of a potential solution. Although still slow, this revision performs the given task on any ZX Spectrum, despite being limited to 32-bit precision.

```1 CLS
2 DIM k(29): DIM q(249)
5 FOR i=4 TO 249: LET q(i)=LN i : NEXT i
6 REM enhancements for the much expanded Spectrum Next:  DIM p(248,249)
7 REM FOR j=4TO 248:FOR i=j TO 249:LET p(j,i)=EXP (q(j)-q(i))*5:NEXT i:NEXT j
15 FOR i=0 TO 9: LET k(i)=240+ i : NEXT i
17 FOR i=10 TO 29: LET k(i)=210+ i : NEXT i
20 FOR w=6 TO 246 STEP 3
21 LET o=w
22 FOR x=4 TO 248 STEP 2
23 IF o<x THEN LET o=x
24 FOR y=10 TO 245 STEP 5
25 IF o<y THEN LET o=y
26 FOR z=14 TO 245 STEP 7
27 IF o<z THEN LET o=z
30 LET o=o+1 : LET m=k(FN f((w+x+y+z),30))
34 IF m<o THEN GO TO 90
40 REM LET s=p(w,m)+p(x,m)+p(y,m)+p(z,m) instead of:
42 LET s=EXP((q(w)-q(m))*5)
43 LET s=EXP((q(x)-q(m))*5)+ s
45 LET s=EXP((q(y)-q(m))*5)+ s
47 LET s=EXP((q(z)-q(m))*5)+ s
50 IF s<>1 THEN GO TO 80
52 LET a=FN f(w*w,m) : LET a=FN f(a*a*w,m)
53 LET b=FN f(x*x,m) : LET b=FN f(b*b*x,m)
55 LET c=FN f(y*y,m) : LET c=FN f(c*c*y,m)
57 LET d=FN f(z*z,m) : LET d=FN f(d*d*z,m)
60 LET u=FN f((a+b+c+d),m)
65 IF u THEN GO TO 80
73 PRINT w;"^5+";x;"^5+";y;"^5+";z;"^5=";m;"^5": STOP
80 IF s<1 THEN m=m-30 : GO TO 34
90 NEXT z: NEXT y: NEXT x: NEXT w
100 DEF FN f(e,n)=e- INT(e/n)*n
```
Output:
```slide rule ready
27^5+84^5+110^5+133^5=144^5
```