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 200 years before it was disproved by the finding of a counterexample in 1966 by Lander and Parkin.
- Euler's (disproved) sum of powers conjecture
- 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.
Lander and Parkin are known to have used a brute-force search on a CDC600 computer restricting numbers to those less than 250.
- Task
Write a program to search for an integer solution to:
x05 + x15 + x25 + x35 == y5
Where all xi's and y are distinct integers between 0 and 250 (exclusive).
Show an answer here.
C++
<lang cpp>#include <iostream>
- include <cmath>
- include <set>
using namespace std;
int main() { const auto MAX = 250; double pow5[MAX]; set<double> pow5s; for (auto i = 1; i < MAX; i++) { pow5[i] = (double)i * i * i * i * i; pow5s.insert(pow5[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 (pow5s.find(sum) != pow5s.end()) { cout << x0 << " " << x1 << " " << x2 << " " << x3 << " " << pow(sum, 1.0 / 5.0) << endl; goto exit;}}}}}
exit:
return 0;} </lang>
- Output:
133 110 84 27 144
ERRE
<lang 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</lang>
- Output:
133 110 84 27 144
F#
<lang fsharp> //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.[249] -> 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)
</lang>
- Output:
"27**5 + 84**5 + 110**5 + 133**5 = 144**5"
Fortran
<lang fortran>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</lang>
- Output:
27 84 110 133 144
Go
<lang go>package main
import ( "fmt" "log" )
func main() { fmt.Println(eulerSum()) }
func eulerSum() (x0, x1, x2, x3, y int) { var pow5 [250]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 }</lang>
- Output:
133 110 84 27 144
J
<lang J> (#~ (= <.)@((+/"1)&.:(^&5)))a.i.(#~ ] -:"1 /:~"1)@:([: ,/ ,"0 _1/)/ 4#,:}.250{.a. 27 84 110 133</lang>
Explanation:
There are 164059875 distinct possibilities to consider here (248 (+ %&(*/) ]) 1 2 3 4x) - so we need about 5GB memory to represent all the arguments using 64 bit integers. That's fairly trivial on some current laptops.
So, <lang J> a.i.(#~ ] -:"1 /:~"1)@:([: ,/ ,"0 _1/)/ 4#,:}.250{.a.</lang> finds all the possibilities for our four arguments and constrains them down to the distinct possibilities at each combining step. We use a character representation for our intermediate results, to limit memory overhead. Since this is a one-off problem, it's good enough to discard sequences which are not sorted.
Then, <lang J> (#~ (= <.)@((+/"1)&.:(^&5)))</lang> 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.
JavaScript
<lang javascript>
var eulers_sum_of_powers = function (iMaxN) {
var aPow5 = []; var oPow5ToN = {};
for (var iP = 1; 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, 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');
</lang>
Julia
<lang 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[1] 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 </lang>
- Output:
A solution is 27^5 + 84^5 + 110^5 + 133^5 = 144^5.
Pascal
slightly improved.Reducing calculation time by temporary sum and early break. <lang pascal>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. </lang>
- 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 6
<lang perl6>constant MAX = 250;
my %p5{Int}; my %sum2{Int};
for 1..MAX -> $i {
%p5{$i**5} = $i;
for 1..MAX -> $j {
%sum2{$i**5 + $j**5} = ($i, $j);
}
}
my @sk = %sum2.keys.sort; for %p5.keys.sort -> $p {
for @sk -> $s {
next if $p <= $s;
if %sum2{$p - $s} {
say (%sum2{$s}[],%sum2{$p-$s}[] X~ '⁵').join(' + ') ~ " = %p5{$p}⁵";
exit;
}
}
}</lang>
- Output:
84⁵ + 27⁵ + 133⁵ + 110⁵ = 144⁵
PHP
<lang php><?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";
?></lang>
- Output:
133^5 + 110^5 + 84^5 + 27^5 = 144^5
Python
<lang python>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())</lang>
- Output:
133**5 + 110**5 + 84**5 + 27**5 == 144**5
The above can be written as:
<lang python>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())</lang>
- 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: <lang python>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()</lang>
- Output:
144 (27, 84, 110, 133)
Racket
<lang scheme>#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 (expt sum 1/5)))
(break))))</lang>
- Output:
(133 110 84 27 144.00000000000003)
REXX
<lang rexx>/*REXX pgm finds an integer solution for: x0^5 + x1^5 + x2^5 + x3^5 =y^5*/ parse arg L H . /*get optional LOW & HIGH values.*/ if L== | L==',' then L= 0 + 1 /*Not given? Then use default. */ if H== | H==',' then H=250 - 1 /* " " " " " */ numeric digits length(H**5)*4 /*be able to use larger numbers. */ H1=H-1; H2=H-2; H3=H-3 /*calculate upper DO loop limits.*/ !.=0 /* [↓] pre─compute 5th powers. */
do pow=L to H; @.pow=pow**5; _=@.pow; !._=1; #._=pow; end
do x0=L to H3; s0= @.x0 /*traipse through all X0 values. */
do x1=x0+1 to H2; s1=s0+@.x1 /* " " " X1 " */
do x2=x1+1 to H1; s2=s1+@.x2 /* " " " X2 " */
do x3=x2+1 to H; s3=s2+@.x3 /* " " " X3 " */
if !.s3 then call results /*Found a solution? Display it.*/
end /*x3*/ /* [↑] the IF uses a boolen val.*/
end /*x2*/
end /*x1*/
end /*x0*/
say "Didn't find a solution."; exit 1 /*──────────────────────────────────RESULTS subroutine──────────────────*/ results: _=left(,10) /*used for spacing between values*/ say 'Found a solution:' say 'x0='x0 _ "x1="x1 _ 'x2='x2 _ "x3="x3 _ 'y='#.s3 exit /*stick a fork in it, we're done.*/</lang> output when using the default inputs:
Found a solution: x0=27 x1=84 x2=110 x3=133 y=144
Ruby
Brute force: <lang ruby>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"}</lang>
- Output:
27**5 + 84**5 + 110**5 + 133**5 = 144**5
Faster version:
<lang ruby>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"}</lang> The output is the same above.
Rust
<lang 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) }</lang>
- Output:
133^5 + 110^5 + 84^5 + 27^5 == 144^5
VBScript
<lang vb>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</lang>
- Output:
133 110 84 27 144
zkl
Uses two look up tables for efficiency. Counts from 0 for ease of coding. <lang zkl>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));
}</lang>
- 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.
<lang zkl>p5,sum2:=D(),D(); 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
}
}</lang> 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