Euler's sum of powers conjecture

From Rosetta Code
Revision as of 16:50, 2 September 2015 by rosettacode>Gerard Schildberger (→‎{{header|REXX}}: added a comment to help explain the DO loop count comments.)
Task
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   CDC 6600   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.

360 Assembly

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. <lang 360asm> 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
        MVC    WC,MASK
        ED     WC,X1              x1
        MVC    BUF+0(8),WC+8     
        MVC    WC,MASK
        ED     WC,X2              x2
        MVC    BUF+8(8),WC+8    
        MVC    WC,MASK
        ED     WC,X3              x3
        MVC    BUF+16(8),WC+8    
        MVC    WC,MASK
        ED     WC,X4              x4
        MVC    BUF+24(8),WC+8     
        MVC    WC,MASK
        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 PL8'250' 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 MASK DC X'40',13X'20',X'212060' CL17 BUF DS CL80

        YREGS  
        END

</lang>

Output:
      27      84     110     133     144


Ada

<lang Ada>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
     Ada.Text_IO.Put(Base'Image(Number));
  end loop;
  Ada.Text_IO.New_Line;
  

end Sum_Of_Powers;</lang>

Output:
 27 84 110 133 144

ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.8.win32

<lang algol68># max number will be the highest integer we will consider # INT max number = 250;

  1. 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;

  1. find the first a, b, c, d, e such that a^5 + b^5 + c^5 + d^5 = e^5 #
  2. as the fifth powers are in order, we can use a binary search to determine #
  3. whether the value is in the table #

BOOL found := FALSE; FOR a TO max number WHILE NOT found DO

   FOR b FROM a TO max number WHILE NOT found DO
       FOR c FROM b TO max number WHILE NOT found DO
           FOR d FROM c TO max number WHILE NOT found DO
               LONG INT sum   = fifth[a] + fifth[b] + fifth[c] + fifth[d];
               INT      low  := d;
               INT      high := max number;
               WHILE low < high
                 AND NOT found
               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</lang> Output:

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

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. <lang c>// Alexander Maximov, July 2nd, 2015

  1. include <stdio.h>
  2. 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; }</lang>

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++

The simplest brute-force find is already reasonably quick: <lang cpp>#include <iostream>

  1. include <cmath>
  2. include <set>

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; } } } } } // not found return false; }

int main(void) { int tm = clock(); if (!find()) cout << "Nothing found!\n"; printf("time=%d milliseconds\r\n", (int)((clock() - tm) * 1000 / CLOCKS_PER_SEC)); return 0; } </lang>

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: <lang> 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())

</lang> 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. <lang> 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; } } } } } // not found return false; } </lang> 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: <lang> 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]; </lang> With this refinement, the exhaustive search up to MAX=1000 takes 16.9 seconds.

Thanks, C guys!


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.

<lang> 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; } </lang>

Thanks, EchoLisp guys!

D

First version

Translation of: Rust

<lang d>import std.stdio, std.range, std.algorithm, std.typecons;

auto eulersSumOfPowers() {

   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[]);

}</lang>

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

<lang d>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.
           }
       }

}</lang>

Output:
144 Tuple!(uint, uint, uint, uint)(27, 84, 110, 133)

Run-time about 0.10 seconds.

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. <lang d>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[0] < b[0] }.uniq.copy(diffs0).length;
   auto diffs = diffs0.assumeSorted!q{ a[0] < b[0] };
   foreach (immutable x4; 4 .. MAX - 1) {
       foreach (immutable x3; 3 .. x4) {
           immutable s2 = pow5[x4] + pow5[x3];
           auto pd0 = diffs[1 .. $].upperBound(Pair(s2, 0));
           PairPtr pd = &pd0[0] - 1;
           foreach (immutable x2; 2 .. x3) {
               immutable sum = s2 + pow5[x2];
               const PairPtr endPtr = &diffs[$ - 1] + 1;
               // This lambda heap-allocates.
               pd = huntForward(pd, endPtr, (in PairPtr p) pure => (*p)[0] < sum);
               if (pd != endPtr && (*pd)[0] == sum && (*pd)[1] < x3) { // Find each solution only once.
                   immutable y = ((*pd)[0] + pow5[(*pd)[1]]) ^^ 0.2;
                   printf("%d %d %d %d : %d\n", x4, x3, x2, (*pd)[1], cast(int)(y + 0.5));
                   if (--how_many <= 0)
                       return true;
               }
           }
       }
   }
   return false;

}


void main() nothrow {

   if (!dPFind(100))
       printf("Search finished.\n");

}</lang>

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.

Run-time about 7.1 seconds.

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. <lang lisp> (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
		

</lang>

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

Works with: Fortran version 95 and later

<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

Translation of: Python

<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

Haskell

<lang haskell>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 ]</lang>
Output:
(27,84,110,133,144)

J

<lang J> require 'stats'

  (#~ (= <.)@((+/"1)&.:(^&5)))1+4 comb 248

27 84 110 133</lang>

Explanation:

<lang J>1+4 comb 248</lang> finds all the possibilities for our four arguments.

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.

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

<lang J>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

)</lang>

Use:

<lang J> find5 ┌─────────────┬───┐ │27 84 110 133│144│ └─────────────┴───┘</lang>

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. <lang java>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</lang> Output:

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

JavaScript

<lang javascript> 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, 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>

Output:
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. <lang jq># Search for y in 1 .. maxn (inclusive) for a solution to SIGMA (xi ^ 5) = y^5

  1. 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 ;</lang>

The task: <lang jq>sum_of_powers_conjecture(249)</lang>

Output:

<lang sh>$ jq -c -n -f Euler_sum_of_powers_conjecture_fifth_root.jq [27,84,110,133,144]</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.

Oforth

<lang 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 ]
             ]
           ]
        ]
     ]

}</lang>

Output:
>eulerSum
[27, 84, 110, 133]

Pascal

Works with: Free 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

Translation of: Python

<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

Translation of: Python

<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

PowerShell

Brute Force Search
This is a very slow algorithm. <lang powershell>#EULER.PS1 $max = 250

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++){

$sum=[Math]::pow($x0,5)+` [Math]::pow($x1,5)+` [Math]::pow($x2,5)+` [Math]::pow($x3,5) $S1=[Math]::truncate([Math]::pow($sum,0.2))

if($sum -eq [Math]::pow($S1,5)){ Write-host $x0 " " $x1 " " $x2 " " $x3 "-" $S1 return }

        }
     }
  }

}</lang>

Output:
133   110   84   27 - 144

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:

Works with: Python version 2.6+

<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

Translation of: C++

<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 (inexact->exact (round (expt sum 1/5)))))
     (break))))</lang>
Output:
(133 110 84 27 144)

REXX

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   (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 5thpowers
  •   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 userID   G. Brougnard}
  •   see if the sum of any four 5th powers is equal to   any   5th power
  •   {all of the above utilizes REXX's   sparse stemmed array hashing   which eliminates the need for sorting.}

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


In essence, the new formula being solved is:       aⁿ   +   bⁿ   +   cⁿ   ==   xⁿ   ─   dⁿ

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 sums match any of the   ≈30k   differences

<lang rexx>/*REXX program finds unique positive integer solution(s) for the equation: ───── ────────────────────────────────────── 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 /* " " " " " " */ numeric digits 1000 /*be able to handle the next expression*/ numeric digits max(9,length(3*H**5)) /* " " " " 3* [H to 5th power]*/ h1=H-1; h2=H-2; h3=H-3 /*calculate the upper DO loop limits.*/ say center(' 'strip(translate(sourceLine(2), ,"*─/"))' ', 79, "─") /*title.*/

                                      /* [↓]  define values of  5th  powers. */

!.=0; do pow=L to H; @.pow=pow**5; _=@.pow;  !._=1; #._=pow; end ?.=0; do j=L+3 to H-1; do k=j+1 to H; _=@.k-@.j; ?._=1; end; end

                                      /* [↑]  define  5th  power differences.*/
  1. =0 /*#: is the number of solutions found.*/ /* [↓] for N=∞ solutions.*/
   do       a=L    to h3;  s0=   @.a  /*traipse through possible  A  values. */   /*◄──done       246 times.*/
     do     b=a+1  to h2;  s1=s0+@.b  /*   "       "        "     B    "     */   /*◄──done    30,381 times.*/
       do   c=b+1  to h1;  s2=s1+@.c  /*   "       "        "     C    "     */   /*◄──done 2,511,496 times.*/
       if ?.s2  then do d=c+1  to H;  s3=s2+@.d  /*find appropriate solution.*/
                     if !.s3  then call results  /*A solution?  Then show it.*/
                     end   /*d*/                 /* [↑]   !.S3  is a boolean.*/
       end                 /*c*/
     end                   /*b*/
   end                     /*a*/

if #==0 then say "Didn't find a solution."; signal done /*────────────────────────────────────────────────────────────────────────────*/ results: _=left(,5); #=#+1 /*_ is spacing between values; bump #.*/ say _ 'solution' #":" _ 'a='a _ "b="||b _ 'c='c _ "d="d _ 'x='#.s3 if #<N then return /*return, keep searching for more sols.*/ done: exit # /*stick a fork in it, we're all done. */</lang> output when using the default inputs:

─────────────────── aⁿ + bⁿ + cⁿ + dⁿ == xⁿ      where  n=5 ───────────────────
      solution 1:       a=27       b=84       c=110       d=133       x=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:

Translation of: Python

<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

Translation of: ERRE

<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.

Translation of: Python

<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