Solving equation a5+b5+c5+d5+e5=f5

The equation A5+b5+c5+d5+e5=f5 has exactly one integer solution satisfying the 0<a≤b≤c≤d≤e≤f≤75. Please write a program to find out the solution:

1 usingSystem;2 3 namespacereversetheexponentiation4 {5     class Program6     {7         Static voidMain (string[] args)8         {9Program P =NewProgram ();Ten p.reversetheexponentiation (); One         } A  -         voidreversetheexponentiation () -         { the             intA, B, C, D, E, F; -              for(F = the; F >0; f--) -             { -                  for(E =1; E <= F; e++) +                 { -                      for(D =1; D <= E; d++) +                     { A                          for(C =1; C <= D; C++) at                         { -                              for(B =1; B <= C; b++) -                             { -                                  for(A =1; A <= B; a++) -                                 { -                                     if(Math.pow (A,5) + Math.pow (B,5) + Math.pow (C,5) + Math.pow (D,5) + Math.pow (E,5) = = Math.pow (F,5)) in                                     { -Console.WriteLine ("A,b,c,d,e,f is: {0},{1},{2},{3},{4},{5}", A, B, C, D, E, F); to                                     } +                                 } -                             } the                         } *                     } $                 }Panax Notoginseng             } -         } the     } +}

View Code

Output:

Note: If the condition becomes 0≤a≤b≤c≤d≤e≤f≤75 there are many solutions, the output is as follows:

Solving equation a5+b5+c5+d5+e5=f5