The equation A**5 + B**5 + C**5 + D**5 + E**5 = F**5 has exactly one integral so

Question

The equation A**5 + B**5 + C**5 + D**5 + E**5 = F**5 has exactly one
integral solutions that satisfies 0 < A <= B <= C <= D <= E <= F <= 75. Write
a program to find the solution. Hint: First, precompute all values of X**5 and
store them in an array. Then, for each tuple (A, B, C, D, E), you will only
need to verify that some F exists in the array. (There are several ways to check
for F, one of which is to use a binary search to check for F. Other methods
might prove to be more efficient.)

Explanation / Answer

//Try to execute in 64 or 32 bit operating system...  

#include <stdio.h>
    #include <limits.h>
    #include<math.h>
    long a[75];
    int binerySearch(long keynum)
    {
      int low=1;
      int high=75;
      int mid;
      do
        {
                mid= (low + high) / 2;
                if ( keynum < a[mid] )
                high = mid - 1;
                else if ( keynum > a[mid])
                low = mid + 1;
                } while( keynum!=a[mid] && low <= high);

                if( keynum == a[mid] )
                {
                        return mid;
                }
                else
                {
                        return -1;
                }
    }
    int main()
    {
        int i,j,k,l,m;
        for(i=1;i<=75;i++)
        {
            a[i]=pow(i,5);
        }
        for(i=1;i<=75;i++)
        {
        for(j=i;j<=75;j++)
                {
                        for(k=j;k<=75;k++)
                        {
                                for(l=k;l<=75;l++)
                                {
                                        for(m=l;m<=75;m++)
                                        {
                                              if(binerySearch(a[i]+a[j]+a[k]+a[l]+a[m])!=-1)
                                        {
                                            printf("Numbers are %d**5+%d**5+%d**5+%d**5+%d**5=%d**5",i,j,k,l,m,binerySearch(a[i]+a[j]+a[k]+a[l]+a[m]));
                                                                                 }
                                        }
                                }
                        }
                }
        }

        return 0;
    }

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