There turns out to be several generalizations; one of them we will call S(n, m),

Question

There turns out to be several generalizations; one of them we will call S(n, m), which asks if there are integer solutions of the form: a n 1 + a n 2 + · · · + a n m = c n . Fermat’s last theorem is just S(n, 2). Show that the generalization of the theorem is in fact not true in general! By this, show that S(n, 4) for n 5 is false, by implementing a C++ program that will find a counterexample; i.e., some positive integers a1, · · · , a4, c and n 5 such that a n 1 + · · · + a n 4 = c n . You can be guaranteed that all of a1, · · · , a4, c 1000, so you don’t have to verify infinitely many possible cases.

write a c++ code to ececute

Explanation / Answer

#include <bits/stdc++.h>

using namespace std;

void testSomeNumbers(int limit, int n)

{

   if (n < 3)

     return;

   for (int a=1; a<=limit; a++)

     for (int b=a; b<=limit; b++)

     {

         // Check if there exists a triplet

                  int pow_sum = pow(a, n) + pow(b, n);

         double c = pow(pow_sum, 1.0/n);

         int c_pow = pow((int)c, n);

         if (c_pow == pow_sum)

         {

             cout << "Count example found";

             return;

         }

     }

     cout << "No counter example within given"

            " range and data";

}

// driver code

int main()

{

    testSomeNumbers(10, 3);

    return 0;

}

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