Fermat's Last Theorem in C++

Fermat’s last theorem in number theory also known as Fermet’s conjecture is a theorem that states that for power n greater than 2. No three values a, b, c satisfy −

          aⁿ + bⁿ = cⁿ

i.e.     if n <= 2,               aⁿ + bⁿ = cⁿ 

          Otherwise,   aⁿ + bⁿ != cⁿ

Example of values for n = 2, 

3, 4, 5 => 3² + 4² = 9 + 16 = 25 = 5². 

5, 12, 13 => 25 + 49 = 169 = 13².

In this problem, we are given three values, L, R, pow denoting range [L, R] and power. Our task is to verify the fermat’s last theorem for the given range and power.

Let’s take an example to understand the problem,

Example 1: 

Input: L = 4, R = 12, power = 2

Output: 5, 12, 13

Example 2: 

Input: L = 4, R = 12, power = 4

Output: No such value found

Solution Approach: 

Here, we will check if the power is greater than 2 or not. If it's greater, print No such value found.  

Else check in the limit if there is a value satisfying the condition aⁿ + bⁿ = cⁿ.

Program to illustrate the working of our solution,

Example

Live Demo

#include <iostream>
#include <math.h>
using namespace std;

void checkFermatsLastTh(int L, int R, int n) {

   if (n >= 3)
   cout<<"No example found!";
   else {
      for (int a = L; a <= R; a++)
      for (int b=a; b<=R; b++)
      {
         int sum = pow(a, n) + pow(b, n);
         double c = pow(sum, 1.0/n);
         int cpowN = pow((int)c, n);
         if (cpowN == sum)
      {
         cout<<"Example found with value : "<<a<<", "<<b<<", "<<c;
         return;
      }
   }
   
   cout << "No example found!";
   }
}

int main() {
   
   int L = 3, R = 15, power = 2;
   cout<<"Run 1 \n";
   checkFermatsLastTh(L, R, power);
   
   L = 5, R = 42; power = 5;
   cout<<"\n\nRun 2\n";
   checkFermatsLastTh(L, R, power);

   return 0;
}

Output −

Run 1
Example found with value : 3, 4, 5
Run 2
No example found!