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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
#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!
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