Challenge: If a number, 'X' is prime then 2^X-1 should also be prime, find at which prime this fails to be true. This uses the trial and error method of finding prime numbers.

Prime.cpp

#include <iostream>
#include <cmath>
#include "Prime.h"

int main()
{
	int x;
	bool Failure = false;

	for (x = 1; Failure != true; ++x)
	{
		if (IsPrime(x))
		{
			std::cout << "Prime: " << x << std::endl;
			if (!IsPrime(pow(2,x)-1))
			{
				std::cout << "Fails on " << x << std::endl;
				Failure = true;
			}
		}
	}

	return 0;
}

Prime.h

#ifndef PRIME_H_24122011
#define PRIME_H_24122011
/* Declare function */
int IsPrime(int number)
{
	/* 1 and 2 are both prime, so this catches them */
	if(number < 3 && number > 0)
		return 1;

	/* Any even number other than 2 is not prime. */
	if(number % 2 == 0)
		return 0;

	/* Declare variables */
	int x;
	int limit = number; // This will make the search faster

	/* Loop while x < limit. */
	for(x = 3; x < limit; x += 2)
	{
	/* If this hits the number isn't prime. */
		if(number % x == 0)
			return 0;

		/* Making the limit number/x instead of just number
			we gain a noticable speed increase */
		limit = number / x;
	}

	/* Number is prime. */
	return 1;
}
#endif