/* test3.cpp
Lucas-Lehmer test for primality of 2^p - 1
If p > 2 is a prime, then 2^p - 1 is prime if
and only if L[p-2] = 0, where the sequence L[i]
is defined as follows: L[0] = 4,
L[i+1] = (L[i]^2 - 2) modulo (2^p - 1)
----------------------------------------------- */
#include <stdio.h>
#include "largeint.h"
void pause();
void timer(int f);
int main() {
int p, i;
LargeInt L, mod;
printf("Enter a prime number: ");
scanf("%d", &p);
timer(0); // start timer
mod.powerTwo(p);
mod = mod - 1; // mod = 2^p - 1
L = 4;
for (i = 2; i <= p - 1; i++) {
if (i % 100 == 0)
printf("%4d\r", i);
L = (L * L - 2) % mod;
}
printf("\n2^%d - 1 is ", p);
if (L == zero)
printf("prime\n");
else
printf("not prime\n");
printf("and the calculation took "), timer(1);
return 0;
}
/* End of File */