Vantieghems Theorem for Primality Test

Vantieghems Theorem is a necessary and sufficient condition for a number to be prime. It states that for a natural number n to be prime, the product of $2^i - 1$ where $0 < i < n$ , is congruent to $n~(mod~(2^n - 1))$ .

In other words, a number n is prime if and only if.

$\prod _{1\leq i\leq n-1}\left(2^{i}-1\right)\equiv n\mod \left(2^{n}-1\right).$

Examples:

For n = 3, final product is (2¹ – 1) * (2² – 1) = 1*3 = 3. 3 is congruent to 3 mod 7. We get 3 mod 7 from expression 3 * (mod (2³ – 1)), therefore 3 is prime.
For n = 5, final product is 1*3*7*15 = 315. 315 is congruent to 5(mod 31), therefore 5 is prime.
For n = 7, final product is 1*3*7*15*31*63 = 615195. 615195 is congruent to 7(mod 127), therefore 7 is prime.
For n = 4, final product 1*3*7 = 21. 21 is not congruent to 4(mod 15), therefore 4 is composite.

Another way to state above theorem is, if $(2^n - 1)$ divides $\prod _{1\leq i\leq n-1}\left(2^{i}-1\right) - n$ , then n is prime.

// CPP code to verify Vantieghem's Theorem 
#include <bits/stdc++.h> 
using namespace std; 
  
void checkVantieghemsTheorem(int limit) 
{ 
    long long unsigned prod = 1; 
    for (long long unsigned n = 2; n < limit; n++) { 
  
        // Check if above condition is satisfied 
        if (((prod - n) % ((1LL << n) - 1)) == 0) 
            cout << n << " is prime\n"; 
  
        // product of previous powers of 2 
        prod *= ((1LL << n) - 1); 
    } 
} 
  
// Driver code 
int main() 
{ 
    checkVantieghemsTheorem(10); 
    return 0; 
} 

Output:

2 is prime
3 is prime
5 is prime
7 is prime

The above code does not work for values of n higher than 11. It causes overflow in prod evaluation.

My Personal Notes

Suggest a Topic

Recommended Posts: