# Vantieghems Theorem for Primality Test

• Difficulty Level : Medium
• Last Updated : 13 May, 2021

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 where , is congruent to In other words, a number n is prime if and only if. 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 divides , then n is prime.

## C++

 // C++ code to verify Vantieghem's Theorem#include 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 codeint main(){    checkVantieghemsTheorem(10);    return 0;}

## Java

 // Java code to verify Vantieghem's Theoremimport java.util.*;class GFG{ static void checkVantieghemsTheorem(int limit){    long prod = 1;    for (long n = 2; n < limit; n++)    {         // Check if above condition is satisfied        if (((prod - n < 0 ? 0 : prod - n) % ((1 << n) - 1)) == 0)            System.out.print(n + " is prime\n");         // product of previous powers of 2        prod *= ((1 << n) - 1);    }} // Driver codepublic static void main(String []args){    checkVantieghemsTheorem(10);}} // This code is contributed by rutvik_56.

## Python3

 # Python3 code to verify Vantieghem's Theoremdef checkVantieghemsTheorem(limit):         prod = 1    for n in range(2, limit):                 # Check if above condition is satisfied        if n == 2:            print(2, "is prime")        if (((prod - n) % ((1 << n) - 1)) == 0):            print(n, "is prime")                     # Product of previous powers of 2        prod *= ((1 << n) - 1)     # Driver codecheckVantieghemsTheorem(10) # This code is contributed by shubhamsingh10

## C#

 // C# code to verify Vantieghem's Theoremusing System;class GFG{  static void checkVantieghemsTheorem(int limit)  {    long prod = 1;    for (long n = 2; n < limit; n++)    {       // Check if above condition is satisfied      if (((prod - n < 0 ? 0 : prod - n) % ((1 << (int)n) - 1)) == 0)        Console.Write(n + " is prime\n");       // product of previous powers of 2      prod *= ((1 << (int)n) - 1);    }  }   // Driver code  public static void Main()  {    checkVantieghemsTheorem(10);  }} // This code is contributed by pratham76.

## Javascript

 
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.

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