Fastest algorithm to find if a BigInteger is a prime number or not? [duplicate]

Answer 1

Check if the integer is a "probable prime." If it's not you know for sure that it's composite and you avoid the slow factorization.

if (!testNumber.isProbablePrime(5)) return false;

Also you only need to make trial divisions only up to the square root of testNumber. If K is composite you know that its least prime factor must be at most sqrt(K).

Answer 2

Some other simple improvements would be to limit your set of possible numbers to only two and odd numbers in your outer loop and also to only iterate up to the square root of "index" (or index / 2 if too hard to calculate) in your inner loop.

Answer 3

Here's an optimized version using testing only up to sqrt(n) and using the Miller-Rabin test (as of Joni's answer):

public boolean returnPrime(BigInteger number) {
    //check via BigInteger.isProbablePrime(certainty)
    if (!number.isProbablePrime(5))
        return false;

    //check if even
    BigInteger two = new BigInteger("2");
    if (!two.equals(number) && BigInteger.ZERO.equals(number.mod(two)))
        return false;

    //find divisor if any from 3 to 'number'
    for (BigInteger i = new BigInteger("3"); i.multiply(i).compareTo(number) < 1; i = i.add(two)) { //start from 3, 5, etc. the odd number, and look for a divisor if any
        if (BigInteger.ZERO.equals(number.mod(i))) //check if 'i' is divisor of 'number'
            return false;
    }
    return true;
}