Prime Factorization Program in Java

Question

I am working on a prime factorization program implemented in Java. The goal is to find the largest prime factor of 600851475143 (Project Euler problem 3). I think I have m

Answer 1

For those answers which use a method isPrime(int) : boolean, there is a faster algorithm than the one previously implemented (which is something like)

private static boolean isPrime(long n) { //when n >= 2
    for (int k = 2; k < n; k++)
        if (n % k == 0) return false;

    return true;
}

and it is this:

private static boolean isPrime(long n) { //when n >= 2
    if (n == 2 || n == 3) return true;

    if (n % 2  == 0 || n % 3 == 0) return false;

    for (int k = 1; k <= (Math.floor(Math.sqrt(n)) + 1) / 6; k++)
        if (n % (6 * k + 1) == 0 || n % (6 * k - 1) == 0) return false;

    return true;
}

I made this algorithm using two facts:

1. We only need to check for n % k == 0 up to k <= Math.sqrt(n). This is true because for anything higher, factors merely "flip" ex. consider the case n = 15, where 3 * 5 = 5 * 3, and 5 > Math.sqrt(15). There is no need for this overlap of checking both 15 % 3 == 0 and 15 % 5 == 0, when we could just check one of these expressions.
2. All primes (excluding 2 and 3) can be expressed in the form (6 * k) + 1 or (6 * k) - 1, because any positive integer can be expressed in the form (6 * k) + n, where n = -1, 0, 1, 2, 3, or 4 and k is an integer <= 0, and the cases where n = 0, 2, 3, and 4 are all reducible.

Therefore, n is prime if it is not divisible by 2, 3, or some integer of the form 6k ± 1 <= Math.sqrt(n). Hence the above algorithm.

--

Wikipedia article on testing for primality

--

Edit: Thought I might as well post my full solution (*I did not use isPrime(), and my solution is nearly identical to the top answer, but I thought I should answer the actual question):

public class Euler3 {

    public static void main(String[] args) {
        long[] nums = {13195, 600851475143L};

        for (num : nums)
            System.out.println("Largest prime factor of " + num + ": " + lpf(num));

    }

    private static lpf(long n) {
        long largestPrimeFactor = 1;
        long maxPossibleFactor = n / 2;

        for (long i = 2; i <= maxPossibleFactor; i++)
            if (n % i == 0) {
                n /= i;
                largestPrimeFactor = i;

                i--;
            }

            return largestPrimeFactor;

    }

}