Fastest way to determine if an integer's square root is an integer

Question

I\'m looking for the fastest way to determine if a long value is a perfect square (i.e. its square root is another integer):

1. I\'ve done it the ea

Answer 1

Newton's Method with integer arithmetic

If you wish to avoid non-integer operations you could use the method below. It basically uses Newton's Method modified for integer arithmetic.

/**
 * Test if the given number is a perfect square.
 * @param n Must be greater than 0 and less
 *    than Long.MAX_VALUE.
 * @return true if n is a perfect
 *    square, or false otherwise.
 */
public static boolean isSquare(long n)
{
    long x1 = n;
    long x2 = 1L;

    while (x1 > x2)
    {
        x1 = (x1 + x2) / 2L;
        x2 = n / x1;
    }

    return x1 == x2 && n % x1 == 0L;
}

This implementation can not compete with solutions that use Math.sqrt. However, its performance can be improved by using the filtering mechanisms described in some of the other posts.