Check if BigInteger is not a perfect square

Question

I have a BigInteger value, let\'s say it is 282 and is inside the variable x. I now want to write a while loop that states:

while b2 isn\'t a perfect square:         


        

Answer 1

private static boolean isSqrt(BigInteger n, BigInteger root)
{
    final BigInteger lowerBound = root.pow(2);
    final BigInteger upperBound = root.add(BigInteger.ONE).pow(2);
    return lowerBound.compareTo(n) <= 0
        && n.compareTo(upperBound) < 0;
}

I tried the above using JavaScript BigInt:

function isPerfectSqrt(n, root) {
  const lowerBound = root**2n;
  const upperBound = (root+1n)**2n
  return lowerBound <= n && n < upperBound;
}

And found it was only about 60% as fast (in Node V8) as the one-liner:

function isPerfectSqrt(n, root) {
  return (n/root === root && n%root === 0n)
}