Check if BigInteger is not a perfect square has code to compute the integer square root of a Java BigInteger. Here it is translated into C#, as an extension method.
public static BigInteger Sqrt(this BigInteger n)
{
if (n == 0) return 0;
if (n > 0)
{
int bitLength = Convert.ToInt32(Math.Ceiling(BigInteger.Log(n, 2)));
BigInteger root = BigInteger.One << (bitLength / 2);
while (!isSqrt(n, root))
{
root += n / root;
root /= 2;
}
return root;
}
throw new ArithmeticException("NaN");
}
private static Boolean isSqrt(BigInteger n, BigInteger root)
{
BigInteger lowerBound = root*root;
BigInteger upperBound = (root + 1)*(root + 1);
return (n >= lowerBound && n < upperBound);
}
Informal testing indicates that this is about 75X slower than Math.Sqrt, for small integers. The VS profiler points to the multiplications in isSqrt as the hotspots.
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