Use fractions (=2 integers) and the Babylonian/Newtonian approach: X(n+1) = (X(n)^2 + N) / (2 * X(n))

Perhaps using binary search will work.

By approximation. Star with a = 1 and check if a*a is > N. If so you return a-1 else a++ and move forward. The fact that here we cannot use the floating point calculation makes everything trivial and not much accurate. Anyway convert this strategy into an accurate one using floating point it's easy as well.