how to check that, is number square of another number, python3

Question

numbers are real long, (for example n=341235129628026803631240389584456), I tried that way: if n**(1/2)-floor(n**(1/2)) == 0: true but program count n**(1/2) like 1.8472550707144556e+16, but i need number without e

Answer 1

You can do it with binary search.
Function f(x) = x² is monotonically increasing for x = 1, 2, ..., k, ... (i.e. natural numbers). Therefore, equation f(m) = m² = n has at most one root (with respect to m), which you can find by binary searching.
It will require O(log³(n)) operations (because there are O(log(n)) iterations of BS, each one requires squaring, which can be done in O(log²(n))).

The implementation may look like this:

def find_if_square(n):
  L, R = 0, n + 1
  while R - L > 1:
    M = (R + L) // 2
    if M * M <= n:
      L = M
    else:
      R = M

  return L * L == n

Here we maintain an invariant: L² &leq; n < R². The distance between L and R decreases until R = L + 1. Then we only need to check whether L² is strictly equal to n (otherwise, it is less, according to our invariant).