From: Sam Kong <sam.s.kong@...>
Date: 2007-02-12T04:30:05+09:00
Subject: What's the correct and fast way to determine if a (gig) number is a perfect square?

Hello,

I'm solving a math problem in Ruby.
I need to determine if a number is a perfect square.
If the number is small, you may do like the following.

def perfect_square? n
  sqrt = n ** 0.5
  sqrt - sqrt.to_i == 0
end

But Float number has limitation on precision.
Thus the function won't work correctly for big numbers like
(123456789123456789).

How would you solve such a case?
It should be fast as well as correct because I will use it repeatedly.

Thanks in advance.

Sam