Except for semicolons, Michael Kleber writes:

[T]here will be at most two [square roots of an integer mod p].

Working mod p, you can still do a^2=b^2; a^2-b^2=0; (a+b)(a-b)=0;

a=b or a=-b, since there are no zero divisors mod p.

There's something different and interesting going on with

square roots in the integers mod n for n not prime,

and especially if n|24.

--Dan