quadratic residue...
1. Prove that is x^2 = a(mod n), then (n-x)^2 = a(mod n)
2. If p is a prime number and both a and b are quadratic residues mod p, prove that ab(mod p) is also a quadratic residue mod p.
2. $\displaystyle x^2\equiv{a}(\bmod.p)$for some natural number $\displaystyle x$ and $\displaystyle y^2\equiv{b}(\bmod.p)$for some natural number $\displaystyle y$ (a and b are quadratic residues modp)
Multiplying: $\displaystyle x^2\cdot{y^2}=(x\cdot{y})^2\equiv{a\cdot{b}}(\bmod .p)$
Thus a·b must be a quadratic residue
It is not that n is 0!
But n belongs to the equivalence class of 0. Put in layman terms, when we are seeing only remainders, n leaves the same remainder as 0 when they are divided by n, so they are equivalent.
To prove that directly from definition(if you are new to congruences)
$\displaystyle n|n(n-2x) \Rightarrow n|(n-x)^2 - x^2 \Rightarrow (n - x)^2 \equiv x^2 ~\text{(mod n)}$