Bounds of GCD for Sum and Difference Congruent Squares

Theorem
Let $x, y, n$ be integers.

Let:
 * $x \not \equiv \pm y \pmod n$

and:
 * $x^2 \equiv y^2 \pmod n$

where $a \equiv b \pmod n$ denotes that $a$ is congruent to $b$ modulo $n$.

Then:
 * $1 < \gcd \left({x - y, n}\right) < n$

and:
 * $1 < \gcd \left({x + y, n}\right) < n$

where $\gcd \left({a, b}\right)$ is the GCD of $a$ and $b$.

Proof
But since $x \not \equiv -y \pmod n$, then:
 * $n \nmid \left({x + y}\right)$

and since $x \not \equiv y \pmod n$, then:
 * $n \nmid \left({x - y}\right)$

Therefore:
 * $\gcd \left({x - y, n}\right) < n$

and:
 * $\gcd \left({x + y, n}\right) < n$

So if $p \mathrel \backslash \left({x - y}\right)$ then:
 * $1 < \gcd \left({x - y, n}\right) < n$

and also there exists $q$ such that:
 * $q \mathrel \backslash n$
 * $q \mathrel \backslash \left({x + y}\right)$
 * $1 < q \le \gcd \left({x + y, n}\right)$

Likewise if $p \mathrel \backslash \left({x + y}\right)$ then:
 * $1 < \gcd \left({x + y, n}\right) < n$

and also there exists $q$ such that:
 * $q \mathrel \backslash n$
 * $q \mathrel \backslash \left({x - y}\right)$
 * $1 < q \le \gcd \left({x - y, n}\right)$