User:Joe/Sandbox

Theorem
If $$x\not\equiv\pm{y}\pmod{n}$$ and $$x^2\equiv y^2\pmod{n}$$ then $$1<\gcd(x-y,n)<n$$ and $$1<\gcd(x+y,n)<n$$.

Proof
Let $$x^2\equiv y^2 \pmod{n}$$

$$\implies n|(x^2-y^2) \implies n|(x+y)(x-y)$$

$$\implies p|(x+y)(x-y)$$ $$ \forall$$ prime divisors $$p$$ of $$n$$

$$\implies p|(x-y)$$ or $$p|(x+y)$$

But since $$x\not\equiv -y \pmod{n}$$, then $$n\not{|}(x+y)$$,

and $$x\not\equiv y \pmod{n}$$, then $$n\not{|}(x-y)$$

$$\therefore \gcd(x+y,n)<n$$ and $$\gcd(x-y,n)<n$$

So if $$p|(x-y)$$ then $$1<p\leq\gcd(x-y,n)$$, also $$\exist q$$ s.t. $$q|n$$ and $$q|(x+y)$$ and $$1<q\leq\gcd(x+y,n)$$

Likewise if $$p|(x+y)$$ then $$1<p\leq\gcd(x+y,n)$$, also $$\exist q$$ s.t. $$q|n$$ and $$q|(x-y)$$ and $$1<q\leq\gcd(x-y,n)$$