Say $n\in \mathbb N$ with n odd and $n|y^2-1$ for some $y \in\mathbb N$. Show that $n=gcd(n,y-1)gcd(n,y+1)$. Also show a counterexample when $n$ is even.
I know that $n|y^2-1=(y+1)(y-1)$ so $n|(y+1)$ and $n|(y-1)$.
Since $n|(y+1)$ we can say $y+1=nx$ for some $x\in \mathbb Z$.
Similarly $n|(y-1)$ we can say $y-1=ny$ for some $y\in \mathbb Z$.
So $y^2-1=(y+1)(y-1)=n^2xy$.
I am having issues connecting this back to what we are trying to show.
First, note that $3 \mid (4^2-1) = 3\cdot 5$, but while $3\mid (4-1)$, we don't have $3\mid (4+1).$ So your sentence beginning with "I know that..." is incorrect.
Second, let's think about the counterexample for when $n$ is even. Just take $n=2$ and $y$ any odd number, say $y=5$. Then $2 \mid (5^2-1) = 4\cdot 6.$ Then compute $\gcd(2,4)\cdot\gcd(2,6) = 2\cdot 2 \neq 4.$ This counterexample should give you some insight as to what's going on in the odd case. The key fact is that $y-1$ and $y+1$ differ by $2$, so the $\gcd(y-1,y+1) =1$ or $2$. So the only prime factor that $y-1$ and $y+1$ could have in common is $2$.
Third, suppose $n$ is an odd prime number $p$. If $p\mid y^2-1$ then either $p\mid (y-1)$ or $p\mid (y+1)$ but not both because of the above paragraph. Then one of $\gcd(p,y-1)$ or $\gcd(p,y+1)$ equals $p$ and the other equals one. So you must have $\gcd(p,y-1)\cdot\gcd(p,y+1)= 1\cdot p = p.$ So the theorem holds for $n$ and odd prime.
Fourth, you should be able to extend this into a nice proof for general odd $n$.
One always has $(y+1,y-1)=1\text { or } 2$ according to $y$ be even or odd respectively. Therefore if $n|(y+1)(y-1)$ then all prime factor of $n$ dividing, say, $(y+1)$ does not divide $(y-1)$. Let $n=PQ$ with $(P,Q)=1$. We have, say, $P|(y+1)$ and $Q|(y-1)$ so $(P,y+1)=P$ and $(Q,y-1)=Q$. The result follows.
As counter example take $4|(11^2-1)=12\cdot10$. In this case $(4,12)\cdot(4,10)=8\ne4$
