Proof: If n is a perfect square, $\,n+2\,$ is NOT a perfect square
"Prove that if n is a perfect square, $\,n+2\,$ is NOT a perfect square." I'm having trouble picking a method to prove this. Would contraposition be a good option (or even work for that matter)? If not, how about contradiction?
Hint: Every perfect square is either $0$ or $1$ modulo $4$.
Suppose $n=m^2$ and $n+2=k^2$. Clearly $k>m$, so $k\ge m+1$. But then $$n+2\ge (m+1)^2=m^2+2m+1\ge m^2+3=n+3$$ a contradiction.
$$n=a^2\,\,,\,\,n+2=b^2\Longrightarrow 2=(n+2)-n=b^2-a^2=(b-a)(b+a)$$
Now check that this is impossible (and $\,b>0\,$ )