I have been told that this proof is incorrect, however, I'm having a hard time seeing the issue.
Prove that for any integer $n > 0$ which is a perfect square, $n+4$ is not a perfect square:
Proof:
$n > 0$
Since $n$ is a perfect square, $a^2=n$
The subsequent perfect square can be written as ${b^2=\left(a+1\right)}^2$
The difference between consecutive squares is then ${|b}^2-a^2|=|2a+1|$
$\forall a\in\mathbb{Z}(|2a+1|\neq 4)$
$\square$
Edit: Thanks everyone, appreciate it :)