Understanding proof of $(m+n+2)^2 \neq 3(mn + 1)$ The problem is: Prove that there are no integers $m$ and $n$ such that:
$$(m+n+2)^2 = 3(mn + 1)$$
I found the explanation for the problem solution, but I still don't understand it.
Here's the solution I found:

If there are numbers $m$ and $n$ such that: $(m+n+2)^2 = 3(mn + 1)$, then $3 
|(m+n+2)^2 \implies 9|(m+n+2)^2$ and $3 | (mn+1)$, that means that numbers $m$ and $n$ cannot be divisible by $3$ and they cannot have the same remainder after dividing by $3$. Let's rewrite $m$ and $n$ as $m = 3p + 1$ and $n = 3q + 2$, but then we have $m+n+2 = 3(p+q+1) + 2$ which is impossible since $3 | m+n+2$

The only thing I don't understand is why $3 | (mn+1)$?
Can someone explain this to me? Thanks.

Alternative proof
$$(m+n+2)^2-3mn-3=0$$
$$\iff  m^2 - m (n -4) + n^2 + 4 n + 1=0 $$
$$\begin{align}\Delta &=-3 n^2 - 24 n + 12\\
&=60 - 3 (n + 4)^2\end{align}$$
$$60 - 3 (n + 4)^2≥0, ~ 60 - 3 (n + 4)^2=T^2, T\in\mathbb Z$$
This means, such $n\in\mathbb Z$ doesn't exist.

Small supplement:
$$\begin{align}0&≤60 - 3 (n + 4)^2\\
&=60 - 3 (|n + 4|)^2\\
&=T^2,~T\in\mathbb Z \end{align}$$
All possible cases are as follows:
$$|n+4|:=1,2,3,4.$$
