Prove or disprove that for any positive integer $m$ there is a positive integer $n$ such that $mn + 1$ is a perfect square.
My reasoning: We're asked to show that $mn + 1 = p^2$ and that given any positive integer $m$ we can find positive integer $n$ where this holds.
Rewriting, we get $$mn = p^2 - 1 = (p - 1)(p + 1)$$
Case #1
Let $m = p - 1$ and $n = p + 1$. For a given $m$ we just take $$n = p + 1 = (p - 1) + 2 = m + 2$$
Note also that since $m > 0$, here we have $n > 0$.
Case #2
Let $m = p + 1$ and $n = p - 1$. Given some value $m$, we should choose $n$ like $$n = p - 1 = (p + 1) - 2 = m - 2$$
Now, this is fine for $m > 2$ but for $m \in \{1,2\}$ we have that the corresponding $n$ is $n \leq 0$. This goes against our requirement that $n > 0$.
Is there an error in this reasoning? To me, looking only at Case #1 seems like a loss of generality.