Prime from a perfect square

Let $Y = (n^2 -1 ) = (n+1) \times (n-1)$

$Y$ can only be a prime number if $(n-1) =1$ or when $n = 2$ and $Y = 3$ .

Let the prime number be $p$ and the square number be $a^2$ . Then $p=a^2-1=(a+1)(a-1)$ In order for the number to be prime, the two factors $(a+1)$ and $(a-1)$ must be equal to $1$ and $p$ . We don't know which is which yet, so test the two cases:

Case 1

$\begin{cases} a+1=1 \\ a-1=p \end{cases}\Rightarrow a=0\Rightarrow p=-1$

But we need $p>2$ , so we reject this case.

Case 2

$\begin{cases} a+1=p \\ a-1=1 \end{cases}\Rightarrow a=2\Rightarrow p=3$

This case works!

Therefore $p=\large \color{#20A900}{\boxed{3}}$ .