TITLE: Primes on the form $p^2-1$
QUESTION [8 upvotes]: Prove that there exists a unique prime number of the form $p^2 − 1$ where $p\geq 2$ is an integer.
I have no idea how to approach the question. any hints will be greatly appreciated

REPLY [6 votes]: This question is asked so frequently on this website that the answer is immediately and unthinkingly coughed up in its shortest form each time.
What I'd like to do is show you one way you might be able to come up with the answer yourself if you don't already know it.
Does $p$ itself have to be prime? I will assume not, so I prefer to use the letter N instead of the letter P. Alright, then, can $n^2 - 1$ be prime?
Suppose $n$ is coprime to $3$. Then $n^2 \equiv 1 \pmod 3$ and therefore $n^2 - 1$ is a multiple of $3$. In every case except $n = -2$ or $2$, we will see that $n^2 - 1$ is a nontrivial multiple of $3$ (since the squares of negative numbers are positive numbers, I won't say anything further about negative numbers in this answer).
Then for $n^2 - 1$ to be prime with $n > 2$, $n$ has to be a multiple of $3$. If $n$ is an odd multiple of $3$, then $n^2$ is odd also but $n^2 - 1$ is an even composite number, that is, a nontrivial multiple of $2$.
Now the only possibility left is  that $n$ is a multiple of $6$. Checking a few small cases, we see a pattern emerge:

$6^2 - 1 = 35 = 5 \times 7$
$12^2 - 1 = 143 = 11 \times 13$
$18^2 - 1 = 323 = 17 \times 19$
$24^2 - 1 = 575 = 5^2 \times 23$ but wait! That's $23 \times 25$.

This suggests $(n - 1)(n + 1)$. Applying FOIL, we get $$(n - 1)(n + 1) = n^2 + n - n - 1 = n^2 - 1.$$