Question

Find all primes $p$ and $q$ such that $p$ divides $q^2-4$ and $q$ divides $p^2-1$.

Answer 1

Suppose that p $\le$ q.
Since q divides (p - 1)(p + 1) and q > p - 1 it follows that q divides p + 1 and hence q = p + 1.
$\therefore p=2$ and $q=3$
On the other hand, if p > q then p divides $(q-2)(q+2)$ implies that p divides q + 2 or q - 2 = 0.
This gives either p = q + 2 or q = 2.
In the former case it follows that that q divides (q + 2)${}^2$ - 1, so q divides 3.
$\therefore p>2,q=2$ and $(p,q)=(5,3)$