Finding all possible values

Question

we have to find all possible prime values $(p,q,r)$ such that

$ pq = r + 1 $

$ 2(p^2+q^2) = r^2 + 1 $

I do not know how to start looking for an answer.

Answer 1

From the second equation we find that $r$ is odd. Hence $r+1$ is even, and one of $p$ and $q$ must be $2$. We may assume that $p=2$. Then we get $2q = r+1$ and $$8+2q^2=r^2+1=(r+1)(r-1)+2=2q(2q-2)+2,$$ implying $2q^2-4q-6=0$ or $(q-3)(q+1)=0$. This has $q=3$ as only prime solution, hence $q=3$ and $r = 2q - 1 = 5$.

In conclusion, there are two solutions: $(p,q,r)=(2,3,5)$ and $(p,q,r)=(3,2,5)$.