Question

Find all triples $(p,q,r)$ of primes such that $pq=r+1$ and $2(p^2+q^2)=r^2+1$, where $p,q,r$ are any natural numbers.

Answer 1

If p and q are both odd, then $r=pq-1$ is even so $r=2$.
But in this case pq $\ge$ 3 $\times$ 3 = 9 and hence there are no solutions.
This proves that either $p=2$ or $q=2$. If $p=2$ then we have $2q=r+1$ and $8+2q^2=r^2+1$.
Multiplying the second equation by 2 we get 2r${}^2$ + 2 = 16 + (2q)${}^2$ = 16 + (r + q)${}^2$.
Rearranging the terms, we have $r^2-2r-15=0$, or equivalently $(r+3)(r-5)=0$.
This proves that $r=5$ and hence $q=3$.
Similarly if $q=2$ then $r=5$ and $p=3$.
$\therefore (p,q,r)=(2,3,5)$ and $(p,q,r)=(3,2,5)$.