Slovenija 2008

First assume $p=q$. Then $47p^3$ must be a perfect square. Since $47p^3$ is divisible by $47$ which is prime, it must also be divisible by $47^2$. This implies that $47$ divides either $p^3$ or $p$. But $p$ is a prime, so this is only possible when $p$ is equal to $47$. Indeed, when $p=q=47$ we have $2p^{2}q+45p{q}^2=47^4$, which is a perfect square.

Now, let $p\ne q$. Since $2p^{2}q+45p{q}^2=pq(2p+45q)$ is a perfect square divisible by $p$, it must also be divisible by $p^2$. So, $p$ divides $q(2p+45q)$ or, since $p$ and $q$ are coprime, $p$ divides $2p+45q$. We conclude that $p$ divides $45q$, so $p$ divides $45$. There are two possible cases: $p=3$ or $p=5$. A similar argument shows that if $q$ divides $2p+45q$, then $q$ divides $2p$, which implies $q=2$. So, $pq(2p+45q)=4p(p+45)$. If $p=3$, then this expression is equal to $4\cdot 3\cdot 48=24^2$. If $p=5$, then it is equal to $4000$ and this is not a perfect square. The only two pairs of primes with the required property are $p=q=47$ and $p=3,q=2$.