, 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 $2p^2q + 45pq^2 = 47^4$, which is a perfect square.

Now, let $p \neq q$. Since $2p^2q + 45pq^2 = pq(2p + 45q)$ 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 $p = 3, q = 2$.