1) $p=q$ result $47p^3=a^2$, so $p=q=47$. 2) Else, must have $2p^2q + 45pq^2=p^2q^2k^2$, so $2p+45q=pqk^2$, so $q$ divide $2$, thus $q=2$, and $p$ divide $45$, thus $p=3,5$, now casework.