1. Find all prime numbers $p$ and $q$ for which the number $2 p^{2} q+45 p q^{2}$ is a perfect square.
I/1. First, let's assume that $p=q$. Then the number $47 p^{3}$ must be a perfect square. Since it is divisible by 47 and 47 is a prime number, it must also be divisible by $47^{2}$, which implies that 47 divides $p^{3}$ and thus 47 divides $p$. Since $p$ is a prime number, it must be equal to 47. Indeed, for $p=q=47$, the number $2 p^{2} q+45 p q^{2}$ is equal to $47^{4}$ and is therefore a perfect square.

Now let $p \neq q$. Since the number $2 p^{2} q+45 p q^{2}=p q(2 p+45 q)$ is a perfect square divisible by $p$, it must also be divisible by $p^{2}$. Thus $p$ divides $q(2 p+45 q)$, or, since $p$ and $q$ are coprime, $p$ divides $2 p+45 q$. From this it follows that $p$ divides $45 q$ or that $p$ divides 45. Therefore, $p=3$ or $p=5$. Similarly, we conclude that $q$ divides $2 p+45 q$, from which it follows that $q$ divides $2 p$ or $q=2$. Thus, $p q(2 p+45 q)=4 p(p+45)$. If $p=3$, this number is equal to $4 \cdot 3 \cdot 48=24^{2}$, and for $p=5$ it is equal to 4000, which is not a perfect square. The only solutions are therefore $p=q=47$ and $p=3, q=2$.[^0]