HMMT-Feb 2020 guts, Problem 11

Find the number of ordered pairs of positive integers $(x, y)$ with $x, y \leq 2020$ such that $3 x^{2}+10 x y+$ $3 y^{2}$ is the power of some prime.

Solution: Solution: We can factor as $(3 x+y)(x+3 y)$. If $x \geq y$, we need $\frac{3 x+y}{x+3 y} \in\{1,2\}$ to be an integer. So we get the case where $x=y$, in which we need both to be a power of 2 , or the case $x=5 y$, in which case we need $y$ to be a power of 2 . This gives us $11+9+9=29$ solutions, where we account for $y=5 x$ as well. $\fbox{29}$.

Answer: 29