, 2022

Solution:

All squares must end with $0, 1, 4, 5, 6$, or $9$, meaning that $p$ must end with $1$ and $9$. Moreover, since all odd squares are $1 \bmod 4$, we know that $p$ must be $1 \bmod 4$. This rules all primes except for $41, 61, 29, 89$. Since $17^2 = 289$, $19^2 = 361$, $23^2 = 529$, $89, 61$, and $29$ all work. To finish, we claim that $41$ does not work. If $100q + 41$ were a square, then since all odd squares are $1 \bmod 8$ we find that $4q + 1 \equiv 1 \pmod{8}$, implying that $q$ is even. But $241$ is not a square, contradiction.

The final answer is $29 + 61 + 89 = 179$.