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 $100 q+41$ were a square, then since all odd squares are $1 \bmod 8$ we find that $4 q+1 \equiv 1(\bmod 8)$, implying that $q$ is even. But 241 is not a square, contradiction. The final answer is $29+61+89=179$.