Problem #9

First, for $p^{2}$ to have $3$ positive factors it must be the case that $p$ is prime, (the $3$ factors then being $1, p$ and $p^{2}$ ).

Next, for $12p + 1$ to have $3$ factors we must have that $12p + 1 = q^{2}$ where $q$ is prime. Now $p = 2$ gives us $12p + 1 = 25 = 5^{2}$ , and since $5$ is prime we have $p = 2$ as one solution.

For $p = 3$ we have $12p + 1 = 37$ which is not the square of a prime, (or of any number for that matter), so $3$ is not a solution.

Now suppose $p$ is any prime greater than $3$ , and that $12p + 1 = q^{2}$ for some prime $q$ , (also greater than $3$ ). Then $12p = q^{2} - 1 = (q - 1)(q + 1)$ . Now clearly $q$ will be odd so both $q - 1$ and $q + 1$ will be even and hence both divisible by $2$ . Also, since $q - 1$ and $q + 1$ are consecutive even numbers, one of them will be divisible by $4$ . Finally, since for any three consecutive integers one of them must be divisible by $3$ , one of $q - 1, q$ or $q + 1$ must be divisible by $3$ . But $q$ is a prime greater than $3$ so cannot be divisible by $3$ , and thus one of $q - 1$ or $q + 1$ must be divisible by $3$ .

This implies that $(q - 1)(q + 1)$ is divisible by $2*4*3 = 24$ , i.e., that $(q - 1)(q + 1) = 24n$ for some integer $n$ , and so $12p = 24n \Longrightarrow p = 2n$ . But we assumed that $p$ was a prime greater than $3$ and hence cannot be divisible by $2$ . We thus have a contradiction, implying that no prime greater than $3$ can be a solution to this problem.

We are thus left with $p = 2$ as the only solution to this problem, and the desired sum is just $\boxed{2}$ .