Prime number and Divisors

Question

Let $p$ be a prime number such that $p^2+12$ has exactly $5$ divisors. What is the maximum value of $p$ ?

I came across this question in a Math Olympiad Competition and had no idea how to solve it

Answer 1

Notice that a number that has 5 divisors must be in the form of $q^{4}$ for some prime $q$. So we get:

$q^{4}=p^{2}+12 \implies (q^{2}-p)(q^{2}+p)=12$

Then do some casework on it.

Answer 2

Using @LordSharktheUnknown's hint that a number with an odd number of divisors must be square

we are then looking for cases where $p^2+12=a^2$ i.e. $a^2-p^2 = 12$

Since $7^2-6^2=13$, this tells you $a^2-p^2 \gt 12$ for $a \ge 7$ and $p \lt a$

so here we must have $a \lt 7$ and thus prime $p \in \{2,3,5\}$. Considering these:

• $2^2+12=16=4^2$ so $p=2$ is indeed a possibility
• $3^2+12=21$, not a square, so $p=3$ is not a possibility
• $5^2+12=37$, not a square, so $p=5$ is not a possibility