Prime is already confusing, now what?

We will see how to get an approach to this problem.

We all know that any number leaves a remainder of either $0,1,2$ on dividing by $3$ .

So, $a \equiv (0,1,2) \pmod{3}$ Then $a^2 \equiv 0,1 \pmod{3}$

Also $a^2 \equiv 0,1 \pmod{4}$

Now, lets come back to problem. For time being, lets consider $p >3$ . So we have either $p \equiv 1 , 2 \pmod{3}$ . That is $p^2 \equiv 1 \pmod{3} \ldots (1)$

Also, primes greater than 3 are odd, so $p^2 \equiv 1 \pmod{4} \ldots (2)$

Now, we notice that $11 \equiv -1 \pmod{3} \ldots (3)$ , and $11 \equiv -1 \pmod{4} \ldots (4)$

Now, from $(1),(3)$ , $p^2+11\equiv 0 \pmod{3}$ And from $(2),(4)$

$p^2+11\equiv 0 \pmod{4}$

Thus, from above results, $p^2+11$ is divisible by $12$ , which has $6$ factors. Also, its obvious that any other multiple of $12$ will lead to higher number of factors. Therefore, the value of $p$ will be $3$

This problem is interesting. Let us prove by contradiction that there are no primes $p \ge 5$ s.t. The expression shown has $6$ divisors.

Suppose that there $\exists$ a prime $p \ge 5$

Then it can be expressed in the form $p = 6k \pm 1$ , where $k \in \mathbb{Z}$

Substituting $p$ we get,

$(6k \pm 1)^2 + 11$

$\Rightarrow$

$36k^2 \pm 12k + 12$

$\Rightarrow$

$12(3k^2 \pm k + 1)$

Notice that $12= 3\times2^2$ has $6$ divisors and the expression $3k^2 \pm k + 1$ has $2$ or more divisors.

Hence there are no primes $p \ge 5$ s.t. $p^2 + 11$ has $6$ divisors.

$\therefore$ The only possibilities are $2$ and $3$ . Checking shows $p=3$ gives $3^2 + 11 = 20$ has $6$ divisors.