Does the source really say "1st Grade" Oh @below that's a nice solution

jasperE3 wrote: Find all prime numbers $p$ for which the number $p^2+11$ has less than $11$ divisors. The answer is $p=2, 3$, which indeed works. Now, consider $p \geq 3$. Then $p$ is odd, so $p^2 \equiv 1 \pmod 4 \Rightarrow 4 \mid p^2+11.$ Since $p$ is prime and not smaller than $3$, it follows that $p^2 \equiv 1 \pmod 3 \Rightarrow 3 \mid p^2+11.$ Therefore $12 \mid p^2+11.$ Since $p^2+11>12$ for $p \geq 3$, it follows that $p^2+11=12k=2^2 \cdot 3 \cdot k$ for some positive integers $k$. It follows that $p^2+11$ has at least $(2+1)(1+1)(1+1)=12$ divisors, contradiction.

$p=5$ also works

wassupevery1 wrote: it follows that $p^2+11=12k=2^2 \cdot 3 \cdot k$ for some positive integers $k$. It follows that $p^2+11$ has at least $(2+1)(1+1)(1+1)=12$ divisors, contradiction. This is wrong. It would be true if $k$ has a prime divisor different from $2$ and $3$, but as pointed out above it happens for $p=5$ that $k=3$ and then the number of divisors is just $9$. But it can be easily fixed: We only need to consider the case $k=2^a \cdot 3^b$. But then $p^2+11$ has $(3+a)(2+b)$ divisors. So $(3+a)(2+b) \le 11$ and hence only $a=2, b=0$ and $a=1,b=0$ and $a=b=0$ and $a=0, b=1$ are possible. This leads to $p^2+11 \in \{48,24,12,36\}$ which only works for $p=5$. So this is the only new solution.