$p$ and $10p^2+11$ are prime numbers. Find $p$

Question

$p$ and $10p^2+11$ are prime numbers. Find $p$

I see that if $p=3$ the $p=3$ and $10p^2+11=101$ are prime numbers.

Answer 1

It's the only case. Otherwise, $p=3k\pm1$, wich implies that $3\mid 10p^2+11$.

Answer 2

Hint: For $p>3$, $10p^2+11$ is divisble by $3$.

Answer 3

This is more for the benefit of others who might come across this question. I figure that as soon as you read $p = 3k \pm 1$ in Jose's answer, you were like "Of course."

So, for the others: Have you tried having Mathematica or Wolfram Alpha give you a bunch of values of $10p^2 + 11$? I get 51, 101, 261, 501, 1221, 1701, 2901, etc. It looks like they're all nontrivial multiples of 3, with the exception of 101.

Indeed if $p \neq \pm 3$, we have $p^2 \equiv 1 \pmod 3$, and $10p^2 \equiv 1 \pmod 3$ also, since $10 \equiv 1 \pmod 3$. Since $11 \equiv 2 \pmod 3$, our intuition about multiples of 3 is confirmed.