If p and $8p^2+1$ are prime numbers then $p=3$?

Question

I'm trying to prove the statement above.

Any hints?

Answer 1

Use the fact you want to prove in previous post: if $3\nmid n$ then $3\mid n^2-1$.

If $p\ne 3$ then $3\mid p^2-1$ so $$3\mid 9p^2-(p^2-1)= 8p^2+1$$ so $8p^2+1=3$ which is impossible. So $p=3$.