Prove that if numbers $p$ and $8p^2+1$ are prime numbers then $8p^2-1$ is also a prime number.

Question

How to prove that if numbers $p$ and $8p^2+1$ are prime numbers then $8p^2-1$ is also a prime number?

Answer 1

Hint: $8p^2+1$ isn't prime very often. Can you find a prime that always divides it (except in one case)?

Answer 2

Yes, if $p\ne 3$ then $p^2\equiv_3 1$ so $q:=8p^2+1 \equiv _3 0$ so $q=3$ which is impossible, so $p=3$...

Answer 3

Prime numbers different from $2,3,5$ are either of the form $6k+1$ or of the form $6k+5$. Insert $p=6k+1$ in $8p^2+1$ and conclude. Then insert $p=6k+5$ in $8p^2+1$ and conclude. Some cases are still there to check them so check them.

Some kind of converse is also true:

If both $8p^2+1$ and $8p^2-1$ are primes then they are twin primes and the sum of two twin primes is always divisible by $12$ if they are not the pair $(3,5)$.

So we have $12 | 16p^2$ which implies $3 | 4p^2$. So $p=3$.