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

Question

I'm trying to prove the statement above.

Any hints?

Consider divisibility by $3$ – Mark Bennet Jun 03 '19 at 20:10 — Mark Bennet, Jun 03 '19 at 20:10

score 2 · Accepted Answer · answered Jun 03 '19 at 20:10

2

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 $p=3$.

answered Jun 03 '19 at 20:10

nonuser

1 Answers1