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If $p$ and $8p^2+1$ is prime number than $$8p^2+2p+1$$ is prime number proof.

I know that prime number can write as a multiply 1 and $8p^2+2p+1$, so if I show that this number is a multiply $a*b$ where $a\not=1$ and $b\not=1$ than that is not a prime number, can someone help me?

nonuser
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Marko Škorić
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    Where did this come from? A priori, I doubt that this is true. – marty cohen Oct 31 '18 at 20:54
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    When you say you have no idea, I don't believe you. You know things about prime numbers. You must have checked a few obvious general things and a few small primes before you came in here and have us this problem. Please tell us about that! – Arthur Oct 31 '18 at 20:56
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    (Repeating the definition of prime is not what I referred to. What happens for $p=2$? What about $3,5$ or $7$? Does it help to look at it modulo $4$ or $6$? These are things you should always try in any problem about primes unless a solution comes to you immediately.) – Arthur Oct 31 '18 at 21:04
  • Questions like this appear in many slight variations - all solvable by exactly the same idea. I chose one as a dupe, but there are many others, e.g. here and here and here. – Bill Dubuque Oct 31 '18 at 21:23

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If $p = 3k\pm 1$ then $8p^2+1 \equiv 0 \pmod3$ so $8p^2+1=3$ which is imposibile. So $p=3$ then $8p^2+2p+1 = 79$

nonuser
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