Question: If $p= n^{2} + 2 $ is a prime number, prove that n is a multiple of 3.

Question

Helo guys!

I have doubts on this question:

"If $p= n^{2} + 2 $ is a prime number, prove that n is a multiple of 3."

I did the test with n=3k (multiple of 3), n=3k+1 and n=3k+2, ($k \in R $)and I tried to show that in the last two cases $p$ was not a prime number and I got to this:

• $p1=(3k+1)^{2} +2 = 9k^{2} + 6k +3 $
• $p2=(3k+2)^{2} +2 = 9k^{2} + 12k +6 $

Why aren't p1 and p2 prime numbers? I can not understand it.

Thanks.

Answer 1

$9k^2+ 6k+ 3= 3(k^2+ 2k+ 1)$, and thus is divisible by $3$.

$9k^2+ 12k+ 6= 3(k^2+ 4k+ 2)$, and thus this is also divisible by $3$.

Answer 2

Hint:
If $n$ is not a multiple of 3 then $n^2 \equiv 1 \pmod{3}.$

What would that say about $n^2 + 2$?

The case of $n=1$ is an exception.