Prove that if P5 is a prime number, then P^2 +2 is never prime?

Question

Prove that if P>5 is a prime number, then P^2 +2 is never prime?

hitaro9 · Answer

These prime proofs confuse me. Can someone help?

ganeshie8 · Answer

If you know fermat's little thm, for all $p\ne 3$ we have: 
$$p^2 \equiv 1 \pmod{3}$$

so 
$$p^2+2\equiv 1+2\equiv 3\equiv 0 \pmod{3}$$

in other words $3|(p^2+2)$ for all $p \ne 3$

hitaro9 · Answer

I don't think we've seen that theorem. Is it difficult to prove without it?

ganeshie8 · Answer

Not at all, just show that $p^2+2$ is divisible by 3 using division algorithm

ganeshie8 · Answer

By division algorithm any integer can be represented in one of the three forms : 3k, 3k+1, 3k+2

since p is a prime > 3, we only get two cases : p = 3k+1, 3k+2

hitaro9 · Answer

Ohhh okay. That's pretty simple then.

ganeshie8 · Answer

case1 : p = 3k+1

p^2 + 2 = (3k+1)^2 + 2 = 9k^2+6k+1+2 = 3(3k^2+2k+1)  which is divisible by 3

ganeshie8 · Answer

you can work the other case similarly :)

hitaro9 · Answer

Yeah, that was a lot easier than I was thinking it was, Thanks a ton.

ganeshie8 · Answer

technically, you can prove ANY divisibility proofs using division algorithm. these get painful once the cases increase but you can see that these always work

hitaro9 · Answer

Right, that makes sense. Again, thank you.