I'm trying to prove that if $p$ is prime,$p\ne3$ then $p^2+2$ is composite. Here's my attempt:
Every number $p$ can be put in the form $3k+r, 0\le r \lt 3$, with $k$ an integer. When $r=0$, the number is a multiple of 3, so that leaves us with the forms $3k+1$ and $3k+2$. The first one will be even when $k$ is odd, and the second one will be even when $k$ is even. So we will see what happens for each form in the case that $k$ is even (for the first form) and $k$ is odd (for the second form):
- $p=3k+1$, $k$ is even
Since $k$ is even, we can write it as $k=2q$ for some $q$. Then
So $p^2+2$ is composite.
- $p=3k+2$, $k$ is odd
Then, $k$ can be written as $k=2q+1$, for some $q$. Then
And again, $p^2+2$ is composite.
Is that a correct proof? Is not the same that comes in the answer books.