Prove that if x is a prime number and it doesn't equal 3, then x squared + 2 can be divided by 3

what is the proof for this?

Let x be a prime number not equal to 3. Then x is not divisible by 3. We know that any number can be written in the form 3k, 3k+1, or 3k+2, for some integer k. Since x is not divisible by 3, it is not of the form 3k, so it is either of the form 3k+1 or 3k+2.

First suppose x=3k+1. Then we have:
x² + 2 = (3k+1)² + 2 = 9k² + 6k + 1 + 2 = 9k² + 6k + 3 = 3(3k² + 2k + 1)
So, x is divisible by 3.

Now suppose instead that x=3k+2. Then we have:
x² + 2 = (3k+2)² + 2 = 9k² + 12k + 4 + 2 = 9k² + 12k + 6 = 3(3k² + 4k + 2)
So again, x is divisible by 3.

Therefore, in each case, x is divisible by 3, as desired.

Hope that helps :)

Edit:
Pauley Morph did use the fact that x is not composite, which follows immediately from x being prime.

x must be of the form x = 3A - 1, x = 3A, or x = 3A + 1

If x is not 3, then it cannot be of the form x = 3A because it would then be a composite number.

So x = 3A ± 1

Then x² + 2 = 9A² ± 6A + 1 + 2 = 9A² ± 6A + 3 = 3(3A² ± 2A + 1)

which is a multiple of 3.

Just to make things clear:
This actually works for any numbers (both composite and prime) that is not dividable by 3.
Don't believe me, try 22,26,125.

Since prime is subset of all numbers then the statement should be also true.