If p##\geq##5 is a prime number, show that p^{2}+2 is composite?

[Math100]

Homework Statement

If p$\geq$5 is a prime number, show that p^{2}+2 is composite.
[Hint: p takes one of the forms 6k+1 or 6k+5.]

Relevant Equations

None.

Proof: Suppose p$\geq$5 is a prime number.
Applying the Division Algorithm produces:
p=6k, p=6k+1, p=6k+2, p=6k+3, p=6k+4 or p=6k+5 for some k$\in\mathbb{Z}$.
Since p$\geq$5 is a prime number,
it follows that p cannot be divisible by 2 or 3 and p must be odd.
Thus, p takes one of the forms 6k+1 or 6k+5.
Now we consider two cases.
Case #1: Suppose p=6k+1.
Then we have p^{2}+2=(6k+1)^{2}+2
=36k^{2}+12k+3
=3(12k^{2}+4k+1)
=3m,
where m=12k^{2}+4k+1 is an integer.
Thus, p^{2}+2 is composite.
Case #2: Suppose p=6k+5.
Then we have p^{2}+2=(6k+5)^{2}+2
=36k^{2}+60k+27
=3(12k^{2}+20k+9)
=3m,
where m=12k^{2}+20k+9 is an integer.
Thus, p^{2}+2 is composite.
Therefore, if p$\geq$5 is a prime number, then p^{2}+2 is composite.

Above is my proof for this problem. Can anyone please verify/review to see if this is correct?

 

[PeroK]

It looks fine. Slicker is
$$p = \pm 1 \ ( mod \ 6) \ \Rightarrow \ p^2 +2 = 3 \ (mod \ 6)$$$$\Rightarrow \ p^2+2 = 3k$$

 

[Math100]

It looks fine. Slicker is
$$p = \pm 1 \ ( mod \ 6) \ \Rightarrow \ p^2 +2 = 3 \ (mod \ 6)$$$$\Rightarrow \ p^2+2 = 3k$$

Thank you.

 

[PeroK]

You should definitely be using modulo arithmetic more. It's not like it's something that takes a lot to learn. It's the sort of thing that some people use before they realize it's a thing that gets taught in maths classes.