How can I prove that one of $n$, $n+2$, and $n+4$ must be divisible by three, for any $n\in\mathbb{N}$ Intuitively it's true, but I just can't think of how to say it "properly".
Take for example, my answer to the following question:

Let $p$ denote an odd prime. It is conjectured that there are infinitely many twin primes $p$, $p+2$. Prove that the only prime triple $p$, $p+2$, $p+4$ is the triple $3,\ 5,\ 7$.

And my solution:

Given an odd integer $n$, between the three integers $n$, $n+2$ and $n+4$, one of them must be divisible by $3$... Three possible cases are $n=3k$, $n+2=3k$, and $n+4=3k$. The only such possible $k$ that makes $n$ prime is $k=1$. In this case, given an odd prime $p$, either $p=3$, $p+2=3$, or $p+4=3$. This would imply that $p=3$, $p=1$, or $p=-1$. The only of these three that is prime is $p=3$, therefore the only three evenly distributed primes are $3$, $5$, and $7$.

Is there a "better" way that I can assert that one of the integers is divisible by 3? This feels too weak.

A somewhat silly answer:
You probably know that $\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}{2}$ and $\displaystyle \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$; if not these can be proven easily via induction. Consider then
\begin{align*}
 \sum_{i=1}^n (i^2 + 3i + 1) &= \frac{n(n+1)(2n+1)}{6} + 3\frac{n(n+1)}{2} + n \\
&= \frac{(2n^3 + 3n^2 + n) + 9(n^2+n) + 6n}{6} \\
&= \frac{2n^3 +12 n^2 + 16 n}{6} \\
&= \frac{n^3 + 6n^2 + 8n}{3} \\
&= \frac{n(n+2)(n+4)}{3}
\end{align*}
On the left hand side, you have a sum of integers, so the right hand side is an integer. Hence $3 | n(n+2)(n+4)$, and since $3$ is prime it divides one of the three factors.
