<s>[INST] 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.
 [/INST] Suppose that n is not divisible by 3.
It means :
n % 3 = {1,2} let call it x
and
n + 2 % 3 = (x + 2) % 3
and n + 4 % 3 = (x + 4) % 3
So if x = 1
n + 2 % 3 = 3 % 3 = 0
n + 4 % 3 = 5 % 3 = 2
n + 2 is divisible by 3
n + 4 is not divisible by 3
else if x = 2
n + 2 % 3 = 4 % 3 = 1
n + 4 % 3 = 6 % 3 = 0
n + 2 is not divisible by 3
n + 4 is divisible by 3
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