# Proof involving twin primes

I have to prove that if $p$ and $p+2$ are twin primes, $p>3$, then $6\ |\ (p+1)$. I figure that any prime number greater than 3 is odd, and therefore $p+1$ is definitely even, therefore $2\ |\ (p+1)$. And if I can somehow prove $3\ |\ (p+1)$, then I would be done. But I'm not sure how to do that. Help!

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Exploit the fact that we can write any prime $p\gt3$ as $p=6n\pm1$. – Workaholic Mar 16 at 16:53

The easy way.

Note that one of $p,p+1,p+2$ must be divisible by $3$, since they are three consecutive numbers, and since $p$ and $p+2$ are prime, that must be $p+1$. We can do the same to show that $p+1$ is divisible by $2$.

Looking modulo $6$.

We can look $\mod 6$. We see that \begin{align} 6k+0\equiv 0\mod 6&\Rightarrow 6|6k+0\\ 6k+1\equiv 1\mod 6&\Rightarrow \text{possibly prime}\\ 6k+2\equiv 2\mod 6&\Rightarrow 2|6k+2\\ 6k+3\equiv 3\mod 6&\Rightarrow 3|6k+3\\ 6k+4\equiv 4\mod 6&\Rightarrow 2|6k+4\\ 6k+5\equiv 5\mod 6&\Rightarrow \text{possibly prime}\\ \end{align}

So for a number to be prime it must be either of the form $6k+1$ or $6k-1$ (equivalent to $6k+5$ since $6k-1=6(k-1)+5$). So if you have two primes, $p$ and $p+2$, then $p=6k-1$ and $p+2=6k+1$ for some $k$; thus, $p+1=6k$ is a multiple of $6$.

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i like this answer. It uses $6$ rather than $3$ and ignores that we have $2$ taken care of already by a different argument. However, I might be misunderstanding something when you write $2|6k + 0$ in the third line. Don't you mean $2|6k + 2$? Similarly for the next two lines where $3$ and $4$ are the remainders.. – Frank Hubeny Mar 16 at 12:52
@FrankHubeny, yes, thank you. That was a result of some not-reading-my-copy-paste-work thourough enough. – vrugtehagel Mar 16 at 12:55

If $p = 3k+1$ then $p+2$ is not prime.So $p=3k+2$ and you are done.

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Note that for every integer $k$, exactly one of $k,k+1$ is divisible by $2$ and exactly one of $k,k+1,k+2$ is divisible by $3$. What do your hypothesis tell you if you put $k=p$?

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