If $p>3$ and $p+2$ are twin primes then $6\mid p+1$ 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!
 A: 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 $6$.
A: If $p = 3k+1$ then $p+2$ is not prime.So $p=3k+2$ and you are done.
A: 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 $k=p$?
A: *

*Consider a triple {m,n,o} of consecutive natural numbers.


*Then, at least one of these numbers must be divisible by 2 - an even number, and at least one of these numbers - even or odd - must be divisible by 3 (not necessarily but possibly the same number).


*The triple must either be odd-even-odd or even-odd-even.


*Since all twin primes are odd primes, only an odd-even-odd triple can contain a twin prime tuple.


*So, because of (2.), for primes > 3, the middle number n between twin primes must be both divisible by 2 and 3.


*In other words, in the triple {p, n, p+2} (| p+2 prime),  n ≡ 0 (mod 6). ■
