Interesting question about twin cousins If $p$ and $p + 2$ are (twin) primes, then $p + (p + 2)$ is divisible by $12$,
where $p > 3$. 
 A: Often if you are asked to show that a number $n$ is divisible by some positive number $N$, a good way to do this is to factor $N = p_1^{a_1} \cdots p_r^{a_r}$ into distinct prime powers and show separately that $n$ is divisible by $p_i^{a_i}$ for all $i$.  (This is clearly necessary, and it is sufficient because $n$ is then divisible by the least common multiple of all these numbers, which is $N$.)  At any rate this sort of local approach of working prime by prime becomes an increasingly important technique as one continues one's study of number theory.
So here I am saying to show separately that $2p+2 = 2(p+1)$ is divisible by $3$ and $4$.  The divisibility by $4$ is especially straightforward: for divisibility by $3$ it is equivalent to show that $3$ divides $p+1$, i.e., $p \equiv 2 \pmod{3}$.  For that, it is enough to rule out the other two possibilities $p \equiv 0 \pmod{3}$ and $p \equiv 1 \pmod{3}$: I leave that to you.
A: To show $p+(p+2)=2p+2$ is divisible by $12$, you can show $p+1$ is divisible by $6$, or $p+1\equiv 0\pmod{6}$, that is, $p\equiv 5\pmod{6}$. Now $p\gt 3$ cannot be equivalent to $0$, $2$, $3$, or $4$ modulo $6$, else it wouldn't be prime. If $p\equiv 1\pmod{6}$, then $p+2\equiv 3\pmod{6}$, which means $p+2$ is not prime, as it is divisible by $3$. So $p\equiv 5\pmod{6}$ as needed.
A: I've seen 4 answers, all good, but no one has suggested working directly on 12. Modulo 12, if $p$ is prime (and greater than 3), then $p$ is 1, 5, 7, or 11 (since those are the only numbers less than and relatively prime to 12). If $p+2$ is also prime, that leaves $p$ being 5 or 11 mod 12, and $p+p+2$ being a multiple of 12. 
If "modulo" is unfamiliar to you, put it this way: $p$ must be of one of the forms $12k+1$, $12k+5$, $12k+7$, $12k+11$ (again, because for any other $m$ less than 12, $m$ will have a common factor of 2 or 3 or more with 12, so $12k+m$ will have that factor and won't be prime). Then for $p+2$ to be prime as well, we're down to $p$ being $12k+5$ or $12k+11$, and then you can calculate $p+p+2$ and see there's a 12 to factor out. 
A: Or to show $6 \mid p+1$ note that any prime $p>3$ is of the form $6k-1$ or $6k+1$. But $p \neq 6k+1$ because if $p =6k+1$, then $p+2 = 6k+3$ which doesn't give a prime always. For example $k=1$ it gives $9$, which is not a prime. Hence $p$ has to be of the form $6k-1$.
A: Although I don't quite know that I understand the purpose of this question, I wanted to answer. This is actually very simple. If we recall that primes > 3 will be either congruent to 1 or 5 $\mod 6$, then we quickly see that all $p$ in this problem are congruent to $5 \mod 6$. How do you get the other factor of 2? One prime will be congruent to $1 \mod 4$, and another $3 \mod 4$. So their sum is divisible by 4 and 6. Therefore, it's divisible by 12.
