# Interesting question about twin cousins

If $p$ and $p + 2$ are (twin) primes, then $p + (p + 2)$ is divisible by $12$, where $p > 3$.

• Is this a question? Are you trying to prove this? I don't understand. – davidlowryduda Jul 4 '11 at 5:23
• And how do cousins enter into it? – Gerry Myerson Jul 4 '11 at 5:43
• @Gerry I think it may be some language thing. I don't know if it was deliberately intended as some sort of pun or something. But in Spanish and (after checking in google) Portuguese the word for cousin is "primo" which is also the word used for prime. I'm guessing that the OP is from Portugal or at least speaks Portuguese, from the website that is mentioned in his profile. – Adrián Barquero Jul 4 '11 at 5:49
• I love it that "interesting" is part of the question. I think the idea is that we get to decide what's interesting by upvoting (or not...) – The Chaz 2.0 Jul 4 '11 at 5:58
• @Gerry: If the parents of the twins were twins, then the twins are also cousins... or something? – Asaf Karagila Jul 4 '11 at 6:02

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.

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.

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.

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 $6k-1$.

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.