$p+1$ is divisible by $3$

Question

Prove that the sum of a pair of twin primes, each greater than $3$, is divisible by $12$

I assume primes are $p$ ,$p+2$ Then their summation is $2(p+1)$ Since $p$ is prime, $p+1$ is divisible by $2$ .hence $2(p+1)$ is divisible by $4$.

Now again I have to show $2(p+1)$ is divisible by $3$ .then $2(p+1)$ will be divided by $12$. But I am getting $p+1$ is divisible by $2$ always

Answer 1

First, $p$ is a prime number greater than $3$, so it can not be divisible by $3$. Hence, either we have $p=1 \;(\text{mod}\; 3)$ or $p=2 \;(\text{mod}\;\;3)$.

In the first case, we then have $p+2=0\;(\text{mod}\;\;3)$. Because $p+2$ also is prime, this is impossible.

Hence, we must have $p=2 \;(\text{mod}\;3)$. It follows that $p+1$ is divisible by $3$.

Answer 2

Since $p$, $p+1$ and $p+2$ are three consecutive integers, one of them must be divisible by $3$. You can rule out $p$ and $p+2$ since they are prime and they are not $3$.