Show that if $p_1$ and $p_2$ are primes such that $p_2 = p_1 + 2$ and $p_1> 3$, then $p_1 + p_2 \equiv 0$ mod $12$

Question

I tried using that $a \equiv a$ mod $m$ and other module properties, but could not solve.

Answer 1

$p_1+p_2=2(p_1+1)$, $p_1$ and $p_2$ are odd implies that $p_1+1$ is even and $2(p_1+1)$ is divisible by $4$. One of $p_1,p_1+1,p_1+2$ is divisible by $3$ it is $p_1+1$.

Answer 2

If $p$ is prime and $p>2$, then $p\equiv\pm1\pmod{12}$ or $p\equiv\pm5\pmod{12}$. So, if both $p$ and $p+2$ are prime, then $p\equiv-1\pmod{12}$ and $p+2\equiv1\pmod{12}$ or $p\equiv-5\pmod{12}$ and $p+2\equiv5\pmod{12}$. In both cases, $p+(p+2)\equiv0\pmod{12}$.

Answer 3

$p_2 \equiv p_1 +2 \pmod 3$.We know $p_1 \equiv 1$ or $p_1 \equiv 2$ modulo $3$ by parity but we must have the latter since $p_2$ is also odd. Then $p_1+p_2 \equiv 2+4 \equiv 0 \pmod 3$. Can you show that $p_1 +p_2 \equiv 0 \pmod 4$ by the same reasoning?

Answer 4

Every integer is congruent, mod $6$, to one of $-2,-1,0,1,2,3$.

Write $p_1 = 6k + r$, where $k \in \mathbb{Z}$ and $r \in \{-2,-1,0,1,2,3\}$.

Since $p_1$ is prime and $p_1 > 3$, it follows that $r$ is odd, and $r \ne 3$.

But we also can't have $r=1$, else $p_2 = 6k + 3$.

Thus, we must have $r=-1$, hence $$p_1 + p_2 = (6k-1) + (6k+1) = 12k$$