GCD of $p+q$ for all pairs of primes $p,q$ of the form $p-q=12n+2x$

Question

I came across this result, and i'm having trouble explaining it. I use it as an argument in a proof, so i need to explain this behavior, in the shortest possible way, and most importantly prove that this is true for all possible pairs of odd prime $p,q$, with $p>q>3$.

Any pair of primes $p,q$ that satisfy $p-q=12n$ also satisfies $p+q\equiv0\pmod2$.
Any pair of primes $p,q$ that satisfy $p-q=12n+2$ also satisfies $p+q\equiv0\pmod{12}$.
Any pair of primes $p,q$ that satisfy $p-q=12n+4$ also satisfies $p+q\equiv0\pmod{6}$.
Any pair of primes $p,q$ that satisfy $p-q=12n+6$ also satisfies $p+q\equiv0\pmod{4}$.
Any pair of primes $p,q$ that satisfy $p-q=12n+8$ also satisfies $p+q\equiv0\pmod{6}$.
Any pair of primes $p,q$ that satisfy $p-q=12n+10$ also satisfies $p+q\equiv0\pmod{12}$.


In other words:

p-q     gcd (p+q)
___________
0       2
2       12
4       6
6       4
8       6
10      12
12      2
14      12
16      6
18      4
20      6
22      12
24      2
26      12
28      6
30      4
32      6
34      12

Answer 1

Here's a way to approach the $12n + 2$ case. I believe you can use this technique for all cases.

All primes are of the form $6k + 1$ or $6k + 5$. That means modulo $12$ all primes must have remainder $1, 5, 7$ or $11$.

When $p - q = 12n + 2$ there are two possibilities. Either $p = 12k + 7$ and $q = 12j + 5$ or $p = 12k + 1$ and $q = 12j + 11$. And in both cases it holds:

$$p + q \equiv 12k + 7 + 12j + 5 \equiv 0 \mod 12$$ $$p + q \equiv 12k + 1 + 12j + 11 \equiv 0 \mod 12$$