For $n$ even, assume
$\large n = \frac{q-p}{6}$
For $a,b \in\mathbb{N}$, primes $p,q$ can be written as
$q = 6a\pm1$
$p = 6b\pm1$
There are $4$ possible cases
Case 1
$n = \frac{(6a-1)-(6b-1)}{6}$, then $n=a-b$
Case 2
$n = \frac{(6a-1)-(6b+1)}{6}$, then $n = a-b-1/3$
Case 3
$n = \frac{(6a+1)-(6b-1)}{6}$, then $n = a-b+1/3$
Case 4
$n = \frac{(6a+1)-(6b+1)}{6}$, then $n=a-b$
For Cases $2$ and $3$, $n$ cannot be even because of $\pm 1/3$.
For Cases $1$ and $4$, both result in $n=a-b$.
Not sure how to proceed from here?