The modulo 6 of primes at certain distances from another prime While doing some statistics on primes pairs for a given gap between them, I noticed this phenomenon:
For each $p$ prime $\ge 7$ and all $q$ prime $> 6p$ such that $(q - 4p)$ is prime, we have the same value for ${q\!\! \mod\!6}$   (either only $1$ or only $5$).
I have verified it through calculations for $p$ between $7$ and $7919$ and for around $1200$ first $q$ values for each $p$.
I am not sure if the above conjecture is true and how hard it is to prove it, can you number theorists take a stab at it ?
Thanks !
 A: All primes $>3$ are either $\pm 1 \pmod 6$.  Let's go case by case.
First, assume $p\equiv 1 \pmod 6$.  Then we have $$q-4p\equiv q-4\pmod 6$$  In that case it is not possible for $q\equiv 1\pmod 6$ as that would give  $$1-4\equiv 3\pmod 6$$  And the "prime" would be divisible by $3$.
Similarly, $$p\equiv -1 \pmod 6\implies q-4p=q+4\pmod 6$$ and here we can't have $q\equiv -1 \pmod 6$ for the same reason, and we are done.
A: It's the direction $(\Rightarrow)$ below in the special case: $\,p,\,q,\,p\!-\!4q\,$ are all primes $ > 3.$
Lemma $\ $ If $\ (p,6)=1=(q,6)\ $ then $\ (p\!-\!4q,6)=1\iff q \equiv -p\pmod{\! 6}$
Proof $\bf\, 1$ $\bmod 6\!:\ p\!-\!4q \equiv (p/q\!+\!2)q\,$ is invertible if $\,p/q = -1,\,$ not if $\,p/q\equiv 1$ 
Proof $\bf\, 2$ $\,\ p,q\,$ coprime to $\,6\,\Rightarrow\,\bmod 6\!:\,\ p,q\equiv \pm1,\,$ so either $\,\color{#c00}{q\equiv -p}\,$ therefore $\, (p\!-\!4\color{#c00}q,\,6) = (5p,6)=(-p,6)=1,\, $ else $\ \color{#0a0}{q\equiv p}\,$ so $\,(p\!-\!4\color{#0a0}q,\,6) = (-3p,6)\neq 1$
