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Inspired by MSE post, I propose the following generalization:

Is the following statement always true?

Consider $n$ and $m$ are non negative Integers. Let $p$ and $q$ are prime with $q=p+6n+2$ then there is no such $m$ gives pair of prime $p'$ and $q'$ with $p'=pq+p+q$ and $q'=p'+6m-2$.

Example: let $n=0$ so choose any twin prime pair, let $p=3,q=5$ and then there is no such $m\ge0$ gives $p'$ and $q'$ both prime.

Source code Pari/GP

for(n=0,10,for(m=0,10,forprime(p=1,10000,forprime(q=p+6*n+2,p+6*n+2,forprime(P=p*q+p+q,p*q+p+q,forprime(Q=P+6*m-2,P+6*m-2,print([n,m,p,q,P,Q])))))))
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    $\begingroup$ If $p = 3$, then $q \equiv 2 \pmod{3}$, so $p' = 23 \equiv 2 \pmod{3}$ and $q' \equiv 0 \pmod{3}$. If $p \equiv 1 \pmod{3}$, then $q \equiv 0 \pmod{3}$ and $q \gt 3$, so it can't be prime. Finally, with $p = 2$, then $q$ is not prime, but if otherwise $p \equiv 2 \pmod{3}$, then $q \equiv 1 \pmod{3}$, so $p' \equiv 2 \pmod{3}$ and $q' \equiv 0 \pmod{3}$. Thus, your statement is always true. In questions like this, you should always first check the smaller modulo values. $\endgroup$
    – John Omielan
    Commented Dec 21, 2020 at 19:12

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Elaborating John Omielan' comment: As $p, q$ both are primes, then, $p$ can't be $0$ or $1$ modulo $3$. Hence, $p$ must have to be $-1 \ \text{modulo} \ 3$.

Now, $p'=q(p+1)+p=p^2+(6n+4)p+(6n+2)$ or $q'=p^2+(6n+4)p+6(n+m)$.

As $p$ is only $-1\ \text{modulo} \ 3$, $q'=1+(1).(-1)+0 =0 \ \text {modulo}\ 3$, hence can't be prime. This means there doesn't exist any such $m$.

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