Inspired by [MSE post](https://math.stackexchange.com/q/3954379/647719), 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])))))))

```


Mapping from prime pairs to non prime pairs
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 $m$.


