Solutions in primes of $\,\frac{p+1}2=q=\frac{r+1}4$ Why the only solutions in primes of
$$\,\frac{p+1}2=q=\frac{r+1}4$$
seem to be $\,(p,q,r)=(3,2,7)\,$ and $\,(p,q,r)=(5,3,11)$?
Said in other terms: "if $\,r\,$ is a prime, different from $\,7\,$ and $\,11$, and $\,r-1=2p\,$ is a semiprime, then $\,r+1=4q\,$ must have at least $\,4\,$ factors $\,[2, 2, s, t]$".
Many thanks.

Note that the conditions imply that $(p,q,r)=(2q-1,q,4q-1)$. If $q>3$ then either $q=3n+1$ or $q=3n-1$ for some $n$.
If $q=3n+1$ then $$r=4q-1=12n+4-1=12n+3$$ is divisible by $3$.
If $q=3n-1$ then $$p=2q-1=6n-2-1=6n-3$$ is divisible by $3$.
