Let me start with a curiosity. The integers $11,13,17,19$ are prime numbers, and $101,103,107,109$ are prime as well. One might wonder whether there is another occurrence where $10^m+1,10^m+3,10^m+7$ and $10^m+9$ are prime numbers. If so, then necessarily $m=2^r$ for some integer $r$ (same argument as for Fermat's numbers).

It is still unknown whether there are infinitely many prime numbers in Fermat's list $2^{2^r}+1$. Presumably, the same question for the list $10^{2^r}+1$ remains open. Likewise, it is not known whether there are infinitely many twin prime numbers $(p,p+2)$. If we impose both constraints, could we expect the question to be more tractable ?

Are there finitely or infinitely many integers $r$ such that $10^{2^r}+\{1,3,7,9\}$ are all primes ? I guess

No. For $r=2$, we have $10001=73\cdot137$. For $r=3$, $p=17$ divides $10^{2^3}+1$.

There is a natural extension of the question, where $10$ is replaced by $n\ge2$ and $\{1,3,7,9\}$ is replaced by a set of representatives of $({\mathbb Z}/n{\mathbb Z})^\times$.

An overwhelming belief is that there are only finitely many such exponents $r$. But I really ask whether there is a proof for some $n$, which involves our current artillery.

mightallow one to produce a proof that at least of $10^{2^{r}} + k$ is composite for all $r \geq 2$. $\endgroup$ – Jeremy Rouse Aug 12 '15 at 15:11