Between Fermat's primes and the twin primes

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.

• While this is a famous unsolved problem, it seems certain that the Fermat primes are finitely many. If $X$ should be prime with probability $1/\log{X}$ (in the naive probabilistic model based on the PNT), then certainly an unbiased set $S \subset \mathbb{N}$ with $\sum_{s \in S} 1/\log{s} < \infty$ should have a finite intersection with the primes. – Vesselin Dimitrov Aug 12 '15 at 9:25
• BTW, for $r=2$ only $7$ and $9$ have the property. For $r=3$, only $7$ has the property. Unfortunately, I have not been able to find tables beyond $10^{13}$, but it would be interesting to see if $10^{16}+\{1,3,7,9\}$ contains any primes at all. – M.G. Aug 12 '15 at 9:32
• Vesselin, key word is unbiased. I am not so sure that it is so for $2^{2^n}+1$. Say, its prime factors a priori belong to arithmetic progression $2^{n+2}x+1$. Probably, we need more careful probabilistic heuristics. – Fedor Petrov Aug 12 '15 at 10:43
• @Fedor: You are right, "unbiased" probably does not apply here; though I still think it should be possible to give a reasonable heuristic taking care of the special property you mention, and predicting finiteness in this problem. (And for all I know, one may never find a prime in $2^{2^n}+1$ beyond $n = 4$.) – Vesselin Dimitrov Aug 12 '15 at 11:15
• For $2 \leq r \leq 1000$, there is always a prime $\leq 11605787$ that divides one of $10^{2^{r}} + k$ for $k \in \{1, 3, 7, 9 \}$. It's also true that $10^{2^{r}} +3$ is a multiple of $7$ if $r$ is even, and $19$ divides $10^{2^{r}} + 3$ if $r \equiv 5 \pmod{6}$. Finding and patching together similar congruences might allow one to produce a proof that at least of $10^{2^{r}} + k$ is composite for all $r \geq 2$. – Jeremy Rouse Aug 12 '15 at 15:11