# A variation on Dirichlet's theorem on arithmetic progressions

Dirichlet's theorem on arithmetic progressions says that if $a$ and $b$ are coprime, then $\{a+bL\}_{L \in \mathbb{N}}$ contains infinitely many prime numbers.

I wonder if the following claim is true:

If $a$ and $b$ are coprime, then $\{a+b^L\}_{L \in \mathbb{N}}$ contains infinitely many prime numbers.

Notice that if my claim is true, then Dirichlet's theorem is true.

Thank you very much!

Edit: After receiving a few helpful comments, perhaps I should change my question to: Is there an additional condition that will make my claim true (a condition on $a$ and $b$)?

• No one knows if, for instance, there are infinitely many primes of the form $2^n+1$, though only a few are known. – lulu Jun 14 '18 at 23:22
• Note: you have to avoid silly counterexamples, like $1+3^n$ is always even. – lulu Jun 14 '18 at 23:23
• We don't even know if there are infinitely many Mersenne primes, i.e. those of the form $2^p-1$ (it's easy to show you need prime $p$ here), which is about the simplest possible. – Chappers Jun 14 '18 at 23:24
• @lulu, thanks for your comments. – user237522 Jun 14 '18 at 23:27
• @Chappers, thanks for your comment. Please, do you claim that if $L$ is not prime then $a+b^L$ is not prime? – user237522 Jun 14 '18 at 23:29

Let $a=4, b=n^4\,$ with $\,n \gt 1\,$, then $a+b^L$ is not a prime for any $L \ge 1\,$ by Sophie Germain's:

$$a+b^L=4 + n^{4L}=(n^{2L}+2+2n^L)(n^{2L}+2-2n^L)$$

• Nice, thank you. (I have added an edit to my question). – user237522 Jun 14 '18 at 23:38
• @user237522 Regarding the edit, an even weaker question is whether there exists any pair $\,a,b\,$ for which the proposition was proved. I don't know the answer to that. – dxiv Jun 14 '18 at 23:54
• ok.. thank you. Please, do you think that there is any hope to say something interesting about primes in $a^{L_1}+b^{L_2}$ (or $a^p+b^q$), where $L_1,L_2 \in \mathbb{N}$ ($p$ and $q$ are prime numbers)? Of course, $a$ and $b$ are assumed to be coprime. – user237522 Jun 14 '18 at 23:56

Here's an alternative view that doesn't try to answer your question but merely points out some observations you could make.

$\Bbb{Z}$ under the open set basis $U(a,b) = a + b\Bbb{N}$ forms a topological ring.

Fixing $a$ you can form a topology at least with $U(b) = a + b^{\Bbb{N}}$ since if $a + b^{\Bbb{N}} \cap a + c^{\Bbb{N}} \neq \varnothing$ say $x \in$ the intersection, then $a + b^n = x = a + c^m$ or $b^n = c^m$. There is a some work involved but you need to conclude that there is another basic open set $a + d^{\Bbb{N}}$ containing $x$ contained in the intersection.

• Thank you, interesting. Is it possible to prove Dirichlet's theorem based on this idea? – user237522 Jun 26 '18 at 10:42
• @user237522 not sure. Idk how Dirichlet's is proved. But in both cases you have a topology on $\Bbb{Z}$. It's only a basic observation I'm afraid. Note, that your case $a + b^{\Bbb{N}}$ may not be a topological ring basis, you'd have to prove that it is, but it's likely not because of addition not being continuous. However, even though the ring operations aren't continuous, other things might be such as certain maps reminiscent of our basis set formula. – EnjoysMath Jul 7 '18 at 15:08
• Thank you very much for the clarification/explanation. – user237522 Jul 8 '18 at 3:43