# Primes in shifted geometric sequence

Call a pair of integers $$(a,b)$$ trivial if it satisfies some simple divisibility condition, like for some prime $$p$$ we have $$p$$ divides both $$a-1$$ and $$b+1$$, or that $$p$$ divides both $$a$$ and $$b$$. This implies that $$a^n+b$$ is always divisible by $$p$$, like $$7^n+2$$ by $$3$$. But there can be more complicated conditions, like $$29^n+26$$, as noted by Julian in a comment. In general, $$(a,b)$$ is trivial if there is a finite collection of primes such that $$a^n+b$$ is always divisible by at least one prime of the collection. (Note: This has been added after Max's answer.)

Is it true that for every $$a\ge 2$$ and $$b$$ integers, if $$(a,b)$$ is non-trivial, then there are infinitely many primes in the sequence $$a^n+b$$?

This would be a generalization of Fermat primes, but I couldn't find anything about this variant. Since for $$b=1$$ we also know that $$n$$ would need to be a power of two for $$a^n+b$$ to be a prime, probably there are counterexamples to my question, but I don't know of any.

What happens if we also suppose $$b\ge 2$$? A trivial calculation would give that the chance of $$a^n+b$$ to be a prime is about $$\frac 1n$$, so there would be infinitely many as $$\sum \frac 1n=\infty$$, unless there are some additional conditions, like in case of $$b=1$$.

• There are other obstructions, e.g. $29^n+26$ is divisible by $5$ when $n$ is odd and is divisible by $3$ when $n$ is even. – Julian Rosen Aug 4 '19 at 13:26

We know that $$78557$$ is a Sierpinski number with the covering set $$S = \{ 3, 5, 7, 13, 19, 37, 73 \}$$, i.e., every integer of the form $$78557\cdot 2^n+1$$ is divisible by an element of $$S$$. Set $$a:=2$$ and $$b:=128100173$$, which form a non-trivial pair and satisfy $$b \equiv 78557^{-1}\pmod{3\cdot 5\cdot 7\cdot 13\cdot 19\cdot 37\cdot 73}.$$ Then the numbers $$a^n+b$$ are never prime as they have the same covering set $$S$$.
• @domotorp Izotov's construction for $k\cdot 2^n+1$ is to take $k$ as a fourth power, say $k=\ell^4$ (and $\ell$ divisible by 5). Then for when $n\equiv2 \pmod4$, use this factorization of $k\cdot 2^n+1$: $$\ell^4\cdot 2^{4j+2} + 1 = (\ell^2\cdot 2^{2j+1} - \ell\cdot 2^{j+1} +1)(\ell^2\cdot 2^{2j+1} + \ell\cdot 2^{j+1} +1)$$ and for other $n$, make sure there is a finite set $S$ of primes that cover. With the same $k$ and $S$, then also $2^n + k$ works. Because again there is a factorization: $$2^{4j+2}+\ell^4 = (2^{2j+1} - 2^{j+1}\cdot\ell + \ell^2)(2^{2j+1} + 2^{j+1}\cdot\ell + \ell^2)$$ – Jeppe Stig Nielsen Aug 1 '20 at 7:27
• (continued) and for $n\not\equiv 2 \pmod4$, again the same set $S$ works. – Jeppe Stig Nielsen Aug 1 '20 at 7:28