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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 $b=1$.

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    $\begingroup$ 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. $\endgroup$
    – Julian Rosen
    Commented Aug 4, 2019 at 13:26
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    $\begingroup$ Also see here: egrove.olemiss.edu/cgi/… $\endgroup$
    – Wolfgang
    Commented Mar 29, 2022 at 14:19

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The answer is No.

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 $S$.

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    $\begingroup$ Covering sets fall under the more complicated conditions for being trivial. I'll try to make the definition more precise. $\endgroup$
    – domotorp
    Commented Aug 5, 2019 at 1:13
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    $\begingroup$ I'm happy to believe you but I would be even happier if someone else could also confirm it, as I don't know the proof. $\endgroup$
    – domotorp
    Commented Aug 5, 2019 at 2:23
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    $\begingroup$ @dom, this might be the Izotov paper: mathstat.dal.ca/FQ/Scanned/33-3/izotov.pdf Bibliographic details: Anatoly S. Izotov (1995). "Note on Sierpinski Numbers". Fibonacci Quarterly. 33 (3): 206. $\endgroup$
    – Gerry Myerson
    Commented Aug 5, 2019 at 12:01
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    $\begingroup$ @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: $$2^{4j+2}+\ell^4 = (2^{2j+1} - 2^{j+1}\cdot\ell + \ell^2)(2^{2j+1} + 2^{j+1}\cdot\ell + \ell^2)$$ $\endgroup$
    – Jeppe Stig Nielsen
    Commented Aug 1, 2020 at 7:27
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    $\begingroup$ (continued) and for $n\not\equiv 2 \pmod4$, again the same set $S$ works. $\endgroup$
    – Jeppe Stig Nielsen
    Commented Aug 1, 2020 at 7:28

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