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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 Aug 4 '19 at 13:26
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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 Aug 5 '19 at 1:13
  • $\begingroup$ @domotorp: Same idea works for Izotov's construction for a Sierpinski number, which does not have a covering set. So, the answer is still No. $\endgroup$ – Max Alekseyev Aug 5 '19 at 2:10
  • $\begingroup$ 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 Aug 5 '19 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 Aug 5 '19 at 12:01
  • $\begingroup$ See also math.stackexchange.com/questions/1683082/… Another reference is oeis.org/A076336 $\endgroup$ – Gerry Myerson Aug 5 '19 at 12:09

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