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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.

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 $b=1$.

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 $b=1$.

Call a pair of integers $(a,b)$ trivial if 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 $3$.

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, but I don't know of any. Are there some other divisibility conditions I've missed? 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 $b=1$.

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.

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 $b=1$.

Call a pair of integers $(a,b)$ trivial if 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 $3$.

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, but I don't know of any. Are there some other divisibility conditions I've missed? What happens if we also suppose $b\ge 2$?

Call a pair of integers $(a,b)$ trivial if 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 $3$.

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, but I don't know of any. Are there some other divisibility conditions I've missed? 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 $b=1$.