Here is a problem I don't even know how to approach: For arbitrary integers a,b,c which satisfy $(a,b)=(b,c)=1$ and $b>1$ there exist infinitely many integers n such that $ab^n+c$ is not prime. Any help with this problem will be appreciated.


closed as off-topic by Matthew Conroy, C. Falcon, Juniven, user91500, user223391 Apr 11 '17 at 21:19

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  • $\begingroup$ What is the notation $(a,b)$ referring to? $\endgroup$ – Myridium Apr 9 '17 at 20:51
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    $\begingroup$ @Myridium: it is the $\gcd$ of $a$ and $b$ $\endgroup$ – Ross Millikan Apr 9 '17 at 20:51
  • $\begingroup$ Hint: What are the possible values of $(a,c)$? See what can be done with that $\endgroup$ – Will Craig Apr 9 '17 at 20:52
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    $\begingroup$ @lhf: No, if $n$ has an odd factor we know $2^n+1$ is not prime. That is an infinite set. $\endgroup$ – Ross Millikan Apr 9 '17 at 20:52

It's easy to show $\gcd(b, ab+c)=1$, from which using Euler's theorem $$b^{\varphi(ab+c)}\equiv 1 \pmod{ab+c} \Rightarrow b^{\varphi(ab+c)+1}\equiv b \pmod{ab+c} \tag{1}$$ or $$ab^{\varphi(ab+c)+1}+c\equiv ab+c \equiv 0 \pmod{ab+c}$$ which means $$ab+c \mid ab^{\varphi(ab+c)+1}+c$$ but, using $(1)$, this is also true for $$ab^{2\varphi(ab+c)+1}+c\equiv ab^{{\varphi(ab+c)+1}}+c \equiv ab+c \equiv 0 \pmod{ab+c}$$ and by induction $$ab^{k\cdot\varphi(ab+c)+1}+c \equiv 0 \pmod{ab+c}, \forall k \geq 1$$ i.e. there will be infinitely many $ab^n+c$ divisible by $ab+c$ and thus not primes.


Suppose $p$ is a prime factor of $ab+c$, then by Fermat's little Theorem, $p$ is a prime factor of $ab^{(p-1)N+1}+c$.


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