# Thread: Infinitely many composite numbers

1. ## Infinitely many composite numbers

Let and be positive integers and let the sequence be defined by and for all non-negative integers . Prove that for any choice of and , the sequence contains infinitely many composite numbers.

How do I do this question? I just started self-learning number theory as recreation so yeah if anyone can show me how to solve it, I would really appreciate it.

2. Originally Posted by usagi_killer
Let and be positive integers and let the sequence be defined by and for all non-negative integers . Prove that for any choice of and , the sequence contains infinitely many composite numbers.

How do I do this question? I just started self-learning number theory as recreation so yeah if anyone can show me how to solve it, I would really appreciate it.
I understand that you are doing mathematics recreationaly and that you just want to see the solution so that you better understand the topic, but that being said I would like to quote the mathematician George Simmons: "Despite our vast advances in mathematics, nothing has replaced the necessity of problem solving."

So, what have you tried?

3. note that the sequence is strictly increasing and also we may assume that $\gcd(a,b) = 1.$ suppose that the claim is false. so there exists some $\ell \in \mathbb{N}$ such that $x_k$ is prime for all $k \geq \ell.$

choose $k \geq \ell$ large enough so that $x_k=p > 2.$ see that $p \nmid a$ and $x_{k+m}=a^mp + (a^{m-1} + \cdots + a + 1)b,$ for all integers $m \geq 1.$ let $a \equiv r \mod p,$ where $1 \leq r \leq p-1.$ now, if $r=1,$

choose $m=p$ and if $r > 1,$ choose $m=p-1.$ see that $p \mid x_{k+m},$ which is a contradiction.

Note: i left a couple of thing for the OP to discover, so Drexel28 wouldn't get too mad at me!