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Let $a$ and $b$ be two relatively prime positive integers, and consider the arithmetic progression $a$, $a+b$, $a+2b$, $a+3b$, . . .

  1. (G. Polya) Prove that there are infinitely many terms in the arithmetic progression that have the same prime divisors.

  2. Prove that there are infinitely many pairwise relatively prime terms in the arithmetic progression.

I was able to solve the first part by showing that the sequence {$ap$, $ap^2$, $ap^3$, . . .} has infinite intersection with the sequence {$a$, $a+b$, $a+2b$, . . .} by considering a prime $p$ $\equiv$ $1$ (mod $b)$ whose existence is guaranteed by Dirichlet.

For the second part, I assumed $k$ terms of the sequence which are pairwise relatively prime and then attempted to show that another pairwise relatively prime term can be generated from the previous $k$ terms by using a construction similar to that of Euclid's. However I wasn't able to get anything meaningful. Does this approach yield the desired result? If so how exactly can we achieve the result?

Arnav
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    I suppose , Second is just the Drichlet Theorem – S.Agera1729 Jan 26 '24 at 08:21
  • Doesn't Dirichlet Theorem say there are infinitely many primes? I mean I know that if we can prove Dirchlet then part 2 is solved, but what about pairwise relatively prime terms? Because I've just started number theory and wasn't able to understand the proof of it. – Arnav Jan 26 '24 at 08:25
  • My first thought when I saw the title (that is essentially question 2), was also Dirichlet's theorem, so your time might be better spend understanding that than searching for a prime of the (acknowledged) weaker statement in the question - it's not given that a weaker statement has a simpler proof. – Henrik supports the community Jan 26 '24 at 09:08
  • Use the elementary "coprime" version of Dirichlet's Theorem here. – Bill Dubuque Jan 26 '24 at 09:08

1 Answers1

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I think we can solve the second part of the question inductively, without using Dirichlet's Theorem.

Let's assume $a+n_1b, a+n_2b, \ ... , a+n_m b $ are some elements of the sequence which are pairwise relatively prime. Since $a$ and $b$ are relatively prime, by Euler's theorem, we have $b \mid a^{\varphi (b)+1} -a,$ where $φ(b)$ denotes Euler's totient function.

Now, consider: $$X=(a+n_1b)^{x_1}(a+n_2b)^{x_2} ... (a+n_m b)^{x_m}+b,$$

such that $x_1, x_2, ...,x_m\geq1$ and $x_1+x_2+...+x_m=(\varphi (b)+1)^k$ for some $k \in \mathbb N. $ It's very easy to verify that $X$ is also a member of the sequence; in other words, $X$ can be written as $a+nb$ for some $n \in \mathbb N$.

Additionally, if $d\mid a+n_ib$ and $d \mid X$, where $1 \leq i \leq m$, then we must have $d \mid b$. So, we must also have $d \mid a$, which implies $d=1$. Hence, $$a+n_1b, a+n_2b, \ ... , a+n_m b, X=a+nb $$ are pairwise relatively prime.

Doing the same with the new sets infinitely many times yields the claim. To start, note that $a+b$ and $a+2b$ are coprime.

We are done.

Reza Rajaei
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