Proof of existence of infinitely many primes that divide the sequence $\{S(k)\}=\sum_{i=1}^{n} a_i^k$ where $\{a_i\}_{i=1}^{n}$ forms an AP

Question

For the sake of a mathjax solution and because of finding an error in my previous solution (which I had posted here for correction), I have edited the whole document

Let $a_i$ for $1 \leq i \leq n$ be a non constant Arithmetic Progression.

Define: $$S(k)=\sum_{i=1}^{n} a_i^k$$

Prove that there are infinitely many primes which divide the sequence $\{ S(k) \} _{k\in \mathbb{N}}$ where $S(k)$ doesn't remain constant.

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The term: infinitely many primes which divide the sequence $\{ S(k) \} _{k\in \mathbb{N}}$ is a bit messy to understand.

Clarification of the statement:

The sequence $\{S(k)\}$ has infinitely many terms different terms since it's not constant. Now the terms are not restricted to being divisible by only a certain number of primes. Like suppose, consider a set of primes: $\mathcal{P}=\{p_1, p_2, \cdots, p_m\}$. Then there will exist a term in the sequence of $\{S(k)\}_{k=1}^{\infty}$ which will get divided by a prime $p_j$ such that $p_j \notin \mathcal{P}$.

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Doubt:

Here (below in the form of images) I have provided half of the solution, as best I could. I think I'm not wrong.

May I get some guidance to complete the second case?

Solution:

First, let us assume to the contrary and let $\mathcal{P}=\{p_1, p_2, \cdots, p_m\}$ denote the set of primes that ONLY divide $S(k)$ and suppose $\text{GCD}(a_1, a_2, \cdots, a_n)=d$. Then we get $a_i=db_i$ such that $\text{GCD}(b_1, b_2, \cdots, b_n)=1$ Choose $p > n$ which means $p \nmid n$.

Fix a prime $p \in \mathcal{P}$ and show that there exists a prime $p \mid S(k)$ such that $p \notin \mathcal{P}$ leading to a condration and hence proving the inexistence of finitely many primes that divide the sequence.

We get two cases:

Case 1: $p \nmid d$

If $p \mid S(k)$, then, $p \mid d^k \sum_{i=1}^n b_i$ and since $p \nmid d$, we see $p \mid \sum b_i$. Set $k=\prod (p_i-1); ~ p \in \mathcal{P}$ From Euler's theorem (the one that uses Totient\Phi function), we get that $b_i \equiv \{0,1\} \pmod p \implies \sum b_i \not\equiv 0 \pmod p$ and hence we get the desired contradiction.

Case 2: $ p \mid d$

(Could not solve)

Answer 1

Let us assume that the prime divisors of $S(k)$ belong to $\{p_1,p_2,\ldots,p_m\}$ for any $k\geq 0$. By Van Der Waerden's theorem, there are arithmetic progressions $k_1,k_2,k_3,\ldots,k_\eta$ of arbitrary length such that $S(k_1),S(k_2),\ldots,S(k_\eta)$ are multiples of some $p_\mu$, and we may assume without loss of generality $k_\eta>p_\mu$. Newton's identities then imply that $a_1,a_2,\ldots,a_n$ are multiples of $p_\eta$. We may define $\tilde{a}_j = a_j/p_\eta$ and repeat the same argument, but since $\mathbb{Z}$ is a UFD an infinite descent is not allowed and the claim is proved.