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Please help me to prove this conjecture:

$$(\forall n,x\in\Bbb N,\forall a\in\Bbb N\setminus\{0,1\})\quad(\gcd(n,x)=1\iff a+\dots+a^n\mid a^x+\dots+a^{nx}).$$

(=> is easy to prove. The question of a possible counterexample of <= is difficult and not a duplicate)

Joel Elsen
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    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Mar 27 '24 at 12:55
  • What makes you think that this conjecture is actually true ? – Peter Mar 27 '24 at 14:22
  • @Peter I brute-force checked for a between 2-100 , x and N each between 2-50. And it held true for every combination checked. (This doesnt Mean it's true, it's just a good reason to believe it's true) – Joel Elsen Mar 27 '24 at 15:13
  • "I came up with a certain number theory idea and proof for it." If you were able to prove it, then it's not a conjecture anymore but a theorem. Are you asking if your proof is correct? If so, then post the proof here. – JRN Mar 27 '24 at 15:36
  • I have rewritten thé question such that it becomes clearer – Joel Elsen Mar 27 '24 at 15:45
  • @AnneBauval Indeed, but if you plugin 0, thé conjecture wont hold. And it's trivial that if a equals 1, that the conjecture doesnt work. I hope this clears it up. Sorry for the unclear language. English is not my first language – Joel Elsen Mar 27 '24 at 15:58
  • @AnneBauval I have changed it. Is it clearer now? – Joel Elsen Mar 27 '24 at 16:13
  • Yes I find it clear now. $\implies$ seems feasible. I have doubts about $\impliedby$. What were your attempts for each direction? – Anne Bauval Mar 27 '24 at 16:19
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    @AnneBauval For the <= arrow I tried to use the geometric series for both and trying to show that the one function could only divide the other if they shared all Roots/solutions if you put {insert their goniometric serie}=0 and thé same for thé other one. Then you would get that the latter has thé same solutions shifted to thé right. (I don't Remember how much). And only if they were coprime they would match up completly(and thus divide). For the other arrow I had no idea either but thé conjecture still holds. (Or maybe I am completly over thinking it) – Joel Elsen Mar 27 '24 at 16:35
  • If one had at least all of the Roots other* – Joel Elsen Mar 27 '24 at 16:42
  • Discussion relevant to the case of $n=3$ here (that recurs often enough). Anyway, its sure looks like cyclotomic polynomials (or rather just $(T^n-1)/(T-1)$) explains this. When $x$ is coprime to $n$, multiplying with it permutes the residue classes modulo $n$, and that's what you need. Like Anne Bauval, I'm not sure about the other direction :-) – Jyrki Lahtonen Mar 27 '24 at 17:35
  • @JyrkiLahtonen Oh yea, that's a lot easier than I had in mind. Haha. Thank you for the help :) – Joel Elsen Mar 27 '24 at 17:48
  • The divsibility follows immediately by Euclid's Lemma, since $,f_n = (x^n-1)/(x-1),$ is a strong divisibility sequence. – Bill Dubuque Mar 27 '24 at 23:06

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