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Let $x > 1$ and let $n$ be a prime. I'm wondering if a characterization of this is known. That is, what are sufficient and necessary conditions for $$ \dfrac{x^n-1}{x-1} = 1 + x + x^2 + \cdots + x^{n-1} $$ to be a prime number? What are these conditions if we restrict $x$ to be a power of a prime? Note that $n$ can not be composite since otherwise it is easy to show that so is $(x^n-1)/(x-1)$. Thanks in advance.

Semiclassical
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user152634
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    I've added the cyclotomic-polynomial tag to your question because, for $n$ prime, that's exactly what your polynomial is. Also, this question is related though narrower than yours. Finally, this paper on arXiv looks relevant. – Semiclassical Aug 02 '14 at 00:24
  • Note that either $p|x-1$ or $n|p-1$, where $p$ is the prime. – N. S. Aug 02 '14 at 00:43
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    @IanMateus: I was wondering if I'd missed something, heh! And for my part, my referencing that arXiv paper may've been hasty: that's about whether $\Phi_n(x)$ is prime in $\mathbb{Z}[x]$ rather than which $x\in \mathbb{Z}$ are mapped to primes by $\Phi_n(x)$. – Semiclassical Aug 02 '14 at 00:46
  • @Semiclassical I don't think there is an easy criterion; look at the Mersenne primes, for instance. This won't be easy (I have already deleted my comment above before I could read yours :-( ) – Ian Mateus Aug 02 '14 at 00:51
  • @ianmateus: Good point. About the most I'm able to reason out is: $\Phi_n(x)$ can't produce primes over the integers if it itself isn't prime in $\mathbb{Z}[x]$. But that's only a negative criterion, and says nothing about any particular choice of integer $x$. – Semiclassical Aug 02 '14 at 01:00
  • Thanks guys. Seems this won't be easy to characterize. BTW this seems related: http://yves.gallot.pagesperso-orange.fr/papers/cyclotomic.pdf

    See Theorem 2.4 there due to Legendre. It would imply that if $\Phi_p(x) = \ell$ is prime for a prime $p$, then $\ell \equiv 1 \pmod{p}$.

     – user152634 Aug 02 '14 at 01:05
  • This implication is immediate from Fermat's little theorem and handling separately the case $x-1\equiv 0\pmod p$. – Ian Mateus Aug 02 '14 at 01:13
  • If ((x^n)-1)/(x-1) = W and g.c.d(a,W) =1 and W is an odd prime {let W(o)= (x^(n-1)-1)/(x-1)} then (a^((W(o) x))-1) is divisible by W; G.c.d( x,W(O)) =1. If you pick 'a' so the order of a mod W is p ( being a prime number) then p divides x or p divides W(O). If p divides W(O) then a^T = 1 mod W where T=(x^(n-1)-1) Is this helpful or correct? – user128932 Aug 14 '14 at 04:55
  • If q is an odd prime and q | n then ($x^q$ -1) | ($x^n$ -1) so ($x^q$ -1)/(x -1) | p therefore q = n. So if ($x^n$ -1)/(x-1) is an element of primes so is n. – user128932 Oct 23 '14 at 05:00
  • If $(x^p-1)$ /(x - 1) = q , p and q being primes (x + 1)| (q -1), assuming p is odd. – user128932 Oct 26 '14 at 03:46
  • Another similar question would be whether every prime can be written in the form $\frac{x^n-1}{x-1}$. It sounds equally hard to me. – ghosts_in_the_code Dec 13 '14 at 15:58

1 Answers1

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Not a complete answer; just some thoughts

It is likely to be very difficult to give a necessary and sufficient condition. For $x=2$ you're asking what are the Mersenne primes so I don't expect a (simple) answer in the near future.

Some observations:

  • Every prime $p$ is of that form: $p=\frac{(p-1)^2-1}{(p-1)-1}$
  • $n$ has to be prime. An alternative but perhaps overpowered way to see this is Zsigmondy's theorem, which shows that $\frac{x^n-1}{x-1}$ has at least $\tau(n)-1$ distinct prime divisors (except for a few exceptions that aren't very interesting here), $\tau(n)$ the number of divisors of $n$.

  • This is related to cyclotomic polynomials. In general, $x^n-1$ factors as $\prod_{d\mid n}\Phi_d(x)$. If $n$ is prime, your question is essentially when $\Phi_n(x)$ is prime. From cyclotomic polynomials we know that

    • If $p$ is a prime divisor of $\Phi_n(x)$, then $p\equiv1\pmod n$ or $p\mid n$.

    This does does however not tell us much if we're looking for a necessary condition, because by Fermat we always have $\frac{x^n-1}{x-1}\equiv1\pmod n$ if $n$ is prime. (Unless $n\mid x-1$, but then it's not hard to see that we need $\frac{x^n-1}{x-1}=n$ which is impossible as $\Phi_n(x)>\Phi_n(1)=n$.)

I don't think elementary arguments can rule out non-trivial possibilities of $x$ or $n$. For example, Birkhoff & Vandiver's proof of Zsigmondy's theorem investigates the (primitive) prime divisors of $\Phi_n(x)$ quite well, but their intermediate results seem

  1. strongest for composite $n$ - "a non-primitive divisor of $x^n-1$" is not a very interesting concept if $n$ is prime
  2. trivial if we require that $\Phi_n(x)$ is prime

and

  1. the final conclusion is simply that there is some primitive divisor (by showing that the primitive part of $x^n-1$ is $>1$), but it does not tell us whether this divisor is prime.

Inspecting the proof a bit closer it does not seem that the prime factorisation of $x$ plays any role in the proof, so I can't tell whether anything changes when we require that $x$ is a prime power. I haven't read other proofs of Zsigmondy's theorem (you can start looking here if you're interested) but there may be one that is more closely related to the factorisation of $x$.

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