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Considering the equation $p^{k+1}-1=(p-1)q^n$, where $p$ and $q$ are primes, $k$ and $n$ are integers such that $k>1$ and $n>0$, is it true that $p?

Thanks in advance.

Edit: it can be shown that $p\lt q$ iff $n\lt k$. I think that whenever $p$ is not a divisor of $k$, then one has $n=\varphi(k)/l_{k}(p)$, where $\varphi$ is Euler's totient function and $l_{k}(p)$ is the order of $p$ in $(\mathbb{Z}/k\mathbb{Z})^{\times}$. I'm not sure this kind of question really fits the "elementary number theory" tag, but I'd be glad if someone could prove this.

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    This amounts to asking whether $q^n = 1+p+p^2+\cdots+p^k$, with $k\gt 1$, $p$ and $q$ primes, implies $p\lt q$.2012-06-29
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    Yes, hence the title.2012-06-29
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    Why not phrase it like that in the body, then?2012-06-29
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    I can't see where the matter is actually. It was just easier to type.2012-06-29
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    @SylvainJulien Does a prime $q$ satisfying the given identity actually exist?2012-06-29
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    Yes. $1+3+9=13$, $1+3+9+27+81=121=11^2$, maybe there are other examples.2012-06-29
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    @SylvainJulien I'm with Arturo - a few extra keystrokes is not much to ask to make a problem easier for your readers to read. After all, you are, by posting a question here, asking a favor from strangers.2012-06-29
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    I didn't think this could lead to a difficulty, sorry.2012-06-29
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    1+5+25=31, so there are examples with $p \neq 3$, but is it always true that $p?2012-06-29
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    One way out could be to prove that the number of divisors of $1+p+...+p^k$ is smaller than the number of divisors of $p^{k+1}$, but I don't know how to achieve this.2012-06-30
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    It seems that $1+p+...+p^{k}\equiv 1\pmod k$. Can one prove this and deduce from this fact that $n$ is a divisor of $\varphi(k)$ through Lagrange's theorem (the order of any element of a finite groupe divides the order of the group)?2012-08-06

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