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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 $p2012-06-29
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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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I think you're looking at some well-known unsolved problems. D10 in Guy, Unsolved Problems In Number Theory, says, in part, "Hugh Edgar asks if there is a solution, other than $1+3+3^2+3^3+3^4=11^2$, of the equation $1+q+q^2+\cdots+q^{x-1}=p^y$ with $p,q$ odd primes and $x\ge5$, $y\ge2$. An important breakthrough in this area is the paper of Reese Scott." I think the reference is to On the equations $p^x-b^y=c$ and $a^x+b^y=c^z$, J Number Theory 44 (1993) 153-165.

EDIT: I'm not sure that the Scott paper has any bearing on the problem.

Hugh Edgar has written about the problem.
1. In The exponential Diophantine equation $1+a+a^2+\cdots+a^{x-1}=p^y$, Amer Math Monthly 81 (1974) 758-759, MR0345906 (49 #10636), he reviews some earlier work and makes some conjectures.
2. In Problems and some results concerning the Diophantine equation $1+a+a^2+\cdots+a^{x-1}=p^y$, Rocky Mountain J. Math. 15 (1985), no. 2, 327–329, MR0823244 (87e:11047), his purpose is "to raise several as yet unsettled questions concerning the equation of the title and to try to induce people to solve these problems."

One should also see Mao Hua Le, On the Diophantine equation $(x^m−1)/(x−1)=y^n$, Acta Math. Sinica 36 (1993), no. 5, 590–599, MR1261944 (95a:11028), where it's shown under various additional conditions that if $(p^m-1)/(p-1)=q^n$ for odd primes $p,q$ then $n$ is the least $r$ such that $q^r\equiv1\pmod p$

Similar results and further progress are in Yann Bugeaud, Maurice Mignotte, Yves Roy, On the Diophantine equation $(x^n−1)/(x−1)=y^q$, Pacific J. Math. 193 (2000), no. 2, 257–268, MR1755817 (2001f:11049). Bugeaud and Mignotte have (at least) two more papers on this equation, but I'm not sure whether they bear on the particular situation when $x,y$ are prime. The paper by these two authors with Guillaume Hanrot, Proc. London Math. Soc. (3) 84 (2002), no. 1, 59–78, MR1863395 (2002h:11027), does look like it might be quite relevant.