# A geometric series equalling a power of an integer

The following problem cropped up whilst considering generalised quadrangles with a product structure, and it boils down to a simple number theoretic problem. Let $s$ be an integer greater than 2 and suppose the geometric series $(s^r-1)/(s-1)$ is a nontrivial power of a positive integer. It seems the following is true:

If $r=3$, then $s= 18$.
If $r=4$, then $s = 7$.
If $r=5$, then $s = 3$.
If $r>5$, there are no solutions.

Does anyone know a proof of this curious property?

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My first instinct (but I really don't know much about number theory) is to try and factorize in terms of cyclotomic integers (corresponding to $r$th roots of unity), as in Lame's attempted proof of FLT. This would only work for certain $r$ though... –  Yemon Choi Mar 17 '11 at 1:01
quid's post says that (A) is conjectured to be true, i.e. there are only three solutions. But I found also the following: (3,2,2,2), (8,3,2,2), (24,5,2,2), (15,4,2,2), (26,3,2,3), (7,2,2,3), (15,2,2,4), (80,3,2,4). Obviously there are lots of solutions. –  Mok-Kong Shen Nov 26 '13 at 15:39
I guess quid meant $n>2$; there are surely infinitely many solutions to $x+1=y^q$... –  Philip van Reeuwijk Nov 26 '13 at 16:27
See mathoverflow.net/questions/39430/…, where I try, though in an incomplete manner, to give an application of the NL equation to the problem of whether odd perfect numbers exist or not. I'm still not quite sure whether such an idea can lead to something interesting though. Hope this helps... –  Sylvain JULIEN Nov 27 '13 at 18:46

This is a well-investigated Diophantine equation known as Nagell--Ljunggren equation (they investigated this equation in the 1920s and 1940s, resp). Indeed, it is conjectured that the three solutions mentioned by the questioner are the only ones; however, it is not even known that the number of solutions is finite, though there are numerous partial result.

Below, I try to give some rough overview of some results that are known, and some references to (recent) articles.

First, I restate the question to bring the notation in line with some sources I quote.

What are the solutions $(x,y,n,q)$ of the equation $$\frac{x^n - 1}{x - 1} = y^q$$ with integers $x,y>1$, $n>2$, $q \ge 2$ ?

As mentioned, in the question, one finds three 'small' solutions $(3,11,5,2)$, $(7,20,4,2)$, and $(18,7,3,3)$. And, the remaining question is:

(A) Are these three solutions all the solutions ?

Or more modestly

(B) Is the number of solutions finite?

As said, even (B) is open; but (A) is conjectured to be true.

By early works of Nagell and Ljunggren it is known that with any of the following conditions there are no other solutions: $q=2$, $n$ a multiple of $3$, $n$ a multiple of $4$, or ($q=3$ and $n$ not $5$ modulo $6$).

Shorey and Tijdeman proved (1976) that the number of solutions is finite with any of the following conditions: $x$ is fixed, $n$ has a fixed prime divisor, $y$ has a fixed prime divisor. Also, Shorey proved that the ABC-conjecture implies that the number of solutions is finite.

There are numerous additional results, imposing various conditions on $x,y,n$ or $q$ (due to Bennet, Bugeaud, Le, Mignotte, and others) for a survey of the state of the art around a decade ago see, e.g., a 2002 survey (in French) of Bugeaud and Mignotte (which was also the main bases for the above written part) available here.

The early results were obtained via passing to certain rings of algebraic integers; later results often used Baker's method (linear forms in logarithms) and results on Diophantine approximation. Some years ago, the solution of Catalan's conjecture (which is on a somewhat similar equation), by Mihailescu that (as far as I understand, very surprisingly) avoided all these types of tools and used instead (only) results on cyclotomic fields/integers, provided a new impetus.

Specifically, it is now known, see Bugeaud and Mihailescu (2007), that

a. for any other solution (so not one of three known ones) the smallest prime divisor of $n$ is at least $29$ and $n$ has at most $4$ prime divisors (counted with multiplicity). Moreover, $n$ is prime if $q=3$. And, if $q\mid n$, then $q=n$.

b. to prove that there are no other solutions, it suffices to show that there is no solution with $n\ge 5$ an odd prime and $q$ an odd prime.

Moreover, related to the latter assertion Mihailescu recently proved (see here and here) various results in the case that $n$ and $q$ are odd primes (saying, in one of the abstracts that methods used in the cyclotomic approach to FLT are used, so Yemon Choi's intuition was very right).

This answer does certainly not give a complete picture (in this format, it would be difficult to give one, and no matter the format, it would be impossible for me); I am aware of various omissions I made, and I am afraid there are many of which I am not aware. The references I mentioned should however allow to retrieve more complete information.

[Note: in case the tex is broken, it is not carelessness; at the moment, for technical reasons, I cannot test it myself.]

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We (the OP and I) are particularly interested in the case when n=4. From the Bugeaud and Mihailescu paper you linked to, it seems that it is known that there are no other solutions when 4 | n and so if I am correct, the little sub-problem we are interested in is actually done. –  Gordon Royle Mar 17 '11 at 23:50
Yes, this is correct and goes back to Nagell--Ljunggren, and is mentioned in this paper. Theorem 1 of B. and M. (this is a. in the answer) gives much more, and in particular allows to exclude for each 1< d <29 (additional) solutions for n that is a (possibly trivial) multiple of d. Btw, this is the precise reference for B. and M.: Math. Scand. 101 (2007), 177-183 ; and the N. and L. papers are quoted there. –  quid Mar 18 '11 at 0:30
Hey quid... we have used this fact in a paper and wish to acknowledge you. Can you please send me your identity (to gordon.royle@uwa.edu.au) if you wish to be acknowledged. –  Gordon Royle Jan 25 '12 at 1:37