A geometric series equalling a power of an integer - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T10:22:38Z http://mathoverflow.net/feeds/question/58697 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58697/a-geometric-series-equalling-a-power-of-an-integer A geometric series equalling a power of an integer John Bamberg 2011-03-17T00:35:03Z 2011-03-17T17:03:29Z <p>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:</p> <p>If $r=3$, then $s= 18$.<br> If $r=4$, then $s = 7$.<br> If $r=5$, then $s = 3$.<br> If $r>5$, there are no solutions.</p> <p>Does anyone know a proof of this curious property?</p> http://mathoverflow.net/questions/58697/a-geometric-series-equalling-a-power-of-an-integer/58715#58715 Answer by quid for A geometric series equalling a power of an integer quid 2011-03-17T03:02:37Z 2011-03-17T17:03:29Z <p>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.</p> <p>Below, I try to give some rough overview of some results that are known, and some references to (recent) articles.</p> <hr> <p>First, I restate the question to bring the notation in line with some sources I quote.</p> <p>What are the solutions $(x,y,n,q)$ of the equation <code>$$\frac{x^n - 1}{x - 1} = y^q$$</code> with integers $x,y>1$, $n>1$, $q \ge 2$ ?</p> <p>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: </p> <p>(A) Are these three solutions all the solutions ?</p> <p>Or more modestly</p> <p>(B) Is the number of solutions finite?</p> <hr> <p>As said, even (B) is open; but (A) is conjectured to be true.</p> <p>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 $6$).</p> <p>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 <a href="http://en.wikipedia.org/wiki/Abc_conjecture" rel="nofollow">the ABC-conjecture</a> implies that the number of solutions is finite.</p> <p>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) <a href="http://www-irma.u-strasbg.fr/~bugeaud/survols.html" rel="nofollow">available here</a>.</p> <p>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 <a href="http://en.wikipedia.org/wiki/Catalan_conjecture" rel="nofollow">Catalan's conjecture</a> (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.</p> <p>Specifically, it is now known, see <a href="http://www-irma.u-strasbg.fr/~bugeaud/travaux/PredaYann1.pdf" rel="nofollow">Bugeaud and Mihailescu (2007)</a>, that</p> <p>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 $q=n$.</p> <p>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.</p> <p>Moreover, related to the latter assertion Mihailescu recently proved (see <a href="http://www.uni-math.gwdg.de/preda/mihailescu-papers/diag.pdf" rel="nofollow">here</a> and <a href="http://www.uni-math.gwdg.de/preda/mihailescu-papers/ndfin.pdf" rel="nofollow">here</a>) 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).</p> <p>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.</p> <p>[Note: in case the tex is broken, it is not carelessness; at the moment, for technical reasons, I cannot test it myself.]</p>