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I've been looking for perfect squares that can be represented as $\sum\limits_{k=0}^{n}p^k$.

Of course, both $n$ and $p$ should be natural numbers larger than $1$.

Searching up to $n=100$ and $p=200$, I found only $2$ cases:

  • $121=11^2=3^0+3^1+3^2+3^3+3^4=\sum\limits_{k=0}^{4}3^k$
  • $400=20^2=7^0+7^1+7^2+7^3=\sum\limits_{k=0}^{3}7^k$

Is there any way to prove that there are no other cases?

barak manos
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  • FWIW, I think these occurrences are rather coincidental and I think you'd be hard pressed to find relevant theory. Also, personally I would say that $121$ and $400$ aren't the only ones since I don't see any heuristic reason why that would be the case. – MT_ Jan 26 '15 at 22:01
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    $p$ an integer or $p$ a prime? – Hagen von Eitzen Jan 26 '15 at 22:07
  • @Soke: Thanks. You are possibly correct in both statements, but I don't see a single mathematical reasoning in your comment to support either one of them. – barak manos Jan 26 '15 at 22:08
  • @HagenvonEitzen: An integer (as specified in the question). I realized that it might be confusing, but I kinda ran out of "natural number denoting" letters that were visibly distinguishable from one another. – barak manos Jan 26 '15 at 22:09
  • @barakmanos I realize, it's just speculation. – MT_ Jan 26 '15 at 22:10
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    Some of thorniest open problems in number theory come about from this interplay between additive number theory and multiplicative number theory. – James47 Jan 26 '15 at 22:35
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    @James47: It reminds me of something I read on some "mathematical novel". I think it was The Music of Primes or Fermat's Last Theorem... Anyway, it said something like - "Mathematics is like a vast ocean of unknown, with little islands of knowledge and nothing that connects between them"... So if I understand you correctly, those two theories that you've mentioned are two separate islands in that ocean. – barak manos Jan 26 '15 at 22:43
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    You can rewrite $\sum_{k=0}^n p^k = \frac{p^{k+1}-1}{p-1}$ – azimut Jan 26 '15 at 23:02
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    see pages 54-57 of Dickson volume 1, https://books.google.com/books?id=FYWn8XWrAI0C&pg=PA54&dq=dickson+fermat+wallis+brouncker&hl=en&sa=X&ei=js_GVIr1OcirogTpy4DYBw&ved=0CCUQ6AEwAQ#v=onepage&q=dickson%20fermat%20wallis%20brouncker&f=false – Will Jagy Jan 26 '15 at 23:37
  • Possibly of interest: http://matwbn.icm.edu.pl/ksiazki/aa/aa64/aa6413.pdf – Barry Cipra Jan 27 '15 at 00:54
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    Definitely of interest: http://arxiv.org/pdf/1312.4037v1.pdf – Barry Cipra Jan 27 '15 at 01:56
  • So by Proposition 1 of the last link the answer is yes. Quite surprising. – Vincent Jan 27 '15 at 03:52
  • I think it's much easier to find primes of this form than squares of this form. In a randomly selected bunch of consecutive integers, the primes are likelier more common than the squares. – Robert Soupe Jan 27 '15 at 05:15
  • @BarryCipra: Yep, definitely of interest. So these are indeed the only $2$ known solutions, out of $4$ known solutions under a slightly more generalized conjecture (of perfect powers instead of perfect squares)... I assume that I will not get a better answer to my question at this point :) – barak manos Jan 27 '15 at 13:20
  • Similar, though not exact duplicates: http://math.stackexchange.com/questions/1098095, http://math.stackexchange.com/questions/372367, http://math.stackexchange.com/questions/379731 – Bart Michels Jan 28 '15 at 14:29

1 Answers1

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Turning a comment into an answer (of sorts)....

Various papers (e.g., this one by Yann Bugeaud and Preda Mihailescu) cite papers by Nagell and Ljunggren as proving there are no perfect squares of the given form other than the ones the OP found.

As for the more general problem of representing an arbitrary perfect power, a recent paper by Michael Bennett and Aaron Levin offers this assessment in its abstract (boldface emphasis added):

The Diophantine equation ${x^n−1\over x-1} = y^q$ has four known solutions in integers $x$, $y$, $q$ and $n$ with $|x|, |y|, q \gt 1$ and $n \gt 2$. Whilst we expect that there are, in fact, no more solutions, such a result is well beyond current technology.

The other two solutions referred to are the perfect cube $7^3$ expressed as $1+18+18^2$ and $1+(-19)+(-19)^2$.

See also the MathOverflow answer by quid.

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Barry Cipra
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  • I thought the article in the link you specified at the comment suggested that the "no perfect squares of the given form" conjecture was still unproved (I must have mistaken it for the more generalized conjecture). Thanks. – barak manos Jan 27 '15 at 16:10
  • @barakmanos, I initially misread the abstract that way too, until I looked carefully at the article's Proposition 1. It's a little unclear how simple or difficult the Nagell-Ljunggren proof for your perfect-square case is; I haven't been able to find it online. (It might be hidden in plain sight in one of the papers; maybe an actual expert can weigh in on the subject.) – Barry Cipra Jan 27 '15 at 17:21
  • Sounds like you're still not 100% sure that it has been proved... – barak manos Jan 27 '15 at 17:28
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    No, the various papers make it clear there's a proof, they just don't reproduce it (that I can see). What I really mean is, there might be some fairly simple argument that takes care of the perfect-square case. – Barry Cipra Jan 27 '15 at 17:33
  • OK, gotcha... And the more generalized (perfect-power) case is yet unproven, right? – barak manos Jan 27 '15 at 17:35
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    Yes, that's how I read things. (Shall we delete the last portion of these comments, to remove clutter no one needs to read?) – Barry Cipra Jan 27 '15 at 17:36