# Integer solutions of $n^k+(n+1)^k+\cdots+(n+m)^k=(n+m+1)^k$

So we all know already that next identities follow: $3^2+4^2=5^2$ $3^3+4^3+5^3=6^3$

So it raises my question: For $(*)n^k+(n+1)^k+...+(n+m)^k=(n+m+1)^k$ are there infinite triples (n,m,k) s.t (*) is satisifed?

Any literature on this question or more general question from this follows?

Thanks.

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See the following question : mathoverflow.net/questions/53134/… Since here you require an additional condition, it might be easier to answer your question. If you fix the exponent, then you're asking for the rational points on an algebraic curve, so it is likely that there will be no solution if the exponent is big enough. –  François Brunault Mar 25 '11 at 15:10
Thanks, I am reading the thread in the link right now. –  Alan Mar 25 '11 at 17:03
Has someone tried to show this? Or is it a good exercise for an thesis article? :-) –  Alan Mar 25 '11 at 17:10
I don't have Guy's Unsolved Problems In Number Theory at hand, but I'm pretty sure this problem is discussed therein. –  Gerry Myerson Mar 25 '11 at 22:04
@Gerry Myreson, I looked at Guy's unsolved problems, it looks more general, but Euler asked are there solutions of the form: $$a_1^k+...+a_{k}^k=b^k$$ I am giving another condition which is strictier, that we have a_i s and b all are arithmetic sequence with difference one (or if you want to look more general than some arbitrary difference, though first let's look at difference 1). Just out of curiosity are there more examples such as: $$3^2+4^2=5^2$$3^3+4^3+5^3=6^3$$And are there infinite such instances? P.S I looked at Guy's second edition, page 139 on sums of like powers. – Alan Mar 26 '11 at 11:01 ## 6 Answers I do not have a definite answer, but in view of the discusion, and since it connects, as requested, the question to problems investigated in the literature, some remarks. I claim that: If an answer to this question is known, then it is of the form that there are infinitely many solutions. This might seem like a strange claim, and I do not claim that the number of solutions is infinite, but here is the reason: Erdős and Moser (in the 50s) conjectured that the equation$$ 1^k + \dots + m^k = (m+1)^k $$has only the solution 1+2=3. Over the decades it was investigated quite a bit, still it is unknown whether the number of solutions to this equation (the special case  n= 1  of the questioner's equation) is even finite. So, even fixing  n = 1  in the equation of the questioner one does not know how to prove that the number of solutions is finite. Thus, the only way an answer to this question can be known, is the one I mentioned, otherwise (and likely) this is an open problem. I already mentioned the names Erdős and Moser; searching for 'Erdős--Moser type equation' will yield various investigations on this and related problems. More specifically, the book 'Unsolved Problems in Number Theory' by Guy, already mentioned by Gerry Meyerson, contains a section on it (namely D7), you might be able to see it on Google books, with various references, in particular to computational work excluding solutions for  m  up to  10^{10^6}  and even beyond that. For more recent work on variants of this problem, see for example, a preprint by MacMillan and Sondow or a paper by Lengyel (scroll to A41). - Thanks, I'm sure it was Guy's section on Erdos-Moser that I had in mind. – Gerry Myerson Mar 27 '11 at 0:39 I'm quite unhappy with the answers I gave until now ; let me try to do better. The first case you mention (k=2 and m=1) falls inside the theory of pythagorean triples ; that is the integer solutions to x^2+y^2=z^2. Some trivial calculations show that the only pythagorean triple which has the more specific form you ask for is the one you gave, with the relation 3^2+4^2=5^2. Then, the more general case you mention falls inside the search of integer solutions to an equation of the form x_1^k+\dots+x_m^k=y^k, where k is fixed, so the equation is polynomial. It is for this general type of problem that I was mentioning that Faltings' theorem applies : if the geometric curve is too complex, then there will be only a finite number of solutions. And hence, only a finite number of solutions of the prescribed form. I tried a few more calculations : for m\geq2 and k=2, I first saw that n could only be 0 or 1 ; then using \sum_{l=0}^Nl^2=\frac{N(N+1)(2N+1)}6, I ruled out all possibilities. For the other cases, I didn't find anything useful. - Can you find any other solutions aside from 1^1+2^1=3^1 and n^0=(n+1)^0? There certainly aren't any others with k<100. You want f(x,k,m)=x^k+(x+1)^k+...+(x+m)^k-(x+m+1)^k=0 with x a positive integer. Let x_{m,k} be the positive real root (if any). Convince yourself that, for k \gt 2, 2k-1 \lt x_{2,k} \lt 2k. Next convince yourself that for fixed m, x_{m,k} is a decreasing function of k and that x_{m,k} \lt 1 if m \gt \frac{3k}{2}. This allows a quick search and the claim above. - "Introduction to elliptic curves and modular forms" by Neal Koblitz discusses at length the pythagorean triples (your first example) More generally, your question falls in the arithmetic geometry realm, and there is a wealth of references. Let me just mention "Arithmetic of elliptic curves" by Joseph Silverman. - Do you know for a fact that such a question or more general one is addressed there? Can you please be more specific as to which chapter it's covered? Thanks in advance. – Alan Mar 25 '11 at 14:26 You can read a few pages of chapter 1 if you search for "Koblitz Introductio to elliptic curves and modular forms" on google books. – Julien Puydt Mar 25 '11 at 14:52 The question seems much harder than the pythagorean triples problem... Now observe that:$$n^k+(n+1)^k+...+(n+m)^k=1^k+2^k+...+(n+m)^k-(1^k+2^k+...+(n-1)^k)$$So you can actually get an expression of this (using Bernoulli numbers). When k and n are fixed, it is polynomial in m of degree k+1 so it equal to (n+m+1)^k only for a finite number of m... You could swap m and n... - The end of the question is much harder ; but the fact that it starts like this makes me think pythagorean triples are already a convenient answer. – Julien Puydt Mar 25 '11 at 15:49 @Aurelien, it can only be a polynomial of degree k in m. @Snark, I am really interested in my question I asked at the end, because the identities (for (3,1,2) and (3,2,3) in my notation (n,m,k)) made me think are these the only such identities out there or are there more. I am not even sure how to check for another such triple, other than by mistake. (-: – Alan Mar 25 '11 at 17:01 No, it is indeed of degree k+1. For example the sum of the first n integers is \frac{n(n+1)}{2}, of degree 2... If you want more formulas (for power 3, 4, ...) see: trans4mind.com/personal_development/mathematics/series/… You can get the coeffs of those polynomials, quoting wikipedia:$$\sum n^m = \frac 1{m+1}\left( B_0n^{m+1}-\binom{m+1}1B_1n^m+\binom{m+1} 2B_2n^{m-1}-\cdots +(-1)^m\binom{m+1}mB_mn\right) Here the $B_m$ are the Bernoulli numbers: en.wikipedia.org/wiki/Bernoulli_number –  Aurelien Mar 25 '11 at 17:24
When $k$ is fixed, you might then be able to bound the possible values of $n$ and $m$... Now if you want some real work, consider Snark's remark... –  Aurelien Mar 25 '11 at 17:27
Ah, OK. But still it doesn't say if there are infinite such triples, right? –  Alan Mar 25 '11 at 17:31

For the more general question, I think : Falting's theorem is what you want to know.

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The link appears to be broken. –  Koundinya Vajjha Mar 25 '11 at 16:00
It was, indeed. –  Julien Puydt Mar 25 '11 at 16:11
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