On powers that are sums of consecutive like powers
 Vandita Patel^{1} and
 Samir Siksek^{1}Email author
Received: 29 July 2016
Accepted: 25 November 2016
Published: 14 February 2017
Abstract
Keywords
Exponential equation Bernoulli polynomial Newton polygonMathematics Subject Classification
Primary 11D61 Secondary 11B681 Background
Theorem 1
If k is odd, then \(\mathcal {A}_{k,r}\) contains all of the odd d: we can take \((x,y,n)=(r(1d)/2,0,n)\). Thus the conclusion of the theorem does not hold for odd k.
2 Some properties of Bernoulli numbers and polynomials
Theorem 2
(Brillhart and Dilcher) The Bernoulli polynomials are squarefree.
2.1 Relation to sums of consecutive like powers
Lemma 2.1
This formula can be found in ([8] Section 24.4), but is easily deduced from the identity (5).
Lemma 2.2
Proof
Lemma 2.3
Let k, r be integers with \(k \ge 2\) and \(r \ne 0\). Let \(q \ge k+3\) be a prime not dividing r such that the congruence \(B_k(x) \equiv 0 \pmod {q}\) has no solutions. Let d be a positive integer such that \({{\mathrm{\upsilon }}}_q(d)=1\). Then Eq. (2) has no solutions (i.e. \(d \notin \mathcal {A}_{k,r}\)).
Proof
Remarks

For \(k \ge 3\) odd, the kth Bernoulli polynomial has known rational roots 0, 1 / 2, 1. Thus the criterion in the lemma fails to hold for all primes q. For even \(k \ge 2\) we shall show that there is a positive density of primes q such that \(B_k(x)\) has no roots modulo q.

The second Bernoulli polynomial is \(B_2(x)=x^2x+1/6\). By quadratic reciprocity, this has a root modulo \(q \not \mid 6\) if and only if \(q \equiv \pm 1 \pmod {12}\). We thus recover the result of Bai and Zhang mentioned in the introduction: if \(q \equiv \pm 5 \pmod {12}\) and \({{\mathrm{\upsilon }}}_q(d)=1\) then (1) has no solutions with \(k=2\).
3 A Galois property of even Bernoulli polynomials
Proposition 3.1
Let \(k \ge 2\) be even, and let \(G_k\) be the Galois group of the Bernoulli polynomial \(B_k(x)\). Then there is an element \(\mu \in G_k\) that acts freely on the roots of \(B_k(x)\).
There is a longstanding conjecture that the even Bernoulli polynomials are irreducible; see for example [3, 4, 14]. One can easily deduce Proposition 3.1 from this conjecture. We give an unconditional proof of Proposition 3.1 in Sect. 5. As noted previously, if k is odd, then \(B_k(x)\) has rational roots 0, 1 / 2, 1, so the conclusion of the proposition certainly fails for odd k.
3.1 A density result
Theorem 3
(Niven) Let \(\{q_i\}\) be a set of primes such that \(\delta (\mathcal {A}^{(q_i)})=0\) and \(\sum q_i^{1}=\infty \). Then \(\delta (\mathcal {A})=0\).
3.2 Proposition 3.1 implies Theorem 1

primes \(q \le k+2\);

primes q dividing r;

primes q dividing the numerator of the discriminant of \(B_k(x)\) (which is nonzero by Theorem 2).
4 The 2adic Newton polygons of even Bernoulli polynomials
Lemma 4.1
 (i)
a horizantal segment joining the points \((0,1)\) and \((k2^s,1)\);
 (ii)
a segment joining the points \((k2^s,1)\) and (k, 0) of slope \(1/2^s\).
Proof
Consider the definition of \(B_k(x)\) in (3). We know that \(B_0=1\), \(B_1=1/2\) and \(B_m=0\) for all odd \(m \ge 3\). From the von Staudt–Clausen theorem, we know that \({{\mathrm{\upsilon }}}_2(B_m)=1\) for even \(m \ge 2\). It follows that the Newton polygon is bounded below by the Horizontal line \(y=1\).
Remarks
Inkeri [12] showed that \(B_k(x)\) has no rational roots for k even. His proof required very precise (and difficult) estimates for the real roots of \(B_k(x)\). Lemma 4.1 allows us to give a much simpler proof of the following stronger result.
Theorem 4
Let k be even. Then \(B_k(x)\) has no roots in \({\mathbb Q}_2\).
Proof
Indeed, suppose \(\alpha \in {\mathbb Q}_2\) is a root of \(B_k(x)\). From the slopes of the Newton polygon segments we see that \({{\mathrm{\upsilon }}}_2(\alpha )=0\) or \(1/2^s\). As \({{\mathrm{\upsilon }}}_2\) takes only integer values on \({\mathbb Q}_2\), we see that \({{\mathrm{\upsilon }}}_2(\alpha )=0\) and so \(\alpha \in {\mathbb Z}_2\). Let \(f(x)=2B_k(x) \in {\mathbb Z}_2[x]\). Thus \(f(\alpha )=0\) and so \(f(\overline{\alpha })=\overline{0} \in {\mathbb F}_2\). However, \(\overline{\alpha } \in {\mathbb F}_2=\{\overline{0},\overline{1}\}\). Now \(f(\overline{0})=\overline{(2B_k)}=\overline{1}\), and from (4) we know that \(f(\overline{1})=f(\overline{0})=\overline{1}\). This gives a contradiciton. \(\square \)
Although Theorem 4 is not needed by us, its proof helps motivate part of the proof of Proposition 3.1.
5 Completing the proof of Theorem 1
5.1 A little group theory
Lemma 5.1
Let H be a finite group acting transitively on a finite set \(\{ \beta _1,\cdots ,\beta _n\}\). Let \(H_i \subseteq H\) be the stabilizer of \(\beta _i\), and suppose \(H_1=H_2\). Let \(\pi \; : \; H \rightarrow C\) be a surjective homomorphism from H onto a cyclic group C. Then there is some \(\mu \in H\) acting freely on \(\{\beta _1,\cdots ,\beta _n\}\) such that \(\pi (\mu )\) is a generator of C.
Proof
Proof of Propostion 3.1 We now complete the proof of Theorem 1 by proving Proposition 3.1. Fix an even \(k \ge 2\), and let L be the splitting field of \(B_k(x)\). Let \(G_k={{\mathrm{Gal}}}(L/{\mathbb Q})={{\mathrm{Gal}}}(B_k(x))\) be the Galois group of \(B_k(x)\). Let \(\mathfrak {P}\) be a prime of L above 2. The 2adic valuation \({{\mathrm{\upsilon }}}_2\) on \({\mathbb Q}_2\) has a unique extension to \(L_\mathfrak {P}\) which we continue to denote by \({{\mathrm{\upsilon }}}_2\). We let \(H={{\mathrm{Gal}}}(L_\mathfrak {P}/{\mathbb Q}_2) \subseteq G_k\) be the decomposition subgroup corresponding to \(\mathfrak {P}\).
From Lemma 4.1 we see that \(B_k(x)\) factors as \(B_k(x)=g(x)h(x)\) over \({\mathbb Q}_2\) where the factors g, h correspond respectively to the segments (i), (ii) in the lemma. Thus g, h have degree \(k2^s\) and \(2^s\) respectively. We denote the roots of g by \(\{\alpha _1,\ldots ,\alpha _{k2^s}\} \subset L_\mathfrak {P}\) and the roots of h by \(\{\beta _1,\ldots ,\beta _{2^s}\} \subset L_\mathfrak {P}\). From the slopes of the segments we see that \({{\mathrm{\upsilon }}}_2(\alpha _i)=0\) and \({{\mathrm{\upsilon }}}_2(\beta _j)=1/2^s\). It clearly follows that h is irreducible and therefore that H acts transitively on the \(\beta _j\). Moreover, from the symmetry (4) we see that \(1\beta _1\) is a root of \(B_k(x)\), and by appropriate relabelling we can suppose that \(\beta _2=1\beta _1\). In the notation of Lemma 5.1, we have \(H_1=H_2\). Now let \(C={{\mathrm{Gal}}}({\mathbb F}_\mathfrak {P}/{\mathbb F}_2)\), where \({\mathbb F}_\mathfrak {P}\) is the residue field of \(\mathfrak {P}\). This group is cyclic generated by the Frobenius map: \(\overline{\gamma } \mapsto \overline{\gamma }^2\). We let \(\pi \; : \; H \rightarrow C\) be the induced surjection. By Lemma 5.1 there is some \(\mu \in H\) that acts freely on the \(\beta _i\) and such that \(\pi (\mu )\) generates C. To complete the proof of Proposition 3.1 it is enough to show that \(\mu \) also acts freely on the \(\alpha _i\). Suppose otherwise, and let \(\alpha \) be one of the \(\alpha _i\) that is fixed by \(\mu \). As \({{\mathrm{\upsilon }}}_2(\alpha )=0\), we can write \(\overline{\alpha } \in {\mathbb F}_\mathfrak {P}\) for the reduction of \(\alpha \) modulo \(\mathfrak {P}\). Now \(\alpha \) is fixed by \(\mu \), and so \(\overline{\alpha } \in {\mathbb F}_\mathfrak {P}\) is fixed by \(\langle \pi (\mu ) \rangle =C\). Thus \(\overline{\alpha } \in {\mathbb F}_2\) and so \(\overline{\alpha }=\overline{0}\) or \(\overline{1}\). Now let \(f(x)=2 B_k(x) \in {\mathbb Z}_2[x]\). Thus \(f(\overline{\alpha })=\overline{0}\). But \(f(\overline{0})=\overline{(2B_k)}=\overline{1}\), and from (4) we know that \(f(\overline{1})=f(\overline{0})=\overline{1}\). This contradiction completes the proof.
Declarations
Open Access
This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Acknowledgements
The firstnamed author is supported by an EPSRC studentship. The secondnamed author is supported by the EPSRC LMF: LFunctions and Modular Forms Programme Grant EP/K034383/1.
Authors’ Affiliations
References
 Bennett, M.A., Patel, V., Siksek, S.: Perfect powers that are sums of consecutive cubes. Mathematika 63, 230–249 (2016)View ArticleGoogle Scholar
 Bennett, M.A., Patel V., Siksek S.: Superelliptic equations arising from sums of consecutive powers, Acta Arithmetica, to appearGoogle Scholar
 Brillhart, J.: On the Euler and Bernoulli polynomials. J. Reine Angew. Math. 234, 45–64 (1969)MathSciNetMATHGoogle Scholar
 Carlitz, L.: Note on irreducibility of the Bernoulli and Euler polynomial. Duke Math. J. 19, 475–481 (1952)MathSciNetView ArticleMATHGoogle Scholar
 Cassels, J.W.S.: A Diophantine equation. Glasgow Math. J. 27, 11–88 (1985)View ArticleMATHGoogle Scholar
 Cassels, J.W.S., Fröhlich, A. (eds.): Algebraic Number Theory. Academic Press, New York (1967)MATHGoogle Scholar
 Dickson, L.E.: History of the theory of numbers, vol. II. Chelsea, New York (1971)Google Scholar
 Dilcher, K.: Chapter 24: Bernoulli and Euler polynomials, in digital library of mathematical functions. http://dlmf.nist.gov/24
 Dilcher, K.: On multiple zeros of Bernoulli polynomials. Acta Arithmetica 134(2), 149–155 (2008)MathSciNetView ArticleMATHGoogle Scholar
 Euler L.: Vollständige Anleitung zur Algebra, vol. 2, St. Petersburg (1770)Google Scholar
 Granville, A.: Arithmetic properties of binomial coefficients. I. Binomial coefficients modulo prime powers. CMS Conf. Proc. Amer. Math. Soc. 20, 253–276 (1997)MathSciNetMATHGoogle Scholar
 Inkeri, K.: The real roots of Bernoulli polynomials. Ann. Univ. Turku. Ser. A I(37), 1–20 (1959)MathSciNetMATHGoogle Scholar
 Niven, I.: The asymptotic density of sequences. Bull. Amer. Math. Soc. 57, 420–434 (1951)MathSciNetView ArticleMATHGoogle Scholar
 Kimura, N.: On the degree of an irreducible factor of the Bernoulli polynomials. Acta Arith. 50, 243–249 (1988)MathSciNetMATHGoogle Scholar
 Soydan G., On the Diophantine equation $$(x+1)^k+(x+2)^k+\cdots +(l x)^k=y^n$$ ( x + 1 ) k + ( x + 2 ) k + ⋯ + ( l x ) k = y n , preprint (2016)Google Scholar
 Stroeker, R.J.: On the sum of consecutive cubes being a square. Compositio Mathematica 97, 295–307 (1995)MathSciNetMATHGoogle Scholar
 Uchiyama, S.: On a Diophantine equation. Proc. Japan Acad. Ser. A Math. Sci. 55(9), 367–369 (1979)MathSciNetView ArticleMATHGoogle Scholar
 Zhang, Z.: On the Diophantine equation \((x1)^k+x^k+(x+1)^k=y^n\). Publ. Math. Debrecen 85, 93–100 (2014)MathSciNetView ArticleMATHGoogle Scholar
 Zhang, Z., Bai, M.: On the Diophantine equation \((x+1)^2+(x+2)^2 + \cdots +(x+d)^2=y^n\). Funct. Approx. Comment. Math. 49, 73–77 (2013)MathSciNetView ArticleMATHGoogle Scholar