On powers that are sums of consecutive like powers
 Vandita Patel^{1} and
 Samir Siksek^{1}Email author
DOI: 10.1007/s4099301600680
© The Author(s) 2017
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
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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
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