Open Access

On powers that are sums of consecutive like powers

Research in Number Theory20173:2

https://doi.org/10.1007/s40993-016-0068-0

Received: 29 July 2016

Accepted: 25 November 2016

Published: 14 February 2017

Abstract

Let \(k \ge 2\) be even, and let r be a non-zero integer. We show that for almost all \(d \ge 2\) (in the sense of natural density), the equation
$$x^k+(x+r)^k+\cdots +(x+(d-1)r)^k=y^n,\quad x,~y,~n \in {\mathbb Z}, \; n \ge 2,$$
has no solutions.

Keywords

Exponential equation Bernoulli polynomial Newton polygon

Mathematics Subject Classification

Primary 11D61 Secondary 11B68

1 Background

The problem of cubes that are sums of consecutive cubes goes back to Euler ([10] art. 249) who noted the remarkable relation \(3^3+4^3+5^3=6^3\). Similar problems were considered by several mathematicians during the nineteenth and early twentieth century as surveyed in Dickson’s History of the Theory of Numbers ([7] p. 582–588). These questions are still of interest today. For example, both Cassels [5] and Uchiyama [17] determined the squares that can be written as sums of three consecutive cubes. Stroeker [16] determined all squares that are expressible as the sum of \(2 \le d \le 50\) consecutive cubes, using a method based on linear forms in elliptic logarithms. More recently, Bennett, Patel and Siksek [2] determined all perfect powers that are expressible as sums of \(2 \le d \le 50\) consecutive cubes, using linear forms in logarithms, sieving and Frey curves. There has been some interest in powers that are sums of k-th powers for other exponents k. For example, the solutions to the equation
$$\begin{aligned} x^k+(x+1)^k+(x+2)^k=y^n, \quad x,~y,~n \in {\mathbb Z}, \; n \ge 2, \end{aligned}$$
have been determined by Zhongfeng Zhang [18] for \(k=2\), 3, 4 and by Bennett, Patel and Siksek [1] for \(k=5\), 6, and similar problems are considered by Soydan [15].
In view of the above, it is natural to consider the equation
$$\begin{aligned} x^k+(x+1)^k+\cdots +(x+d-1)^k=y^n,\quad x,~y,~n \in {\mathbb Z}, \; n \ge 2 \end{aligned}$$
(1)
with k, \(d \ge 2\). This was studied by Zhang and Bai [19] for \(k=2\). They show that if q is a prime \(\equiv \pm 5 \pmod {12}\) and \({{\mathrm{\upsilon }}}_q(d)=1\) then Eq. (1) has no solutions for \(k=2\); here \({{\mathrm{\upsilon }}}_q(d)\) denotes the q-adic valuation of d. It follows from a standard result in analytic number theory (as we shall see later) that the set of d for which there is a solution with \(k=2\) has natural density 0. We prove the following generalization to all even exponents k.

Theorem 1

Let \(k\ge 2\) be even and let r be a non-zero integer. Write \(\mathcal {A}_{k,r}\) for the set of integers \(d \ge 2\) such that the equation
$$\begin{aligned} x^k+(x+r)^k+\cdots +(x+(d-1)r)^k=y^n,\quad x,~y,~n \in {\mathbb Z}, \; n \ge 2, \end{aligned}$$
(2)
has a solution (xyn). Then \(\mathcal {A}_{k,r}\) has natural density 0; by this we mean
$$\begin{aligned} \lim _{X \rightarrow \infty } \frac{ \# \{d \in \mathcal {A}_{k,r} \; :\; d \le X\}}{X} =0. \end{aligned}$$

If k is odd, then \(\mathcal {A}_{k,r}\) contains all of the odd d: we can take \((x,y,n)=(r(1-d)/2,0,n)\). Thus the conclusion of the theorem does not hold for odd k.

2 Some properties of Bernoulli numbers and polynomials

In this section we summarise some classical properties of Bernoulli numbers and polynomials. These are found in many references, including [8]. The Bernoulli numbers \(B_k\) are defined via the expansion
$$\begin{aligned} \frac{x}{e^x-1}=\sum _{k=0}^\infty B_k \frac{x^k}{k!} \, . \end{aligned}$$
The first few Bernoulli numbers are
$$\begin{aligned} B_0=1, \quad B_1=-1/2, \quad B_2=1/6, \quad B_3=0, \quad B_4=-1/30, \quad B_5=0, \quad B_6=1/42. \end{aligned}$$
It is easy to show that \(B_{2k+1}=0\) for all \(k \ge 1\). The \(B_k\) are rational numbers, and the von Staudt–Clausen theorem asserts that for \(k \ge 2\) even
$$\begin{aligned} B_k+\sum _{(p-1) \mid k} \frac{1}{p} \; \in \; {\mathbb Z}\end{aligned}$$
where the sum ranges over primes p such that \((p-1) \mid k\).
The k-th Bernoulli polynomial can be defined by
$$\begin{aligned} B_k(x)=\sum _{m=0}^k \left( {\begin{array}{c}k\\ m\end{array}}\right) B_m x^{k-m}. \end{aligned}$$
(3)
Thus it is a monic polynomial with rational coefficients, and all primes appearing in the denominators are bounded by \(k+1\). It satisfies the symmetry
$$\begin{aligned} B_k(1-x)=(-1)^k B_k(x), \end{aligned}$$
(4)
the identity
$$\begin{aligned} B_k(x+1)-B_k(x)=k x^{k-1}, \end{aligned}$$
(5)
and the recurrence
$$\begin{aligned} B_k^\prime (x)=k B_{k-1}(x). \end{aligned}$$
(6)
Whilst all the above results have been known since at least the nineteenth century, we also make use of the following far more recent and difficult theorem due to Brillhart [3] and Dilcher [9].

Theorem 2

(Brillhart and Dilcher) The Bernoulli polynomials are squarefree.

2.1 Relation to sums of consecutive like powers

Lemma 2.1

Let r be a non-zero integer and k, \(d \ge 1\). Then
$$\begin{aligned} x^k+(x+r)^k+\cdots +(x+r(d-1))^k= \frac{r^k}{k+1} \left( B_{k+1}\left( \frac{x}{r}+d\right) -B_{k+1}\left( \frac{x}{r}\right) \right) . \end{aligned}$$

This formula can be found in ([8] Section 24.4), but is easily deduced from the identity (5).

Lemma 2.2

Let \(q\ge k+3\) be a prime. Let a, r, d be integers with \(d \ge 2\), and \(r \ne 0\). Suppose \(q \mid d\) and \(q \not \mid r\). Then
$$\begin{aligned} a^k+(a+r)^k+\cdots +(a+r(d-1))^k \equiv r^k\cdot d \cdot B_k(a/r) \pmod {q^2}. \end{aligned}$$

Proof

By Taylor’s Theorem
$$\begin{aligned} B_{k+1}(x+d)=B_{k+1}(x)+d \cdot B_{k+1}^\prime (x) + \frac{d^2}{2} B_{k+1}^{(2)}(x)+\cdots +\frac{d^{k+2}}{(k+2)!} \cdot B_{k+1}^{(k+2)}(x). \end{aligned}$$
It follows from the assumption \(q \ge k+3\) that the coefficients of \(B_{k+1}(x)\) are q-adic integers. Thus the coefficients of the polynomials \(B_{k+1}^{(i)}(x)/{i!}\) are also q-adic integers. As \(q \mid d\) and \(q \not \mid r\) we have
$$\begin{aligned} B_{k+1}\left( \frac{a}{r}+d\right) - B_{k+1}\left( \frac{a}{r}\right) \equiv d \cdot B_{k+1}^\prime (a/r) \pmod {q^2}. \end{aligned}$$
The lemma follows from (6) and Lemma 2.1. \(\square \)

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

Suppose \((x,y,n)=(a,b,n)\) is a solution to (2). By Lemma 2.2,
$$\begin{aligned} r^k \cdot d \cdot B_k(a/r) \equiv b^n \pmod {q^2}. \end{aligned}$$
However, the hypotheses of the lemma ensure that the left-hand side has q-adic valution 1. Thus \({{\mathrm{\upsilon }}}_q(b^n)=1\) giving a contradiction. \(\square \)

Remarks

  • For \(k \ge 3\) odd, the k-th 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 \(B_2(x)=x^2-x+1/6\). 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 long-standing 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

Let \(\mathcal {A}\) be a set of positive integers. For X positive, define
$$\begin{aligned} \mathcal {A}(X)=\# \{d \in \mathcal {A}: d \le X \}. \end{aligned}$$
The natural density of \(\mathcal {A}\) is defined as the limit (if it exists)
$$\begin{aligned} \delta (\mathcal {A})=\lim _{X \rightarrow \infty } \frac{\mathcal {A}(X)}{X}. \end{aligned}$$
For a given prime q, define
$$\begin{aligned} \mathcal {A}^{(q)}=\{d \in \mathcal {A}\; : \; {{\mathrm{\upsilon }}}_q(d)=1\}. \end{aligned}$$
We shall need the following result of Niven ([13] Corollary 1).

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

We now suppose Proposition 3.1 and use it to deduce Theorem 1. Let \(k \ge 2\) be an even integer. Write \(G_k\) for the Galois group of the Bernoulli polynomial \(B_k(x)\). Let \(\mu \in G_k\) be the element acting freely on the roots of \(B_k(x)\) whose existence is asserted by Proposition 3.1. By the Chebotarev density theorem ([6] Chapter VIII) there is a set of primes \(\{q_i\}_{i=1}^\infty \) having positive Dirichlet density such that for each \(q=q_i\), the Frobenius element \({{\mathrm{Frob}}}_q \in G_k\) is conjugate to \(\mu \). We omit from \(\{q_i\}\) (without affecting the density) the following:
  • primes \(q \le k+2\);

  • primes q dividing r;

  • primes q dividing the numerator of the discriminant of \(B_k(x)\) (which is non-zero by Theorem 2).

As \(\mu \) acts freely on the roots of \(B_k(x)\), it follows that the polynomial \(B_k(x)\) has no roots modulo any of the \(q_i\). Now let \(\mathcal {A}=\mathcal {A}_{k,r}\) be as in the statement of Theorem 1. By Lemma 2.3, if \({{\mathrm{\upsilon }}}_{q_i}(d)=1\) then \(d \notin \mathcal {A}\). It follows that \(\mathcal {A}^{(q_i)}=\emptyset \). By Theorem 3, we have \(\delta (\mathcal {A})=0\) as required.

4 The 2-adic Newton polygons of even Bernoulli polynomials

Lemma 4.1

Let \(k \ge 2\) be even and write \(k=2^s t\) where t is odd and \(s \ge 1\). The 2-adic Newton polygon of \(B_k(x)\) consists two segments:
  1. (i)

    a horizantal segment joining the points \((0,-1)\) and \((k-2^s,-1)\);

     
  2. (ii)

    a segment joining the points \((k-2^s,-1)\) 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 \(y=-1\).

We shall need to make use of the following result of Kummer (see [11]): if p is a prime, and u, v are positive integers then
$$\begin{aligned} \left( {\begin{array}{c}u\\ v\end{array}}\right) \equiv \left( {\begin{array}{c}u_0\\ v_0\end{array}}\right) \left( {\begin{array}{c}u_1\\ v_1\end{array}}\right) \pmod {p}, \end{aligned}$$
where \(u_0\), \(u_1\) are respectively the remainder and quotient on dividing u by p, and likewise \(v_0\), \(v_1\) are respectively the remainder and quotient on dividing v by p. Here we adopt the convention \(\left( {\begin{array}{c}r\\ s\end{array}}\right) =0\) if \(r<s\). Applying this with \(p=2\) we see that
$$\begin{aligned} \left( {\begin{array}{c}k\\ 2^s\end{array}}\right) = \left( {\begin{array}{c}2^s t\\ 2^s\end{array}}\right) \equiv \left( {\begin{array}{c}t\\ 1\end{array}}\right) \equiv t \equiv 1 \pmod {2}. \end{aligned}$$
Thus the coefficient of \(x^{k-2^s}\) in \(B_k(x)\) has 2-adic valuation \(-1\). Since the constant coefficient of \(B_k(x)\) also has valuation \(-1\), we obtain the segment (i) as part of the Newton polygon. We also see that for \(0<v<2^s\),
$$\begin{aligned} \left( {\begin{array}{c}k\\ v\end{array}}\right) \equiv 0 \pmod {2}, \end{aligned}$$
and so the valuation of the coefficient of \(x^{k-v}\) is \( \ge 0\). Finally the coefficient of \(x^k\) is \(B_0=1\) and so has valuation 0. This gives segment (ii) and completes the proof. \(\square \)

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

Let \(m=\#C\) and write \(C=\langle \sigma \rangle \). Consider the subset
$$\begin{aligned} C^\prime =\{ \sigma ^r \; : \; \gcd (r,m)=1\}; \end{aligned}$$
this is the set of elements that are cyclic generators of C, and has cardinality \(\varphi (m)\), where \(\varphi \) is the Euler totient function. As \(\pi \) is surjective we see that
$$\begin{aligned} \# \pi ^{-1}(C^\prime )=\frac{\varphi (m)}{m} \cdot \# H. \end{aligned}$$
(7)
As H acts transitively on the \(\beta _i\), the stabilizers \(H_i\) are conjugate and so have the same image \(\pi (H_i)\) in C. If this image is a proper subgroup of C, then take \(\mu \) to be any preimage of \(\sigma \). Thus \(\pi (\mu )=\sigma \) is a generator of C, and moreover, \(\mu \) does not belong to any of the stabilizers \(H_i\) and so acts freely on \(\{\beta _1,\cdots , \beta _n\}\), completing the proof in this case. Thus we suppose that \(\pi (H_i)=C\) for all i. It follows that
$$\begin{aligned} \# \pi ^{-1} (C^\prime ) \cap H_i= \frac{\varphi (m)}{m} \cdot \# H_i = \frac{\varphi (m)}{m} \cdot \frac{\# H}{n}, \end{aligned}$$
(8)
where the second equality follows from the Orbit-Stabilizer Theorem. The lemma states that there is some element \(\mu \) belonging to \(\pi ^{-1}(C^\prime )\) but not to \(\cup H_i\). Suppose otherwise. Then
$$\begin{aligned} \pi ^{-1}(C^\prime ) \subseteq \bigcup _{i=1}^n H_i, \end{aligned}$$
and therefore
$$\begin{aligned} \pi ^{-1} (C^\prime )=\bigcup _{i=1}^n \pi ^{-1}(C^\prime ) \cap H_i. \end{aligned}$$
(9)
Now (7), (8) and (9) together imply that the \(\pi ^{-1}(C^\prime ) \cap H_i\) are pairwise disjoint. This contradicts the hypothesis that \(H_1=H_2\) completing the proof. \(\square \)

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 2-adic 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 \(k-2^s\) and \(2^s\) respectively. We denote the roots of g by \(\{\alpha _1,\ldots ,\alpha _{k-2^s}\} \subset L_\mathfrak {P}\) 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 first-named author is supported by an EPSRC studentship. The second-named author is supported by the EPSRC LMF: L-Functions and Modular Forms Programme Grant EP/K034383/1.

Authors’ Affiliations

(1)
Mathematics Institute, University of Warwick

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