In this paper we consider the Diophantine equation \begin{align*}b^k
+\left(a+b\right)^k &+ \cdots + \left(a\left(x-1\right) + b\right)^k=\\ &=d^l +
\left(c+d\right)^l + \cdots + \left(c\left(y-1\right) + d\right)^l,
\end{align*} where $a,b,c,d,k,l$ are given integers. We prove that, under some
reasonable assumptions, this equation has only finitely many integer solutions.Comment: This version differs slightly from the published version in its
  expositio

Bazsó, A.

Kreso, D.

Luca, F.

Pintér, Á.

English

arXiv

In this paper, we consider the Diophantine equation bk +(a+b)k + ··· + (a(x-1) + b)k= dl + (c+d)l + ··· + (c(y-1) + d)l, where a,b,c,d,k,l are given integers with gcd (a,b) = gcd (c,d) = 1, k ą l. We prove that, under some reasonable assumptions, the above equation has only finitely many solutions

Bazsó, András

Kreso, Dijana

Luca, Florian

Pintér, Ákos

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GLASNIK MATEMATICˇKIVol. 47(67)(2012), 253 – 263ON EQUAL VALUES OF POWER SUMS OF ARITHMETICPROGRESSIONSAndra´s Bazso´, Dijana Kreso, Florian Luca and A´kos Pinte´rUniversity of Debrecen, Hungary, Technische Universita¨t Graz, Austria andUniversidad Nacional Auto´noma de Me´xico, MexicoAbstract. In this paper, we consider the Diophantine equationbk + (a+ b)k + · · ·+ (a (x− 1) + b)k == dl + (c+ d)l + · · ·+ (c (y − 1) + d)l ,where a, b, c, d, k, l are given integers with gcd(a, b) = gcd(c, d) = 1, k 6= l.We prove that, under some reasonable assumptions, the above equationhas only finitely many solutions.1. Introduction and resultsFor a positive integer n ≥ 2, let(1.1) Ska,b (n) = bk + (a+ b)k+ · · ·+ (a (n− 1) + b)k.It is easy to see that the above power sum is related to the Bernoullipolynomials Bk(x) in the following way:(1.2)Ska,b (n) =akk + 1([Bk+1(n+ba)−Bk+1]−[Bk+1(ba)−Bk+1]),where the polynomials Bk(x) is defined by the generating seriest exp(tx)exp(t)− 1=∞∑k=0Bk(x)tkk!2010 Mathematics Subject Classification. 11B68, 11D41.Key words and phrases. Diophantine equations, exponential equations, Bernoullipolynomials.253254 A. BAZSO´, D. KRESO, F. LUCA AND A´. PINTE´Rand Bk+1 = Bk+1(0). For the properties of Bernoulli polynomials which willbe often used in this paper, sometimes without special reference, we refer to[7, Chapters 1 and 2]. We can extend Ska,b for every real value x as(1.3) Ska,b (x) =akk + 1(Bk+1(x+ba)−Bk+1(ba)).We denote by C[x] the ring of polynomials in the variable x with complexcoefficients. A decomposition of a polynomial F (x) ∈ C[x] is an equality ofthe following formF (x) = G1(G2(x)) (G1(x), G2(x) ∈ C[x]),which is nontrivial ifdegG1(x) > 1 and degG2(x) > 1.Two decompositions F (x) = G1(G2(x)) and F (x) = H1(H2(x)) are saidto be equivalent if there exists a linear polynomial ℓ(x) ∈ C[x] such thatG1(x) = H1(ℓ(x)) and H2(x) = ℓ(G2(x)). The polynomial F (x) is calleddecomposable if it has at least one nontrivial decomposition; otherwise it issaid to be indecomposable.In a recent paper, Bazso´, Pinte´r and Srivastava ([1]) proved the followingtheorem about the decomposition of the polynomial Ska,b (x) defined above.Theorem 1.1. The polynomial Ska,b (x) is indecomposable for even k. Ifk = 2v − 1 is odd, then any nontrivial decomposition of Ska,b (x) is equivalentto the following decomposition:(1.4) Ska,b (x) = Ŝv((x+ba−12)2).Proof. This is [1, Theorem 2].Using Theorem 1.1 and the general finiteness criterion of Bilu and Tichy([2]) for Diophantine equations of the form f(x) = g(y), we prove the followingresult.Theorem 1.2. For 2 ≤ k < l, the equation(1.5) Ska,b(x) = Slc,d(y)has only finitely many solutions in integers x and y.Since the finiteness criterion from [2] is based on the ineffective theoremof Siegel, our Theorem 1.2 is ineffective. We note that for a = c = 1, b = d = 0our theorem gives the result of Bilu, Brindza, Kirschenhofer, Pinte´r and Tichy([3]).Combining a result of Brindza [5] with recent theorems by Rakaczki ([8])and Pinte´r and Rakaczki ([6]), for k = 1 and 3 we obtain effective statements.ON EQUAL VALUES OF POWER SUMS OF ARITHMETIC PROGRESSIONS 255Theorem 1.3. For k = 1 and l /∈ {1, 3, 5}, the equation(1.6) S1a,b(x) = Slc,d(y)implies max(|x|, |y|) < C1, where C1 is an effectively computable constantdepending only on a, b, c, d and l.In the exceptional cases l = 3, 5 one can give some values for a, b, c, dsuch that the corresponding equations possess infinitely many solutions. Forexample, if k = 1, a = 2, b = 1, l = 3 or l = 5, c = 1, d = 0 we havex2 = 1 + 3 + · · ·+ 2x− 1 = 13 + 23 + · · ·+ (y − 1)3orx2 = 1 + 3 + · · ·+ 2x− 1 = 15 + 25 + · · ·+ (y − 1)5,respectively. These equations have infinitely many integer solutions, see [9].Theorem 1.4. For k = 3 and l /∈ {1, 3, 5}, the equation(1.7) S3a,b(x) = Slc,d(y)implies max(|x|, |y|) < C2, where C2 is an effectively computable constantdepending only on a, b, c, d and l.2. Auxiliary resultsIn this section, we collect some results needed to prove Theorem 1.2.First, we recall the finiteness criterion of Bilu and Tichy ([2]). To do this, weneed to define five kinds of so-called standard pairs of polynomials.Let α, β be nonzero rational numbers, µ, ν, q > 0 and ρ ≥ 0 be integers,and let ν(x) ∈ Q[x] be a nonzero polynomial (which may be constant).A standard pair of the first kind is (xq , αxρν(x)q) or switched, (αxρν(x)q ,xq), where 0 ≤ ρ < q, gcd(ρ, q) = 1 and ρ+ deg ν(x) > 0.A standard pair of the second kind is (x2, (αx2 + β)ν(x)2) or switched.Denote byDµ(x, δ) the µ-th Dickson polynomial, defined by the functionalequationDµ(z + δ/z, δ) = zµ + (δ/z)µor by the explicit formulaDµ(x, δ) =⌊µ/2⌋∑i=0dµ,ixµ−2i with dµ,i =µµ− i(µ− ii)(−δ)i.A standard pair of the third kind is (Dµ(x, αν), Dν(x, αµ)), wheregcd(µ, ν) = 1.A standard pair of the fourth kind is(α−µ/2Dµ(x, α),−β−ν/2Dν(x, β)),where gcd(µ, ν) = 2.A standard pair of the fifth kind is ((αx2 − 1)3, 3x4 − 4x3) or switched.256 A. BAZSO´, D. KRESO, F. LUCA AND A´. PINTE´RThe following theorem is the main result of [2].Theorem 2.1. Let R(x), S(x) ∈ Q[x] be nonconstant polynomials suchthat the equation R(x) = S(y) has infinitely many solutions in rationalintegers x, y. Then R = ϕ ◦ f ◦ κ and S = ϕ ◦ g ◦ λ, where κ(x), λ(x) ∈ Q[x]are linear polynomials, ϕ(x) ∈ Q[x], and (f(x), g(x)) is a standard pair.The following lemmas are the main ingredients for the proofs of Theorems1.3 and 1.4.Lemma 2.2. For every b ∈ Q and rational integer k ≥ 3 with k /∈ {4, 6}the polynomial Bk(x) + b has at least three zeros of odd muliplicities.Proof. For b = 0 and odd values of k ≥ 3 this result is a consequence ofa theorem by Brillhart ([4, Corollary of Theorem 6]). For non-zero rational band odd k with k ≥ 3 and for even values of k ≥ 8 our lemma follows from[6, Theorem] and [8, Theorem 2. 3], respectively.Our next auxiliary result is an easy consequence of an effective theoremconcerning the S-integer solutions of so-called hyperelliptic equations.Lemma 2.3. Let f(x) be a polynomial with rational coefficients and withat least three zeros of odd multiplicities. Further, let u be a fixed positiveinteger. If x and y are integer solutions of the equationf(xu)= y2,then we have max(|x|, |y|) < C3, where C3 is an effectively computableconstant depending only on u and the parameters of f .Proof. This is a special case of the main result of [5].Let c1, e1 ∈ Q∗ and c0, e0 ∈ Q.Lemma 2.4. The polynomial Ska,b(c1x + c0) is not of the form e1xq + e0with q ≥ 3.Lemma 2.5. The polynomial Ska,b(c1x+ c0) is not of the forme1Dν(x, δ) + e0,where Dν(x, δ) is the ν-th Dickson polynomial with ν > 4, δ ∈ Q∗.Before proving the above lemmas, we introduce the following notation.PutSka,b(c1x+ c0) = sk+1xk+1 + skxk + · · ·+ s0,andc′0 =ba+ c0.ON EQUAL VALUES OF POWER SUMS OF ARITHMETIC PROGRESSIONS 257We havesk+1 =akck+11k + 1,(2.8)sk =akck12(2c′0 − 1),(2.9)sk−1 =akck−1112k(6c′20 − 6c′0 + 1), k ≥ 2,(2.10)and for k ≥ 4,(2.11) sk−3 =akck−31720k(k − 1)(k − 2)(30c′40 − 60c′30 + 30c′20 − 1).Proof of Lemma 2.4. Suppose that Ska,b(c1x+ c0) = e1xq + e0, wherewe have q = k+1 ≥ 3. It follows that sk−1 = 0, so 6c′20 − 6c′0+1 = 0. Hence,c′0 /∈ Q, which is a contradiction.Proof of Lemma 2.5. Suppose that Ska,b(c1x + c0) = e1Dν(x, δ) + e0with ν > 4. Thensk+1 = e1,(2.12)sk = 0,(2.13)sk−1 = −e1νδ,(2.14)sk−3 =e1(ν − 3)νδ22.(2.15)From (2.8), (2.12) and (2.9), (2.13), respectively, it follows that(2.16) e1 =aν−1cν1νand c′0 =12.In view of (2.10), substituting (2.16) together with k = ν − 1 into (2.14), weobtain(2.17) −aν−1cν−21 (ν − 1)24= −aν−1cν1νδν,which implies(2.18) c21 =ν − 124δ.Similarly, comparing the forms (2.11) and (2.15) of sk−3 with the substitutionsk = ν − 1 and (2.16), we obtain(2.19)7aν−1cν−41 (ν − 1)(ν − 2)(ν − 3)5760=aν−1cν1(ν − 3)νδ22ν,which implies(2.20) c41 =7(ν − 1)(ν − 2)2880 δ2.258 A. BAZSO´, D. KRESO, F. LUCA AND A´. PINTE´RAfter substituting (2.18) into (2.20), we obtain 7(ν − 2) = 5(ν − 1), whichimplies ν = 9/2, a contradiction.One can see that the condition ν > 4 is necessary. Indeed,S22,1(x) =43x3 −13x =43D3(x,112),andS32,1(x) = 2x4 − x2 = 2D4(x,18)−116.3. Proofs of the TheoremsProof of Theorem 1.3. Using (3), one can rewrite equation (6) ascll + 1(Bl+1(y +dc)−Bl+1(dc))=12ax2 +(b−a2)xor8acll + 1(Bl+1(y +dc)−Bl+1(dc))= 4a2x2 + 8a(b−a2)x= (2ax+ 2b− a)2 − (2b− a)2.Then our result is a simple consequence of Lemmas 2.2 and 2.3.Proof of Theorem 1.3. Following Theorem 1.1, we haveS3a,b(x) =a34(x+ba−12)4−a38(x+ba−12)2+a4 − 16a2b2 + 32ab3 − 16b464a.Using the above representation, we rewrite equation (7) as64aSlc,d(y) = (2ax+2b−a)4− 4a2(2ax+2b−a)2+a4− 16a2b2+32ab3− 16b4or64aSlc,d(y) + 3a4 + 16a2b2 − 32ab3 − 16b4 = (X − 2a2)2,where X = (2ax + 2b − a)2. As in the previous case, Lemmas 2.2 and 2.3complete the proof.Proof of Theorem 1.2. If the equation (5) has infinitely many integersolutions, then by Theorem 2.1 it follows that Ska,b(a1x + a0) = ϕ(f(x)) andSlc,d(b1x + b0) = ϕ(g(x)), where (f, g) is a standard pair over Q, a0, a1, b0, b1are rationals with a1b1 6= 0 and ϕ(x) is a polynomial with rational coefficients.Assume that h = degϕ > 1. Then Theorem 1.1 implies0 < deg f, deg g ≤ 2,ON EQUAL VALUES OF POWER SUMS OF ARITHMETIC PROGRESSIONS 259and since k < l, we have deg f = 1, deg g = 2. In particular, k + 1 = hand l + 1 = 2h, so l = 2k + 1. Therefore, if l 6= k + 1, we then must haveh = degϕ = 1 and l = 2k + 1.Condition k 6= 1 implies k ≥ 2 and since l = 2k + 1, it follows that l ≥ 5.Since deg f = 1, there exist f1, f0 ∈ Q, f1 6= 0, such that Ska,b(f1x+f0) = ϕ(x),soSka,b(f1g(x) + f0) = ϕ(g(x)) = Slc,d(b1x+ b0).As g(x) is quadratic, by making the substitution x 7→ (x− b0)/b1, we obtainthat there are c2, c1, c0 ∈ Q, c2 6= 0, such thatSka,b(c2x2 + c1x+ c0) = Slc,d(x).Since degSka,b(x) = k+ 1 ≥ 2 and c2 6= 0, we have a decomposition of Slc,d(x)which is equivalent to S((x+b/a−1/2)2) for some S ∈ Q[x] with deg S = k+1,according to Theorem 1.1. Therefore, there exists a linear polynomial l(x) inC[x] such thatc2x2 + c1x+ c0 = l((x+ b/a− 1/2)2)and S(x) = Ska,b(l(x)). Hence, there are A,B ∈ C, A 6= 0, such thatc2x2 + c1x+ c0 = A(x + b/a− 1/2)2 +B.Clearly, this implies that A,B ∈ Q andSka,b(A(x+ b/a− 1/2)2 +B)= S2k+1c,d (x).By the linear substitution x 7→ x− b/a+ 1/2, we obtain(3.21) Ska,b(Ax2 +B) = S2k+1c,d (x− b/a+ 1/2).Thus, we have an equality of polynomials of degree 2k + 2 ≥ 6. We calculateand compare coefficients of the first few highest monomials participating inthe above polynomials. The coefficients of the polynomial in the right–handside above are easily deduced by setting c1 = 1, c0 = −b/a + 1/2 in (2.8),(2.9), (2.10) and (2.11). Therefore, if we denoteS2k+1c,d (x− b/a+ 1/2) = r2k+2x2k+2 + · · ·+ r1x+ r0,andc′0 =dc−ba+12,260 A. BAZSO´, D. KRESO, F. LUCA AND A´. PINTE´Rthen the coefficients are:r2k+2 =c2k+12k + 2,r2k+1 =c2k+12(2c′0 − 1),r2k =c2k+1(2k + 1)12(6c′20 − 6c′0 + 1),r2k−2 =c2k+1(2k + 1)2k(2k − 1)720(30c′40 − 60c′30 + 30c′20 − 1).On the other hand, the coefficients sk+1, sk, . . . s0 for the polynomial Ska,b(x)can be found by setting c1 = 1, c0 = 0 in (2.8), (2.9), (2.10) and (2.11). SinceSka,b(Ax2 +B) =k+1∑m=0smm∑i=0(mi)(Ax2)iBm−i,it follows that if we putSka,b(Ax2 +B) = t2k+2x2k+2 + · · ·+ t1x+ t0,thent2k+2 =akAk+1k + 1,t2k+1 = 0,t2k = akAkB +akAk2(2(ba)− 1),t2k−1 = 0,t2k−2 =akk2Ak−1B2 +akk2Ak−1B(2(ba)− 1)+akk12Ak−1(6(ba)2− 6(ba)+ 1).Now we compare the coefficients. Comparing the leading coefficients yields(3.22)akAk+1k + 1=c2k+12k + 2, so 2akAk+1 = c2k+1,and2ca=c2k+2ak+1Ak+1.Therefore,k+1√2ca∈ Q.ON EQUAL VALUES OF POWER SUMS OF ARITHMETIC PROGRESSIONS 261If a and c do not fulfill the above condition, we are through, otherwise weproceed. Comparing the coefficients of index 2k + 1, we getc2k+12(2c′0 − 1) = 0,so c′0 = 1/2, which impliesdc=ba.If the coefficients a, b, c and d do not satisfy the last property above, thenwe eliminate the possibility degϕ > 1. Therefore, we proceed with the casewhere a, b, c and d do satisfy this property. Comparing the next coefficientsand using (3.22), we obtain(3.23)ba−12= −112A(2k + 1)−B.Comparing the coefficients of index 2k − 2 and using c′0 = 1/2, we getakk2Ak−1B2 +akk2Ak−1B(2(ba)− 1)+akk12Ak−1(6(ba)2− 6(ba)+ 1)=78·c2k+1(2k + 1)2k(2k − 1)720.By using also (3.22) and simplifying, we obtainB22+B2(2(ba)− 1)+112(6(ba)2− 6(ba)+ 1)=7(4k2 − 1)A21440.By using also (3.23), the last relation above can be transformed intoB22+B(−112A(2k + 1)−B)+12(−112A(2k + 1)−B)2−124=7A2(4k2 − 1)1440.After simplification, we obtainA2(k − 3)(−2k − 1) = 15.For k ≥ 3, the expression in the left–hand side above is negative or zero,which is a contradiction. If k = 2, then A2 = 3, which contradicts the factthat A ∈ Q. Therefore there are no rational coefficients a, b, c, d, A and Bsuch that (3.21) is fulfilled, which implies that degϕ = 1.262 A. BAZSO´, D. KRESO, F. LUCA AND A´. PINTE´RNow, we haveSka,b(a1x+ a0) = e1f(x) + e0 and Slc,d(b1x+ b0) = e1g(x) + e0,where 0 6= e1, e0 ∈ Q. Further, we have deg f = k + 1 and deg g = l+ 1.In view of the assumptions on k and l, it follows that the standard pair(f, g) cannot be of the second kind, and with the exception of the case (k, l) =(3, 5), of the fifth kind either.If it is of the first kind, then one of the polynomials Ska,b(a1x + a0) andSlc,d(b1x + b0) is of the form e1xq + e0 with q ≥ 3. This is impossible byLemma 2.4.If (f, g) is a standard pair of the third or fourth kind, we then haveSlc,d(b1x + b0) = e1Dν(x, δ) + e0 with ν = l + 1 ≥ 5 and δ ∈ Q∗, whichcontradicts Lemma 2.5 or k = 2, l = 3. In this case Theorem 1.4 gives aneffective finiteness result.Now returning to the special case (k, l) = (3, 5), by using formula (2.10)for k = 3 it is easy to see that S3a,b(c1c+c0) = e1(3x4−4x3)+e0 is impossible,see the proof of Lemma 2.4.Acknowledgements.The authors are grateful to the referee for her/his careful reading andhelpful remarks.The research was supported in part by the Hungarian Academy ofSciences, by the OTKA grants K75566, K100339, NK101680, NK104208and by the TA´MOP 4.2.1./B-09/1/KONV-2010-0007 project implementedthrough the New Hungary Development Plan cofinanced by the EuropeanSocial Fund and the European Regional Development Fund. Dijana Kresowas supported by the Austrian Science Fund (FWF): W1230-N13 and NAWIGraz.References[1] A. Bazso´, A´. Pinte´r and H. M. Srivastava, A refinement of Faulhaber’s Theorem con-cerning sums of powers of natural numbers, Appl. Math. Lett. 25 (2012), 486–489.[2] Y. F. Bilu and R. F. Tichy, The Diophantine equation f(x) = g(y), Acta Arith. 95(2000), 261–288.[3] Y. F. Bilu, B. Brindza, P. Kirschenhofer, A´. Pinte´r and R. F. Tichy, Diophantineequations and Bernoulli polynomials (with an Appendix by A. Schinzel), CompositioMath. 131 (2002), 173–188.[4] J. Brillhart, On the Euler and Bernoulli polynomials, J. Reine Angew. Math. 234(1969), 45–64.[5] B. Brindza, On S-integral solutions of the equation ym = f(x), Acta Math. Hungar.44 (1984), 133–139.[6] A´. Pinte´r and Cs. Rakaczki, On the zeros of shifted Bernoulli polynomials, Appl.Math. Comput. 187 (2007), 379–383.[7] H. Rademacher, Topics in Analityc Number Theory, Springer-Verlag, Berlin, 1973.ON EQUAL VALUES OF POWER SUMS OF ARITHMETIC PROGRESSIONS 263[8] Cs. Rakaczki, On some generalizations of the Diophantine equation s(1k +2k + · · ·+xk) + r = dyn Acta Arith. 151 (2012), 201–216.[9] J. J. Scha¨ffer, The equation 1p + 2p + 3p + · · · + np = mq , Acta Math. 95 (1956),155–189.A. Bazso´Institute of MathematicsMTA-DE Research Group ”Equations, functions and curves”Hungarian Academy of ScienceUniversity of DebrecenH-4010 Debrecen, P.O. Box 12HungaryE-mail : bazsoa@science.unideb.huD. KresoInstitut fu¨r Mathematik (A)Technische Universita¨t GrazSteyrergasse 30, 8010 GrazAustriaE-mail : kreso@math.tugraz.atF. LucaMathematical Center UNAMUNAM Ap. Postal 61–3 (Xangari)CP 58 089, Morelia, Michoaca´nMexicoE-mail : fluca@matmor.unam.mxA´. Pinte´rInstitute of MathematicsMTA-DE Research Group ”Equations, functions and curves”Hungarian Academy of ScienceUniversity of DebrecenH-4010 Debrecen, P.O. Box 12HungaryE-mail : apinter@science.unideb.huReceived : 7.11.2011.Revised : 27.12.2011.

On equal values of power sums of arithmetic progressions

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