GENERAL SOLUTIONS OF SUMS OF CONSECUTIVE CUBED INTEGERS EQUAL TO SQUARED INTEGERS 5 1 VLADIMIRPLETSER 0 2 n Abstract. All integer solutions (M,a,c) to the problem of the sums of M a consecutive cubed integers (a+i)3 (a>1, 0≤i≤M−1) equaling squared J integers c2 are found by decomposing the product of the difference and sum of the triangular numbers of (a+M−1) and (a−1) in the product of their 4 greatest common divisor g and remaining square factors δ2 and σ2, yielding 2 c=gδσ. Further, the condition that g must be integer for several particular andgeneralcasesyieldgeneralizedPellequationswhosesolutionsallowtofind ] T allinteger solutions(M,a,c)showingthat thesesolutionsappear recurrently. In particular, it is found that there always exist at least one solution for the N cases of all odd values of M, of all odd integer square values of a, and of all h. evenvaluesofM equaltotwiceanintegersquare. t Keywords: Sums of consecutive cubed integers equal to square integers ; a Quadratic Diophantine equation ; Generalized Pell equation ; Fundamental m solutions;Chebyshevpolynomials [ 1 MSC2010 : 11E25 ; 11D09 ; 33D45 v 8 9 1. Introduction 0 ItisknownsincelongthatthesumofM consecutivecubedpositiveintegersstarting 6 0 from1equalsthesquareofthesumoftheM consecutiveintegers,whichitselfequals . the triangular number of the number of terms M, 1 △M 0 M M 2 M(M +1) 2 5 (1.1) i3 = i = = 2 1 ! 2 △M : Xi=1 Xi=1 (cid:18) (cid:19) v for M Z+. The question whether this remarkable result can be extended to i ∀ ∈ X other integer values of the starting point, i.e. whether the sum of consecutive r cubed positive integers starting from a=1 is also a perfect square a 6 M−1 (1.2) (a+i)3 =c2 i=0 X has been addressedby severalauthorsbut has receivedso far only partialanswers. Withthenotationofthispaper,Lucasstated[15]thattheonlysolutionsforM =5 area=0,1,96and118(missing the solutiona=25,seefurther Table1),andthat there are no other solutions for M = 2 than a = 1. Aubry showed [1] that a solution for M = 3 is a = 23, c = 204, correcting Lucas’ statement that there are no other solution for M = 3 than a = 1. Other historical accounts can be found in [7]. Cassels proved [3] by using the method of finding all integral points on a given curve of genus 1 y2 = 3x x2+2 (with x = a+1, y = c in this paper notations), that the only solutions for M = 3 are a = 1,0,1 and 23. Stroeker (cid:0) (cid:1) − obtained [37] complete solutions for 2 M 50 and M = 98, using estimates of ≤ ≤ 1 GENERAL SOLUTIONS OF SUMS OF CONSECUTIVE CUBED INTEGERS EQUAL TO SQUARED INTEGERS2 lower bound of linear forms in elliptic logarithms to solve elliptic curve equations of the form Y2 = X3 + dX where d = n2 n2 1 /4, X = nx + n(n 1)/2, − − Y = ny (with n = M, x = a, y = c in this paper notations). The method (cid:0) (cid:1) reported, although powerful, appears long and difficult and caused some problems for the cases M = 41 and 44. Stroeker remarked also that M = a = 33 with c = 2079 = 33 63. This is not the single occurrence of M = a, as it occurs also × for M =a=2017,124993,7747521,...(see [24, 25]). One of the reasons that these previous attempts to find all solutions (M,a,c) to (1.2) were only partially successful was most likely due to the approach taken to startthesearchforsolutionsforsinglevaluesofM,onebyoneandinanincreasing order of M values. The method proposed in this paper is to tackle the problem in a different way and instead of looking at each individual values of M one by one, to consider the problem in a more global approach by comparing different sets of known solutions, and instead of listing solutions in increasing order of M values, to look at two other parameters, δ and σ, defined further. This new beginning then leads to a more classical approach using general solutions of Pell equations, that allows to find all solutions in (M,a,c) of (1.2) for all possible cases. Note that Pell equations were already used previously by various authors (e.g. Catalan [4,5],Cantor[2],Richaud[34])inthe19thcenturyinattemptstosolvethepresent problem, albeit without reaching a complete resolution of the problem. The approach proposed in this paper includes three steps. Step1inSection2isbasedonthedecompositionoftheproductofthedifference∆ andthe sum Σ ofthe two triangularnumbers of(a+M 1)and (a 1) insimple − − factors,g,δ,σ where g =gcd(∆,Σ), allowingto find generalexpressionsofa andc in function of M,δ and σ that are always solutions of (1.2). In step 2 in Section 3, some conditions on δ and σ are explored to obtain three particular cases of solutions yielding specific expressions of M in function of k, k Z+, including the case of M taking all odd positive integer values. Section 4 ∀ ∈ recalls some basics on Pell equation solutions to introduce the third step. Instep 3 inSection 5,a generalsolutionin (M,a,c) is found for allvalues ofδ and σ,basedonsolutionsofsimpleandgeneralizedPellequationsinvolvingChebyshev polynomials,allowingtofindallthesolutionsin(M,a,c)to(1.2). Asanalternative to step 3, recurrence relations are deduced in Section 6. Section 7 summarizes all findings. Trivial solutions are not considered here. For instance, M 0 is meaningless; for ≤ M = 1, the only solution is obviously a = α2, yielding (M,a,c) = 1,α2,α3 ; therefore,weconsideronlyM >1. Ifa<0,therearenosolutionsifM <(1 2a); (cid:0) − (cid:1) if M = (1 2a), the only solution is (M,a,c) = ((1 2a),a,0); if M > 1 2a, − − − the solutions are (M,a,c) = M,(1 a), 2 2 , i.e. a shift of the − △M+a−1−△(−a) first term from a negative va(cid:16)lue a to a po(cid:16)sitive value (1 a)(cid:17)(cid:17)with a reduction of − the number of terms (M +2a 1). If a = 0 or 1, the classical solutions (1.1) − are (M,a,c) = M,0, 2 or M,1, 2 . Therefore, we limit our search to △M−1 △M solutions for M >1 and a>1. (cid:0) (cid:1) (cid:0) (cid:1) GENERAL SOLUTIONS OF SUMS OF CONSECUTIVE CUBED INTEGERS EQUAL TO SQUARED INTEGERS3 2. Step 1: A General theorem Theorem1. For δ Z+, σ,a,M,c Z+,κ Q+,suchasδ <σ,gcd(δ,σ)=1, ∀ ∈ ∃ ∈ ∈ κ=(σ/δ)>1, M−1 (2.1) (a+i)3 =c2 i=0 X holds if M κ2 1 +1+ M2(κ4 1)+1 (2.2) a = − − 2 (cid:0) (cid:1) p κM (2.3) c = κ2M + M2(κ4 1)+1 2 − Proof. For a,M,c,∆,Σ,g,∆′,Σ(cid:16)′,δ,σ,Cp Z+, i,k Z∗,(cid:17)κ Q+, with M > 1, ∈ ∈ ∈ 0 i M 1, κ=(σ/δ)>1, the sum of cubes of M consecutive integers (a+i) ≤ ≤ − for i=0 to M 1 can be written successively as − M−1 a+M−1 a−1 (2.4) (a+i)3 = i3 i3 − i=0 i=0 i=0 X X X (2.5) = △2a+M−1−△2a−1 (2.6) = ( a+M−1 a−1)( a+M−1+ a−1) △ −△ △ △ (2.7) = (∆)(Σ) where ∆= a+M−1 a−1 and Σ= a+M−1+ a−1, i.e. the difference and the △ −△ △ △ sum of the triangular numbers of (a+M 1) and (a 1), with obviously ∆<Σ, − − that can also be written M 1 (2.8) ∆ = M a+ − 2 (cid:18) (cid:19) M(M 1) (2.9) Σ = a2+a(M 1)+ − − 2 Let g =gcd(∆,Σ), yielding ∆=g∆′ and Σ=gΣ′. For (2.1) to hold, c2 =g2∆′Σ′ ′ ′ and as ∆ and Σ are coprimes and their product must be square, both must be integer squares, i.e. ∆′ =δ2 and Σ′ =σ2, with gcd(δ,σ)=1 and δ <σ, yielding (2.10) c=gδσ From (2.8) and (2.9), one has then respectively 2gδ2 (2.11) 2a+M 1 = − M (2.12) = 4gσ2 (M2 1) − − where the + sign is taken in front of thpe square root in (2.12) as 2a+M > 1. Solving for g yields then M (2.13) g = κ2M + M2(κ4 1)+1 2δ2 − (cid:16) p (cid:17) Replacing in (2.11) or (2.12) yields then directly (2.2) and in (2.10) yields directly (2.3). (cid:3) GENERAL SOLUTIONS OF SUMS OF CONSECUTIVE CUBED INTEGERS EQUAL TO SQUARED INTEGERS4 3. Step 2: Three Particular Solutions and a Generalization For which values of M does (2.1) hold ? Answers can be found for at least three particular cases. Other general approaches are developed further. Table 1 gives for 1 < M 45 and 1 < a < 105, the first values of M,a and ≤ associated ∆,Σ,g,δ,σ and c values. For M and a < 105, there are 892 solutions (M,a,c), given in [26, ?], such that (4.1) holds. One observes very easily that: (i) all odd values of M have at least one entry (in bold in Table 1) with g = 2 (M−1)/2 = M2 1 /4,δ =2M andσ = 2M2 1 ,yieldingc=M M2 1 2M2 1 /2 △ − − − − with a=M3 (3M 1)/2; −(cid:0) −(cid:1) (cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) (ii) those odd values of a equal to odd integer squares have at least one entry (e.g. forM =17inTable1)withg =(a 1)/8,δ =(2a 1)andσ =(2a+1),yielding − − c=(a 1) 4a2 1 /8 with M =(√a 1)(2a 1)/2 − − − − (iii) those even values of M equal to twice an integer square have at least one (cid:0) (cid:1) entry (in italics in Table 1) with g = M M2 1 /2, δ = 1, σ = M/2, yielding − c=M M2−1 M/2/2 with a=△M−(cid:0)1. (cid:1) p These three casescanbe generalizedrespectivelyto allodd values ofM,to allodd integer(cid:0)square v(cid:1)aplues of a = (2k+1)2, and to all even values of M equal to twice integer squares in the following Theorem 2. k Z+, δ,σ,M,a,c Z+ such that (2.1) holds : ∀ ∈ ∃ ∈ (i) if σ = δ2 2 /2, with − (cid:0) (cid:1) (3.1) M = (2k+1) (3.2) a = (2k+1)3 (3k+1) − (3M 1) (3.3) = M3 − − 2 (3.4) c = 2k(k+1)(2k+1)(8k(k+1)+1) M M2 1 2M2 1 (3.5) = − − 2 (cid:0) (cid:1)(cid:0) (cid:1) (ii) if σ =δ+2, with (3.6) a = (2k+1)2 (3.7) M = k(8k(k+1)+1) (√a 1)(2a 1) (3.8) = − − 2 k(k+1) (3.9) c = 4(2k+1)4 1 2 − (a 1) 4(cid:16)a2 1 (cid:17) (3.10) = − − 8 (cid:0) (cid:1) GENERAL SOLUTIONS OF SUMS OF CONSECUTIVE CUBED INTEGERS EQUAL TO SQUARED INTEGERS5 Table 1. First values of M,a,∆,Σ,g,δ,σ and c for 1 < M 45 ≤ and 11 and (3.11) M = 2k2 (3.12) a = k2 2k2 1 − M(M 1) (3.13) = (cid:0) − (cid:1) 2 (3.14) c = k3 4k4 1 − M(cid:0)M2 1(cid:1) M (3.15) = − 2 2 (cid:0) (cid:1)r Proof. For k,δ,σ,g,a,M >1,c Z+, q Q+, for (2.1) to hold: ∈ ∈ (i) let σ = δ2 2 /2. Replacing in (2.13) yields − (cid:0) M2(cid:1) δ2 (3.16) g = δ4 4δ2+4+ δ8 8δ6+8δ4+16δ2 2 +16 8δ4 − s − (cid:18)M2 − (cid:19) ! Forthepolynomialinδ underthe squarerootsigntobeasquare,let(δ/M)2 2= − 2, i.e. δ = 2M, yielding δ8 8δ6 + 8δ4 + 32δ2 + 16 = δ4 4δ2 4 2, giving − − − g = M2 δ2 4 /4δ2 = M2 1 /4 where δ was replaced by 2M. As g Z+, M − − (cid:0) (cid:1)∈ cannot be even and must be odd, i.e. M =2k+1 k Z+, yielding g =k(k+1), (cid:0) (cid:1) (cid:0) (cid:1) ∀ ∈ δ = 2M = 2(2k+1), σ = 2M2 1 = 8k(k+1)+1. Replacing in (2.2) and (2.3) − yields directly (3.2) to (3.5). (ii) Let σ =δ+2. Replacing in (2.13) yields M (3.17) g = M(δ+2)2+ M2 (δ+2)4 δ4 +δ4 2δ4 r − ! (cid:16) (cid:17) For the polynomial in δ under the square root sign to be a square, let M = qδ and δ = 8q(q +1)+1 with q Q+. It yields then q2 (δ+2)4 δ4 +δ2 = ∈ − r δ2 4q(δ+1), giving g = q(q+1)/2. As g Z+, q Z+(cid:16)and let q = k,(cid:17)yielding − ∈ ∈ g = , δ = 16 +1, σ = 16 +3, M = kδ = k(16 +1)=(3.7). Replacing k k k k △ △ △ △ in (2.11) and (2.10) yields (3.6) and (3.9). Replacing k in function of a from (3.6) yields also (3.8) and (3.10). (iii) Let δ =1 and σ =k with k >1. Then (2.13) reads M k2M + M2(k4 1)+1 − (3.18) g = (cid:16) p 2 (cid:17) whichtakesintegervaluesifM =2k2,yielding g =k2 4k4 1 =M M2 1 /2. Replacing in (2.11) and (2.10) yields directly (3.12) to (3.15−). − (cid:3) (cid:0) (cid:1) (cid:0) (cid:1) The case (i) of Theorem 2 confirms the statement of Stroeker ([37], p. 297) about alloddvalues ofM (in boldin Table 1)havinga solutionto (2.1). The first50000 values of (M,a,c) for this case (i) are given in [28, 29] For the case (ii) of Theorem2, Table 2 givesthe first five values of M,a,g,δ,σ and c. The first 50000 values of (M,a,c) for this case (ii) are given in [30, 31]. Asthereexistothertripletsofvaluesof(M,a,c)suchthat(2.1)holdswithM =kδ forthesamevalueofg =∆ ,thecase(ii)ofTheorem2canbegeneralizedtoother k values of a as follows. Table 3 shows some of these values for 1 k 2 and ≤ ≤ 11, such that (2.1) holds n n n n n n ∀ ∈ ∃ ∈ if σ = δ2 +(δ2 1)/2k(k+1), with n n n− (3.25) p Mn = kδn (3.26) = k Un (2k+1)2 Un−1 (2k+1)2 − δh+1(cid:16) (cid:17) (cid:16) (cid:17)i n (3.27) a = n 2 Un (2k+1)2 Un−1 (2k+1)2 +1 − (3.28) = (cid:16) (cid:17) 2 (cid:16) (cid:17) k(k+1)((2k(k+1)+1)δ2 1) (3.29) c = δ n− n n 8 r k(k+1)U (2k+1)2 2n (3.30) = 2(cid:16) (cid:17) Proof. For k,n,δ ,σ ,g,M ,a ,c Z+, M >1, let n n n n n n ∈ σ = δ2 +(δ2 1)/2k(k+1). Replacing in (2.13) yields n n n− p g = Mn (2k(k+1)+1)δn2 −1 M + 2δ4 2k(k+1) n n (cid:18)(cid:18) (cid:19) (2k(k+1)+1)δ2 1 2 (3.31) M2 n− δ4 +δ4 vu n (cid:18) 2k(k+1) (cid:19) − n! n u For the polynomial intδ under the square root sign to be a square,let M =kδ , n n n yielding after simplification g = k(k+1)/2 = . Replacing further in (2.2) k △ and (2.10) yields (3.27) and (3.29). Replacing further δ by (3.23) yields (3.26) n and (3.28). Replacing δ and σ in (2.10) yields (3.30), noting that δ σ = n n n n U2 (2k+1)2 U2 (2k+1)2 =U (2k+1)2 as can be shown by replac- n − n−1 2n ing(cid:16)Un2(x) an(cid:17)d Un2−1(x(cid:16)) in functi(cid:17)on of Ch(cid:16)ebyshev p(cid:17)olynomials of the first kind, respectivelyT (x) andT (x) (see e.g. [36])andsimplifying appropriately. (cid:3) 2n+2 2n For the case (iii) of Theorem 2, the first 50000values of (M,a,c) are given in [32]. Thereexistalsoothertripletsofvaluesof(M,a,c)suchthat(2.1)holdswithδ =1 andσ =k. Table4showssomeofthesevaluesfor11,Y >0, Z+). Among the five methods 1 1 ∈ listed by Robertson [35] to find the fundamental solution (X ,Y ), the classical 1 1 method based on the continued fraction expansion of the quadratic irrational √D introduced by Lagrange [12] is at the core of several other methods. It can be summarized as follows. One computes the rth convergent (p /q ) of the continued r r fraction [α ;α ,...,α ,α ,...] of √D which becomes periodic after the following 0 1 r r+1 term α =2α if √D is a quadratic surd or quadratic irrational(i.e. √D / Z+) r+1 0 ∈ and with α = √D i.e. the greatest integer √D . The terms p and q can be 0 i i ≤ found by the rejcurreknce relations (4.5) pi =αipi−1+pi−2 , qi =αiqi−1+qi−2 with p−2 = 0,p−1 = 1,q−2 = 1,q−1 = 0. The fundamental solution is then (X ,Y )=(p ,q ) if r 1(mod2) or (X ,Y )=(p ,q ) if r 0(mod2). 1 1 r r 1 1 2r+1 2r+1 ≡ ≡ For the general case of N = 1, the generalized Pell equation (4.1) can have either 6 no solution at all, or one or several fundamental solutions (X ,Y ), and all integer 1 1 solutions, if they exist, come on double infinite branches that can be expressed in function of the fundamental solution(s) (X ,Y ) and (X , Y ). Several authors 1 1 1 1 − (see e.g. [12, 6, 20, 19, 35, 16, 17, 11] and references therein) discussedhow to find the fundamental solution(s) of the generalized Pell equation, based on Lagrange’s GENERAL SOLUTIONS OF SUMS OF CONSECUTIVE CUBED INTEGERS EQUAL TO SQUARED INTEGERS10 methodofcontinuedfractionswithvariousmodifications(seee.g. [22]),andfurther how to find additional solutions from the fundamental solution(s). The method indicated by Matthews [17] based on an algorithm by Frattini [8, 9, 10] using Nagell’s bounds [20, 18], will be used further. Noting now (x ,y ) the fundamental solutions of the related simple Pell equation f f (4.3), the other solutions (X ,Y ) can be found from the fundamental solution(s) n n by n (4.6) X +√DY = X +√DY x +√Dy n n 1 1 f f ± (cid:16) (cid:17)(cid:16) (cid:17) for a proper choice of sign [35], or by the recurrence relations ± (4.7) Xn = xfXn−1+DyfYn−1 (4.8) Yn = xfYn−1+yfXm−1 that can also be written as (4.9) Xn = 2xfXn−1 Xn−2 − (4.10) Yn = 2xfYn−1 Yn−2 − or in function of Chebyshev’s polynomials of the first kind Tn−1(xf) and of the second kind Un−2(xf) evaluated at xf (see [23]) (4.11) Xn = X1Tn−1(xf)+DY1yfUn−2(xf) (4.12) Yn = X1yfUn−2(xf)+Y1Tn−1(xf) For N = η2 an integer square, the generalized Pell equation (4.1) admits always ′ ′ integer solutions. The variable change (X ,Y ) = ((X/η),(Y/η)) transforms the generalizedPellequationinasimplePellequationX′2 DY′2 =1whichhasinteger ′ ′ − solutions (X ,Y ). The integer solutions to the generalizedPell equation can then n n ′ ′ be found as (X ,Y ) = (ηX ,ηY ), or from ((4.9),(4.10)) with (X ,Y ) = (1,0) n n n n 0 0 and (X ,Y )=(ηx ,ηy ), yielding simply 1 1 f f (4.13) X = ηT (x ) n n f (4.14) Yn = ηyfUn−1(xf) (whichisalsovalidforthesimplePellequation(4.2)withη =1). Notehoweverthat not all solutions in (X,Y) may be found in this way (see e.g.[39]) and, depending on the value of D, other fundamental solutions may exist. 5. Step 3: A More General Approach ThethreecasesofTheorem2cannowbegeneralizedasshowninthenexttheorem, that includes also the general method to find all solutions (M,a,c) such that (2.1) holds.

General solutions of sums of consecutive cubed integers equal to squared integers

Similar General solutions of sums of consecutive cubed integers equal to squared integers