On 11 Sep 1997 rusin@vesuvius.math.niu.edu (Dave Rusin) wrote: > Recall we are looking for a set of consecutive integers the sum of > whose cubes is a cube. Here's the "right" notation: let a be the > number of integers, b the sum of the first and last. Then we have > an integer point on the surface a b (a^2+b^2-1) = c^3; conversely, > any integer point determines such a set of cubes, provided a and > b are of opposite parity... The "torsion points" I mentioned are > those in the family a= u^3, b = 3( (u^2-1)/3 )^2. ...it's helpful > to arrange the known solutions (not those in the infinite family) > as follows. Write a=p(a')^2 and b=q(b')^2, where p and q are > square-free... We could expand on this a little by characterizing the solutions in terms of the mutually coprime components of the three factors on the left side of your formula A B (A^2 + B^2 - 1) = K^3 (1) where I'll stipulate that A is odd and B is even. If we define x = gcd( A, A^2+B^2-1 ) y = gcd( A, B ) z = gcd( B, A^2+B^2-1 ) g = gcd( A, B, A^2+B^2-1 ) then x,y,z are pairwise coprime, and we have pairwise coprime integers a,b,c such that A = axyg B = byzg (A^2 + B^2 - 1) = cazg If we substitute for A and B into the right hand relation, it's clear that g must equal 1, and we have (axy)^2 + (byz)^2 - 1 = caz (2) Also, substituting into (1) gives (abc)(xyz)^2 = K^3 (3) Letting G denote the greatest common divisor of abc and xyz, there must be coprime integers u,v such that abc = Gu^3 xyz = Gv^3 (4) and so K = Guv^2. Following is a list of these parameters for all the solutions with A,B less than 30000. This includes two solutions that haven't been mentioned before. A B a b c x y z G u v ---- ---- --- --- ----- --- --- --- ---- --- --- * 3 8 1 1 3 3 1 8 3 1 2 25 4 5 1 32 5 1 4 20 2 1 25 20 5 1 256 1 5 4 20 4 1 25 36 5 3 32 5 1 12 60 2 1 49 20 7 1 20 7 1 20 140 1 1 49 630 7 3 13310 1 7 30 210 11 1 * 75 64 5 8 81 15 1 8 120 3 1 99 120 3 1 55 11 3 40 165 1 2 99 1210 3 11 49130 3 11 10 330 17 1 * 125 192 125 8 2187 1 1 24 24 45 1 153 578 3 17 59582 3 17 2 102 31 1 153 1444 3 19 544 51 1 76 3876 2 1 * 343 768 343 16 14739 1 1 48 48 119 1 833 288 7 3 68 119 1 96 1428 1 2 * 1323 512 7 64 1331 189 1 8 56 22 3 * 1331 4800 1331 40 206763 1 1 120 120 451 1 * 2197 9408 2197 56 555579 1 1 168 168 741 1 * 3267 1000 11 125 4913 297 1 8 11 85 6 * 4913 27648 4913 32 912673 1 1 864 32 1649 3 6591 7200 2197 15 595508 1 3 160 60 689 2 *12675 2744 65 343 107811 195 1 8 195 231 2 13923 19942 3 13 14042 357 13 118 547638 1 1 14161 17408 119 32 2248091 7 17 32 3808 131 1 *21675 4096 85 512 238521 255 1 8 2040 172 1 ? ? 33124 70225 91 53 100 364 1 1325 482300 5 1 63001 85340 251 1 33094240 1 251 340 85340 5 1 The two new [A,B] solutions are [6591,7200] and [13923,19942]. The second of these is interesting because it shows that [49,20] is not the only solution with u=v=1. The solutions with asterisks are those where either A or B is the cube of some integer m, so they are members of the infinite family of "torsion" solutions. Over the range covered in this table these torsion solutions all share the property that two of the three parameters x,y,z are cubes, and the third is either (m^2 - 1) or 3(m^2 - 1), depending on whether or not v is divisible by 3. It's interesting that for the entire list of solutions, including the two larger solutions with A,B greater than 30000 found by Rusin and Buddenhagen, the parameter v is always 1, 2, 3, or (in one case) 6. What is the smallest solution with v not equal to one of these four values? Also, notice that the only solutions in this table with v = 3 or 6 are members of the torsion set, so if we eliminate those, all the remaining solutions have v = 1 or 2. What is the smallest non-torsion solution with v greater than 2? Not surprisingly, the value of G is always greater than 1 in the above table, because if we had G=1 then equations (4) would imply that each of the numbers a,b,c and x,y,z is a cube, and so equation (2) would have the form [(a'x'y')^2]^3 + [(b'y'z')^2]^3 = [(c'a'z')]^3 + 1 (5) where the primed variables are cube roots of the corresponding unprimed variables. This is a restricted case of the "near-miss" solution of Fermat's equation for cubes, which is interesting because it's another case where there is a known infinite family of solutions, as well as a semingly plentiful supply of "sporadic" solutions. In other words, the equation X^3 + Y^3 = Z^3 + 1 (6) has infinitely many solutions because of identities such as (1 + 9m^3)^3 + (9m^4)^3 = (9m^4 + 3m)^3 + 1 (7) but there are other solutions as well, and it doesn't seem to be known if all the solutions are members of some infinite family. In any case we can certainly show that (5) has no solution of the form (7), because that would require 1 + 9m^3 to be a square, which implies an integer r such that 9m^3 = r^2 - 1 = (r+1)(r-1) Note that r+1 and r-1 cannot both be divisible by 3, so one of them is divisible by 9 and the other is coprime to 3. Thus, if r is even we can choose its sign to give integers s,t such that r+1 = 9s^3 r-1 = t^3 Subtracting one from the other gives 2 = 9s^3 - t^3 which is impossible (mod 9) where 0,+1,-1 are the only cubes. On the other hand, if r is odd then we know m is even, and so 2 divides one of (r+1),(r-1) and 4 divides the other. This gives two possibilities r+1 = 18s^3 , r-1 = 4t^3 or r+1 = 36s^3 , r-1 = 2t^3 In the first case we can substract the two equations to give 2 = 18s^3 - 4t^3, which is the same as 1 = 9s^3 - 2t^3, clearly impossible (mod 9). In the second case we have 2 = 36s^3 - 2t^3, which is the same as 1 + t^3 = 18s^3. This factors as (1+t) (1-t+t^2) = 18s^3 and each factor is divisible by 3 if and only if t=2 (mod 3), so we have an integer q such that t=3q+2 and the above equation becomes (q+1) (1 + 3q + 3q^2) = 2 s^3 From this it's clear that q must be odd, so we have integer k such that q = 2k-1, which gives k(12k^2 - 6k + 1) = s^3 The factors are coprime so we have integers m,n such that k = m^3 12k^2 - 6k + 1 = n^3 Solving the second of these for k gives __________________ ___________ 6 +- / 36 - 48(1 - n^3) 3 +- / 12n^3 - 3 k = ------------------------- = ------------------ 24 12 To yield a rational result there must be an integer h such that h^2 = 12n^3 - 3 = 3(4n^3 - 1) which implies that 4n^3 - 1 must be of the form 3d^2 for some integer d. Thus we have 3d^2 + 1 = 4n^3 But notice that if we define the rational number f = (3d-1)/2 we have f^2 + f + 1 = (9/4)d^2 + (3/4) = 3n^3 Furthermore, we have the convenient identity (f+2)^3 + (1-f)^3 = 9(1 + f + f^2) so we can combine this with the previous relation to give (f+2)^3 + (1-f)^3 = (3n)^3 This is Fermat's equation for cubes, which of course has no rational solutions. So we've shown that if G=1 in our original problem then there must be an integer solutions of equation (6), but it can't be of the form (7). Therefore, it would have to be one of the other (much less dense) parametric solutions, or a sporadic solution. __________________________________________________________________ | MathPages /*\ http://www.seanet.com/~ksbrown/ | | / \ | |____________/"Truth for any man is that which makes him a man."___|============================================================================== [I just stumbled upon this Math Reviews item which is clearly of some related interest! Also, see Problem 6552 in Amer Math Monthly v 97 #7 -- djr] ============================================================================== 96i:11038 11D85 11D25 Stroeker, R. J.(NL-ROTT-E) On the sum of consecutive cubes being a perfect square. (English. English summary) Special issue in honour of Frans Oort. Compositio Math. 97 (1995), no. 1-2, 295--307. _________________________________________________________________ In this paper, the author describes a systematic method that answers the following problem: Given $n\in{\bf N}$, find all integers $x$ for which $x\sp 3+(x+1)\sp 3+\cdots+(x+n-1)\sp 3$ is a perfect square. Clearly, for $x=1$ and any $n$, this sum is a perfect square but, if one is interested in exhausting all such $x$ (for a given $n$) he or she can find only sparse results in the literature and, surely, no result of a general character. The author makes the observation that the above problem can be answered provided that one can explicitly calculate all integral points on the (already extensively investigated) elliptic curve $Y\sp 2=X\sp 3+d\sb nX$, where $d\sb n={1\over4}n\sp 2(n\sp 2-1)$. Recently, the author and the reviewer elaborated a method [Acta Arith. 67 (1994), no. 2, 177--196; MR 95m:11056] for finding all integral points on a Weierstrass model of an elliptic curve. (Independently, J. Gebel, A. Petho and H. G. Zimmer [Acta Arith. 68 (1994), no. 2, 171--192; MR 95i:11020] developed a similar method.) The realization of this method is heavily based on a recent estimate of S. David for linear forms in elliptic logarithms [Mem. Soc. Math. France (N.S.) No. 62 (1995), 143 pp.]. The advantage of the method is that, once the generators for the Mordell-Weil group of the corresponding elliptic curve are known, a number of clear steps, independent of any ad hoc arguments, lead to the explicit determination of all integral points. On the other hand, its uniform character permits one to deal with many curves simultaneously. The author takes advantage of this feature and utilizes successfully the method in order to answer explicitly the initial problem for all $n$ from 2 to 50 and for $n=98$. His strategy, in its essential lines, is independent from those particular values of $n$ and, surely, can be applied to other values of $n$ as well. The paper is very neatly written and its style is attractive. Reviewed by Nikos Tzanakis © Copyright American Mathematical Society 1996, 1997