From: ksbrown@seanet.com (Kevin Brown) Newsgroups: sci.math Subject: Sum of Consecutive Cubes Equals a Cube Date: Tue, 02 Sep 1997 05:29:21 GMT The famous "cannonball stacking" problem of Lucas (1875) requires a sum of consecutive squares, beginning with 1, equal to a square. The only non-trivial solution is 1^2 + 2^2 + 3^3 + ... + 24^2 = 70^2 However, as discussed in Laurent Beeckmans' article in the May 94 Mathematical Monthly, if we allow ANY set of k consecutive squares (not necessarily beginning with 1) there are solutions for k = 1, 2, 11, 23, 24, 33, 47,etc. For each of these we have infinitely many sequences of k consecutive squares whose sum is a square. For example, with k=24 we not only have the above sequence, we also have 9^2 + 10^2 + 11^2 + ... + 32^2 = 106^2 20^2 + 21^2 + 22^2 + ... + 43^2 = 158^2 etc. In general, the sum of the 24 squares beginning with m^2 is a square for m = 1, 9, 20, 25, 44, 76, ... These values can be determined by solving the corresponding Pell equation, since the sum of k consecutive squares beginning with m^2 is (k/6)(6m^2 - 6m + 6mk + 1 - 3k + 2k^2) Setting this equal to an arbitrary square N^2 gives, for any fixed k, a Pell equation in m and N. For example, with k=24 we have 24m^2 + 552m + 4324 = N^2 Solving this for m gives ____________ -138 + / 6N^2 - 6900 m = -------------------- 12 so there must be an integer q such that 6N^2 - q^2 = 6900 There are six infinite families of solutions [N,q], whose smallest members are [70,150] [106,246] [158,378] [182,438] [274,666] [430,1050] For any of these six fundamental solutions [N0,q0] the rest of the corresponding infinite family is given by _ _ _ _ _ _ | Nj | | 5 2 |j | N0 | | | = | | | | |_ qj _| |_ 12 5 _| |_ q0 _| Equivalently, both the sequences Nj and qj satisfy the second order recurrence s[j] = 10 s[j-1] - s[j-2]. For any q the corresponding value of m is (q-138)/12. If we go on to consider sums of higher powers, it appears that there are no sums of two or more consecutive 4th powers equal to a 4th power, or in general sums of two or more consecutive nth powers equal to an nth power for any n>3. [Can anyone supply a proof, reference, or counter-example?] This leaves the case of cubes, for which there are certainly some solutions, but they don't reduce to simple Pell equations so they aren't as easy to analyze as in the case of squares. The sum of k consecutive cubes begging with m^3 is (k/4)(2m+k-1)(2m^2 - 2m + 2km - k + k^2) Setting this quantity equal to a cube N^3, the only solutions in small integers that I've found are k m N --- --- ---- 3 3 6 4 11 20 20 3 40 20 15 70 25 6 60 49 291 1155 64 6 180 99 11 330 99 556 2805 125 34 540 153 213 1581 153 646 3876 288 273 2856 343 213 2856 512 406 5544 It's interesting that the values of k include the squares 4, 25, 49, and 64, and the cubes 125 and 343. Also there are pairs of solutions for k=20, 99, 153, and also with m=3, 11, 6, and 213. It's also interesting that the cube 2856^3 is expressible as a sum of consecutive cubes in two distinct way. Beyond those in the above table, what is the next cube expressible as a sum of (two or more) consecutive cubes? _________________________________________________________________ | MathPages /*\ http://www.seanet.com/~ksbrown/ | | / \ | |____________/Yet seemed it winter still, and, you away, _________| As with your shadow I with these did play. ============================================================================== From: James Buddenhagen Newsgroups: sci.math Subject: Re: Sum of Consecutive Cubes Equals a Cube Date: Fri, 05 Sep 1997 00:18:42 -0500 Kevin Brown wrote: [...] > This leaves the case of cubes, for which there are certainly some > solutions, but they don't reduce to simple Pell equations so they > aren't as easy to analyze as in the case of squares. The sum of k > consecutive cubes begging with m^3 is > > (k/4)(2m+k-1)(2m^2 - 2m + 2km - k + k^2) > > Setting this quantity equal to a cube N^3, the only solutions in small > integers that I've found are > > k m N > --- --- ---- > 3 3 6 > 4 11 20 > 20 3 40 > 20 15 70 > 25 6 60 > 49 291 1155 > 64 6 180 > 99 11 330 > 99 556 2805 > 125 34 540 > 153 213 1581 > 153 646 3876 > 288 273 2856 > 343 213 2856 > 512 406 5544 > > It's interesting that the values of k include the squares 4, 25, 49, > and 64, and the cubes 125 and 343. Also there are pairs of solutions > for k=20, 99, 153, and also with m=3, 11, 6, and 213. It's also > interesting that the cube 2856^3 is expressible as a sum of > consecutive cubes in two distinct way. > > Beyond those in the above table, what is the next cube expressible as > a sum of (two or more) consecutive cubes? Your equation: (k/4)(2m+k-1)(2m^2 - 2m + 2km - k + k^2) = N^3 with k as parameter is a family of elliptic curves of positive rank (over Q). I don't see any way to make use of this for your problem. Some searching with k a square gave a few new values: (k,m,N)= ( 64^2, 8790, 175440 ) (k,m,N)= ( 119^2, 1624, 249424 ) (k,m,N)= ( 251^2, 11170, 1962820 ). --Jim Buddenhagen(jbuddenh@swbell.net) [newsreader-fodder deleted -- djr] ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Sum of Consecutive Cubes Equals a Cube (long response...) Date: 5 Sep 1997 06:39:21 GMT Summary: Elliptic curves strike again Kevin Brown recently asked for consecutive sets of integers whose cubes sum to another cube. He included a small table of known solutions and asked if there are others. This turns out to suggest a number of computational problems in elliptic curves, which I spent a couple of days thinking about. I found one infinite family of solutions, but no other new solutions. Indeed, there is some evidence here to suggest that Brown's list (as amended) could be complete. This is rather a long article, so I've divided it into sections with headers. To get the ball rolling, let's muse with Brown on the possible generalizations of this avenue of research: >If we go on to consider sums of higher powers, it appears that there >are no sums of two or more consecutive 4th powers equal to a 4th >power, or in general sums of two or more consecutive nth powers equal >to an nth power for any n>3. [Can anyone supply a proof, reference, >or counter-example?] Counter-example: (-2)^5 + (-1)^5 + (0)^5 + (1)^5 + (2)^5 = 0^5 Only half a smiley here, because this is relevant in the case of cubes. As Brown noted, the sum of k consecutive cubes beginning with m^3 is (k/4)(2m+k-1)(2m^2 - 2m + 2km - k + k^2). Thus the sum of consecutive cubes is another cube precisely when the Diophantine solution N^3 = (k/4)(...) has an integer solution. As in the displayed equation, we clearly have a solution whenever m<0, with k = -2m+1 (and N=0). There are more trivial examples too: we could take any m<0 and use (k,N) = (-2m,m) or (k,N) = (-2m+2,-m+1). Are these trivial examples, together with Brown's list, the only solutions? I. ALGEBRAIC GEOMETRIC VIEWPOINT AND NOTATION Let me adjust the perspective in two ways. First, rather than looking for points (k,m,N) on an algebraic surface in 3-space, let me hold k fixed (but arbitrary) and look for points (m,N) on a curve in 2-space. This just slices up our search space into curves. One may also discuss the "generic" curve, that is, we can consider the curve defined over the polynomial ring Z[k]. Second, let me enlarge my search to look not only for integer points (m,N) on the curve, but for rational points. Any rational point on this curve gives k cubes of integers _in arithmetic progression_ whose sum is a cube; perhaps this is also interesting. Then for _every_ k, the trivial examples in the previous paragraph give me three points on each of these curves: (m,N) = ( (1-k)/2, 0 ), (-k/2, -k/2), and (1 - k/2, k/2). (Equivalently, these give three points on the generic curve over Q(k).) Very well then, we have now a cubic equation in two variables over the rationals; the equation has a rational point (say the point P0=( (1-k)/2, 0 ), just to pick one). The solution set is then an elliptic curve! Indeed, it can be rendered into normal form by introducing new variables x = 2k(k^2-1) N/(2m+k-1) y = k^2(k^2-1)^2/(2m+k-1) which satisfy (*) y^2 = x^3 + D where D = -k^4(k^2-1)^3. Equation (*) defines a well-known family of elliptic curves. (See e.g. the section IX.7 in Silverman's first book.) We will write E_k for the set of rational points satisfying (*). (Just to be precise here, each point (m,N) on the original curve determines a point (x,y) on the new one E_k, unless m = (1-k)/2 (and then N=0). Conversely, each rational point (x,y) corresponds to a rational solution (m,N) unless y=0 (and then x=-D^(1/3); this is possible iff k is a cube).) The other two "trivial" rational points on our first curve define points on E_k, namely P1=(k^2(k^2-1), k^2(k^2-1)^2) and its negative. (Recall that an elliptic curve is a group, and that in particular the negative of (x,y) is (x,-y).) II. GOALS AND OVERVIEW OF RESULTS Brown's original question is now, "are there any (rational) points on any E_k for which the corresponding coordinates m and N are integers?". (I believe he intended k, m, and N to be positive, but that's really irrelevant.) Observe that if m is integral, then so is N (assuming as throughout this article that all variables are rational). Thus we need only check the integrality of m. Our basic approach is to find the group of all rational points on E_k and then ask which of these satisfy the appropriate integrality conditions. We know that E_k contains at least the (infinite) cyclic group < P1 >; the question is whether or not there's anything else. I believe one can prove the "generic" curve (the curve over Q(k)) to have rank 1 and no torsion; that is, the only solutions to (*) with each of x and y expressed as rational functions of k are those obtained as multiples of P1 on this curve. There is a homomorphism from the generic curve to each of its specializations E_k; in general, one expects this map to be an isomorphism for "typical" k. The case k=5 is "typical" in this way: The group E_5 has rank 1; there is no torsion; the known point P1=(150,1800) is a generator. The points of E_5 are then easily enumerated: we have points 2P1=(825/4, 2925/8), 3P1=(22744/49, -3278528/343), 4P1=(81240225/2704, -732245324175/140608), 5P1=(2070265047000/2555403601, 2954483842534545600/129178207434151),... and their negatives. We can compute the corresponding (m,N): P1 -> m=-3/2, N=5/2 2P1 -> m= 230/13, N=440/13 3P1 -> m=-282083/102454, N=-298515/102454 4P1 -> m=-6513346782/3254423663, N=-1126531120/3254423663 etc. The negatives are related in a natural way, e.g. -P1 -> m=-5/2, N=-5/2 Each of these corresponds to a set of four "consecutive" cubes summing to another cube, just as Brown requested, except for the obvious fact that they are not cubes of _integers_ in general. For example, the data from P1 itself implies (-3/2)^3 + (-3/2+1)^3 + ... + (-3/2+4)^3 = (5/2)^3 It seems clear from the data that the denominators of the m (and N) for n.P1 grow with n, so we expect in particular no integer solutions to the original problem with k=5. A similar analysis applies for the "typical" k. One easily evaluates the multiples 2P1, 3P1, ... of our "trivial" solution P1, and notices quickly that the corresponding m are not integral except in the trivial cases arising from P1 itself and its negative (which are in fact only integral if k is even). Since for a "typical" k, the curve E_k contains only < P1 >, this means there are no solutions to Brown's original problem for this k. So we only have solutions to that problem when k is not "typical", allowing E_k to have "exceptional" points. There are five ways this can happen, listed here in order of deviation from "typical" behaviour: A) E_k = , indeed, but other multiples of P1 have integral m B) E_k is indeed cyclic but contains < P1 > as a proper subgroup C) E_k is indeed of rank 1 but has a nontrivial torsion subgroup D) is indeed infinite but not of finite index in E_k E) E_k isn't even an elliptic curve of rank at least 1. To decide whether or not these events occur is difficult in general, but for individual k each may be resolved. I ran some computations on a few small values of k; the results are summarized below, as well as some more general remarks. (All calculations were performed with the Maple suite APECS. In some cases they depend on widely-believed conjectures, so I suppose they are, technically, incomplete.) I'll discuss the five problems in reverse order. III. WHAT IF E_k LOOKS NOTHING LIKE THE GENERIC CURVE? (Problem (E) ) We will eventually show that the point P1 has infinite order in E_k for k>2, so at least the specialization map from the generic curve to E_k is an injection. For k=0 and k=1, the curve is E: y^2 = x^3. This is singular (i.e. it's not an elliptic curve at all) and has infinitely many rational points. This of course is predictable: it's not hard to have a sum of k=0 or k=1 consecutive cubes be a cube... For k=2, the rank of E: y^2=x^3-432 is zero, and indeed the only points on E are (x,y)=(12, +-36). But _of course_ there are no nontrivial rational points, as Fermat's Last Theorem for exponent 3 allows only trivial solutions to x^3+(x+1)^3 = z^3 ! By the way, there is a natural isomorphism between the curves with a given k and its negative, so in the sequel we will assume k >= 0. (I believe there is also value to considering other automorphisms of the original surface, but I haven't the tools to do so. For example, the pair of equations 3^3 + 4^3 + 5^3 = 6^3 (-2)^3 + (-1)^3 + ... + 2^3 + 3^3 + 4^3 + 5^3 = 6^3 illustrates an automorphism given by (k,m,N) -> (k+2m-1, 1-m, N). This automorphism swaps the trivial curves E_0 and E_1 with the sets of trivial points P0 and P1 on the other E_k. ) IV. CURVES OF RANK GREATER THAN 1 (Problem (D)) The case k=3 is the first one to illustrate this possibility. The rank of the group is 2, and the combination Brown reported leads to a point P2=(36,72) on the curve which together with P1=(72,576) generates E_3. The same is true of Brown's examples with k=20, 25, 49, and 288. When the curve E_k has other generators P2, P3, ... then of course there are more points to check for integrality. For example, in the cases k=20 and k=99, the two values Brown listed correspond to the points P2 and P1-P2 of E_k (each a group of rank 2); no other small linear combinations yielded integral m. The case k=153 is more interesting; here E_k has rank 3 and the two points found by Brown are independent of each other and of P1. No other small linear combination of the three gives integral m. It is not necessarily true that when the rank is larger than one, an exceptional solution will result. For example, the curve E_6 is also of rank 2; an additional generator is the point P2=(385,1225) corresponding to m=31/2 and N=33. Neither this point nor any small combination a.P1 + b.P2 appears to yield an integral coordinate m except of course a=(+-1) and b=0. (P2 itself is a close miss, though, and represents the interesting equality 31^3 + 33^3 + 35^3 + 37^3 + 39^3 + 41^3 = 66^3 .) The integer values of k for which E_k has rank greater than 1 are k=6, 13, 14, 16, 19, 20, 21, 25, 27, ... All these have rank 2; there are probably values of k for which the rank is much greater than 2, although as I understand it, the fact that the generic rank is 1 should mean that the "average" rank over all integers k should be 2 or less. Whether these curves of higher rank will provide infinitely many examples of points with _integer_ m is not at all clear to me. On the one hand there seem to be plenty such curves E_k, but on the other hand it seems very rare to attain integral m . V. PERFECT CUBES GIVE TORSION (Problem (C) ) Brown observed >It's interesting that the values of k include the squares 4, 25, 49, >and 64, and the cubes 125 and 343. [and 512] It's more appropriate to say "...cubes 64, 125, 343, and 512, and squares 4, 25, and 49", since the squares appear to be accidents but the cubes are not: Theorem: The torsion subgroup of E_k for k > 1 is of order 2, if k is a perfect cube of order 3, if k=2 zero otherwise Proof. We've already considered the case k=2 in section III. Clearly if k=u^3 we have the point (u^4(u^6-1), 0) in E_k, necessarily of order 2. Now we need to show there is never any other torsion. Elements (x,y) of order 2 in a curve in Weierstrass form must have y=0, which in our case means -D=k^4(k^2-1)^3 must be a cube. Elements P of order 3 have the property that the x-coordinates of P and 2P must be equal. Using standard formulae for 2P, we find this requires x^3=-4D and then y^2=-3D. The first requires that k have the form k=2w^3 for some w; then the second forces 3(4w^6-1) to be a square. But the curve Y^2 = 3(4X^3-1) has only trivial rational points (with X=1), so w=1 and then k=2. For P to have order 4 requires the y-coordinate of 2P to be zero; this translates to the vanishing of a quadratic polynomial in x^3/D, a polynomial which however has no rational roots, so no such (x,y) exists. Likewise for P to have order 5 or 7 requires x^3/D to be (obviously rational and) a root of an irreducible polynomial of degree greater than 1. Details left to the reader. Now use Mazur's theorem that the exponent of the torsion subgroup of an elliptic curve has exponent dividing 8*9*5*7. QED The upshot of this theorem is that we'll almost never have additional points to consider in E_k besides the ones we otherwise knew about, except when k is a cube. Those cases are interesting, however. As it turns out (see Sect. I) the torsion points P_T do not correspond to a pair (m,N). However, the points P1+P_T do, and a brief computation shows they have coordinates k = u^3 m = (u^3-2*u^2-4*u-4)*(u-1)/6 N = u*(u^2-1)*(u^2+2)/6 These expressions are integral unless u is a multiple of 3. This accounts for Brown's exceptional values with k=64, 125, 343, and 512. (Note that it's hard to spot a pattern for cubes which does not apply to u=0, u=1, u=2, u=3, or u=6 if you only check as far as u=8!) This adds infinitely many points to Brown's table, beginning with k=1000, m=1134, N=16830 but I would argue that this family is not "exceptional", and so its members are removed from Brown's table. There are groups with both torsion and a rank larger than 1, e.g. k=27; of course this means there are more points to check for integrality. VI. RANK-1 GROUPS WITH EXTRA POINTS (Problem (B) ) One of the difficulties in Elliptic Curve computations is that even when the rank is successfully deduced, it's not easy to find generators of the group. Even if the right _number_ of independent points is found, there's no guarantee that the subgroup they generate isn't a proper subgroup of all of E (of finite index). For any specific group, one _can_ compute an upper bound on the sizes of the x and y which much be searched (once the rank is known), but that's often impractical. In particular, even if E_k has rank 1, we don't really know P1 is a generator; for example E_k could well be a cyclic group < P > where P is some point with P1 = 2.P This is a concern particularly if we intend to search for solutions systematically -- we need to check the multiples n.P for integrality, not just n.P1 -- but in the end may not affect our results, since even if such a P exists, if it (and all odd multiples) have non-integral m, they don't correspond to solutions of our original problem about consecutive cubes. (It's not true, necessarily, that P itself will have integral coordinates even if, say, P1=2.P does -- doubling in an elliptic curve does not uniformly make the coordinates more complex. For example, the point P1=(9900,980100) on E_10 has a double 2.P1=(2700,99900) with smaller coordinates! (Its m is not integral, however.) ) No examples have been found of an E_k in which P1 was a multiple of another point P in E_k -- integral or otherwise. It's not hard to write down the equations determining the coordinates m and N of a point P with 2.P = P1, for example. The question is whether there exist _integer_ (or even _rational_) such m and N. I did test for k=3..500 and found that no such rational m exists. I suppose there is no "half" of P1 for any k but I didn't try to prove this. Likewise, one should be concerned that there be points P with 3.P=P1 and so on. For a specific curve of rank 1, it's not too hard to verify in this way that P1 is really a generator; but I don't know how to prove this for all k. (Of course, if the rank of the curve is greater than 1, we must worry more generally about whether there exist generators of lower height than a given independent set.) VII. INTEGRAL ELEMENTS IN < P1 > ITSELF (Problem (A) ) If the whole of the elliptic curve is known through its generators P1, P2, ... (and possibly a torsion element P_T), then in principle we must check all linear combinations a.P1 + b.P2 + ... (+ P_T) to see if m is integral. Now, in general, I believe the theory of heights on elliptic curves is sufficient to show that there will be only a finite number of such integral combinations, and indeed that the number of digits in the denominator of m is roughly proportional to |a|+|b|+... In particular, if E_k really is just , then there should be only a few multiples n.P1 in E_k which have integral m. In the examples I tried, the denominators become larger than 1 as soon as n>1 -- in general, but not universally. When k=4, the point 2.P1 still has integral m, and accounts for the remaining element in Brown's table. To what extent is this possible more frequently? I worked through the formulas for doubling on the elliptic curve, starting with a point whose m and N coordinates are known. It's not pretty; for example, the new N coordinate is 2 2 / N2 = - (k - 1) (k + 1) (5 k + 18 m k - 9 k - 18 m + 18 m + 4) N / / 2 2 2 3 4 (2 + 144 m k - 324 m k - 180 m k + 41 k - 36 m - 18 k - 36 k + 11 k 3 2 2 3 4 3 2 + 216 k m + 180 m k + 72 m k + 108 m - 216 m + 144 m ); the new m coordinate has the same denominator. In particular, we can apply this formula to our one generator P1; each curve E_k has a rational point 2.P1 whose coordinates are 2 (k - 1) (k + 1) (k + 8) N {N2 = 2 --------------------------, 2 4 - 8 - 20 k + k 2 3 (k - 1) k (- 28 k + 8 - 8 k + k ) m2 = - 1/2 ----------------------------------} 2 4 - 8 - 20 k + k Well, can these be integral? The answer is "yes" if k=4, but I can prove that's the only case. Here's how. I computed the resultant of the numerator and denominator of m2; it's 69657034752. That means there are integral polynomials f(k) and g(k) with f(k)*numer + g(k)*denom = 69657034752. In particular, if there is a k for which m2 is an integer, the denominator divides the numerator, and thus divides 69657034752 = 2^17*3^12. There are only finitely many such divisors, and for each only at most 4 k's with -8 - 20k + k^4 equal to that divisor. Guess what: the only rational solutions are k=+-1, +-2, and +-4. So for every curve E_k, 2.P1 is already non-integral (if k>4). One can presumably do a similar analysis to show 3.P1 is never integral, and so on, but this doesn't seem very productive. Since in general the denominators of n.P1 increase with n, it would appear we never get integral points this way except if n=1 (or k=4). Of course, if the rank of E_k is greater than 1, it's quite conceivable that both a new generator P2 and its double 2.P2 have integral m. I have not observed this to occur. (It's hard enough to find a P2 which itself has integral m, let alone asking 2.P2 to do the same!) VIII. CONCLUSION So what _are_ the consecutive sets of integer cubes which sum to a cube? We have identified several types of solutions: 1. The trivial solutions -- m=(1-k)/2 (so N=0) for odd k, and m=-k/2 or -k/2+1 for even k 2. The torsion solutions -- m=(u^3-2*u^2-4*u-4)*(u-1)/6 when k=u^3 3. Other elements in the subgroup generated by 1. and 2. (Conjecturally only k=4, m=11) 4. Other elements of rank-1 curves (None such? see Sect. VI) 5. Other generators of rank >1 curves (and linear combinations thereof) as for k=3, 20, 25, 49, 99, 153, 288 Types 3. and 4. may well provide no more solutions, but I suppose I consider type 5. cases more likely to yield additional solutions. (There are of course the negatives of these points on elliptic curves; they don't lead to _positive_ m). This was more of a fun problem than I expected. dave ============================================================================== From: ksbrown@seanet.com (Kevin Brown) Newsgroups: sci.math Subject: Re: Sum of Consecutive Cubes Equals a Cube (long response...) Date: Sat, 06 Sep 1997 06:06:08 GMT On 5 Sep 1997 rusin@vesuvius.math.niu.edu (Dave Rusin) wrote: > .... > VIII. CONCLUSION > So what _are_ the consecutive sets of integer cubes which sum to a > cube? We have identified several types of solutions: > 1. The trivial solutions -- m=(1-k)/2 (so N=0) for odd k, and > m=-k/2 or -k/2+1 for even k > 2. The torsion solutions -- m=(u^3-2*u^2-4*u-4)*(u-1)/6 when k=u^3 > 3. Other elements in the subgroup generated by 1. and 2. (Conjecturally > only k=4, m=11) > 4. Other elements of rank-1 curves (None such? see Sect. VI) > 5. Other generators of rank >1 curves (and linear combinations thereof) > as for k=3, 20, 25, 49, 99, 153, 288 > Types 3. and 4. may well provide no more solutions, but I suppose I > consider type 5. cases more likely to yield additional solutions. Splendid analysis! I see that Jim Buddenhagen has confirmed the existence of more solutions of type 5. Interestingly he found more cases where k is a square (119^2 and 251^2). By the way, I notice that the same problem was treated by C. Pagliani back in 1830, prior to the development of the modern machinery of elliptic curves. He proceeded by noting that the sum of the cubes of x+1, x+2,...,x+m is s = m(y+1)(y^2 + 2y + m^2)/8 where y = 2x+m. If we set m = 8n^3 then s is the cube of n(y+4n^2) if y=0 or y satisfies the equation 3 (4n^2 - 1) y = 2 (32n^6 - 24n^4 + 1) Letting v denote 2n, the above is equivalent to the statement that (x+1)^3 + (x+2)^3 + ... + (x+v^3)^3 = [ vx + v^3 (v+1)/2 ]^3 if 6x = (v^2 - 1)^2 - 3(v^3 + 1), which implies that x is an integer for any integer v not divisible by 3. For example, the cases v = 2, 4, and 10 give [-2^3 + -1^2 + 0^3 + 1^2 + 2^2 +] 3^3 + 4^3 + 5^3 = 6^3 6^3 + 7^3 + 8^3 + 9^3 + ... + 69^3 = 180^3 1134^3 + 1135^3 + ... + 2133^3 = 16830^3 So this corresponds to the infinite family of "torsion solutions" identified by Dave Rusin. Beyond these, it seems that even today we have no way of finding solutions other than by searching, possibly with guidance from the algebraic factorization of the sum. ============================================================================== Date: Sat, 6 Sep 1997 13:09:27 -0500 (CDT) From: Dave Rusin To: ksbrown@seanet.com Subject: Re: Sum of Consecutive Cubes Equals a Cube (long response...) Newsgroups: sci.math Just thought I'd add that the equation to be solved can be presented in a really pretty form: N^3 = x y (x^2+y^2-1) (where x and y are sum and difference of the terms at the ends of the string of integers to be cubed) Also, I've found that it _is_ possible, as I suggested, to prove that for each k, there are only finitely many points on the curve E_k whose "m" coordinate is integral. The proof is not effective in the sense that you can't put an upper bound on the m's to be searched for each k, although I think it's possible to pre-compute an upper bound on the _number_ of m's which will work. I see Jim B. has also posted the infinite family of "torsion" points, but his comment (to the effect that the corresponding E_k must then have rank at least 2) is incorrect, since I don't think he realized this new point does not add another generator _of infinite order_. dave ============================================================================== From: no-cet1-spam@cam.ac.uk (Chris Thompson) Newsgroups: sci.math Subject: Re: Sum of Consecutive Cubes Equals a Cube Date: 7 Sep 1997 23:57:01 GMT In article <340b9897.41266384@news.seanet.com>, ksbrown@seanet.com (Kevin Brown) writes |> |> It's interesting that the values of k include the squares 4, 25, 49, |> and 64, and the cubes 125 and 343. and Dave Rusin and Jim Buddenhagen have pointed out that the cubes are explained but the squares, if this is a real effect, are not. In article <5uo9eq$5uo$1@gannett.math.niu.edu>, rusin@vesuvius.math.niu.edu (Dave Rusin) writes |> |> (I believe there is also value to considering other automorphisms of |> the original surface, but I haven't the tools to do so. For example, |> the pair of equations |> 3^3 + 4^3 + 5^3 = 6^3 |> (-2)^3 + (-1)^3 + ... + 2^3 + 3^3 + 4^3 + 5^3 = 6^3 |> illustrates an automorphism given by (k,m,N) -> (k+2m-1, 1-m, N). If one looks at Kevin Brown's table with these negative-m entries added, the squares look even more prevalent k m N k m N --- --- ---- --- --- ---- 3 3 6 8 -2 6 *CU *SQ 4 11 20 25 -10 20 *SQ 20 3 40 25 -2 40 *SQ 20 15 70 49 -14 70 *SQ *SQ 25 6 60 36 -5 60 *SQ *SQ 49 291 1155 630 -290 1155 *CU 64 6 180 75 -5 180 99 11 330 120 -11 330 99 556 2805 1210 -555 2805 *CU 125 34 540 192 -33 540 153 213 1581 578 -212 1581 153 646 3876 1444 -645 3876 *SQ 288 273 2856 833 -272 2856 *CU 343 213 2856 768 -212 2856 *CU 512 406 5544 1423 -405 5544 and various other k's are 2*square (288) or 3*square (75,192,768). Is there something real going on here, or is the Strong Law of Small Numbers in operation again? BTW, it's nice to see some real mathematics in sci.math for once... Chris Thompson Email: cet1 at cam.ac.uk (edit From field in obvious way if replying) ============================================================================== From: ksbrown@seanet.com (Kevin Brown) Newsgroups: sci.math Subject: Re: Sum of Consecutive Cubes Equals a Cube Date: Mon, 15 Sep 1997 00:33:03 GMT
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

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