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 + 3^2 + ... + 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. Dickson's "History of the Theory of Numbers" discusses the more general problem of finding a cube that is equal to the sum of the cubes of consecutive elements of any arithmetical progression. The special case of consecutive cubes is discussed on page 582. C. Pagliani (1830) treated the problem of finding 1000 consecutive integers the sum of whose cubes is a cube. Notice 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). This implies that x is an integer if v is not divisible by 3. For example, the cases v = 2, 4, and 10 give 3^3 + 4^3 + 5^3 = 6^3 6^3 + 7^3 + ... + 69^3 = 180^3 1134^3 + ... + 2133^3 = 16830^3 So this gives an infinite family of solutions. However, not all sums of consecutive cubes equal to a cube are members of this family. Note that the sum of k consecutive cubes beginning 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 first several solutions in integers 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 easy to see that the solutions with k equal to a cube (such as k=64, 125, 343, and 512 in the above list) are members of the infinite family found by Pagliani. But the remaining solutions are more interesting. Note that the values of k include the squares 4, 25, 49, and 64. 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 ways. Dave Rusin approached the problem from the standpoint of elliptic curves and classified several types of (possible) 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 He speculated that Types 3 and 4 may provide no more solutions, but that Type 5 might be more likely to yield additional solutions. The "torsion solutions" correspond to the infinite family of solutions found by Pagliani. 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. Jim Buddenhagen searched for solutions with k equal to a square, and found the cases k = 119^2 and k = 251^2. It was also pointed out by Dave Rusin that if we let A denote the number of consecutive cubes and B denote the sum of the first and last cubes, then we have A=k and B=2m+k-1, which can be substituted into the basic equation (k/4)(2m+k-1)(2m^2 - 2m + 2km - k + k^2) = N^2 and, multiplying both sides by 8, we have A B (A^2 + B^2 - 1) = K^3 (1) where K = 2N. Note that this equation appears symmetrical in A and B, but since A+B = 2(m+k)-1 is odd, it's clear that A and B have opposite parity. Rusin notes that the "torsion points" are given by A= u^3, B = 3( (u^2-1)/3 )^2, and goes on to classify the other solutions in terms of the greatest common divisor of A and B. 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 (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) = cxzg 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 = cxz (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 (2+f)^3 + (1-f)^3 = 9(1 + f + f^2) so we can combine this with the previous relation to give (2+f)^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.

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