On New Year's Eve, 1994, an announcement appeared on the Internet stating that The Sunday Telegraph of London had offered a cash prize (450 pounds sterling) to the first person to send them a solution in co-prime positive integers greater than 100 of the equation (A^3/B^3) + (C^3/D^3) = 6 Just for fun I emailed them the first such solution that came to mind: A = 792220572662549608190252926112121617686087939438245\ \665806051608621113641830336450448115419524772568639 C = 677959805103821424723263992665061838773573375138707\ \37934706199386093375292356829747318557796585767361 B = D = 436066841882071117095002459324085167366543342937477\ \344818646196279385305441506861017701946929489111120 The offer turned out to be legitimate, and the Telegraph actually did send me L450 ($706.50). Problems of this type were "solved" in principle by Viete, Fermat, Euler, etc. In fact, Diophantus dealt with similar equations. However, it seems Legendre once stated that this particular equation had no solutions, although it's not clear why he thought so. In any case, Lame' pointed out that (17/21)^3 + (37/21)^3 = 6. (The Telegraph's stipulation that the integers be greater than 100 was clearly intended to exclude this "easy" solution.) Of course, given one solution, the tangent-chord method, or some variation, yields infinitely many other solutions. As Poincare conjectured and Mordell proved, all the rational points on a curve such as x^3 + y^3 = 6 can be generated by tangent and chord constructions applied to a finite set of points. Do we know the set of points that generate all solutions in this particular case? We know that ALL the rational solutions are based on Lame's point (17/21,37/21), because the group of rational points on this curve has rank 1 (following the proof of Mordell's theorem). The smallest solution in positive integers (after 17,37,21) is A = 1498088000358117387964077872464225368637808093957571271237 C = 1659187585671832817045260251600163696204266708036135112763 B,D = 1097408669115641639274297227729214734500292503382977739220 The next smallest is the one I sent to the Telegraph, and as noted above there are infinitely many others. This raises the question of what led Legendre to believe there were NO solutions? Interestingly, it can _almost_ be proved by descent that the equation can have no solutions. Starting from scratch, we observe that (AD)^3 + (BC)^3 = 6(BD)^3 implies that D^3 divides B^3 and that B^3 divides D^3, so we have B=+-D and the equation can be written in the form A^3 + B^3 = 6C^3 where A,B,C are integers. Since A^3 + B^3 is even, whereas A and B cannot both be even, they must both be odd, so we have co-prime integers u,v such that A=u+v and B=u-v. Making these substitutions gives u(u^2 + 3v^2) = 3C^3. Clearly u is divisible by 3. Also, since u and (u^2+3v^2)/3 are co-prime, they must both be cubes, so we have integers a,b such that u = 27a^3 and u^2 + 3v^2 = 3b^3. From the theory of binary quadratic forms it's known that all the solutions of the latter equation are of the form u = 9y(x^2 - y^2) v = x(x^2 - 9y^2) b = x^2 + 3y^2 where gcd(x,y)=1. Noting that u must be of the form 27a^3 we have 3a^3 = y(x+y)(x-y) This shows that 3 must divide exactly one of the quantities y, (x+y), or (x-y). In the first case it follows that (y/3), (x+y), and (x-y) must all be cubes. Setting (y/3) = q^3 (x+y) = r^3 (x-y) = s^3 we have (r)^3 + (-s)^3 = 6(q)^3, which is a solution of the original equation in absolutely smaller integers. Therefore, if this were the only possible case, we would be forced to conclude that no solution is possible (by the principle of "infinite descent"). Legendre may have been thinking along these lines when he stated the equation was insoluable. However, if y is not divisible by 3, then 3 divides either (x+y) or (x-y), in which case we arrive at (s)^3 + 2(q)^3 = 3(r)^3 or (r)^3 + 2(-q)^3 = 3(s)^3 respectively. At this point the "descent" argument no longer applies, i.e., a solution to this equation does not necessarily imply an absolutely smaller solution. It follows that every solution of A^3 + B^3 = 6C^3 can be reduced down through a sequence of successively smaller triples (q,r,s) that satisfy the equation r^3 +s^3 = 6q^3 until reaching a triple that satisfies instead the equation r^3 + 2s^3 = 3q^3. The smallest solution (r,s,q) of the latter equation is (1,1,1), and the tangent line through that point gives (-5,4,1). Every solution of the Telegraph's problem that I've seen so far can be reduced to one of these two irreducible cases. However, there are other solutions that reduce to, for example, (655,-488,253), which is on the tangent line through (-5,4,1). Conversely, given a solution of either r^3 + s^3 = 6q^3 or r^3 + 2s^3 = 3q^3 we can generate the solution of A^3 + B^3 = 6C^3 "above it" by working backwards. For example, in the latter case we would compute y=s^3 and x=3q^3 - s^3, and then the "higher" solution of A^3+B^3=6C^3 is given by A = 9y(x^2-y^2) + x(x^2-9y^2) B = 9y(x^2-y^2) - x(x^2-9y^2) Has anyone tried constructing a "family tree" of all the solutions of A^3 + B^3 = 6C^3 generated from the irreducible solutions of the related equation r^3 + 2s^3 = 3q^3 ?

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