From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: tantalising teaser Date: 10 Dec 1999 16:56:56 GMT Newsgroups: sci.math Keywords: integers satisfying two Pellian equations (integer points on ECs) [This is a repost since our newsfeed was down for a while and this article seems not to have propagated. Since it was first posted I have checked for solutions to y^2 = 74 t^4 + 1850 with numerator and denominator of t up to e^12=162753, finding that none exist. Indeed none exist assuming a parity conjecture on the rank of the Jacobian of this curve.] In article <8267ut$ga2$1@nnrp1.deja.com>, ray steiner wrote: > Note that x= 49 is a square. >Also x-1=48 is 3 times a square >and x+1 = 50 is twice a square. >Show that 49 is the only >natural number with this >property. [...] >I recently read that the first >problem has been solved and >that the solution "will >appear", but so what! In this case we wish to solve for integers x,y,z x-1=3y^2 x+1=2z^2 x=w^2 Eliminating x gives w^2 - 1 = 3 y^2, w^2 + 1 = 2 z^2. Diophantine problems with a few variables and lots of squares tend to be right at the edge of what's easily solved with current understanding. In this case it's two equations in three unknowns, which geometrically describes a curve in 3-space; the equations being quadratic, this is an intersection of two conic sections, making it an elliptic curve. That means on general principles that there will be, um, a knowable set of rational solutions and a finite set of integer ones. What gets you a paper is actually showing that a particular finite set of integer solutions is complete; depending on the particular equations involved that often requires some ingenuity (some would say 'trickery'). The same remarks apply to this addendum if k and m are fixed: >More generally, when is >x^2+1 k times a square >and x^2-1 m times a square? We have to solve w^2 - 1 = m y^2, w^2 + 1 = k z^2 [I'll not use x which was interpreted differently before] There can be NO integer solutions if there are no rational ones. That will happen, for example, if there are no real solutions (did you exclude the possibility that k < 0 ?) or no solution modulo a prime (e.g. if k is divisible by any prime congruent to 3 mod 4 then there can be no integral solution). On the other hand, if these two conditions are met, then it is known that there _are_ rational solutions at least to w^2 + 1 = k z^2. Of course there are rational solutions to the other equation, a general expression for them being (w, y) = (-1,0) or w = (m+t^2)/(m - t^2), y = 2*t/(t^2-m) (t rational, t^2 <> m). Substituting into the other equation, we find there are simultaneous _rational_ solutions to the pair of Pellian equations for each t for which (2k)( t^4 + m^2) is a square. Again it might be that there are (real or) p-adic reasons why there are no rational solutions (e.g. if m=3, k=13 then there are no rational solutions because there are no 3-adic solutions; this seems to happen more commonly than not with small m and k) but this time it's not so easy to offer a converse: there are quartic polynomials which take on square values in each completion of the rationals but are never actually a rational square. For example, is there a rational solution when m=5, k=37? I don't know! Sort of frustrating... (Once _a_ rational solution is found, the curve can be reduced to Weierstrass Normal Form and attacked using various techniques which generally, but not universally, will produce a description of all rational solutions. For example, the case m=3, k=2 mentioned at the outset describes a curve with rank 1 and torsion Z/2 x Z/2. Some "small" solutions have |t| = 2, 7/20, 1702/1551, 4653/1702, ... corresponding to |w| = 7, 1249/1151, 10113607/4319999, ...) For combinations of {k,m} which have rational points, it's then a reasonable question to ask what the integer points are, but just offhand I don't know how I would approach that question in any generality. Where will the case m=3, k=2 appear? dave ============================================================================== [Some more details and data --djr] We seek combinations [k,m] for which there are rational points on (2k) ( t^4 + m^2) = y^2. Simple transformations of variables allow us to assume k, m are squarefree. It is clear that there are modular (i.e. p-adic) obstructions in some cases, e.g. if k is divisible by a prime =3 mod 4 or if gcd(k,m)>2. On the other hand, there are occasions in which there are no local obstructions and nonetheless no rational points. Finally, of course, there are combinations (k,m) for which rational points do exist; we look to see how many there are (but do not check for integrality). Data checked for squarefree k<=36, m<=100. (Next up: k=37, 41, 53, 58, 61, 65, 73, 74, 82, 85, 89, 97, ...) In all cases, APECS able to find points fairly rapidly. In a few cases need to issue e.g. RkNC(100) to make it look harder, or to issue Mest() to have it try harder to get an upper bound on rank. In some cases, L-series and parity conjectures are used to pin down rank. We observe APECS is sensitive to presentation since the 2-torsion isn't cyclic. (Notified Connell; he says he will fix that.) Let's consider the theory here. If (2k)(t^4 + m^2) = z^2 then x=(2k)(t^2) and y=2ktz are coordinates of a point on y^2=x(x^2 + 4D) where D=(km)^2, and obviously x is 2k times a square. On the other hand, the set of values of x modulo squares, as (x,y) ranges over the points of this curve other than (0,0), forms a group which is known to be contained in <-1> x Prod(

, p | 2D ); it happens to be naturally isomorphic to E/phi(E') where phi:E'->E is the isogeny from the curve y^2=x(x^2-D) given by phi(x,y)=((y/x)^2, y(D+x^2)/x^2), and in particular is a homomorphic image of E/2E, an elementary-abelian 2-group of order 2^(rank(E)). (With an extra factor of 2 if k=m=1, since in that case the torsion of E is Z/4Z, and is not in the image of phi; in all other cases, torsion is Z/2Z, contained in im(phi).) So the existence of rational solutions to (2k)(t^4 + m^2) = z^2 is equivalent to 2k being an element of E/phi(E'). (In the special case k=2, we are looking for rational t with t^4 + m^2 = (z/2)^2 for some z. Of course one such solution is t=0!) If a solution exists, then the curve is birationally equivalent to the curve y^2=x(x^2-D) itself. This curve always has exactly four torsion points, which correspond to the four possible sign changes in (x,z). In order for there to be more rational solutions, we require rank(E) to be positive. This is automatic when there is a solution, if k <> 2, as noted above. The jacobian of the quartic is exactly this curve y^2=x(x^2-D), which is the (k*m)-twist of y^2=x^3-x. See Silverman p 309 et seq for more information. Case I: No rational points at all because of local obstruction. Applies to all cases not listed below. Case II: Only finitely many rational points. As noted above, the existence of a rational solution on the quartic gives the elliptic curve positive rank, disallowing this case, except in the cases m=k=1 and those cases with k=2 and rank=0. The latter occurs when m = 1, 2, 5, 6, 13, 21, 22, 29, 33, 34, 37, 38, 41, 53, 57, 61, 65, 66, 70, 73, 85, 86, 89, 93 (some of theses -- 41, 73, 89 -- required Ein(0,0,0,-(k*m)^2,0) rather than Ein(0,0,0,4*(k*m)^2,0). ) The solutions are of course (t,z)=(0,+-2m) and the two points at infinity. Case III: No rational points at all because rk(E)=0 Again, the existence of a point on the quartic implies the jacobian ought to have positive rank, except for the combinations in Case II. We find comparatively few cases in which the rank is zero apart from the situations in Case I which have a p-adic restriction. In order to show the rank is zero, we must rely on APECS's procedures, some of which assume established conjectures. Rank of E is zero in these cases: k=1, m=17,73,89,97 (and thus m=1, k=17,73, 89,97) using second descent In some cases second descent shows rank=0 but ONLY starting from Ein(0,0,0,4*(k*m)^2,0): likewise (k,m)=(5,37), (13,37), (26,58), (29,61) (This may be thought of as a weakness in APECS but is not an error!) (k,m)=(10,62) uses descent, Mestre, & parity conjecture to get rank=0 (k,m)=(29, 69) works similarly if Mest(5000) called manually; else apparent non-vanishing of L(1) is used to get rank=0. Likewise [29, 77] (Mest(4500)) (k,m)=[34, 53] rank is only found to be zero by nonvanishing of L-function. Case IV: No rational points because E/phi(E') does not include 2k. (k,m)=(26,34): second descent limits E/2E to rank 1, and we find the generator of E'/2E' is not in image(phi'), i.e. E/phi(E')=1, so obviously does not include {2k}. Exactly so with (k,m)=[17,47], (34, 26), This time we find cases in which 2nd descent only works starting with Ein(0,0,0,-(k*m)^2,0) ! k = 34: m= 19, 43 ,47, 59, 94 Again the conclusion is the same: E/phi(E')=1, does not contain 2k (&rk=1) k=34, m=83: Using Ein(...-(km)^2,0) easily shows rank <=1 but fails to find point easily; using Ein(...4(km)^2,0) finds a point without too much work: x=(-34)*(165/19)^2 on former curve. Still, E/phi(E')=1. (k,m)=(10,82): This time 4(km) easily gives rank=1. Still, E/phi(E')=1. (k,m)=(17,89): Now both descents find rank=2. Both E/phi(E') and E'/phi(E) contribute one generator. E/phi(E')=<2*17*89> isnt 34 ! (k,m)=(34,89): Again both descents work, rank=2. Now E/phi(E')=1. In the remaining cases, there ARE rational solutions, indeed small solutions are quickly found by APECS, and then transformation is made to the curve y^2=x(x^2-D^2). The size of the solution set is the same as the size of the rational points on the elliptic curve itself: (Z/2)^2 torsion and a free rank of 1 or 2 (nothing greater seen): Case V: Rank=1 (unconditionally, with assists to find first rational pt.) k=1, m= 7, 23, 31, 47, 71, 79 [need t=161] k=2, m= 3, 7, 10, 11, 14, 15, 19, 23, 26, 30, 31, 35, 39, 42, 43, 46, 47, 51, 55, 58, 59, 62, 67, 71, 74, 78, 79, 83, 87, 91, 94, 95 k=5, m= 3, 43, 67, 83 [need t=213/13] k=10, m= 2, 22, 38 k=13, m= 11, 19, 59 [need t=4289/425] 67, 83 k=17, m= 15, 55, 87 k=26, m=2, 6, 86 k=29, m=3, 11, 19, 43 [need t=206787/2563] k=34, m=67 Case VI: Rank=2: (unconditionally, using second descent as needed) k=1, m= 41 k=2, m= 17, 69, 77, 82, 97 k=5, m= 13, 53, 93 k=10, m= 58 k=13, m= 5, 85 k=17, m= 33 k=26, m= 46 k=29, m= 21, 37 k=34, m= 1, 13, 38, 86 Sample: k=5, m=3 corresponds to the case {x^2-1=3y^2, x^2+1=5z^2}. Solutions resulting from smallest elements of corresponding elliptic curve (rank=1) are for x= [2, 698/671, 124429898/13190689, 6470326198450802/5462872634963521, 154796675829869145981594002/107879777271702406762612799, 1501957082163564236627871511434743411498/437038783361465466891780172562053299551] Don't know if x=2,y=z=1 is the only integral solution. Maple codes look like these: kdjr:=29:mdjr:=0: mdjr:=mdjr+1: xxdj:=quar(2*kdjr,0,0,0,2*kdjr*mdjr^2); while mobius(mdjr)=0 or (type(xxdj,integer) and xxdj<>0) do mdjr:=mdjr+1: xxdj:=quar(2*kdjr,0,0,0,2*kdjr*mdjr^2);od:mdjr,xxdj, type(xxdj,integer); #when point found, use this -- but don't bother if xxdj=0 ! Quar(2*kdjr,0,0,0,2*kdjr*mdjr^2);Rk();