From: Dave Rusin Date: Fri, 19 Jun 1998 14:18:27 -0500 (CDT) To: greil@muc.de Subject: Re: the "hyperbolic" FLT Right, the equation is sextic, not quadratic, but you have already introduced a transformation to make it quadratic. I didn't look closely at it until today, so I see now that the analysis is incomplete, but you can get part-way. From a^2 + b^2 =c^2 * (1 + a^2 * b^2) divide by (abc)^2 to get (1/bc)^2 + (1/ac)^2 = (1/ab)^2 + 1, so you only need [1/bc, 1/ac, 1/ab] to be a point on the quadratic surface x^2+y^2=z^2+1. I _can_ find all rational solutions to this, since it _is_ a quadratic with obvious rational points. The set of solutions is given as the set of all triples [x,y,z] of the form [(n^2-m^2+1)/D, 2m/D, 2n/D] where D=n^2-m^2-1; m and n are arbitrary along with a small set of "extraordinary" points [1,m,m] and [1,m,-m], m arbitrary. These may be found with the technique described in the article I suggested to you yesterday. Unfortunately, to recover a,b,c you must have xyz a rational square. This always happens for points of the form [1,m,m] (giving points a=m,b=1,c=1); it never happend for points of the form [1,m,-m]; and for the other family of points it happens iff 4mn(n^2-m^2+1)(n^2-m^2-1) is a rational square. The collection of such pairs (m,n) may be found by considering a surface, e.g. the surface given by the quartic equation n^4 + m^4 - 2m^2n^2 - 1 = mn p^2 which, it may be shown, is equivalent to the surface you started with! So really no progress has been made. Your hope (I imagine) is that the set of rational solutions can be given by specifying a, b, and c as rational functions of some parameters u and v. Not all surfaces have this property; the ones which do are called _rational surfaces_ (or actually a slightly broader class called _unirational surfaces_) There are probably people who can tell you whether or not your surface is rational; I'm not one of them! I guess it's likely that the surface is not rational, so no simple formula for all the rational points on it would be possible. Then what would you want? This is part of Algebraic Geometry (perhaps specifically "Arithmetic Algebraic Geometry"); see e.g. index/14-XX.html dave ============================================================================== From: Dave Rusin Date: Tue, 30 Jun 1998 16:52:48 -0500 (CDT) To: greil@muc.de Subject: Re: the "hyperbolic" FLT (LONG) I've been clearing my mailbox and gave some more attention to your question regarding the equation a^2 + b^2 = c^2 * (1+a^2*b^2). I can't quite remember how you came upon this equation or what you wanted to know about it, but I can make some observations. Permit me to summarize them -- at least for my own benefit! You might just want to read the conclusions in (5). (1) INTEGRAL SOLUTIONS. The equation is equivalent to (1-c^2)(a^2+b^2) = c^2(a^2-1)(b^2-1) so that if |c| > 1 then the left side is negative (unless a=b=0) forcing one of |a| and |b| to be greater than one and one to be less than one. In particular, if |c| > 1, then if a and b are to be _integral_, one must be zero (and the other equals +-c ). If |c|=1, then it is necessary and sufficient that one of a or b is +-1 as well. If c=0, then a=b=0. So all the _integral_ solutions to your equation are of the form (0,c,c), (c,0,c), (0,-c,c), or (-c,0,c) for some (integer) c or of the form (1,b,1), (1,b,-1), (a,1,1), (a, -1,1). Let's call these "trivial" solutions; then all integral solutions are trivial. (2) SMALL SOLUTIONS There are however nontrivial rational solutions. One may define the height of a rational number x/y to be max(|x|,|y|) and the height of a solution to the original equation to be max(ht(a),ht(b),ht(c)). Then one may check for small solutions by brute force. For example, the nontrivial solutions with a=i/j, b=k/l, c=m/n sorted by max(|i|,|j|,|k|,|l|,|m|,|n|) are minor variations of the points [4/13, 16/21, 4/5] [7/19, 19/22, 17/19] [11/24, 19/22, 10/11] [11/24, 11/23, 11/17] [1/18, 11/24, 6/13] [14/27, 16/21, 6/7] [1/9, 14/27, 9/17] [3/14, 9/19, 15/29] [9/32, 11/24, 8/15] [7/32, 4/7, 17/28] [3/14, 9/32, 6/17] [7/22, 19/33, 11/17] [4/33, 7/32, 1/4] [3/14, 7/23, 13/35] [1/21, 7/19, 13/35] [4/7, 3/5, 29/37] [2/23, 3/29, 5/37] [1/13, 1/9, 5/37] [7/23, 11/23, 23/41] [1/21, 1/18, 3/41] [1/13, 4/13, 13/41] [11/45, 3/5, 25/39] [4/33, 11/45, 3/11] [3/37, 7/46, 5/29] [8/49, 7/22, 5/14] [8/49, 3/14, 7/26] etc. (The "variants" are discussed with "torsion points", below.) (3) EMBEDDED CURVES There are many ways to obtain infinitely many rational solutions to the initial equation. One could for example hold a fixed (e.g. at any of the values indicated above) and then find infinitely many pairs (b,c) which lie on the corresponding elliptic curve. We shall do this instead with c held fixed in the next section. (It seems more natural to fix c than a because of the symmetry of the equation, but using the two ideas alternately will find many rational points quickly.) Also, rather than holding one of the coordinates fixed, one may allow all to vary, that is, one may look at other curves embedded in the surface. There are even curves of genus zero in there, e.g. one may take 2 4 2 (t - 3) t t - 9 t - 3 {a = 2 -------------, b = 1/8 ------, c = - 1/2 ------} 4 2 2 t + 2 t + 9 t t which corresponds to 4 2 6 4 2 t - 14 t + 9 t - 7 t - 21 t + 27 X = - 1/2 --------------, Y = 1/2 ----------------------. 2 3 t t With this parameterization we may for example find points on the surface with c = 1/4, 4/11, 5/11, 3/13, 8/13, 1/15, 13/20, 11/21, 7/23, 12/23, 11/24... This parameterization traces out the curve defined by a=c/(2+c^2), b^2=(c^4+3*c^2)/4. Another similar parameterization results from the intersection of the surface with the surface a=8*c/(1+6*c+c^2) and leads to the values c = -(7t^2+1)/(7+t^2), e.g. c=1/4, 1/7, 2/11, 7/22, 11/23, 14/23... As far as I can see there is no rational map (a,b,c): Q^2 -> surface, that is, the surface is not unirational. (4) COORDINATE ELLIPTIC CURVES Not only are there are infinitely many nontrivial rational solutions, there are infinitely many having certain fixed values of c. Solving for a^2 shows we need a^2=(b^2-c^2)/(b^2c^2-1) (if c=+-1, there are also the trivial solutions with b=+-1 ). For fixed c this gives a rational value of a only if (b^2-c^2)(b^2c^2-1) is a square; that is, for fixed c, the solutions correspond to points on the curve V^2 = (c^2) b^4 - (c^4+1) b^2 + (c^2) other than those with b=0 or b=+-c; here V = a*(b^2*c^2-1) . Now, those trivial solutions imply that this curve _does_ have rational points; using say (b,V) = (0,c) to be a "point at infinity" this curve becomes an elliptic curve, E(c) say. We may express this curve in Weierstrass normal form. I used the Maple package APECS to do the work: 2 3 4 2 4 4 4 he Weierstrass form is Y = X + (-c - 1) X - 4 X c + 4 c (c + 1) and the transformations are 4 c (X - c - 1) b = 2 -------------- Y 2 3 4 2 2 8 4 c (-Y + 2 X - 4 c X - 4 X + 2 c X + 4 X c + 2 X) V = ------------------------------------------------------- 2 Y with inverse transformation c (V + c) X = 2 --------- 2 b 2 4 2 2 c (-2 c V - 2 c + c b + b ) Y = -2 ------------------------------ 3 b Again, the point is that (for fixed c) the curve has the canonical form Y^2 = cubic in X, so the techniques of Elliptic Curves may be readily applied. For example, we may search for torsion points, and find the torsion group to have order exactly 8; it consists of the identity element (the point at infinity), the points [X, Y] = 2 2 4 4 [-2 c , 0], [2 c , 0], [c + 1, 0], of order 2, and the points [2, 2 c - 2], 4 4 6 2 4 6 2 [2, 2 - 2 c ], [2 c , 2 c - 2 c ], and [2 c , -2 c + 2 c ] of order 4. Each leads either to "points at infinity" or to "trivial" points in the original surface. (More generally, the group of 16 transformations of the original surface generated by the four transformations (a,b,c) -> (-a,b,c), (a,-b,c), (b,a,c), or (1/a,1/b,c) corresponds to endomorphisms of the elliptic curve generated by addition of torsion elements, and negation. We simply let this group act on the origin, (a,b,c)=(c,0,c), to get the whole torsion subgroup.) So our goal is to find values of c which give curves E(c) with more points than just torsion. Note that E(-c) = E(c), so we may as well assume c > 0. E(c) and E(1/c) are also isomorphic, since (a,b,c) is a solution to the original equation iff (a, 1/b, 1/c) is; thus we may restrict to c<1. (The curves with c=0 and c=+-1 are singular, and contain only "trivial" points.) Well, for each fixed c, it is fairly easy to find the structure of the the elliptic curve E(c). (More precisely, there are algorithms which can analyze the curve, and often terminate quickly with a complete description of E(c), but which may sometimes terminate by reporting a failure to determine the complete structure.) I have no idea how to describe the situation for _all_ c, but I can report the situation for "small" c. First, we find that for many c, the curve has rank 0, that is, it consists only of the 8 torsion points; in these cases there are no nontrivial points on the original surface having this value of c. This situation occurs for c =1/2,1/3,2/3,3/4,1/5,2/5,3/5,... I found that rank(E(c))=0 for all fractions c=c1/c2 with 0 < c1 < c2 < 13 except those discussed below. (I also found rank(E(1/N))=0 for N=13,14,16,17,18,19.) One case I attempted (c=5/8) had rank 0 but this could only be detected using Mestre's bound on rank, since Sha(E(c)) is nontrivial and even a second 2-descent failed. Second, it may be that the rank of E(c) is one. In this case, there is a generator P such that all rational points on E(c) are of the form m*P + T for some integer m and some torsion element T. It suffices to compute the points (a,b,c) corresponding to P, 2P, 3P, ... since we have already discussed the effect of adding torsion and negating. Unfortunately I did not determine without amiguity this generator P in any case except the smallest (c=1/4); it is conceivable that the point P used is actually a multiple m'*P' of some other element of E(c). I would be very surprised if this were true, however; I suspect the given P really is a generator. An explicit formula may be written in terms of a,b,c only, which returns the coordinates a',b',c' of the point obtained by "doubling" P=(a,b,c). It's terrible. The formulas for tripling (etc.) or addition in E(c) are even worse. These are the cases with rank(E(c))=1 among fractions with c1 To: Dave Rusin Subject: Re: the "hyperbolic" FLT (LONG) Dear Dave, > I've been clearing my mailbox and gave some more attention to your > question regarding the equation > a^2 + b^2 = c^2 * (1+a^2*b^2). Thank you very much for your study of my equation and your communication of your observations and results! I'm really interested in. Please let me apologize for the next ten days from responding, because of the finish of a software development project where I'm working, with the "usual circumstances" of such a situation ... I could send you in the meantime a copy of an email with some considerations to this equation which are elementary and straigtforward. This refers to a preprint of Prof. Abraham A. Ungar, "The Hyperbolic Pythagorean Theorem in the Poincare Disc Model of Hyperbolic Geometry". which appears as a note in one of the next issues of "American Math Monthly". Many greetings! Anton --------------------------- copy ---------------------------------------------- A working hypothesis of Prof. Ungar is to regard fundamental analogies between hyperbolic and Euclidean geometry. The above paper with the discovery of the new (old!) form of the Pythagoras invites to "play" with that principle in related situations. In the context of the "classical" Pythagoras of the Euclidean geometry, there exist the (integer) "Pythagorean triples", like 3^2 + 4^2 = 5^2, 5^2 + 12^2 = 13^2, 20^2 + 21^2 = 29^2, ... The set of these (integer) triples can be generated in a well known manner. Only a little more difficult is the task, to find all triples of rational numbers (unrestricted to their seize!) who fullfill the "Euclidean" Pythagorean theorem. I tried to find all rational numbers q, 0 < q < 1, who fullfill the "hyperbolic Pythagorean theorem" in the form of Prof. Ungar: a^2 (+) b^2 = c^2, 0 < a,b,c < 1 |---------------------------------------------------------------------------| | This is the "little problem" I'd like to add to the note of Prof. Ungar. | |---------------------------------------------------------------------------| (The restriction 0 < a,b,c < 1 in the hyperbolic case is the analogon to no restriction in the Euclidean case, because the first restriction defines the model of the Poincare unit disk for the hyperbolic plane *within* the Euclidean plane; but within the hyperbolic plane the resp. equation is has the same "degree of freedom"). It seems to be a "nice" problem with a "round solution". This I only can suppose after some investigations. I don't have a proper solution. May be this is a well known and solved problem. A little computer programm led to a list of (obviously infinite) such triples. In the appendix some examples are listed. The program uses the presentation of the rational numbers by the "Farey Tree" or "Stern-Brocot Tree" ( --> Graham/Knuth/Patashnik, "Concrete Mathematics"). Thus I produced all rational numbers q (0 < q <1) on the first n levels of the tree and then checked for all pairs among them, if the relations of the "Euclidean" or "hyperbolic" Pythagoras are fullfilled. An observation over the first 12 levels of the Stern-Brocot tree shows that the "Euclidean" relation is more frequently fullfilled than the "hyperbolic" one. This is plausible when setting a := p1/q1, b:= p2/q2, c:= p3/q3 and analyzing the equations implied by the resp. Pythagoras: 1. In the Euclidean case, each rational triple can be derived from a unique integer Pythagorean triple. 2. In the hyperbolic case, for each rational triple there are *two* integer Pythagorean triples involved, which interact in a restrictive way. This explains the fewer solutions in the hyperbolic case, following the enumeration in the above algorithm. A transformation of the hyperbolic problem is encouraging that there is a "nice" solution: a^2 (+) b^2 = c^2 (a^2 + b^2) / (1 + a^2*b^2) = c^2 a^2 + b^2 - c^2 = (abc)^2 with new variables: x := a / (abc) = 1/(bc) y := b / (abc) = 1/(ac) z := c / (abc) = 1/(ab) and xyz = 1 / (abc)^2, x,y,z rational and x,y,z > 1 the problem is equivalent to find all rational points of the one-sheeded hyperboloid x^2 + y^2 - z^2 = 1 in R^3, where the rational Cartesian coordinates (x,y,z) have to fullfill the additional feature: xyz is the square of a rational number! ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ------------------------------------------------------------------------------- Appendix: ------------------------------------------------------------------------------- "RPT": rational Pythagorean triple "PT": integer Pythagorean triple Each RPT is based (= "derived from") the associated 1 PT (Euclidean case) 2 PT's (hyperbolic case) ------------------------------------------------------------------------------- Euclidean Plane: some Rational Pythagorean Triples -------------------------------------------------- RPT: (1/2,2/3,5/6) PT: (3,4,5) RPT: (1/2,3/8,5/8) PT: (4,3,5) RPT: (3/4,5/7,29/28) PT: (21,20,29) // norm(hypothenuse) > 1 ! RPT: (3/4,2/5,17/20) PT: (15,8,17) RPT: (4/5,3/5,1/1) PT: (4,3,5) RPT: (4/5,1/3,13/15) PT: (12,5,13) RPT: (3/5,4/7,29/35) PT: (21,20,29) RPT: (3/5,1/4,13/20) PT: (12,5,13) RPT: (5/8,1/3,17/24) PT: (15,8,17) RPT: (4/7,3/7,5/7) PT: (4,3,5) RPT: (1/3,1/4,5/12) PT: (4,3,5) RPT: (3/8,1/5,17/40) PT: (15,8,17) Hyperbolic Plane: some Rational Pythagorean Triples (a la A. Ungar) --------------------------------------------------- RPT: (7/8,15/26,218/233) PT1: (182,120,218) PT2: (208,105,233) RPT: (17/21,9/20,389/447) PT1: (340,189,389) PT2: (420,153,447) RPT: (16/21,4/13,4/5) PT1: (208,84,4) PT2: (273,64,5) RPT: (3/5,4/7,29/37) PT1: (21,20,29) PT2: (35,12,37) RPT: (5/8,34/55,41/50) PT1: (275,272,41) PT2: (440,170,50) RPT: (5/8,3/14,74/113) PT1: (70,24,74) PT2: (112,15,113) RPT: (4/7,7/32,17/28) PT1: (128,49,17) PT2: (224,28,28) RPT: (19/33,9/16,85/111) PT1: (304,297,85) PT2: (528,171,111) RPT: (8/15,7/26,233/394) PT1: (208,105,233) PT2: (390,56,394) RPT: (2/5,11/24,73/122) PT1: (48,55,73) PT2: (120,22,122) RPT: (7/15,4/13,109/197) PT1: (91,60,109) PT2: (195,28,197) RPT: (11/24,9/32,8/15) PT1: (352,216,8) PT2: (768,99,15) RPT: (8/19,5/21,193/401) PT1: (168,95,193) PT2: (399,40,401) RPT: (3/8,5/14,58/113) PT1: (42,40,58) PT2: (112,15,113) RPT: (7/23,3/14,13/35) PT1: (98,69,13) PT2: (322,21,35) RPT: (9/32,3/14,6/17) PT1: (126,96,6) PT2: (448,27,17) -------------------------------------------------------------------------------