From: Paul Bruckman Newsgroups: sci.math.num-analysis Subject: extending Fermat's Last Theorem Date: Fri, 24 Jul 1998 01:13:40 -0700 I have asked this question before in this newsgroup, but I was being too general, and I will now make my question more specific. Consider the equation x^3 + y^3 = z^3, to be solved in algebraic integers x,y,z in Q(Sqrt(5)). That is, x,y and z are to be of the form pa+qb+r, where p,q,r are rational integers, a = (1+Sqrt(5))/2, b = (1-Sqrt(5))/2 . I seek solutions. I know that there is at least one non-trivial solution, namely with x = a+4, y = b+4 , z = 6 . In this sense, FLT is "false", if we extend it to allow algebraic integers of Q(Sqrt(5)) (and no doubt false if we require x, y, z to be algebraic integers in some other common quadratic field). We may take specific instances, e.g. requiring z only to be a rational integer ; note that the form for x and y in any event reduces to x = p'+aq', y = r'+bs', for some integers p', q',r', s', since a+b=1. The proposed equation : x^3 + y^3 = z^3 , with x = p+aq, y=p+bq, p,q and z rational integers may be put into the Diophantine form : 2p^3 + 3p^2q + 9pq^2 + 4q^3 = z^3 . This looks awful, but fortunately can be put into a more attractive form, namely : 2z^3 = (2q+p)^3 + (q+p)^3 + p^3 + (-q+p)^3 Note that this requires 2z^3 to be the sum of four cubes, the cube roots of which are to be in arithmetic progression. As previously noted, we have one solution with z=6, taking p=4, q=1 (since 6^3 = 5^3 + 4^3 + 3^3). The questions become : how many solutions are there, is there an upper bound on solutions, and can we find an infinite sequence of solutions? The corresponding cases for the exponent being > 3 must have finitely many solutions, as someone in this newsgroup has already pointed out (due to Faltings' Theorem). The cubic case seems to have appeal as a problem waiting to be solved. Any suggestions towards the solution of the above equation or of the more general case where x, y and z are all in Q(Sqrt(5)) but are not rational integers would be welcome. Paul Bruckman ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math.num-analysis Subject: Re: extending Fermat's Last Theorem Date: 26 Jul 1998 06:00:31 GMT Followup-To: sci.math In article <6p9il8$sgg$1@news-1.news.gte.net>, Paul Bruckman wrote: >The proposed equation : x^3 + y^3 = z^3 , with x = p+aq, y=p+bq, p,q >and z rational integers may be put into the Diophantine form : > >2p^3 + 3p^2q + 9pq^2 + 4q^3 = z^3 . ... >Note that this requires 2z^3 to be the sum of four cubes, the cube roots >of which are to be in arithmetic progression. As previously noted, >we have one solution with z=6, taking p=4, q=1 >(since 6^3 = 5^3 + 4^3 + 3^3). The questions become : how many >solutions are there, is there an upper bound on solutions, and can we >find an infinite sequence of solutions? Infinitely many solutions. Homogeneous equations in three variables define (projective) curves over the rationals. When the degree is three, these curves are (usually) "elliptic" curves (not ellipses, BTW) and admit a careful analysis. In this case, we can substitute 3 2 X - 1575 + 225 Y p = 1/15 ------------------- z (2 Y + 1) X 3 -4 X + 3375 q = 1/15 ------------ z (2 Y + 1) X and obtain a one-to-one correspondence between relatively-prime solutions (p,q,z) to the original equation, and rational solutions (X,Y) to 2 3 Y + Y = X - 844 There are infinitely many such rational points, and they form a cyclic group with your point as generator. (Your point p=4, q=1, z=6 becomes X=10, Y=-13 in these coordinates.) What this means is that the points may be written P_n with n a nonzero integer; P_1 = [10,-13], P_2 = [124, 1380], P_(-n) = [X, -1-Y] if P_n = [X,Y], and the remaining P's are found recursively by drawing the line through P_(-1) = [10,12] and P_(-n); it meets the curve shown in precisely one new point, which is taken to be P_(n+1).) Changing back to p-q-z coordinates, I get (besides the trivial solution [p,q,z]=[-1,2,0], which I used as the "point at infinity") the following points, my P_1, P_(-1), P_2, P_(-2), etc.: [4, 1, 6] [5, -1, 6] [1493, -2761, 1860] [-1268, 2761, 1860] [3561647, 6210602, 13610574] [9772249, -6210602, 13610574] [5970871773676, -7206070204127, 9826055642040] [-1235198430451, 7206070204127, 9826055642040] [-9919623053517464957, 45203693464083879145, 58327764169410959934] [35284070410566414188, -45203693464083879145, 58327764169410959934] [56387705867458121335713215543, -32921137187930482842826371886, 76735807224652508480701925580] [23466568679527638492886843657, 32921137187930482842826371886, 76735807224652508480701925580] etc. The list continues indefinitely, and yes, a few obvious patterns continue. There are of course other solutions obtained by multiplying p,q, and z by a fixed integer, but none others. It's perhaps not surprising that you didn't find other integer solutions if you were looking only for positive integer solutions p,q; the next smallest is P_3 = [3561647, 6210602, 13610574]; next smallest is the last one shown. Calculations courtesy the Maple package APECS. For elliptic curves in general see index/14H52.html Since this is number theory, not numerical analysis, I have set followups to a different newsgroup. dave ============================================================================== Date: Thu, 30 Jul 1998 23:13:20 -0700 From: Paul Bruckman To: Dave Rusin Subject: Re: extending Fermat's Last Theorem TaDave Rusin wrote: > > In article <6p9il8$sgg$1@news-1.news.gte.net>, > Paul Bruckman wrote: > > >The proposed equation : x^3 + y^3 = z^3 , with x = p+aq, y=p+bq, p,q > >and z rational integers may be put into the Diophantine form : > > > >2p^3 + 3p^2q + 9pq^2 + 4q^3 = z^3 . > ... > >Note that this requires 2z^3 to be the sum of four cubes, the cube roots > >of which are to be in arithmetic progression. As previously noted, > >we have one solution with z=6, taking p=4, q=1 > >(since 6^3 = 5^3 + 4^3 + 3^3). The questions become : how many > >solutions are there, is there an upper bound on solutions, and can we > >find an infinite sequence of solutions? > > Infinitely many solutions. > > Homogeneous equations in three variables define (projective) curves over > the rationals. When the degree is three, these curves are (usually) "elliptic" > curves (not ellipses, BTW) and admit a careful analysis. > > In this case, we can substitute > 3 > 2 X - 1575 + 225 Y > p = 1/15 ------------------- z > (2 Y + 1) X > > 3 > -4 X + 3375 > q = 1/15 ------------ z > (2 Y + 1) X > > and obtain a one-to-one correspondence between relatively-prime solutions > (p,q,z) to the original equation, and rational solutions (X,Y) to > 2 3 > Y + Y = X - 844 > There are infinitely many such rational points, and they form a cyclic group > with your point as generator. (Your point p=4, q=1, z=6 becomes X=10, Y=-13 > in these coordinates.) What this means is that the points may be written > P_n with n a nonzero integer; P_1 = [10,-13], P_2 = [124, 1380], > P_(-n) = [X, -1-Y] if P_n = [X,Y], and the remaining P's are found > recursively by drawing the line through P_(-1) = [10,12] and P_(-n); > it meets the curve shown in precisely one new point, which is taken to be > P_(n+1).) > > Changing back to p-q-z coordinates, I get (besides the trivial solution > [p,q,z]=[-1,2,0], which I used as the "point at infinity") the following > points, my P_1, P_(-1), P_2, P_(-2), etc.: > > [4, 1, 6] > [5, -1, 6] > [1493, -2761, 1860] > [-1268, 2761, 1860] > [3561647, 6210602, 13610574] > [9772249, -6210602, 13610574] > [5970871773676, -7206070204127, 9826055642040] > [-1235198430451, 7206070204127, 9826055642040] > [-9919623053517464957, 45203693464083879145, 58327764169410959934] > [35284070410566414188, -45203693464083879145, 58327764169410959934] > [56387705867458121335713215543, -32921137187930482842826371886, > 76735807224652508480701925580] > [23466568679527638492886843657, 32921137187930482842826371886, > 76735807224652508480701925580] > > etc. The list continues indefinitely, and yes, a few obvious patterns > continue. There are of course other solutions obtained by multiplying > p,q, and z by a fixed integer, but none others. It's perhaps not > surprising that you didn't find other integer solutions if you were > looking only for positive integer solutions p,q; the next smallest is > P_3 = [3561647, 6210602, 13610574]; next smallest is the last one shown. > > Calculations courtesy the Maple package APECS. > > For elliptic curves in general see > index/14H52.html > > Since this is number theory, not numerical analysis, I have set followups > to a different newsgroup. > > dave Thank you, Dave, for your very informative response. No doubt, this problem can be extended in similar fashion to cover the more general (disguised) Diophantine equation : x^3 + y^3 = z^3 , where ALL of x,y and z are integers in Q(Sqrt(5)) (with at least one not a rational integer). Regards--Paul Bruckman ============================================================================== From: Dave Rusin Date: Fri, 31 Jul 1998 07:24:43 -0500 (CDT) To: aardwolf@gte.net Subject: Re: extending Fermat's Last Theorem Yes, the whole of your particular problem can be treated with elliptic curves. The Fermat equation describes a planar curve; when the exponent is 3, it's a curve of genus 1 (an elliptic curve). (This may be given a topological interpretation: scale the problem to x^3+y^3=1; now view this as describing a set of pairs of complex numbers x & y; the solution set has complex dimension 1, i.e., real dimension 2. Well, orientable real surfaces of dimension 2 are easy to characterize -- look at the number of "holes"; that's the genus. The complex equation x^3+y^3=1 describes a torus.) With suitable transformations, the Fermat curve may be expressed as Y^2 = X^3-27/4. When you ask for solutions to the Fermat equation with exponent 3 having x and y in a certain field, you are simply asking for points on this elliptic curve over that field. Well, the arithmetic theory of elliptic curves is pretty well understood; while it's usually presented for curves over the rationals, there's essentially no change for curves over number fields and indeed one may generalize to curves over function fields etc. Silverman's book (the first one) is very appropriate here. In your case, your question seems to be, what are all the rational points of the curve x^3+y^3=1 over Q(sqrt(5)). The answer is: I don't really know. By general theory, the set of these points will be a finitely generated Abelian group; here the group operation is a little murky at this level, but the "chord and tangent" process I mentioned in my post is pretty close (If P and Q are on the curve Y^2=X^3-27/4, the third point of intersection of this curve and the line joining P and Q is the _negative_ of "P+Q"; you get P+Q itself by changing the sign of the Y coordinate.) Cassels has an easy-going book on elliptic curves which clarifies some of this. It should be possible to compute the rank and torsion subgroup of this Abelian group but I'm not willing to try this by hand. As I showed in my post, there are infinitely many points on this curve (even of the special type you proposed before) so we know the group is of positive rank. Elements of order 2 occur when X is a root of X^3-27/4; there are none in this field. Elements of order 3 occur when X=0,3, or a root of X^2+3X+9; only the case X=3 occurs in this field, and these points correspond to the trivial solutions {x,y}={0,1} on the Fermat curve. Elements of order 5 require (X/3)^3 to be a root of -1/5+5*x2-3*x2^2-19*x2^3+x2^4, but this polynomial has no roots in this field. Elements P of order 7 or 9 have 8P=+-P and so require one of x0^3+9*x0^2-27, 7*x0^6+27*x0^3+729, x0^6-9*x0^5+81*x0^4-54*x0^3+243*x0^2+729 x0^9-27*x0^8+648*x0^6+1458*x0^5+2187*x0^3-19683*x0^2-19683, x0^18-3807*x0^15-264627*x0^12+37870092*x0^9-393797781*x0^6+688747536*x0^3+ 387420489, x0^18+27*x0^17+729*x0^16+1296*x0^15+16038*x0^14-78732*x0^13+424278*x0^12 -866052*x0^11+3188646*x0^10+2794986*x0^9+9034497*x0^8-57395628*x0^7 -20726199*x0^6+71744535*x0^5+387420489*x0^4-86093442*x0^3 -387420489*x0^2+387420489 to vanish, but Maple reports that each of these polynomials is in fact irreducible over your field. As far as I know no one knows how many primes you'd have to check for torsion over this field. (Over Q it is known that there can only be 2-, 3-, 5-, and 7-torsion.) Based on these calculations so far, it looks pretty likely that the torsion subgroup is just Z/3, accounting just for the trivial solutions to the Fermat question. As you might deduce from this search for points of finite order, it seems to me the more interesting question is not the complete determination of the structure of the curve over _this_ field, but rather the question, what are the other fields Q(sqrt(m)) (say) for which the curve x^3+y^3=1 has positive rank, or torsion. As far as I know the tools for answering this question do not exist. (We could answer it for each m in turn, but unless we see a particular pattern we're sunk for the general case.) dave ==============================================================================