From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: Challenge: x^4 + y^4 = 17 Date: 5 Sep 1999 05:22:04 GMT Newsgroups: sci.math In article <7qp8n2$379@news2.aero.org>, L. Andrew Campbell wrote: > >From a colloqium talk (Victor Flynn, Univ. of Luverpool, >speaking at USC on "Rational Points on Curves"), I learned >that the problem of determining all the rational solutions >of x^4 + y^4 = 17 has the best and greatest stumped, despite >its surface simplicity. There are four obvious solutions: >(+/-1,+/-2) and (+/-2,+/-1). It is conjectured that those are >the only solutions. Have at it. Gee, that's odd. I thought this was a done deal. Here's the web page http://www.math.uiuc.edu/Algebraic-Number-Theory/0068/index.html : On Serre's Equation X^4+Y^4=17 Z^4, by Samir Siksek In this paper we settle the problem proposed by Serre of showing that the only rational points on the curve D: X^4+Y^4=17 Z^4 are the eight obvious ones. To do this we construct a curve C which is an unramified cover of D defined over K=Q(i), and show that every rational point on D lifts to some K-rational point on C. We then use the method of Dem'yanenko to bound the heights of the K-rational points on C which we are eventually able to enumerate. This enables us to recover the rational points on D. * Serre.dvi (79360 bytes) [1997 Jul 22] * Serre.dvi.gz (33107 bytes) * Serre.ps.gz (107819 bytes) _________________________________________________________________ Samir Siksek There's a description of the problem in Cassels, J. W. S.; Flynn, E. V. Prolegomena to a middlebrow arithmetic of curves of genus $2$. London Mathematical Society Lecture Note Series, 230. Cambridge University Press, Cambridge, 1996. ISBN: 0-521-48370-0 but I don't think the programme outlined there has much in common with Siksek's method. dave ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: Challenge: x^4 + y^4 = 17 Date: 6 Sep 1999 15:14:22 GMT Newsgroups: sci.math In article <37d28c2d.3086868@news.demon.co.uk>, John R Ramsden wrote: >another possible approach would be to start with >general parametrization[s] of x^4 + y^4 = z^4 + t^4 and show >that there are no integer solutions with, say, |z.t| = 2. > >There are such parametrizations, although I don't know if they >have been shown to be general (probably not). Perhaps one could >work with an easier case, like x^4 + y^2 = t^2 + z^2, and show >that with |t|, |z| = 1, 4 one cannot have |y| a perfect square. Well, whatever you can make work is fine by me, but: The equation x^4+y^4=17 describes a _curve_; algebraic curves are reasonably well understood. In particular, Falting's theorem (previously known as the Mordell Conjecture) assures us that this curve has at most a finite number of rational points. It's the non-effectiveness of the proof which is frustrating in this case. On the other hand, x^4 + y^4 = z^4 + t^4 describes a projective _surface_; I would have to say surfaces are less well understood, and in particular the situation regarding algebraic points is much less clear. The easiest case would be when there is a parameterization of solutions which uses _two_ variables (well, three, if the parameterization is also homogeneous). For irreducible surfaces, this can only happen when the parameterization yields _all_ rational points on the surface. More commonly, though, a parameterization only traces out a curve on the surface, leaving open the possibility that there are other rational points. (Indeed, just offhand I can't think of any irreducible surface for which it has been proved that all rational solutions are those given by parameterizations, except when the whole surface is rational or a product of two curves.) Just for comparison, the surface x^4 + y^4 + z^4 = t^4 is known to have infinitely many rational points, but we don't "know where they are"; we don't know whether there is any parameterized family of solutions, for example. The other equation x^4 + y^2 = t^2 + z^2 is at first blush even worse, since it is not homogeneous, and therefore must be regarded as an _affine_ variety of dimension _three_. But it's nonetheless easy to find the points on this one: writing the equation as 1 + (y/x^2)^2 = (t/x^2)^2 + (z/x^2)^2 we see that the solutions may be easily obtained from the solutions to 1 + u^2 = v^2 + w^2 and this one is a rational surface: the rational points are (u,v,w)= ( (-s^2+1-w^2)/(2*s), (s^2+1-w^2)/(2*s), w ) for arbitrary w, s. Taking, say, s=3,w=3 gives the point (u,v,w)=(-17/6,1/6,3) which corresponds to (x,y,t,z)=(6*a,-102*a^2,6*a^2,108*a^2) for arbitrary a. But I don't see how that's particularly useful for the original problem. dave ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: Challenge: x^4 + y^4 = 17 Date: 8 Sep 1999 22:50:46 GMT Newsgroups: sci.math In article , ray steiner wrote: > If we want to solve x^4+y^4=17, what would be wrong with factorising the > LHS over Q(sqrt(i)) and proceeding from there? It is known that Z(sqrt(i)) > is a Euclidean domain, so maybe there is some hope!? And again I say unto you, if you can make it work, have at it. But I don't really quite know what is being proposed. By positivity it's clear that the only _integer_ solutions are {|x|,|y|}={1,2}. So the problem is to find the _rational_ solutions. How, then, do we use the properties of the ring Z[sqrt(i)]? For comparison, let me note that, say, the equation x^4-239y^4=17 can be analyzed this way: in this case it is _not_ so obvious that integer solutions would have to be small, so that even determining the integer solutions is of interest. Well, in this case one does just what was suggested: pass to a suitable extension field in which the left side factors, and then, well, do something. Here's what you could do: observe that x=4, y=1 is one solution, and therefore 4 - w has norm 17, where w is a root of x^4-239=0. You want other elements x - y w whose norm is also 17, and so you find yourself looking at the problem of finding a unit u in this ring by which to multiply 4 - w to get x - y w. What makes this challenging is that, although the ring has plenty of units, you don't in general expect a unit times 4 - w to give a ring element which is a linear combination of ONLY 1 and w; the ring is a rank-4 module over the integers, so there will usually be some spilling over into the other two generators. This is indeed how one approaches problems of this sort: given an irreducible form f(X,Y) of degree greater than 2, one analyzes the units in the splitting field of f and notices that it is impossible to contain the products ( a + b w )*(unit) to a rank-2 subgroup except for a bounded set of units; then an exhaustive search can find the remaining solutions. These are Thue equations. Note that the conclusions are very different when solving f(X,Y)=c for a form of degree 2 ( Pell equations). But I digress: there's no particularly easy way I can see to find all the _rational_ solutions to x^4 + y^4 = 17. It would be handy if, for example, there were few rational solutions to z^2 + y^4 = 17, but this equation determines an elliptic curve of rank two (and torsion Z/2Z) -- i.e., it has lots of rational solutions. We know on general principles (Falting's theorem) that there are only finitely many rational solutions to x^4+y^4=17, but no real idea just how many. dave ============================================================================== From: "Everett W. Howe" Subject: Re: Challenge: x^4 + y^4 = 17 Date: Mon, 06 Sep 1999 00:35:04 -0700 Newsgroups: sci.math L. Andrew Campbell wrote: > From a colloqium talk (Victor Flynn, Univ. of Luverpool, > speaking at USC on "Rational Points on Curves"), I learned > that the problem of determining all the rational solutions > of x^4 + y^4 = 17 has the best and greatest stumped, despite > its surface simplicity. There are four obvious solutions: > (+/-1,+/-2) and (+/-2,+/-1). It is conjectured that those are > the only solutions. Have at it. and Dave Rusin responded: > Gee, that's odd. I thought this was a done deal. Here's the web page > http://www.math.uiuc.edu/Algebraic-Number-Theory/0068/index.html : This problem, and Siksek's paper, came up in conversation when I was visiting with Flynn a few weeks ago. Flynn said that there is a computational error in Siksek's paper that destroys his argument. Siksek constructs a curve C and an elliptic curve E, both defined over a number field K, and in order for his argument to work the rank of the group of morphisms (modulo constant morphisms) from C to E must exceed the rank of the group E(K). Flynn told me that Siksek's calculation of one of these ranks (I forget which one) is incorrect. -- Everett ________________________________________________________________________ Everett Howe Center for Communications Research username: however 4320 Westerra Court domain-name: alumni.caltech.edu San Diego, CA 92121 ============================================================================== From: Dave Rusin Subject: Re: Challenge: x^4 + y^4 = 17 Date: Mon, 6 Sep 1999 09:42:03 -0500 (CDT) Newsgroups: sci.math To: however@alumni.caltech.edu Hi Everett, Thanks for responding. I wondered why the paper had not gone beyond the preprint stage. Does Flynn hold out much hope for the method he and Cassels outlined in their book? I seem to recall I tried to flesh it out and found it promising but much too cumbersome for the casual reader like me. dave ============================================================================== From: "Everett W. Howe" Subject: Re: Challenge: x^4 + y^4 = 17 Date: Wed, 8 Sep 1999 13:54:36 -0700 (PDT) Newsgroups: [missing] To: Dave Rusin Hi Dave, Flynn said that this was a very frustrating problem --- every method that he's tried has failed to work, but *just barely*. As Flynn and Cassels note in their book, their attempt at solving the problem using covering curves failed because the rank of the Jacobian of a certain genus-2 curve was too large. Theoretically, one could still use the covering curve idea, but *not* make their reduction to a genus-2 curve. Apparently, though, using Chabouty on the higher-genus curves that come up is too difficult at the moment. Flynn said that the other method mentioned in their book --- the one that involved calculating the Mordell-Weil group of an elliptic curve over a number field --- is also too difficult to carry through. He mentioned some other avenues of attack as well, but I can't remember all of the details. -- Everett ________________________________________________________________________ Everett Howe Center for Communications Research however@alumni.caltech.edu 4320 Westerra Court http://www.alumni.caltech.edu/~however San Diego, CA 92121