Newsgroups: sci.math From: jacobson@oucsace.cs.ohiou.edu (E.L.K.S) Subject: Solutions needed to weird equation Date: Wed, 4 Oct 1995 02:20:53 GMT The following weird equation really has come up in some research I am involved with. I seek *all* integer solutions a, b to: a^6 + 5(a^4)b + 6(a^2)(b^2) + b^3 = 1 So far, extensive computer analysis has given only the following solutions: a = 1, b = -5 a = 1, b = -1 a = 1, b = 0 a = 0, b = 1 a = -1, b = -5, a = -1, b = -1, Prove for me, please, that these are all of the solutions. -- =============================================================== T. Eliot, Top Bard. E-mail: jacobson@oucsace.cs.ohiou.edu -- Il est trop tard, maintenant, il sera toujours trop tard. ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Solutions needed to weird equation Date: 7 Oct 1995 07:59:44 GMT In article , E.L.K.S wrote: >The following weird equation really has come up in >some research I am involved with. I seek *all* >integer solutions a, b to: > >a^6 + 5(a^4)b + 6(a^2)(b^2) + b^3 = 1 > >So far, extensive computer analysis has given >only the following solutions: (a=0, a=+-1) >Prove for me, please, that these are all >of the solutions. Others have pointed out that the problem is difficult and that there may well be no other solutions. I'd just like to clarify some of the details. First of all, this equation describes a curve of genus 2, and so by Falting's theorem, there are only a finite number of solutions with a and b _rational_, let alone integral. Secondly, there is a two-sheeted covering of this curve over an elliptic curve (genus 1) obtained by setting x=a^2 and y=b, yielding the curve x^3 + 5 x^2 y + 6 x y^2 + y^3 = 1. Now Falting's theorem does not apply (the conclusion is in fact false, as we shall see), but there is still Siegel's theorem, stating that there are only finitely many _integral_ points on the curve. In general, it is extremely difficult to turn this finiteness result into a concrete theorem listing all integral points on a curve. However, there are exceptions, and they are not limited to the cases in which there are in fact finitely many rational points. For example, for some curves of the form y^2=x^3 + D one can indeed find with proof the set of all integral points even when the curve is of rank at least 1. (I recall this is in one of those Springer GTM books on elliptic curves -- Silverman's maybe?) Of course it would be ideal if there were only finitely many rational points on this curve, but that turns out not to be true. Here's how I analyzed this. Following standard techniques for cubics with a rational point, I make the substitutions 56 yy - 252 xx - xx yy x:= - 6 ---------------------- 2 3 - 3136 + 84 xx - xx 3 1568 + 168 yy - 672 xx - 12 xx yy - xx y:= --------------------------------------- 2 3 - 3136 + 84 xx - xx (as a sequence of more reasonable substitutions, of course) and get the cubic into a normal form of 2 3 yy = xx + 784 (actually the minimal form of this curve is y^2+y=x^3+12). Now, it is easy enough to feed this curve into, say, pari-gp and find that there is a torsion subgroup of order 3 generated by the point P=(0,28) (in this xx,yy-coordinate system) and no other torsion. I had to turn to John Cremona's program mwrank to discover that the curve's rank is indeed one, and that a generator is the point Q=(-7,-21). These points have (x,y) coordinates P=(3,-2) and Q=(-2,1) respectively. It's too bad for the original poster that the rank wasn't just zero, for then we could easily compute all the rational points of the elliptic curve and, eliminating those points with x not a square, find all the rational points on the original curve. However, we at least have an efficient _algorithm_ for looking for any other rational points on the original curve. One need only run through the 6 sequences of points nQ nQ+P nQ+2P n(-Q) n(-Q)+P n(-Q)+2P for n=0,1,2,..., compute the original coordinate x at each of these points, and look for perfect squares. You need only add Q or -Q to the points created successively, and that's equivalent to the chord-and-tangent process, which may be done right on the non-normalized cubic. I ran through n=25 or so just now, generating points with hundreds of decimal digits (normal behaviour on an elliptic curve) and found no new squares to report, although at this hour I won't won't swear by my programming. William C Waterhouse suggested in another article using a different map from the genus-2 curve to an elliptic curve: >Here's an approach that is not guaranteed but might happen >to work. Neglecting the case with a=0, set x = b/a^2 and >y = 1/a^3. These satisfy the equation > > x^3 + 6x^2 + 5x + 1 = y^2. > >If this elliptic curve happens to have only finitely >many rational points, you could check to see which of them >have y of the form 1/a^3. This curve, in minimal form, is y^2=x^3-7x+7, and also has Mordell-Well rank one, with generator (1,1). (That's (1,-1) in Waterhouse's coordinates.) Possibly a way to rule out the existence of other points on the genus-2 curve is to look at other rational maps to elliptic curves, hoping for one onto a curve of rational rank 0. Perhaps someone else can elighten me as to the feasibility of such an approach in general. I would be interested in hearing what problem motivated this sextic in the first place. dave ============================================================================== Newsgroups: sci.math From: jacobson@oucsace.cs.ohiou.edu (E.L.K.S) Subject: Re: Solutions needed to weird equation Date: Sun, 8 Oct 1995 00:50:24 GMT >E.L.K.S wrote: >>The following weird equation really has come up in >>some research I am involved with. I seek *all* >>integer solutions a, b to: >> >>a^6 + 5(a^4)b + 6(a^2)(b^2) + b^3 = 1 >> >>So far, extensive computer analysis has given >>only the following solutions: > >(a=0, a=+-1) > >>Prove for me, please, that these are all >>of the solutions. > > >Others have pointed out that the problem is difficult and that there may >well be no other solutions. I'd just like to clarify some of the details. [original poster (E.L.K.S) snips some insightful analysis] > >I would be interested in hearing what problem motivated this sextic >in the first place. > >dave Sure, I'd be glad to let you in on it. I am currently involved in joint work studying the non-uniform distribution of two-term recurrence sequences modulo powers of 2. We have discovered a classification method for these distributions that depends on the exact power of 2 that divides b*U_5 - 1, U_6, and U_7 - 1. (Here, U_{n+1} = a*U_n + b*U_{n-1}, and the sequence begins U_0 = 0, U_1 = 1). Our theorem's basic condition then is that this exact power of 2 dividing each of these terms actually makes sense. Thus one has to separately examine each of the sequences for which b*U_5 - 1 = 0, U_6 = 0, and U_7 - 1 = 0. The first two equations turned out to be perfectly straightforward. The last equation, that U_7 - 1 = 0, is the one that generated the polynomial on which the post was based. Thus the real question is: Which two-term recurrence sequences (starting 0, 1) have U_7 = 1? Thank you very much for the time and effort you put into your response to this post. I find your apporach fascinating, though it definitely stretches my year's worth of struggling with Hartshorn's tome in grad school way back... --Eliot [sig deleted - djr] ============================================================================== From: ksbrown@ksbrown.seanet.com (Kevin Brown) Newsgroups: sci.math Subject: Re: Solutions needed to weird equation Date: Sun, 08 Oct 1995 03:17:17 GMT E.L.K.S wrote: > The following weird equation really has come up in some research I > am involved with. I seek *all* integer solutions a, b to: > > a^6 + 5(a^4)b + 6(a^2)(b^2) + b^3 = 1 rusin@washington.math.niu.edu (Dave Rusin) wrote: > I would be interested in hearing what problem motivated this sextic > in the first place. E.L.K.S wrote: > Sure, I'd be glad to let you in on it. I am currently involved > in joint work studying the non-uniform distribution of two-term > recurrence sequences modulo powers of 2. We have discovered > a classification method for these distributions that depends on > the exact power of 2 that divides b*U_5 - 1, U_6, and U_7 - 1. > (Here, U_{n+1} = a*U_n + b*U_{n-1}, and the sequence begins > U_0 = 0, U_1 = 1)... Thus the real question is: Which two-term > recurrence sequences (starting 0, 1) have U_7 = 1? Notice that the coefficients of U_n are the nth diagonal of Pascal's triangle. Thus, for U_7 the coefficients are taken from the 7th diagonal as shown below 1 1 1 1 2 1 1 3 3 [1] 1 4 [6] 4 1 1 [5] 10 10 5 1 [1] 6 15 20 15 6 1 If we put x=a^2/b and set m equal to (n-1)/2 or (n-2)/2 (depending on whether n is odd or even) we have U_n[a,b] = (1/b^m) P_n[x] The polynomials P_n[x] have some interesting properties. For example, a linear fractional transformation az+b z -> ------ cz+d is periodic with a period n iff the quantity (a+d)^2 s = ------- (ad-bc) is the negative of a root of P_n[x]. (The parameter "s" is simply the squared trace of the transformation.) Also, note that the m roots of P_n[x] are simply -4cos^2(j PI/n), j=1,2,..,m. This shows that these are essentially Tchebychev polynomials. Another interesting thing about these polynomials is that -P[-y^2] always factors into two conjugate polynomials in y. For example, (y^6 - 5y^4 + 6y^2 - 1) = (y^3+y^2-2y-1) (y^3-y^2-2y+1) = [(y^3-2y) + (y^2-1)] [(y^3-2y) - (y^2-1)] = [y^3-2y]^2 - [y^2-1]^2 From this we can see that the original poster's equation can be written in a somewhat "Pellish" form [a(a^2-2b)]^2 - b [a^2-b]^2 = 1 so any higher solutions (a,b) must be such that the ratio a(a^2-2b)/(a^2-b) is a convergent of the continued fraction of sqrt(b). This connection to continued fractions is interesting because every general linear fractional transformation is conjugate to one of the form z -> 1 + (1/sz), where s is now the negative of the squared trace defined above (so s is a root of P_n[x] if the transformation has a period of n). It follows that the iterates of this transformation are the convergents of a simple continued fraction. For example, beginning with z_0 the 3rd iterate is given by (1/s) z_3 = 1 + ------------- (1/s) 1 + ----------- (1/s) 1 + -------- z_0 Of course, there is also a simple closed-form expression for z_n. There are other useful properties of P_n[x], such as the fact that P_n[x] is divisible by P_d[x] for every divisor d of n. Other "slices" through Pascal's triangle (e.g., taking every other element from a horizontal row) give families of polynomials with similar properties. ============================================================================== {Incredibly, I seem not to have saved the resolving post, which recognized this as a Thue equation, and thus having a computably bounded number of solutions -- in fact, none not already known. For practical solutions of the Thue equation see e.g. a paper eith that title by N Tzanakis and BMM de Weger, J Number Thoery 31 (1989) 99-132. -- djr} =============================================================================