From - Sun Jan 11 15:12:59 1998 From: cohen@wiener.math.ethz.ch (Prof. Henri Cohen) Newsgroups: sci.math.research Subject: a^m+b^n=c^p (was: Re: Sum of two powers = Square) Date: 9 Jan 1998 17:56:01 GMT Organization: Math Department, ETHZ, Zurich, Switzerland The following comes from a thread on sci.math and sci.math.num-analysis, but I thought it would also be worthwhile for sci.math.research readers. Apologies to Fritz, Noam or Don if you are reading this. From: cohen@wiener.math.ethz.ch (Prof. Henri Cohen) Subject: Re: Sum of two powers = Square Path: wiener!cohen Newsgroups: sci.math.num-analysis,sci.math Distribution: local Followup-To: References: <34B520CD.B748A1F2@hotmail.com> <34B618E0.BDE098AA@hotmail.com> Organization: Math Department, ETHZ, Zurich, Switzerland Keywords: Let me from memory post all that one knows about the general equation a^m+b^n=c^p in integers. One of the main world experts on the subject is F. Beukers, but I would be surprised if he read sci.math! 1) If a, b, c are NOT coprime, you can trivially construct an infinite number of solutions, even when m, n, p are distinct. For example, assume (almost random example) that you want a solution with m=3, n=5, p=7. Choose n1=84, n2=90 (see why?). Then, since 1+2=3, you obtain a=2^28.3^30, b=2^17.3^18, c=2^12.3^13 as a solution. So assume from now on that a, b, c are coprime (by pairs or globally, this is equivalent). To exclude trivial cases, we assume in addition a, b and c nonzero, and we identify solutions coming from trivial sign changes. We distinguish 3 cases. 2) The hyperbolic case: this is when 1/m+1/n+1/p<1. By a simple counting argument, one EXPECTS that there is only a finite number of solutions. That this is the case is a very deep theorem only proved 3 years ago by Darmon and Granville: they were able to reduce the problem to Faltings's result on the Mordell conjecture, but the reduction is highly non trivial because the equation is not homogeneous. Computer calculations over a very large range have given exactly 13 (I think) solutions (counting as a single one 1^m+2^3=3^2), some of them quite large, and all of them with one of the exponents equal to 2. It is conjectured that there are only finitely many in total (Darmon- Granville only prove that it is finite for FIXED m,n,p). In fact very possibly there are no more than those discovered, or perhaps 1 or 2 more. I have the list at home for those interested (as well as all the references I do not give), but go back home only in 3 weeks! 3) The parabolic case: this is when 1/m+1/n+1/p=1. This is the easiest, and in a sense has been completely treated by Fermat. It includes for example a^3+b^3=c^3, a^4+b^4=c^2, etc... Here the answer is simple: there are no solutions apart from the universal 1^m+2^3=3^2, here with m=6. 4) The elliptic case: this is when 1/m+1/n+1/p>1, and includes the equation a^4+b^3=c^2 considered in this thread. A counting argument shows that it SHOULD have an infinite number of solutions. This is indeed the case, and has been proved by F. Beukers. However his result is much more spectacular, and answers positively the question asked by P. Lounesto: For given elliptic (m,n,p), the set of solutions is COMPLETELY described by a FINITE family of polynomial parametric solutions, where the polynomials are homogeneous (of known degree) with 2 variables. These families do NOT intersect, except trivially, so the complete solution set is known. The question of finding explicitly these polynomials is another matter. However, the elliptic case boils down to only a small number of cases. Let (u,v,w) be (m,n,p) ordered so that u<=v<=w. We can exclude the trivial case where u=1, so assume u>=2. Then we have the following cases: u=2, v=2, w>=2 : this can be called the dihedral case of order w. u=2, v=3, w=3: the tetrahedral case u=2, v=3, w=4: the octahedral (or cubic) case u=2, v=3, w=5: the icosahedral (or dodecahedral) case. The relation with platonic solids is not a coincidence as I will explain. Anyway, to summarize, Zagier and Beukers have found the complete parametric set in all the elliptic cases, except for the icosahedral case. In the latter, already 22 independent parametric families are known, but there may be many more. In particular, the complete parametric solution of the equation of the thread (a^4+b^3=c^2) is known, but the explicit form is at home (you know, the margin of this post is too small....) Now let me explain what all this has to do with the platonic solids by taking the example of the thread a^4+b^3=c^2. We are going to do a little magic, but believe me when I tell you the magic works also in the other platonic cases. Of course the solution using Pell's equation is perfectly acceptable, but does not generalize to the other cases, while this one does. a) Draw in space a regular octahedron symmetric with respect to the z=0 plane, whose interection with the z=0 plane is the square formed by the 4 fourth roots of unity (+1, -1, +i, -i). This is not essential, but just to fix the picture, which evidently I am unable to draw in ASCII. Take one of the vertices not on the z=0 plane as pole and project all the vertices from there (the standard stereographic projection). Let z_k be the 6 points (in the projective plane P_1(C), but if you don't know what this is don't worry) thus obtained: they simply are +1,-1,+i,-i,0,infinity. Now call f the two variable homogeneous polynomial having the z_k as simple roots: here this is simply x*y*(x^4-y^4) (infinity corresponds to the root y=0). So f(x,y)=x*y*(x^4-y^4). Now more magic: let h(x,y) be the Hessian of f, i.e. fxx*fyy-fxy*fxy where fx, fy, etc... denotes derivation with respect to x, y, etc... Here we see that it is divisible by (-25) so we set in fact h to be equal to (-1/25) times the Hessian, explicitly h(x,y)=x^8+14*x^4*y^4+y^8. Let j(x,y) be the Jacobian of f and h, i.e. fx*hy-fy*hx. This is divisible by (-8) so we set in fact j equal to (-1/8) times the Jacobian, explicitly j(x,y)=x^12-33*y^4*x^8-33*y^8*x^4+y^12. The punchline and final piece of magic: -108*f^4+h^3=j^2 !!! Now this isn't our equation a^4+b^3=c^2, but it is pretty close, and it is easy to get rid of the annoying -108: for instance, set formally x=s*4^(1/4), y=t*(-3)^(1/4). Then, since h and j involve only x^4 and y^4, only integral coefficients will change, while in f^4 an extra factor of -12 will come out, and (-12)*(-108)=6^4. Thus, we have an explicit parametrization for a^4+b^3=c^2: a=6*s*t*(4*s^4+3*t^4) b=16*s^8-168*s^4*t^4+9*t^8 c=64*s^12+1584*s^8*t^4-1188*s^4*t^8-27*t^12 I seem to remember that there is at least another one. Easy exercise: do the same for the tetrahedron. Slightly more difficult exercise: for the icosahedron. Enjoy! Henri Cohen - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - An article by Frits Beukers (University Utrecht Dept. Math. Preprint no. 912 April 1995) discusses the equation x^5 + y^3 = z^2, sort of related to what you're asking. Since this equation is basically an icosahedral syzygy, you can use geometry to study it. Jerry Shurman [SMTP:jerry@reed.edu]