Re: Re: Help with algorithm to find rational

*To*: mathgroup at smc.vnet.net*Subject*: [mg105266] Re: [mg105207] Re: [mg105080] Help with algorithm to find rational*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Wed, 25 Nov 2009 23:00:18 -0500 (EST)*References*: <200911201139.GAA03433@smc.vnet.net> <4B0B1AE9.1000004@wolfram.com> <200911241049.FAA00779@smc.vnet.net>

Tito Piezas wrote: > Hello Daniel, > > Thanks for the reply. It had funny side comments, btw. :-) > > Anyway, this problem arose in the context of trying to solve the multi-grade > eqn, > > x1^k + x2^k + ... + x6^k = y1^k + y2^k + ... + y6^k, for k = 2,4,6,8 > > One way to solve this is to use the highly symmetric form, > > (pa+qb+c)^k + (pa+qb-c)^k + (qa-pb+d)^k + (qa-pb-d)^k + (ra+sb)^k + > (sa-rb)^k = > (pa-qb+c)^k + (pa-qb-c)^k + (qa+qb+d)^k +(qa+pb-d)^k + (ra-sb)^k + > (sa+rb)^k (Eq.1) > > where {c^2, d^2} = {ta^2+ub^2, tb^2+ua^2}. > > Note how "b" has just been negated in the RHS. For some constants > {p,q,r,s,t,u}, this can have an _infinite_ number of non-trivial solns as > the quadratic conditions on {c,d} imply an elliptic curve. This has only > three known families: > > 1. {p,q,r,s,t,u} = {1, 3, 2, 8, 45, -11} (Piezas, 2009, for k = 2,4,6,8,10} > 2. {p,q,r,s,t,u} = {1, 10, 1, 11, -27/5, 248/5} (Wroblewski, 2009) > 3. {p,q,r,s,t,u} = {2, 5, 4, 6, -11/5, 64/5} (Wroblewski, 2009) > > One can assume p < q without loss of generality. I found the first one > using one approach, while Wroblewski found the next two using a numerical > search (something I should have done!). Solving for {c,d} by taking square > roots, Eq. 1 is already true for k = 2. By expanding for k = 4,6, one can > linearly express {t,u}, in terms of {p,q,r,s}, so those 4 are the only true > unknowns. For k = 8, one gets a palindromic eqn of the form, > > P1a^4 + P2a^2b^2 + P1b^4 = 0 > > To make this true, one should find {p,q,r,s} such that P1 = P2 = 0. These > are 15-deg eqns so to resolve them is horrendous. But we can use a trick to > simplify matters. Note how r = np for n = {2,1,2} in the identities above. > And since the system is homogeneous, it can be set s = 1 without loss of > generality. So let, > > {r,s} = {pn, 1} > > The trick of making r = pn makes p have only even powers, so let p = z^(1/2) > to reduce the degree even further. We end up with two 5th degs in z which is > manageable. Let, > > Factor[Resultant[P1, P2, z]] > > and, after just a short while, Mathematica spits out an irreducible 22-deg > eqn solely in n and q, call this E22. (Disregard the trivial linear factors > in n,q.) > > Let n = 1, and E22 has one non-trivial linear factor, q = 10/11. > Let n = 2, and E22 has two non-trivial linear factors, q = {3/8, 5/6} > > So far, I haven't been able to find another rational n that does not involve > mere transposition of the known {p,q,r,s}. (For example, let n = 1/10, and > it gives q = -1/11). > > Question: So what's the easier route: a) find n such that E22 has > a linear factor?, or b) just brute-force search for non-trivial > {p,q,r,s} that makes Eq.1 true for k = {2,4,6,8}? > > Tip 1: To speed up things, one can assume p < q. > Tip 2: If Eq.1 = 0 for k = 12, then those {p,q,r,s} are trivial. > > Anyone can find a 4th infinite family? Any help is appreciated. > > - Titus About the only pointer I can give is that the method I outlined is not likely to be of significant help, unless you get lucky and for some prime most substitutions in one variable give no solution in the other. The more I look at it, the more that method appears to be brute force, albeit well disguised. I should have caught that earlier (like, before I posted...). Daniel Lichtblau Wolfram Research

**References**:**Help with algorithm to find rational roots of a bivariate equation?***From:*TPiezas <tpiezas@gmail.com>

**Re: Help with algorithm to find rational roots of a***From:*Tito Piezas <tpiezas@gmail.com>