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Re: Re: Help with algorithm to find rational


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


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