Axel Vogt schrieb:   
   >   
   > On 05.04.2014 20:49, dwelz@web.de wrote:   
   > >   
   > > Maybe you can do this faster than Derive 6.10. I want to eliminate the   
   > > two variables x1, x2 from the three polynomial equations   
   > >   
   > ...   
   > >   
   > > Thus I started GROEBNER_BASIS([lhs1, lhs2, lhs3], [x1, x2, x3, x4]) on   
   > > Derive, where lhs1, lhs2, lhs3 are the left-hand sides of the three   
   > > equations; the computation is still running. However, I am interested   
   > > only in the basis elements not involving x1 and x2.   
   >   
   > Using Maple:   
   >   
   > lhs1 respectively lhs2 are of degree (3,5) resp. (3,6)  in (x1,x2).   
   >   
   > If I use 'eliminate' w.r.t. {x1, x2} I get three solutions for each   
   > of the systems, allowing x2 to be chosen and all are (formally) not   
   > polynomials but rational algebraic (roots and divisions).   
   >   
   > Finally one can feed x2 by solving lhs3 = x2^2 + x4^2 - 2.   
   >   
   > Due to degree(x2) = 5 in lhs1 a sqrt (formally) would survive.   
   >   
   > So I have doubts one can do what you want to achieve.   
   >   
   > And my dull understanding for 'eliminate' is to project a system   
   > onto remaining coordinates [partially uniquely]).   
      
   Thanks for contributing to the solution of an important problem of the   
   21st century. Here is a lower-degree example of what I am trying to do:   
      
   GROEBNER_BASIS([x1^2*(4*x2^2 - 4*x4^2 - 1) - 4*x1*x2*(x2^2 + 4*x3*x4   
   - 3*x4^2) + x2^4 - 2*x2^2*(2*x3^2 - 6*x3*x4 + 3*x4^2) + x3^2*(4*x4^2   
   + 1) - 4*x3*x4^3 + x4^4, - 2*(4*x1^2*x2*x4 + x1*(2*x2^2*(2*x3 - 3*x4)   
   - x3*(4*x4^2 + 1) + 2*x4^3) + 2*x2*(x2^2 + 2*x3*x4 - x4^2)*(x4 - x3)),   
   x2^2 + x4^2 - 1], [x1, x2, x3, x4])   
      
   for which Derive 6.10 returns in no time at all:   
      
   [x3^4*(256*x4^4 + 288*x4^2 + 81) - x3^3*(512*x4^5 + 416*x4^3 + 72*x4)   
   + x3^2*(384*x4^6 + 200*x4^4 + 32*x4^2 + 9) - x3*(128*x4^7 + 32*x4^5   
   + 16*x4^3 + 4*x4) + 16*x4^8 + 4*x4^4, x2^2 + x4^2 - 1, x1*(4*x4^5   
   + 2*x4^3 - 6*x4) - x2*(x3^3*(256*x4^4 + 288*x4^2 + 81)   
   - x3^2*(384*x4^5 + 312*x4^3 + 54*x4) + x3*(192*x4^6 + 104*x4^4   
   + 20*x4^2 + 9) - 32*x4^7 - 8*x4^5 - 4*x4^3 - 6*x4), x1*(75*x3 + 8*x4^3   
   - 38*x4) - x2*(x3^3*(512*x4^2 + 288) - x3^2*(768*x4^3 + 192*x4)   
   + x3*(384*x4^4 - 8*x4^2 + 82) - 64*x4^5 + 20*x4^3 - 38*x4),   
   x1*x2*(4*x4^3 + 6*x4) + x3^3*(256*x4^4 + 288*x4^2 + 81)   
   - x3^2*(384*x4^5 + 312*x4^3 + 54*x4) + x3*(192*x4^6 + 104*x4^4   
   + 20*x4^2 + 9) - 32*x4^7 - 8*x4^5 - 4*x4^3 - 6*x4, 45*x1^2   
   - x1*x2*(16*x4^2 + 60) - x3^3*(1024*x4^3 + 576*x4) + x3^2*(1536*x4^4   
   + 384*x4^2 - 45) - x3*(768*x4^5 - 16*x4^3 + 44*x4) + 128*x4^6   
   - 40*x4^4 + 16*x4^2 + 15]   
      
   Only the first element does not involve x1, x2; in factored form it   
   reads:   
      
   (x3^2*(16*x4^2 + 9) - x3*(16*x4^3 + 4*x4) + 4*x4^4)*   
   (x3^2*(16*x4^2 + 9) - x3*(16*x4^3 + 4*x4) + 4*x4^4 + 1)   
      
   I want to know in particular if the bivariate elements for the   
   higher-order problems possess a similar structure.   
      
   Martin.   
      
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