I have a system of equations for algebraic curve given by the zero locus of some polynomial encoded in the system of equations (I want to eliminate variable z and get algebraic curve in terms of x and y, C12 and C22 can be assumed to be integers and even positive if that helps):

Eliminate[{y == z^(2 + 2 C12)/(-x^2 + z^2), 
1 == ((x - z) (x/z)^(2 C12) z^(-2 - 2 C12 + 2 C22) (x + z))/(-1 + 
z^2)}, {z}]

Eliminate::ifun: Inverse functions are being used by Eliminate, so some solutions may not be found; use Reduce for complete solution information.

However Mathematica can't eliminate z.

When I use Solve I get:

Solve::nsmet: This system cannot be solved with the methods available to Solve.

Maple gives some answers but they aren't very useful, for instance they don't give algebraic curve that I want.

Is there any smart way to do what I want?

  • $\begingroup$ You are raising z to a symbolic power so this is not an algebraic curve. Even assuming it is positive and integral does not make it one (it becomes a parametrized family of such curves). $\endgroup$ – Daniel Lichtblau Mar 13 '17 at 15:25
  • $\begingroup$ Yes, that's what I want. To obtain a family of curves indexed by C's, given by some zero locus polynomial of y and x where their powers depend on C's, but not in the form of this system but in the form of zero locus of some polynomial. Is this possible? $\endgroup$ – Caims Mar 13 '17 at 15:44
  • $\begingroup$ Offhand I do not know how to do that. There has been some work on Groebner bases parametrized by exponents, but I don't think it ever went far enough to do this. $\endgroup$ – Daniel Lichtblau Mar 13 '17 at 18:07
  • $\begingroup$ Can you give me some reference? $\endgroup$ – Caims Mar 13 '17 at 18:10
  • $\begingroup$ 1 2 3 4 $\endgroup$ – Daniel Lichtblau Mar 13 '17 at 20:17

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