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I have the following system of equations,

1+x+y+z==0, 1+x*y+y*z+x*z==0

which I want to solve in the extension field of GF(2), the algebraic closure of GF(2) for example. There is a parametric solution of these equations in terms of the parameter s as x=1+s, y=1+$\omega$ s, z=1+$\omega^2$s where $s$ is the parameter and $\omega^2+\omega+1=0$. Is there a way for Mathematica to give such parametric solutions to an incomplete system of polynomial equations?

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If you substitude {x -> 1 + s, y -> 1 + \[Omega] s , z -> 1 + \[Omega]^2 s} into your equation you can solve for s,\[Omega]

eqn = {1 + x + y + z == 0 && 1 + x*y + y*z + x*z == 0} 
/. {x -> 1 + s,y -> 1 + \[Omega] s , z -> 1 + \[Omega]^2 s} ;
Solve[eqn, {s, \[Omega]}] 
(*{{s -> 1/2 (-3 - Sqrt[5]), \[Omega] -> 1/2 (3 - Sqrt[5])}, 
{s ->1/2 (-3 + Sqrt[5]), \[Omega] -> 1/2 (3 + Sqrt[5])}}*)
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  • $\begingroup$ thanks but actually my question is about getting the parametric solution and not to find solution for s in reals $\endgroup$ – cleanplay Apr 8 '19 at 22:03
  • $\begingroup$ @cleanplay My result shows that your suggested parametric solution is only valid for special values s, \[Omega] $\endgroup$ – Ulrich Neumann Apr 9 '19 at 10:52
  • $\begingroup$ I think that is because you chose $\omega$ to have solution in reals or complex numbers, while it is an element of $GF(4)$ $\endgroup$ – cleanplay Apr 10 '19 at 1:38

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