Solve a set of non linear Equations on Galois Field I have the following set of equations:
$$M_{1}=\frac{y_1-y_0}{x_1-x_0}$$
$$M_{2}=\frac{y_2-y_0}{x_2-x_0}$$
$M_1, M_2,  x_1, y_1, x_2, y_2,$ are known and they are chosen from a $GF(2^m).$ I want to find  $x_0,y_0$
I ll restate my question.
Someone chose three distinct x0,x1,x2, as well as y0,y1,y2, then computed M1, M2, and finally revealed M1,M2,x1,y1,x2,y2, but not x0,y0 to us.All the variables are chosen from a Galois Field.
I want to recover the unknown $x_0,y_0.$ Is it possible to accomplish that?
If a set of nonlinear equations have been constructed with the aforementioned procedure e.g.
$$M_1=\frac{k_1-(y_0+(\frac{y_1-y_0}{x_1-x_0})(l_1-x_0))}{(l_1-x_0)(l_1-x_1)}$$
$$M_2=\frac{k_2-(y_0+(\frac{y_1-y_0}{x_1-x_0})(l_2-x_0))}{(l_2-x_0)(l_2-x_1)}$$
$$M_3=\frac{k_3-(y_0+(\frac{y_1-y_0}{x_1-x_0})(l_3-x_0))}{(l_3-x_0)(l_3-x_1)}$$
$$M_4=\frac{k_4-(y_0+(\frac{y_1-y_0}{x_1-x_0})(l_4-x_0))}{(l_4-x_0)(l_4-x_1)}$$
where $x_0,y_0 x_1,y_1$ are the unknown GF elements.
 Can I recover the unknown elements?
My question was if the fact that the set of equations is defined on a Galois Field imposes any difficulties to find its solution. 
If not I suppose that the set can be solved. Is this true?
Has mathematica or matlab any package that will help me to verify it?
When I tried to solve a system similar to the one above posted I found out that ${x_i}^{2}, 0\leq i \leq 2$ has come up. 
I think that I should have to compute the square root of the x0. 
Is it possible in GF?
 Is it also possible to compute the $\sqrt[1/n]{x_0}$?
 A: The first system can be solved in the usual way, provided the "slopes" $M_i$ are distinct. Solve each for the knowns $y_k$, $k=1,2$ and subtract. You can then get to
$$x_0=\frac{M_2x_2-M_1x_1-y_2+y_1}{M_2-M_1},$$
and then use one of the equations you already formed with this $x_0$ plugged in to get $y_0.$ Since this method only uses addition/subtraction multiplication/(nonzero)division it works in any field, in particular in your Galois field.
A: I think tha the second system can be solved lik the first one. More specifically,
The second system can be solved too. If it is setted $\frac{y_1-y_0}{x_1-x_0}=a_1$ and $a_0=y_0$ the system of equations can be rewritten as
$$M_1=\frac{k_1-(a_0+a_1(l_1-x_0))} {(l_1-x_0)(l_1-x_1)}$$
$$M_2=\frac{k_2-(a_0+a_1(l_2-x_0))} {(l_2-x_0)(l_2-x_1)}$$
$$M_3=\frac{k_3-(a_0+a_1(l_3-x_0))} {(l_3-x_0)(l_3-x_1)}$$
$$M_4=\frac{k_4-(a_0+a_1(l_4-x_0))} {(l_4-x_0)(l_4-x_1)}$$
 where $M_1, M_2, M_3,M_4, k_1,l_1,k_2,l_2,k_3,l_3,k_3,k_4$ are known.
Then the 4 unknowns are the $x_0,x_1, a_0, a_1$.
The system of equations is linear, and if there is a solution, the system can be solved.
Afterwards, by   $$\frac{y_1-y_0}{x_1-x_0}=a_1$$   $$a_0=y_0$$ the  $y_0, y_1$ can be recovered. True or FAlse????
