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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 $x_0,y_0$

Does the previous set of equations is solvable?

And more...

If I have the following set of equations:

$$M_1=\frac{k_1-(y_0+(\frac{y_1-y_0}{x_1-x_0})(l_1-x_0))}{(l_1-x_0)(l_1-x_1)}$$

$$M_2=\frac{k_2-(y_0+(\frac{y_1-y_0}{x_1-x_0})(l_2-x_0))}{(l_2-x_0)(l_2-x_1)}$$

$$M_3=\frac{k_3-(y_0+(\frac{y_1-y_0}{x_1-x_0})(l_3-x_0))}{(l_3-x_0)(l_3-x_1)}$$

$$M_4=\frac{k_4-(y_0+(\frac{y_1-y_0}{x_1-x_0})(l_4-x_0))}{(l_4-x_0)(l_4-x_1)}$$

where $x_0,y_0 x_1,y_1$ are the unknown GF elements.

As Dilip Sarwate clarified the set of equations is constructed by someone who " chose three distinct x0,x1,x2, as well as y0,y1,y2, then computed M1, M2, and finally revealed $M_1,M_2,x_1,y_1,x_2,y_2$ but not $x_0,y_0$ to us" i.e. it is known that the system has solution.

My question was: Can I recover the $x_0, y_0$ or in the second set of equations can I recover $x_0, x_1, y_0, y_1$ and generally in nonlinear sets to recover the respective $x_i, y_i$ by the provided info, on a GF? and the main point of my question: Does the fact that the set of equations is defined on a Galois Field impose any difficulties to find its solution?

I was not sure that it is possible to compute the solution of the problem with the aforementioned parameters on a Galois Field.

As Dilip Sarwate stated in his answer the solution of the previous problem can be recovered for linear and nonlinear equations.

Herc11
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2 Answers2

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There have been extensive comments by the OP on this question as well as a related one and its answers and the consensus don't seem to be converging at all to anything sensible.

Broadly speaking, the field in which we are operating influences the answer to the question of whether a system of equations has solutions or not to some extent, but not in the ways that the OP thinks it does. Whether we are operating in a prime field or an extension of a prime field (what the OP calls gf or GF) has relatively little to do with the matter. In particular, for linear equations, the general theory of linear equations over a field usually has more to say about the matter than the identity of the field.

As Henrick Hellström has pointed out in a comment on the question, the first set of equations in the OP's question can be converted into a pair of linear equations in the unknowns $x_0$ and $y_0$, say $$\begin{align} a_{11}x_0 + a_{12}y_0 &= b_1\\ a_{21}x_0 + a_{22}y_0 &= b_2\end{align}$$ where the $a_{ij}$ and the $b_k$ are functions of the known quantities $M_1, M_2, x_1, y_1, x_2, y_2$. I refuse to calculate and state explicitly what the $a_{ij}$ and the $b_k$ are in terms of the known quantities. I will differ from Henrick slightly, though, in that I do not think that these equations are solvable for each and every choice of $M_1, M_2, x_1, y_1, x_2, y_2$. This has nothing to do with whether these quantities belong to GF$(2^m)$ or GF$(p^m)$ for $p > 2$ or GF$(p)$ and everything to do with basic linear equation theory: the matrix might be singular for some choices of $M_1, M_2, x_1, y_1, x_2, y_2$ in which case we get multiple solutions rather than a unique solution, or the matrix might be singular and the equations might be inconsistent in which case no solution exists.

All this, however, as well as the OP's questions about other equations is irrelevant in the context of Shamir's secret sharing scheme which is referenced in the other question but not in this one but which I believe is the reason for these questions.

If someone chose three distinct $x_0, x_1, x_2$, as well as $y_0, y_1, y_2$, then computed $M_1$, $M_2$, and finally revealed $M_1, M_2, x_1, y_1, x_2, y_2$, but not $x_0, y_0$ to us, then we can find $x_0$ and $y_0$ from the given relationships, and inconsistency is not an issue. Note the requirement that $x_0, x_1, x_2$ be distinct which allows the unnamed someone to avoid the heartbreak of division by $0$ when computing $M_1$ and $M_2$.

Similar remarks apply to the nonlinear equations in the OP's question as well.

Community
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Dilip Sarwate
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The first question is solvable:

$$x_0=x_1+(y_0-y_1)M_1^{-1}$$ $$x_0=x_2+(y_0-y_2)M_2^{-1}$$

So $$y_0=\frac{M_1M_2(x_2-x_1)+M_2y_1-M_1y_2}{M_2-M_1}.$$

And about the second one, we have a similar method.

Meysam Ghahramani
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