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Oct
11
comment Why does Shamir's Secret Sharing Scheme need a finite field?
So, finite fields are employed because it is guaranteed that the solution of the problem can be found? In rings R,Q is not possible even if the computations are tranformed to avoid computers' prblems like overflow and underflow?
Jul
25
comment Solve a system of non linear equations over GF
Are my questions that strange???
Jul
25
comment Solve a system of non linear equations over GF
Lagrange interpolation does not need n-th roots, but if you try to solve the problem posted in the question I think that they may be needed. Once again, Can I compute roots $\sqrt[1/n]{x}$?
Jul
25
comment Solve a system of non linear equations over GF
Ok. The same stands for e.g.$\sqrt[3]{x_0}$ or more generally $\sqrt[n]{x_0}$ ? I am asking because in some cases the system may have 10 unknonws and 10 equations. Thus, powers such as ${x_0}^{10}$ will come up and the computation of $\sqrt[10]{x_0}$ will be needed.
Jul
25
comment Solve a system of non linear equations over GF
When I tried to solve a system of quadratics I found out that ${x_0}^{2}$ has come up. I think that I should have to compute the square root of the x0. Is it possible in GF?
Jul
24
revised Solve a system of non linear equations over GF
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Jul
22
revised Polynomial multiplication and division in2^128
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Jul
22
revised Solve a system of non linear equations over GF
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Jul
22
revised Solve a system of non linear equations over GF
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Jul
20
comment Why does Shamir's Secret Sharing Scheme need a finite field?
If the set of equations is constituted by polynomials of n-degree and there are $$2*n$$ such polynomials over R the set has a solution. I am thinking that over GF the set cant be solved because it is no possible to find $$(x)^{1/n}=???$$.
Jul
20
revised Solve a system of non linear equations over GF
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Jul
20
revised Solve a system of non linear equations over GF
added 1 characters in body; edited title
Jul
20
revised Solve a system of non linear equations over GF
edited title
Jul
20
comment Why does Shamir's Secret Sharing Scheme need a finite field?
Sorry I didn't make clear that the polynomials of the are of second degree not linear...I get confused...
Jul
20
comment Why does Shamir's Secret Sharing Scheme need a finite field?
Ok.The equations of question 9294 represent 4 polynomials where for each polynomial I know one point and the leading coefficient. I know that these 4 polynomials (quadraticks) have two common points. If I was working in R I could find their intersection points by solving that set. Is it possible in GF?there is a solution as i know that the 4 quad. have two common points. Sorry if i getting you tred...
Jul
20
comment Why does Shamir's Secret Sharing Scheme need a finite field?
But I would like to know if any set of e.g 8 unknows and 8 equations can be solved...
Jul
20
comment Why does Shamir's Secret Sharing Scheme need a finite field?
So, It can be solved?i.e If I know that the set is defined in such a way that there is a solution, this solution can be computed? the set had 4 equations does this change anything?If it is a bigger set e.g. 6 eq?Its the same?
Jul
19
comment Why does Shamir's Secret Sharing Scheme need a finite field?
Your answer about the roots of $$x^2-1$$ puzzled me. If I ave a set of equations let's say with 3 equations and 3 unknowns, where the knowns and unknown variables are GF elements I cannot be sure that I ll find the right answer?It depends on the irreducible polynomial? I am referring to GF generated by ir. polynomials
Jul
19
awarded  Commentator
Jul
19
comment Why does Shamir's Secret Sharing Scheme need a finite field?
Thanks Dilip for your answer. Have you got any advice for this problem? crypto.stackexchange.com/questions/9294/…