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It is 135=2mod19 which is right. (Always I think..)
I verified it with the calculator. It seems right!
true i think
You first multiply then tou do modulo p. The division, you first find the reverse of the number mod p and then multiply
I do not know how square root is computed.
The operations are not like those o f standard arithmetic. If mathematica does the solving may has their definitions. GF elements are not integers they seem like integers. Different properties
ok. but this is not the problem.the problem is to have the definition of addition multiplication division square root etc on GF.You cannot use integer arithmetic
That right! All the problem is define over GF.
I do not know if what I have is helpful. All I have is the abiltiy to select the coef. of the polynomial and to evaluate it to some x_i s.
Hmm...I can create a polynomial and pre select its leading coefficient. Also I can pselect random points from it.
By I m not able to constrct polynomials which have intersection points...
This was very helpful to me
http://www.dragonwins.com/domains/getteched/crypto/polynomial_arithmetic_in_gf%28p%5En%29.htm
Is this helpful? I ll try to find more...
http://mathworld.wolfram.com/FiniteField.html
If all the parameteres of the problem are defined correctly I think that there will be no rpoblem. But I remind you that GF elements are not numbers the look like number. I used polynomial basis representation when I worked over GF @^128.
Yes, its the same problem tha we were discussing in the previous post only that the elements now are selected by a GF(2^128).
In GF(p) all the operations are executed modulo p.
In a order to construct an extension field GF(2^128) an irreducible polynomial is employed in order the elements of the field to be generated. Then all the operations are executed modulo this ir. pol.
I think that I can provide you with points of polynomials over GF(2^128). I can also choose the leading coefficient. But I think that in order the resuls to be right both of us have to employ the same irreducible polynomial.
I mean that if it is easy to "feed in" Mathematica with the irreducible polynomial, the required points the leading coefficient I can give you this info.
I thought that you can transform the set of equations to linear. Now I think that is not true so the square roots etc... will stay on but i do not think that this i s aproblem to mathematica.
I think that my post #194 is wrong. If you substitute the differences (x_i-x_0) with variables, the list of unknown variables become larger and larger...
Any results?
May mathematica has library for GF but i am nto familiar with it.
the initial points will be lements of the gf e.g. (x3,y3)=(2,7) etc. The problem is how can you implement add, multiplication and division to have the right results?
Thats is what I am asking....
Multiplications addtion and division over the gf is not as standard opeerations. More generally all the operations executed over GF are performed modulop. e.g 5+15=20modp=3mod17.
The algorithms for the operations are different thats why all the results end up to b.e Gf elements
Yes!