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You used GF operations didnt you?
Ok I ll chek it. But it is impossible if the paraneteres of the problem are set correctly to return non integer values.
No, a0 a1 a2 are the newton coef. the formula tha associates the newton coef. with polynomial's coef is:
But I dont think that you need the above formula.
The only thing is to solve the set of the 4 a2's.
I thought so...
Is it easy for you to pass all that information to Mathematica?
x2 y2_s are the coordinates of the points of the parabolas. In specific, these are the known points (plus the A_2 s) that we have in our disposal to solve the system.
Here is what I have. I hope that I didnt make any mistake while copying the numbers. Note tha they are in Hex.
I give you the common x_0 y_0 x_1 y_1 a_0 a_1.
and then the different x_2 y_2 a_2
X_0={0xffbbac00aadd00ad04ddaaffccddddcc} <--this is the right
X_1={0x0add0000ffffcc00015f00ffccdd00dd}
Y_0={0x144a00005566aa031111233423050066}
Y_1={0xac1243476566cc0fff11043423450066}
A[0]=Y_0
A[1]={0xb15e102aad1db8e4e224d46c7f715c22}
----1---------
X_2={0x142200ffddddcc00ccddddccdd0000cd}
Y_2={0x21dffa476566baff111123342355ff66}
A[2]={0x08a95bc0baa4cfef4ba9a3b40af2e6c5}
-----2-------
X_2={ 0x142200ffddddcc00ccddddccdd000000}
Y_2= Y_2={0x21dffa476566baff111123342355ff00}
A[2]= 0x09530aa1f9a3e1245bb246c66f273c5a
-----3-------
X_2={ 0x142200ffddddcc00ccddddccdd000011}
Y_2= Y_2={0x21dffa476566baff111123342355ff11}
A[2]=0x03f7c3839906d4a6bee8ef3fea254596
--4---
X_2={ 0x142200ffddddcc00ccddddccdd000022}
Y_2= {0x21dffa476566baff111123342355ff22}
A[2]=0x8dbd3b684dde8c4beb6a9662439766dc
The irreducible polynomial 0x0180000000000000000000000000000043
Then I can provide you with the points and lead . coef etc... I mean that It might be a problem because you have to perform all the operations based on the irreducible problem.
Yes. this is true that why I told you that there will be problems with 2^128.
Although, If we use 2^128 I can provide you the points and the coef. of course not the graphs.
(Graphs cannot be obtained for GFs beacuse GF elementes are not numbers. )
But then is there a way to solve the set?
I think that there will be no problem as 2^128 is an extension field. I ve read it somewhere...
As you wish. I used 2^128 but as I said I dont know if you can solve the set afterwards...
I think that you will find it difficult to find 4 quadraticks with 2 intersection points over GF{113}. I can construct polynomials over the GF{2^128}. I might be able to do it over 2^32. In 2^32 can you solve the set?i.e. has the Mathematica the appropriate equations?
I thought that it was clear...All the parameteres of the problem are GF elemets . The set of equations is defined on GF you can not mix GF elements with integers. The problem will have no sense. The points are GF elements too.
You can only choose by the set of GF{0,..,122}
No, all the coefficients have to be elements of the GF. There must not be non GF elements.
You can mix integeres with GF elements. It does not make any sense.
I see in the polot that you have negative numbers. Is it for represenative reasons?
Maybe your parabola' do not intersect at two points. This is too likely to happen.
The choice of same x's would a problem if wou were solving the problem over real numbers?
Dιδ you used x=3 for creating the parabolas?
How did you construct the parabolas?Can you be sure that they have 2 intersection points? Are the intersection points known?
What went wrong?? mod113
What do you mean?They have to return as elements. If not something is wrong...
You can use Newton Interpolation i.e the formulas for a that I have posted eralier to compute the polynomials.
What I mean is that you can preselect the common x_0 y_0 x_1 y_1 for the four polynomials. Then You can use the postedd formulas for recovering a_0 a_1. Then for each polynomial you can select a different x_3 y_3. Then by the posted formula again you can compute a_3.
After that tou will have 4 polynomials and the set of equations is ready.
The problem is to "pretend" that you do not know the 2 intersection points x_0x_1 y_0 y_1 and try to solve the previous system.
If the set is small maybe is a good choce to use a bigger p.
why?
Ok!!!
Yes. I want to know if we can find the intersection points of the polynomial as syou didi wi th real numbers
Yes try a small prime number. I think if it is solvable for small p it will be solavble for larger p's..pick 113 i think it is prime