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Thanks again for your time.
My mistake but if you have already solved the problem what do you want me to do?
I do not agree, on the other forum they urged you to try to formulate this geometrically, you now say that it is imaginary. If we can not check graphically or algebraically how will you ever know you are right?
First of all, I prove it for a system of 2 linear equations.
How does it prove that there is a common 2 points of intersection for all 4 quadratics?
In order to be sure that the system has a solution I construt the system of equations to have a common point the x0 y0. i.e. I define the intersection point x0y0 and then I pretend that I do not know it..
You are right about that. We have x0,x1,x2 and y0,y1,y2 so what is there to solve for?
Hmmm. The equations were solved on GF 2^128. Want you try to solve it? i think that you can because you do not ser right the problem.
Those are the equations for the slope of of the line between (x0,y0) (x1,y1) and (x0,y0)(x2,y2).
What do you mean?
Okay, but isn't that just the slope?
Ok for the set of linear equations that I said, I solved it.
That is what they teach in school and it is the biggest myth in the entire world. There are many cases where theory does not match the numbers. Theoreticians can not get numbers! Computational people can! Math is split into pure and applied and they do not even speak the same language.
The first a_1 and the second a_1.
You did not offend me. The people on that site that made such an assertion without providing any worked example did. In computation we want numbers not concepts and jargon.
Ok. I understand that I can not give you clear answers because I also do not understand many things.