I m telling you another one time that I am not sure If I can recover a known solution.

Ofcourse we can check it algebraically. You can not geometrically formulate Gfs. They have distinct elements etc...

Try to draw f(x) where x and f(x) take values from 1,2,3,4,5,6,7. Will it be consistent?

]]>As for the 4 quadratics, I told you that I used Newton Interpolation to compute the coefficients of the polynomials.

My mistake but if you have already solved the problem what do you want me to do?

There are no lines that intercept only in our imagination.

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?

I have devoted alot of time to a problem without a solution, coffeemath might be able to assist you better over at the SE. I have done everything I can, good luck.

]]>As for the 4 quadratics, I told you that I used Newton Interpolation to compute the coefficients of the polynomials. It is a basic problem of linear algebra (polynomial reconsruction. I used two common points for all 4 quadratics. What does it mean common? That these are their intersection points. When the problem was defined over real numbers you solved it.

The problem is if you can find the solution in GFafter you know that there is a solution.

There can be no graphical illustration of the problem beacuse we work on GF not on real numbers.

There are no lines that intercept only in our imagination.

If the system is not linear I can not compute roots etc..thats the problem!!!

]]>You can not prove even for 1 set of quadratics by words and theorems that you have not quoted. Please produce your answer and let me see if it meets at two points.

You have no numerical evidence, not a single example, no theorems to quote, at this point you do not even have a speculative conjecture.

Please show the solution and prove it.

]]>That was the problem I posted yesterday. But in that problem the eq. were quadratics...So I used to construct it, two points x0x1 y0 y1. Then by knowing M1..M4 and the four unique X2 y2 for each polynomial andby pretending that you dont know the x0x1 y0 y1 you had to solve the system.

You said the system that hadn t got a solution in GF. As i have constructed the system I was sure that a solution existed.

Thats why I was sure that the system has a solution!!!! Because I knew the solution.

I was not sure If the solution can be found.

I said all that many times.

]]>Had you used a small m on the left I would have spotted them quicker.

]]>They are advanced in mathematics. The know that if theory is correct, numbers are never a problem.

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.

Now, to have any hope of doing one of these in a convincing way the field will have to be smaller than 2^128. These are 40 digit numbers, too big for a test.

]]>They are advanced in mathematics. The know that if theory is correct, numbers are never a problem.

Our problem is that we are not sure, we do not know the theory.

]]>What is M1 and M2?

]]>I thought that you read my post where this set was described in detail and the solution is provided by another user user. This is a system of 2 linear equations. If we want to be mathematically correctly there is no geometrical representation of this problem in the GF.

I did not want to offend you, if I did.

]]>