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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

How does it prove that there is a common 2 points of intersection for all 4 quadratics?

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.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

First of all, I prove it for a system of 2 linear equations.

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!!!

*Last edited by Herc11 (2013-07-24 22:58:46)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

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.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

Thanks again for your time.

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?

*Last edited by Herc11 (2013-07-24 23:14:00)*

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