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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,262

Hi;

A Galois Field is a finite set, what is p?

**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

p is a prime number. The GF is constituted by the elements 0,1...p-1.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,262

Yes, I know bit to compute one we would have to assign a value to p.

**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

Hmmm ok. I used an extension field GF 2^128 I used a polynomial basis representation and a pentanomial irreducible polynomial for the generation of the field/. But I think that if the system can be solved on GF(p) will be solved at GF(2^128) .

So we can assume p=17.Why not?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,262

Now let me get our definitions synchronized. When they say solve over the Reals for an equation they mean the roots ∈ R. When they say over the rationals that means the roots are ∈ Q. For the integers, the roots are ∈ Z.

Now to solve your set of equations over GF(17) = {1,2,3,...16} that would mean the roots of the equations would be a0,a1,a2... ∈ {1,2,3,4,...16}

**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

Yes!if the system is sovlable the intersection points will be defined and there will be x0,y0...x2y2={0,...,16}. Remember that the operations are not the same as on real...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,262

Remember what you were solving for, the a0,a1,a2... these are the roots of the equations we set up. The way I understand it the roots are the quantities that have to be ∈ GF(17). That means the a0,a1,a2... ∈ GF(17)

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

As the problem was set the a0,a1 a2 where knowns as the x3 y3 all the otheres wew unknowns.Alll the variables should be Gf(17) elements.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,262

That is what I mean all the variables that we solve for have to be ∈ GF(17)

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

Yes!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,262

My feeling is that no matter how large you pick p to be it is unlikely that the variables will be elements of it.

Be that as it may be, what can I do for you?

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

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,262

What about if the equations have no solutions? The restriction of modulo arithmetic might mean no solutions.

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

Thats is what I am asking....

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,262

We will only need to find one example of when there are no solutions.

How do you plan to handle the initial points? They are all lattice points now?

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

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?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,262

There are limits to what can be computed. The only thing that is required is that Mathematica be able to solve modulo p - 1. I will begin to investigate what it can do.

What we need is a problem. I will set one up using GF(17).

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

May mathematica has library for GF but i am nto familiar with it.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,262

It can solve lots of equations using modular arithmetic.

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

Any results?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,262

Not yet. I am going to have to find my notes on the original problem. I have forgotten everything about it. I will post immediately when I have something.

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

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...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,262

Still putting together the notes of the last problem. Once that is done I will have some idea whether or not I can do what you ask.

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

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.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,262

I hope not. Hopefully, it will just do what is required but even if it does it will take some time to find a counterexample.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

**Online**