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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 94,792

Hi;

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

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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: 94,792

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

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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: 94,792

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}

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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: 94,792

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)

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**Herc11****Member**- Registered: 2013-06-19
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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: 94,792

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

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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: 94,792

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?

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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: 94,792

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

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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: 94,792

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?

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**If it ain't broke, fix it until it is.**

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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: 94,792

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).

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**If it ain't broke, fix it until it is.**

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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: 94,792

It can solve lots of equations using modular arithmetic.

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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: 94,792

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.

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**If it ain't broke, fix it until it is.**

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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: 94,792

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.

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**If it ain't broke, fix it until it is.**

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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: 94,792

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.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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