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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

No, I chose the points later. I can find different points on the parabolas. It will just mean reprogramming it.

The choice of same x's would a problem if wou were solving the problem over real numbers?

That is a good question and I do not know how the simultaneous set will react when the same x's are chosen for all the points. Remember, I only solved 2 problems before this one over the Reals. That is not a lot. I do know that on those problems I chose different x's.

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

Maybe your parabola' do not intersect at two points. This is too likely to happen.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

I am sure they intersect at 2 points. Here is a plot of 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

I see in the polot that you have negative numbers. Is it for represenative reasons?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

The coefficients of the parabolas are negative integers. For instance one of them is y = 10 - 14 x + 5 x^2. Only the leading coefficient needs to be an element of GF(113).

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

No, all the coefficients have to be elements of the GF. There must not be non GF elements.

You can mix integeres with GF elements. It does not make any sense.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

That I did not know. You say that all the coefficients of the four parabolas must be over GF(113) not just the first one?

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 it was clear...All the parameteres of the problem are GF elemets . The set of equations is defined on GF you can not mix GF elements with integers. The problem will have no sense. The points are GF elements too.

You can only choose by the set of GF{0,..,122}

*Last edited by Herc11 (2013-07-22 22:48:38)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Okay, I will setup up the problem to be like that. I will post when I have it done.

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 you will find it difficult to find 4 quadraticks with 2 intersection points over GF{113}. I can construct polynomials over the GF{2^128}. I might be able to do it over 2^32. In 2^32 can you solve the set?i.e. has the Mathematica the appropriate equations?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

I think it will be impossible to find 4 quadratics over GF(113). You want to go larger?

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 you wish. I used 2^128 but as I said I dont know if you can solve the set afterwards...

*Last edited by Herc11 (2013-07-22 23:00:15)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

2^128 is not a prime.

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 there will be no problem as 2^128 is an extension field. I ve read it somewhere...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

One more thing. 2^128 is a 40 digit number. That means coefficients will be very large and graphing because of scaling will be difficult.

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. this is true that why I told you that there will be problems with 2^128.

Although, If we use 2^128 I can provide you the points and the coef. of course not the graphs.

(Graphs cannot be obtained for GFs beacuse GF elementes are not numbers. )

But then is there a way to solve the set?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

I do not think solving is a problem at all. Forming a set of problems to solve has been troublesome. I will try with 2^128.

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

Then I can provide you with the points and lead . coef etc... I mean that It might be a problem because you have to perform all the operations based on the irreducible problem.

*Last edited by Herc11 (2013-07-22 23:22:55)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Okay, provide me with what you have.

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

Here is what I have. I hope that I didnt make any mistake while copying the numbers. Note tha they are in Hex.

I give you the common x_0 y_0 x_1 y_1 a_0 a_1.

and then the different x_2 y_2 a_2

X_0={0xffbbac00aadd00ad04ddaaffccddddcc} <--this is the right

X_1={0x0add0000ffffcc00015f00ffccdd00dd}

Y_0={0x144a00005566aa031111233423050066}

Y_1={0xac1243476566cc0fff11043423450066}

A[0]=Y_0

A[1]={0xb15e102aad1db8e4e224d46c7f715c22}

----1---------

X_2={0x142200ffddddcc00ccddddccdd0000cd}

Y_2={0x21dffa476566baff111123342355ff66}

A[2]={0x08a95bc0baa4cfef4ba9a3b40af2e6c5}

-----2-------

X_2={ 0x142200ffddddcc00ccddddccdd000000}

Y_2= Y_2={0x21dffa476566baff111123342355ff00}

A[2]= 0x09530aa1f9a3e1245bb246c66f273c5a

-----3-------

X_2={ 0x142200ffddddcc00ccddddccdd000011}

Y_2= Y_2={0x21dffa476566baff111123342355ff11}

A[2]=0x03f7c3839906d4a6bee8ef3fea254596

--4---

X_2={ 0x142200ffddddcc00ccddddccdd000022}

Y_2= {0x21dffa476566baff111123342355ff22}

A[2]=0x8dbd3b684dde8c4beb6a9662439766dc

The irreducible polynomial 0x0180000000000000000000000000000043

*Last edited by Herc11 (2013-07-23 23:58:01)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Hi;

A question:

Are the x2,y2 the coordinates of the points on the parabolas?

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

x2 y2_s are the coordinates of the points of the parabolas. In specific, these are the known points (plus the A_2 s) that we have in our disposal to solve the system.

*Last edited by Herc11 (2013-07-23 22:16:02)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Hi;

Okay, thanks for the clarification.

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

Is it easy for you to pass all that information to Mathematica?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Nothing is easy with Mathematica. It is a real bear. Its strengths are also its weaknesses. Thousands of commands makes everyone go wow but it is difficult to remember them all.

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

Combinatorics is Algebra and Algebra is Combinatorics.

**Online**