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

true i think

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,771

Here is another one:

We get x = 5, which corresponds to solving that equation over GF(19)

**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

I verified it with the calculator. It seems right!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,771

I changed post #252, so please check again.

**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

It is 135=2mod19 which is right. (Always I think..)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,771

Okay, then for the sake of argument because we have nothing else to go on let's assume Mathematica can solve your equations over the GF you want. This is a pragmatic decision, if we assume it can not we are done right here. Let's proceed with the assumption and see what the heck happens.

**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

Ok!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,771

Now, I can continue to ask the remaining questions.

GF(p) what is p going to be? I would very much like to have a p smaller than 2^128. Is this possible?

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 try a small prime number. I think if it is solvable for small p it will be solavble for larger p's..pick 113 i think it is prime

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,771

This is what we agree on:

So the leading coefficients of each of the 4 given polynomials will be ∈ GF. The points given will have x's and y's that are ∈ GF. Now you are requiring the points of intersection of the 4 quadratics to also be ∈ GF?

A Gf(113) will be used.

Agreed?

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. I want to know if we can find the intersection points of the polynomial as syou didi wi th real numbers

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,771

I understand the problem well enough to work on it. I am currently a little busy with 4 other problems and will start modifying the cubic code that I used to solve the other problems. This will take some time.

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

Ok!!!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,771

So far I have found it very difficult to even construct 4 quadratics and 2 points of intersection all of them over Z let alone 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

why?

*Last edited by Herc11 (2013-07-22 07:54:45)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,771

It is a little early to say. But it does suggest they are rare. That means that getting an answer to them in the required form is very rare.

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

If the set is small maybe is a good choce to use a bigger p.

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

You can use Newton Interpolation i.e the formulas for a that I have posted eralier to compute the polynomials.

What I mean is that you can preselect the common x_0 y_0 x_1 y_1 for the four polynomials. Then You can use the postedd formulas for recovering a_0 a_1. Then for each polynomial you can select a different x_3 y_3. Then by the posted formula again you can compute a_3.

After that tou will have 4 polynomials and the set of equations is ready.

The problem is to "pretend" that you do not know the 2 intersection points x_0x_1 y_0 y_1 and try to solve the previous system.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,771

I can do all that. The trouble is the answers are not coming back as integers let alone 1 - 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

What do you mean?They have to return as elements. If not something is wrong...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,771

I think I have a method now that is working to generate the problems. I will go back to work on the solutions.

Half way through with it. I need a little sleep, see you later hopefully with a full solution.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,771

Hi;

I have 4 parabolas that intersect at two points. The leading coefficients are 5, 22 17, 29. I have 4 points on those 4 parabolas, 1 on each parabola.

They are:

(3,13), (3,47), (3,37), (3,61)

You can see that so far all the conditions have been met. Now I must calculate from this information the points of intersection modulo 112.

The first idea to solve it failed completely. I think I see what went wrong.

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

What went wrong?? mod113

*Last edited by Herc11 (2013-07-22 22:02:03)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,771

That will be a problem when I get to that.

What went wrong first is the choice of points. You notice the x's are all 3. This caused the system to not have a unique solution. Why? I do not know! I will have to adjust the test problem to use all different or at least some 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

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

Dιδ you used x=3 for creating the parabolas?

How did you construct the parabolas?Can you be sure that they have 2 intersection points? Are the intersection points known?

*Last edited by Herc11 (2013-07-22 22:19:53)*

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