Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

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

Hi bobbym, the problem that you had solved (on real numbers) was linear?

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

That I do not know offhand but I tend to disagree.

But what I do think is that if you can not find a solution in the reals then there is no solution in any other subset of the Reals.

So far looking at what you posted elsewhere you kept much information from me. But that is water under the bridge.

**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.**

Offline

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

I didnt know that this information would be helpful. From my point of I retstated the same question in a way that would be more familiar to you.

""So far looking at what you posted elsewhere you kept much information from me. But that is water under the bridge.""

What was that helpful to you?

As the problem was set it surely has a solution. I stated this many times. The problem is if there is a way to found the solution on GF.

"But what I do think is that if you can not find a solution in the reals then there is no solution in any other subset of the Reals."

This has no meaning as it is not the same system of equations. The latter system of equations is defined on GF so there is no relation to real.

*Last edited by Herc11 (2013-07-24 19:20:47)*

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Was anyone else able to provide a solution?

My point is no solution could be found in other set other than the reals.

When we started you convinced me there were many such solutions so all I had to do was find one that did not work. As I began solving these I began to think there were not many solutions. Now, I do not believe there is even one such solution.

**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.**

Offline

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

No. only one answer there was.

The author of this answer said that if the problem is constructed as I stated it can be solved for linear or even non linear systems in GF.

By the way, according to my calculations the system is always linear.

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

No. only one answer there was.

I read several answers. One of them said it was unsolvable as you posted it. I am inclined to agree.

The author? Who is he and are you understanding what he meant correctly?

**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.**

Offline

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

You read several comments. No I am not sure that I understood correctly what he wrote. But I think that I understood.

http://crypto.stackexchange.com/questions/9294/solve-a-system-of-non-linear-equations-over-gf

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Broadly speaking, the field in which we are operating influences the answer to the question of whether a system of equations has solutions or not to some extent, but not in the ways that the OP thinks it does. Whether we are operating in a prime field or an extension of a prime field (what the OP calls gf or GF) has relatively little to do with the matter. In particular, for linear equations, the general theory of linear equations over a field usually has more to say about the matter than the identity of the field.

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

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

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

yes.so?

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

First of all he says

but not in the ways that the OP thinks it does

There is something wrong with your understanding of the problem and how it is related to the so called GF.

Whether we are operating in a prime field or an extension of a prime field (what the OP calls gf or GF) has relatively little to do with the matter.

A prime field should have been adequate to answer this question. It was not. I tried GF(113) Gf(2^16), Gf(2^32) all without a single solution that I could find.

This might mean that maybe the author is incorrect. Does he have a working example to prove his assertions?

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

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

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

but not in the ways that the OP thinks it does

I thought that the filed plays a significant role to the solution of the system. And he says that it plays a role but not in the ways that I think.

There were many mistekes at the problem you tried to solve.

You chose negative integers that they do not exist in GF.

The operations returned non integers, which is not possible if the operations executed correctly.

you ve mixed real with GF.

In the problem where I provided you with a_0 a_1 a_2 and the intersection points etc. I used polynomial basis representation with a ceratin irreducoible polynomial.

You controlled the data as they were integers.

It was more than certain that no correct solution will ever be returned.

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

You chose negative integers that they do not exist in GF.

I did not choose anything that is what the multiplications from your formula produced. I asked what you wanted to do next on those coefficients. I would have used a modulo on them to bring them into the set. You had no answer.

You controlled the data as they were integers.

Your set or any prime field is nothing but a subset of the integers. If there was no intersection over the integers how can there be one over a subset of the integers?

I have recommended to you a site where they would further answer your questions. They provided no solution either. Maybe there isn't one! This is the possibility you keep avoiding.

I have asked you to produce any solution or to have your author produce one. I would like to see it.

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

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

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

I used the field 2^128 and also polynomial representation basis. Therefore, addition multiplication and division is not defined as it were on real numbers. I have tested it many times that these operations never return negative elements. It is not possible.

In math stack they all said that the first problem where linear equation are empolyed, there is a solution on GF. Do you think that you can solve it?

My problem is that I dont have an algorithm to solve the second problem. I have the definitions of the operations but I

cannot solve a system of 4 equations with 4 unknowns. I am not sure that there is a solution but I think tha there is. I do

I do not avoid any possibility but I think that believing that there is no solution is the easy but unhopefully the wrong way.

*Last edited by Herc11 (2013-07-24 20:24:29)*

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

I have tested it many times that these operations never return negative elements

Modulo would have brought them into your field. There still would not have been a solution.

Do you think that you can solve it?

Solve what? I do not like when people give offhand information like "this is a linear system" but provide no solution. How is that a representation of 4 quadratics? What am I solving for?

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

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

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

Ok. I understand that I can not give you clear answers because I also do not understand many things.

I thought that you read my post where this set was described in detail and the solution is provided by another user user. This is a system of 2 linear equations. If we want to be mathematically correctly there is no geometrical representation of this problem in the GF.

I did not want to offend you, if I did.

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

You did not offend me. The people on that site that made such an assertion without providing any worked example did. In computation we want numbers not concepts and jargon.

What is M1 and M2?

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

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

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

The first a_1 and the second a_1.

They are advanced in mathematics. The know that if theory is correct, numbers are never a problem.

Our problem is that we are not sure, we do not know the theory.

*Last edited by Herc11 (2013-07-24 20:52:55)*

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

They are advanced in mathematics. The know that if theory is correct, numbers are never a problem.

That is what they teach in school and it is the biggest myth in the entire world. There are many cases where theory does not match the numbers. Theoreticians can not get numbers! Computational people can! Math is split into pure and applied and they do not even speak the same language.

Now, to have any hope of doing one of these in a convincing way the field will have to be smaller than 2^128. These are 40 digit numbers, too big for a test.

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

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

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

Ok for the set of linear equations that I said, I solved it.

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Okay, but isn't that just the slope?

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

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

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

What do you mean?

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Those are the equations for the slope of of the line between (x0,y0) (x1,y1) and (x0,y0)(x2,y2).

Had you used a small m on the left I would have spotted them quicker.

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

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

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

Hmmm. The equations were solved on GF 2^128. Want you try to solve it? i think that you can because you do not ser right the problem.

*Last edited by Herc11 (2013-07-24 22:15:19)*

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

You are right about that. We have x0,x1,x2 and y0,y1,y2 so what is there to solve for?

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

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

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

In order to be sure that the system has a solution I construt the system of equations to have a common point the x0 y0. i.e. I define the intersection point x0y0 and then I pretend that I do not know it..

That was the problem I posted yesterday. But in that problem the eq. were quadratics...So I used to construct it, two points x0x1 y0 y1. Then by knowing M1..M4 and the four unique X2 y2 for each polynomial andby pretending that you dont know the x0x1 y0 y1 you had to solve the system.

You said the system that hadn t got a solution in GF. As i have constructed the system I was sure that a solution existed.

Thats why I was sure that the system has a solution!!!! Because I knew the solution.

I was not sure If the solution can be found.

I said all that many times.

Offline