Friday, September 12, 2008


Latest idea, solving 2 variable Diophantine equations

After pondering the TSP situation for a while I decided to let that subject drop for a while, as it incubated, and wandered off to do other things, but one day found myself pondering the 3 variable Diophantine equation of the form

c_1*x^2 + c_2*xy + c_3*y^2 = c_4*z^2 + c_5*zx + c_6*zy

And I figured out this theorem about it, and noticed that with z=1, I had

c_1*x^2 + c_2*xy + c_3*y^2 = c_4 + c_5*x + c_6*y

and a way to simplify from that to an equation of the form

A(x+y)^2 - B(x+y) + C = w^2


A =(c_2 - 2c_1)^2 + 4c_1*(c_2 - c_1 - c_3)

B = 2(c_2 - 2c_1)*(c_6 - c_5) + 4c_5*(c_2 - c_1 - c_3)


C = (c_6 - c_5)^2 - 4c_4*(c_2 - c_1 - c_3)

so I found a way to simplify any 2 variable Diophantine equation to a simpler one, as if you find integers w and x+y, such that the second is true, you can then solve for x and y directly, so you solve the first, if it has solutions.

But if the second does not have solutions then neither does the first!

For those wondering about solutions, with Diophantine equations solutions must be integers only.

So, for instance, with x^2 - 2y^2 = 1, x=17 and y = 12 work because 17^2 - 2(12)^2 = 1.

I found other tidbits along the way like that given the Pell's Equation:

x^2 - 2y^2 = 1

you automatically have a solution to the negative Pell's Equation:

z^2 - 2(x+y)^2 = -1

so you also can immediately get that 29 is a solution for x+y, and then find that z=41, as

41^2 - 2(29)^2 = -1.

And yes, I'm talking about these things with math people but so far in arguments they are just saying I have nothing new!!!

So here we go again. I say I found something nifty and people jump out of the woodworks to claim it's not.

Maybe Patricia or that Cranmer guy have comments this time?

Ok, so what good is the result?

Well, for physics people it could mean some explanations for physics stuff, but I'm not totally sure.

I'm just a guy who has ideas and the professionals in these fields blow me off, so I end up posting about them.

If you program the mathematics above you might want to go to my math blog where I have a complete theory, which includes an idea for determining when solutions can exist and solving using what I call Diophantine chains.

And yes, it is frustrating to me that no matter what I can prove it seems that established people who I've seen time and time again betray their academic credentials just get to act like normal, go to class, teach their students, collect their paychecks and government grants—while I'm stuck begging for attention for mind-blowing, revolutionary research on newsgroups.

The system is broken. It is hostile to amateur researchers. And the gatekeepers have just locked the doors and thrown away the key.

So I get to deal with people who are often very wrong about the details of my research, but who know that the status quo is to disagree with me, so they do.

Proof is not enough. These class wars are pushing the limits as the people who are at the top feel comfortable with things as they are.

If it were up to them, humanity wouldn't need to learn anything new at all, as what more do they need anyway?

They already rule the world.
An amazing result that still absorbing the full implications of, as before, the state of the art in this area was the use of far more complicated techniques to simplify to a simpler equation and then to solve it!

To get a better picture of what I mean, here is an example, where if you wish you can go to other sources to see how it is done with the other techniques known. And yes, for me it is kind of wild to be playing with techniques I just invented about a week ago:

To keep things easy for me, I'll use

x^2 + 2xy + 3y^2 = 4 + 5x + 6y

so I have

c_1 = 1, c_2 = 2, c_3 = 3, c_4 = 4, c_5 = 5, and c_6 = 6

so next I need to calculate

A = (c_2 - 2c_1)^2 + 4c_1*(c_2 - c_1 - c_3) = -8

B = 2(c_2 - 2c_1)(c_6 - c_5) + 4c_5*(c_2 - c_1 - c_3) = -40


C = (c_6 - c_5)^2 - 4c_4*(c_2 - c_1 - c_3) = 33

and I have then the new quadratic Diophantine:

(2A(x+y) - B)^2 - 4A*S^2 = B^2 - 4A*C

which is

(-16(x+y) + 40)^2 + 32S^2 = 2656

and dividing off 16, I have

(-4(x+y) + 10)^2 + 2S^2 = 166.

Which has a solution at S=9, giving

-4(x+y) + 10 = +/- 2

and trying the positive first gives x+y = 2, while the negative gives x+y = 3.

Trying the first case, x = 2-y and plugging that into the equation gives

(2-y)^2 + 2(2-y)y + 3y^2 = 4 + 5(2-y) + 6y

which is

4 - 4y + y^2 + 4y - 2y^2 + 3y^2 = 4 + 10 - 5y + 6y

which is

2y^2 - y - 10 = 0,

so y = (1 +/- sqrt(1 + 80))/4 = (1+/-9)/2 = -2 as the other case is a fraction.

Then x=4, so I can try x=4, y=-2, with

x^2 + 2xy + 3y^2 = 4 + 5x + 6y

and get

16 + 2(4)(-2) + 3(-2)^2 = 4 + 5(4) + 6(-2)

which is 12 = 12, so they balance out as they must. I'll leave the second solution to the reader. Notice there are only two.

So the equation turned out to be easy to solve using these techniques.

<< Home

This page is powered by Blogger. Isn't yours?