Latest idea, solving 2 variable Diophantine equations

[JSH]

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

where

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)

and

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.


James Harris

 

[JSH]

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

where

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)

and

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.

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

and

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.


James Harris

 

[JSH]

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

and

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
Gives.


(-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.

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.

Here's another example as I had yet another advance using this theory
and came up with this simple relation you can easily play with when
you're bored.

Using the math in my prior posts you can get the general result that
whenever you have

x^2 + Dy^2 = F

you have a connected equation

z^2 + D(x+y)^2 = F*(D+1)

where finding a solution for it gives you a solution for the first,
you can immediately figure out how to get an infinite number of
solutions with D=-2 and F=1, the classical Pell's equation, as then
you have

x^2 - 2y^2 = 1, followed by

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

and the next in the series is

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

so you just get this flipping back and forth, and with one solution at
the start you can get the solutions that follow, so with x=3, and y=2,
you have next that

z^2 - 2(5)^2 = -1, so z^2 = 49, so z=7, and then you have

w^2 - 2(3+2+7)^2 = 1, so w^2 - 2(144) = 1, so w^2 = 289 and w=17.

And you can do that forever.

So like if you're bored you can play with the numbers and watch the
patterns play out.

But you may wonder, why should I keep bugging your newsgroup with
these results?

Well, remember, I came here first to try and get feedback on my Class
Viewer for Java open source project. That didn't go so well.

THEN I put forward some ideas I had for the Traveling Salesman Problem
which got me more attention but from your newsgroup cops trying to
shoot down the idea and then lying about their success.

So now I'm showing you another result to give you some sense of the
length and breath of the reality of my research and discoveries,
mainly for later.

Yup. I want my own computer science company (among other companies)
so part of what I'm doing now is evaluating the community.

Oh, one of your newsgroup cops, Cranmer, wandered over into math
newsgroup threads of mine until I guess he got bored or realized he
was out of his depth.

But where's your Patricia? She claimed interest in math, before.

The TSP idea is not dead either. I've just put it on the back-burner
while I pondered other things.

I am a busy person.


James Harris

 

[Patricia Shanahan]

JSH wrote:
....

But where's your Patricia? She claimed interest in math, before.

....

Well, for the last couple of weeks I've been in England for a family
emergency.

I don't plan to spend any time looking at anything else you do unless
and until you produce a working Java implementation of your claimed TSP
algorithm. A proof of correctness would also be nice.

Patricia

 

[JSH]

JSH wrote:

...> But where's your Patricia?  She claimed interest in math, before.

...

Well, for the last couple of weeks I've been in England for a family
emergency.

Oh, sorry to hear that...

I don't plan to spend any time looking at anything else you do unless
and until you produce a working Java implementation of your claimed TSP
algorithm. A proof of correctness would also be nice.

Patricia

That research is on the back-burner. I'm at 2 variable quadratic
Diophantine equations at the moment.

___JS