## Sunday, December 30, 2012

### Considering a behavior

One of the things that helped me a lot recently was a result of mine with a famous equation, where finally I could look at something that seemed to explain a lot for me. So for background information for those interested in my research and issues around it, I want to talk about this result as objectively as possible. So deliberately I'll try to stay away from emotion, and focus on the facts.

The famous equation is x2 - Dy2 = 1, which according to various web sources has been known for over a thousand years. Typically integer solutions for x and y, given positive integer D, are discussed in regard to it, for instance: 32 - 2(2)2 = 1

However, the size relative to D of the smallest set of positive non-zero integers which will give a solution, which is called the fundamental solution, can vary extensively.

Current accepted number theory about the equation x2 - Dy2 = 1  cannot explain the size in general of the fundamental solution.

My own research proves that D is key, and if D is a prime or two times a prime, then solutions will be the largest, with exceptions, when D-1, D+1, D-2, or D+2 is a square, as x is forced to equal 1 or -1 mod D, especially if D-1 has small prime factors, and if 4 is also factor they will be even bigger still.

That is, x-1 or x+1 must have D as a factor, if D is a prime or twice a prime, which forces a larger solution unless D-1, D+1, D-2 or D+2 is a square. And when a larger solution is forced, if D-1 has small prime factors, solutions will be even larger. And then if 4 is also a factor they are the largest.

For example there is a large fundamental solution at D = 61.

The smallest positive non-zero x and y that will work are x = 1766319049 and y = 226153980.

17663190492 - 61(226153980)2 = 1

Currently accepted number theory cannot tell you why that solution is so large, as my research is to my knowledge not yet accepted.

Here notice that x = -1 mod 61, as: 1766319049 + 1 = (28956050)(61)

From my research it follows that the solution is so large because x = -1 mod 61, D-1, D+1, D+2, and D-2 are not square, while D-1 = 60, which has the first three primes as factors, as 4(3)5) = 60, and because 4 is a factor as well.

So I can explain why D=61 is so large and my explanation is backed by a mathematical proof which has been public on this blog since September 2011.

Also I directly contacted various mathematicians last year, giving them the rules. Usually I went to some effort to find number theorists so that the people contacted would presumably have an interest in this result.

Also one of the mathematicians I knew of from searches on the equation commonly called "Pell's Equation" by mathematicians as he had a book listed on Amazon.

Contacts were successful to some extent in that I got emailed replies from several mathematicians at first, but later, no additional emails garnered replies.

To my knowledge there has been no acknowledgement of this result by established mathematicians.

The equation is very famous and presumably is taught yearly in number theory courses.

As far as I know, mathematicians do not explain the size of the fundamental solutions at all.

Remarkably you can find papers on the subject, however, for instance:

"ON THE SIZE OF THE FUNDAMENTAL SOLUTION OF PELL
EQUATION" by ETIENNE FOUVRY.

It is a 29 page paper and I looked through it trying to find an example with an actual D. There was none. I also scanned through looking for an explanation for the fundamental solution size. I saw none, but I'm not a mathematician.

Another reference to a paper where I could only see the abstract:

"The size of the fundamental solutions of consecutive Pell equations" by Michael J. Jacobson and Hugh C. Williams

There the abstract indicates, well I'll just quote part of it:

"...We also provide some heuristic reasoning which suggests that there exists an infinitude of values of D for which $\rho(D) \gg \sqrt{D} \log \log D / \log D$, and that this is the best possible result under the Extended Riemann Hypothesis. Finally, we present some numerical evidence in support of this heuristic."

I could present more examples but I think those two are indicative.

Looking over what I can see on the web, modern mathematicians cannot explain the size of the fundamental solution to the equation x2 - Dy2 = 1.

But also, now a year since I informed some of them, they have yet to acknowledge the solution.

That is the behavior under consideration in this post for background.

James Harris

## Friday, December 28, 2012

### Hidden help?

One of the reasons I like noting I'm not a mathematician is routinely I make dumb mistakes with the math, and oddly enough I've been corrected often through the years by math people. I say "math people" to include those who may not be actual mathematicians, versus trying to figure out who is, and who is not.

And recently I had an erroneous argument which I thought proved that x2 - 34y2 = -1 had no integer solutions using some recent math finds of mine. That was pointed out to me in a comment to a post which I've yanked. It was a solid comment, very helpful, and now it's gone and I'm worrying to myself, shouldn't I acknowledge the person some kind of way?

Now it has been an issue for years to what extent I should give credit for any such help, and what happened is that for a long time I was arguing with people who were mostly insulting. And the way I see it, if someone is calling you names or insulting your intelligence, while they point out an error, why bother giving them any credit?

So I fell out of the habit.

So here's a post to acknowledge LOTS of help through the years from critics! People who for whatever reason pointed out when I made mistakes. And with my techniques, which involve brainstorming--i.e. often I JUMP to a huge conclusion very quickly as I throw up ideas--that has been often.

Erroneous argument is YANKED, and I'm puzzling as usual how I had some stupid math mistakes. But often it's easy enough to explain--you really WANT to believe something that is just not true. And your mind tells you what you wish to hear.

Thankfully someone pointed out my error, and thanks! And even thanks to all the others through the years, even the ones who were insulting, as you not only helped me, you saved me time from not needing to worry about acknowledging you.

Oh yeah, making mistakes SUCKS!!! I hate it, because then you have to clean up the mess. Here thankfully someone helped me early before I went too far with a mistake as I was getting ready to really get excited with what I thought was a pivotal result.

James Harris

## Wednesday, December 26, 2012

### Disruptive mathematics?

A recent discovery of mine is a modular solution for x2 - Dy2 = F.

With a non-zero integer N for which a residue m exists where--m2 = D mod N, and r, any residue modulo N for which Fr-1 mod N exists then:

2x = r + Fr-1 mod N and 2my = r - Fr-1 mod N

It is easy to derive too. I use (x+my)(x-my) = F mod N, where m2 = D mod N.

But the reduced form of what is called a binary quadratic Diophantine equation is supposedly well studied by mathematicians (I'm not a mathematician) but I haven't seen this simple solution. That's scary.

It's so easy I still puzzle myself about why it took me so long to notice it, but it has implications in some important areas where I hesitate to mention bigger ones, but for instance xy = T is a binary quadratic Diophantine equation.

That is, integer factorization is covered in this area, and can be considered to be simply solving a binary quadratic Diophantine equation. But the supposed difficulty of integer factorization for certain very large numbers is a reason for using it in web encryption. And, for instance, if you go to a site with "https" your computer sends something called a public key, which is worthless as a security system if someone could just quickly factor it.

I've now put a hold on research of mine which I worry might lead to a general way to solve binary quadratic Diophantine equations, and note that I'm not sure I could have achieved it. But mathematicians aren't acknowledging my results! It'd be irresponsible of me to continue further in such a situation.

Most people may not realize that mathematicians see their field as free from disruptions from sweeping changes. Unlike people in just about every other discipline out there they see mathematics as an ever growing mountain of research, where their job is to build on what was done before--not overturn it.

To give an example the way I see it, as I got an undergraduate degree in physics, imagine if a group of people said that human transportation was by horse and buggy, and all we could do is build better and better horse drawn carriages? What if you drove a car by them and they just refused to acknowledge it existed?

And airplanes? Are you kidding? They'd probably laugh and tell you if vehicles ever fly, it will be if horses can first.

That to me is like this situation.

NO one else thinks like mathematicians.

For most people society is advancing, which means that some old ideas get overturned as better ways of doing things are found. But mathematicians pride themselves on the belief that no such thing happens in their field, and they refuse to acknowledge information that shows otherwise.

The breakdown of their worldview, however, in this area could have shattering repercussions for the world. While I am not saying my ideas definitely lead to a general way to solve binary quadratic Diophantine equations, there is enough done already for sensible people to have concerns.

But mathematicians are confident in their worldview and they don't approach these equations in this way!

If they turn out to be wrong, imagine, the news headlines blaring about a sudden collapse of the system used for encryption. Major companies around the world scrambling for some way to figure out how to continue doing business. Big tech giants find their continued existence questioned by pundits. And that's just on the corporate view without considering the country security issues. Or issues of personal security for most people.

And what would mathematicians say?

I suspect they'd be befuddled. Maybe even pathetic at that point. Insistent that these kinds of things just don't happen in their field.

And no one would care then.

I'm not interested in having anything to do with a new financial crisis that could take the world by surprise, but my not continuing research in this area doesn't mean someone else isn't.

But in my experience, I can assure you that mathematicians will calmly inform you that there is no point in worrying at all.

James Harris

## Sunday, December 23, 2012

### Concepts in binary quadratic Diophantine equations

While I am NOT a mathematician I have found myself playing around with intriguing ideas around what are called binary quadratic Diophantine equations and thought it would be a good idea to explain the basics as I know them.

First off binary quadratic Diophantine equations are when you look for integer solutions to equations like:

c1x2 + c2xy + c3y2 = c4 + c5x + c6y

Here x and y are the unknowns to be figured out. An example of such an equation is:

x2 + 2xy + 3y2 = 4 + 5x + 6y

where I've simply used,  c1 = 1, c2 = 2, c3 = 3, c4 = 4, c5 = 5, c6 = 6.

Such equations are also called binary quadratic forms, and in general can be reduced to a more basic binary quadratic form, like:

u2 - Dv2 = C

Here the letters used don't matter, of course, and in this case u and v represent the unknowns.

With my simple example above that reduction to the simpler form using my own research gives:

(-4(x+y) + 10)2 + 2s2 = 166

So to fit into the form above:

u = -4(x+y) + 10 and v = s, while D = -2, and C = 166.

The reason to reduce to a simpler form is to aid in finding solutions. And in this case I can find the answers rather simply.

Subtracting 2s2 from both sides the equation above is:

(-4(x+y) + 10)2 = 166 - 2s2 = 2(83 - s2)

Running through possible odd s's I notice that s=9 works to give -4(x+y) + 10 = 2 or -2, so x+y = 2, or x+y = 3. And x = 4, y = -2, or x = 5, y = -2 work.

The letters do not matter so the general form can also be written as x2 - Dy2 = F.

That form has a modular solution for x and y.

Given x2 - Dy2 = F where all variables are non-zero integers:

With a non-zero integer N for which a residue m exists where--m2 = D mod N, and r, any residue modulo N for which Fr-1 mod N exists then:

2x = r + Fr-1 mod N and 2my = r - Fr-1 mod N

It is derived here.

That result gives solutions to x2 - Dy2 = F mod N.

If modular arithmetic is unfamiliar to you, I have my own short introduction.

And that covers enough of the basic concepts of binary quadratic Diophantine equations to help in understanding my posts on the subject.

James Harris

## Thursday, December 06, 2012

### Surprising modular discovery progression

After years with my own mathematical results I was surprised to come across something which to me is bizarrely simple for the 21st century, as how can this approach be new? But I haven't found it elsewhere yet, while I keep looking. And I'll put it up in a bit, and then walk down the progression from it to a general modular solution.

So I found out recently that a certain well-known equation could be solved modularly:

A modular solution to x2 - Dy2 = 1 is:

2x = r + r-1 mod D-1 and 2y = r -  r-1 mod D-1

Proof:

x2 - Dy2 = 1 mod D-1, and D mod D-1 = 1, so:

x2 - y2 = 1 mod D-1

Which is: (x+y)(x-y) = 1 mod D-1, and letting r be any residue with an inverse modulo D-1:

x+y = r mod D-1 and x-y = r-1 mod D-1, so:

2x = r + r-1 mod D-1 and 2y = r -  r-1 mod D-1

Proof complete.

That one is easy and short so I decided to use more of a traditional format. Math people can get really weird about how you present things. It's their way, or...well it's just THEIR way. And they get years of training at colleges and universities for a particular format. I didn't get that training in their format. I was a physics student. But I digress.

Next step is to generalize beyond modulo D-1.

Given n such that: n2 = D mod p, where p is a prime number, and r, any residue modulo p:

2x = r + r-1 mod p and 2y = n-1(r - r-1) mod p

gives solutions for x2 - Dy2 = 1 mod p.

And THAT can be generalized to modulo N.

But this approach can be taken still further to something even more general.

With x2 - Dy2 = F where all variables are non-zero integers:

Given a non-zero integer N for which a residue m exists where--m2 = D mod N, and r, any residue modulo N for which Fr-1 mod N exists then:

2x = r + Fr-1 mod N and 2my = r - Fr-1 mod N

Which of course gives you the original result, with F = 1 and N = D-1.

Or the modulo p result with F = 1, N = p.

Of course the simple reality I'm exploiting is that in modular arithmetic a non-zero D is always a square!

Proof: Let N = D - m2, or N = m2 - D, for any non-zero integer m, then D is a quadratic residue modulo N.

To me it is a very surprising progression, with simple solutions which can actually solve these equations explicitly if x and y exist less than N, though I'm not suggesting it as a practical technique in these forms.

Why such a simple, basic modular arithmetic concept is not part of what I'm seeing on math websites about these equations is a puzzle to me.

James Harris