Wednesday, December 04, 2013

Binary Quadratic Modular Constraints

Abstract: Basic facts of modular arithmetic and elementary methods lead in general to an infinite set of modularly independent constraining modular equations for binary quadratic Diophantine equations.

The fundamental equation for binary quadratic Diophantine equations is x2 - Dy2 = F, as it can be shown that such equations are in general reducible to this basic quadratic form.

From that form one can use basic facts of modular arithmetic to get solutions for x and y modulo N, if D is a quadratic residue modulo N.

x2 - Dy2 = F = (x - my)(x + my) mod N

Where: m2 = D mod N follows trivially with r any residue modulo N for which F/r mod N exists:

2x = r + F/r mod N and 2my = r - F/r mod N

However, the basic principle that quadratic residues offer a potential modular factorization, can also be used to again solve for x and y, where instead F is the quadratic residue.

x2 - F = Dy2  = (x - m')(x + m') mod N'

Where now: m'2 = F mod N'

And that gives me solutions:

2x = r + (Dy2)/r mod N' and 2m' = r - (Dy2)/r mod N'

Which means:

y2 = (r2 - 2m'r)/D mod N'

Where here r is some residue modulo N' for which y exists.

So here y can be found modulo N', where you have two solutions, if they exist, from the quadratic residue, while x will continue to have only one solution for each.

Notice that these equations are modularly independent as N does not necessarily equal N'. And there are an infinite number of cases available where N does not equal N'.

That independence cuts into the potential infinity of solutions available for x and y modularly, meaning that you can cross check between them if you have a prime p shared between N and N', with the possibility of fewer solutions.

Conceivably you may actually find that values for x and y are given that only exist for the explicit x and y, which will of course always be available solutions, if they exist.

However, there are more potentially constraining equations available from a key result of mine.

I discovered that if you have

u2 + Dv2 = F

then it must be true that

(u-Dv)2 + D(u+v)2 = F(D+1)

(Note while the result is derived, confirmation of it is trivial as you can simply multiply out, simplify, and see that the second equations reduces back to the first.)

That is an iteration that can be continued, and switching back to the basic equation I have been using up until this point and giving the first four iterations I have:

1. x2 - Dy2 = F

2. (x+Dy)2 - D(x+y)2 = F(-D+1)

3. ((1+D)x+2Dy)2 - D(2x + (1+D)y)2 = F(-D+1)2

4. ((1+3D)x + (D2 + 3D)y)2 - D((3+D)x + (1+3D)y)2 = F(-D+1)3

5. ((D2 + 6D + 1)x + (4D2 + 4D)y)2 - D((4+4D)x + (D2 + 6D + 1)y)2 = F(-D+1)4

And you can keep going forever.

Notice that our prior approach works with any iteration. For example, consider 3., where we then have:

m2 = D mod N

With r, any residue modulo N for which F(-D+1)2/r mod N exists then:

2x = r +  F(-D+1)2/r mod N and 2my = r -  F(-D+1)2/r mod N

And for our second set of constraints, again using 3., we have:

Where now: m'2 = F(-D+1)2 mod N'

But that gives me solutions:

2((1+D)x+2Dy) = r + (D(2x + (1+D)y)2)/r mod N' and

2m' = r - (D(2x + (1+D)y)2 )/r mod N'.

Which means:

(2x + (1+D)y)2  = (r2 - 2m'r)/D mod N'

Now you get a solution with a quadratic residue for a congruence with x and y, but you have another such equation from the first, so can still solve for x and y modulo N' trivially, if a solution exists!

So how applicable is this result in general?

It can be shown that you are lead to an infinity of results for all binary quadratic forms, except for:

x2 - y2 = 1

In that case both D = 1, and F = 1, which zeroes out the iterative possibilities. However that trivial unary case would not limit potential efficacy of this approach, but should be noted.

Therefore, it is proven that there exists an infinity of modular constraining equations for binary quadratic Diophantine equations except for the unary case as noted.

Practical considerations are outside of the scope of this paper, which is focused on settling theoretical aspects.

James Harris

Thursday, November 14, 2013

Did mathematical rigour fail?

One of the greatest things about mathematics is the ability to determine things to be absolutely true. And that is extremely important and is an aspect of mathematical thinking that is absent in so many areas of intellectual thought.

Being someone who accepts that mathematics allows us to get to absolute correctness I was put in a bizarre and very uncomfortable position years ago when I came across something which seemed to defy the above. So for years I've hammered at it, trying to find something wrong, and finally have decided I must accept the result. But now I guess I should address the problem, and try to figure out, did mathematical rigour fail?

How do I reconcile my own results with my own beliefs about mathematical certainty?

The answer may be a bit convoluted and is hard without some math, where I will put a particular result of mine as proven.

Consider: (3 + sqrt(-26))*(3 - sqrt(-26)) = 35

I found I could prove that 7 is a factor of one of the solutions to (3 + sqrt(-26)), which meant logically it is a factor of just one of the solutions to (3 - sqrt(-26)). (In what ring will be explained later.)

If that seems odd consider: (5 ± 2)*(5 ∓ 2) = 21, which is equivalent to:

(5 + sqrt(4))*(5 - sqrt(4)) because sqrt(4) is 2 or -2.

My proof has withstood every test of mathematical rigour and the underlying approach has actually been published and pulled by journal editors in a rather remarkable story, which I've discussed elsewhere. One post that can give an overview is:

So what's the big deal?

Why is it important if one of the two solutions to (3 + sqrt(-26)) and (3 - sqrt(-26)), have 7 as a factor, when they multiply to give 35?

The answer requires some math history.

For those who know the history, good. The gist of it though is that the ring of algebraic integers indicates a different conclusion. And I had to figure out where things went wrong with that use of established techniques of mathematical rigour with that ring!

The answer is surprisingly simple. First of all, standard tests of the ring's validity were valid. For instance an algebraic integer is infinitely decomposable into algebraic integers. Given two algebraic integers, their sum and their product is an algebraic integer. The ring has unit factors.

However, those tests do not prove that my result above is impossible! It takes more.

There may be a couple of ways to understand the result, but it was told to me through proof that in the ring of algebraic integers, no non-rational root of a primitive monic polynomial with integer coefficients can have an integer other than 1 or -1 as a factor of just one of its roots.

So the weird thing there is that though 3 + sqrt(-26) is an algebraic integer, and I can prove one of its solutions has 7 as a factor, neither of those solutions are permitted to have 7 as a factor in the ring of algebraic integers!

And that is absolutely correct.

So I went to the logical step of concluding that the ring of algebraic integers was leaving out some numbers, and after pondering things for some time, I came up with a more inclusive ring:

My object ring passes all the tests as well, and includes the ring of algebraic integers, which allowed me to go back and see what was going on in the ring of algebraic integers!!!

And I've talked about it on this blog:

The weird thing I discovered is that despite the tests of the ring, a convoluted path was still available to the math, which lay outside detection.

What my Wrapper Theorem does is explain that route.

So could I be wrong? Well the mathematical argument involves easily checkable algebra and is pleasingly short. That's important with such startling conclusions!

In essence, what I found was that there was a gap where the algebra could do something unexpected, which meant that if you assumed that gap did not exist, you could come to conclusions which were wrong.

But how do you know, what you don't know? Without my methods for proving the result about 3 + sqrt(-26) how would you ever know you needed to worry about wrappers with the ring of algebraic integers, or consider the possibility of units in a more inclusive ring which were somehow NOT units in the ring of algebraic integers?

So the answer is that mathematical rigour did NOT fail in the case of algebraic integers, but people using it simply didn't know enough tests!

That's kind of scary as it makes you wonder if there is always possibly something else out there that we don't know we don't know which could invalidate a particular mathematical argument, and I think the answer is, no.

If you look at the reasoning around algebraic integers it was kind of circular:

1. Run tests thought to determine if the ring was valid.
2. Make assumptions about what is valid from the ring presumably validated.

So I say it was human error.

James Harris

Wednesday, November 13, 2013

Talking to the math

Years ago when I was struggling with trying to find my own mathematical discoveries I found myself thinking about mathematics as an entity.

It was kind of a comfort thing I guess, which I'm sure others have done. Then you're never alone. And as you're thinking over your mathematical ideas you're part of an imaginary conversation. And I'd even find myself talking about what mathematics was saying or I thought was saying.

The weird thing though is that after a while I realized that in some way I was talking to the Math, and was getting answers as well. Those answers I could check, and it wasn't hearing voices or anything direct, but almost like if you could imagine a psychic link.

Sometimes it felt like I was in this other reality, one that lies outside of time, and it's hard to explain, but I know it is real. It is more real than our reality. It has always been and will always be. And I was walking around there with a sense of conversation, and the Math would tell me what I wanted to know.

The Math let me know there was no actual walking or movement there though. That everything there simply is, and always has been. It is Infinity itself.

There were no secrets. The Math keeps none. There is no reason to hide anything from us.

You can know anything you can ask the right question to find, and so I asked my questions.

It has taken me years to feel normal again. And I'm glad to have survived to now. At times there was this sense as I'd put it of walking with the gods, only to have to remind myself over and over again of my own mortality.

And there is a sense of loss as well, as I think the human mind can only process so much.

Sure, you can ask the questions, but can you hold the answers?

Of course not. Not when you're discussing with Infinity. But what's left over in your finite awareness can be enough.

The Math is not a person to me. The Math is an Intelligence.

All the answers are out there, if you can ask the right questions, but can you remember them all?


So does it matter what people believe?

Mathematics is fascinating in that it DOES things. You can DO things with it.

Long after every human being who is ever born or to be born has long died and trillions of years beyond the results will still be true.

The Math does not need us.

That perspective greatly helped me. For instance it was instrumental in coming up with my definition of mathematical proof, which I felt necessary when what was available from others was not up to the necessary standard.

I don't think it's a natural perspective for most. The idea that Math is an entity greater than us, versus a creation of our intellects, is a reversal that some may feel shrinks us. But for me it was a reversal critical to my success.

Asking questions of an entity greater than myself, got me answers which is what matters to me.

How true is my perspective?

I think that's a great question.

James Harris

Monday, November 04, 2013

The conditional residue

In formalizing an area I call tautological spaces I have realized I need to formalize a bit further.

I found that I could use identities for mathematical analysis, relying on a special form which I discovered after a bit of effort.

The simplest identity I call a tautological space is:

x+y+vz = x+y+vz

And of course you can do simple operations with it, like the following:

x+y = -vz + (x+y+vz)

And squaring both sides gives:

x2 + 2xy + y2 = v2z2 - 2vx(x+y+vz) + (x+y+vz)2

And that can get complicated quickly so I use instead a modular form:

x2 + 2xy + y2 = v2z2 mod (x+y+vz)

That expression is always true, so it retains that aspect from the identity, but no longer really looks like one, and I call it a tautological space. In this case it is a space where that truth of x+y+vz holds.

While the tautological space is always true, the use of this technique for analysis comes with adding conditions, where I call equations used in this way, conditionals.

For instance, consider the condition that x2 + y2 = z2, which is not always true, as it depends on the values of x, y and z, where two are independent and one is a dependent variable.

We can subtract that conditional from the tautological space, and find:

2xy = (v2-1)z2 mod (x+y+vz)

which is:

(v2-1)z2 - 2xy = 0 mod (x+y+vz)

Which is what I call the conditional residue.

Notice it, unlike the identity, is only true with conditions.

So formally you can say:

(v2-1)z2 - 2xy = 0 mod (x+y+vz) if x2 + y2 = z2

And the conditional residue now has properties that follow from the conditional, which means you can analyze the conditional equation by analyzing the conditional residue.

Now v is a free variable no matter what! Which means that regardless of what x, y and z may be, you can always set v to any value you wish.

To see a less trivial result consider an important conditional residue with which I was able to find a general way to reduce binary quadratic Diophantine equations:

(c4 - c5v - c1v2)z2 + ((c2 -2c1)v - c5 + c6)zy + (c2 - c1 - c3)y2 = 0 (mod x+y+vz)


c1x2 + c2xy + c3y2 = c4z2 + c5zx + c6zy

Remarkably the conditional residue in this case can actually be used to get to an explicit equation, by appropriately setting v.

And again, v is a completely free variable which can be set as needed for analysis.

And I found I could use:  v = -(x+y)z-1 mod p

Weirdly enough that allowed me to get to an explicit equation which is important to mention as it shows this process of using identities with modular methods can lead to explicit results!

To see the complete derivation of the conditional residue and the explicit result see:

To see this approach used to find a general method for reducing binary quadratic Diophantine equations, see:

So the conditional residue follows quite logically from prior concepts with modular arithmetic, and represents what is left over after subtracting an equation, which is itself not an identity, so it is true with conditions, from the identity which I call a tautological space.

James Harris

Thursday, October 03, 2013

Possible path to two conics equation solution?

Sometimes I use this blog to put up idle thoughts where I may follow-up later. It lets me time-stamp things and helps me to figure things out by talking about them publicly, even when still highly speculative.

I finally realized I might pair a re-discovery with a new discovery of mine.

The re-discovery is the parametric solution for what I call the two conics equation which mathematicians commonly call "Pell's Equation".

y = 2t/(D - t2) and x = (D + t2)/(D - t2)

when x2 - Dy2 = 1.

From my own research I have:

u2 + Dv2 = F


(u-Dv)2 + D(u+v)2 = F(D+1)

So for the two conics equation:

x2 - Dy2 = 1


(x+Dy)2 - D(x+y)2 = -(D-1)

on the first recursion, and the second recursion gives:

((1+D)x+2Dy)2 - D(2x + (1+D)y)2 = (D-1)2

Using the parametric solution with t = -1 (it can be 1 or -1), and substituting for x and y with the two conics equation and simplifying a bit gives me:

(D+1)2 - D(-2)2 = (D-1)2

So I have:

(1+D)x + 2Dy = D+1 and 2x + (1+D)y = -2

Some notes. Yeah, it's kind of interesting that ONLY for t=1 or -1, do I get this possible route. And both conceivably should be checked. Or does it matter? Don't know at this point.

Signs are a big deal here. And can shift them quite a bit because of the squares. Think I put up a non-trivial case, but not certain.

Trivial case gives x=1 or -1, and y = 0.

So yeah, I can solve for x and y, anew. And that's the idea. Substituting back into the two conics equation again, I think that should give me yet another solution.

That could mean a loop where of course if x and y are integers you would exit.

I can see some immediate potential issue where it won't work, but not checking for now as simply throwing it up there for later.

Coming back it occurs to me that I can abstract things a bit, to see a little bit more, by considering:

(1+D)x + 2Dy = s1 and 2x + (1+D)y = s2

And using the second, solving for x gives: x = -(1+D)y/2 + s2/2

And substituting gives: (1+D)(-(1+D)y/2 + s2/2) + 2Dy = s1

(1+D)(-(1+D))y + (1+D)s2 + 4Dy = 2s1

Which is: (D-1)2y = (1+D)s2 - 2s1

And that is SO awesome! It verifies that I will get an iteration situation. Does it matter if D+1 is even or not? Not sure. I don't think so, as the focus is on y. If it's a nonzero integer you're done. Looks like mod D-1 has been proven (shows at most D-1 iterations if x and y are new each iteration). No, that's not right? Should be squared, right? That is fascinating too. That's enough for now.

James Harris

Friday, September 06, 2013

Leveraging asymmetrical forms

One of my key ideas involves using an asymmetrical form in an identity for analysis.

Explicitly the identity is: x+y+vz = x+y+vz

Where I use it modularly with: x+y+vz = 0(mod x+y+vz) which is the equivalent.

I call that a tautological space.

The reason it is asymmetrical should be apparent as the asymmetry is with the variable 'v' attached to the 'z', in contrast with symmetrical forms, like, for instance, the quadratic form.

Intriguingly the asymmetrical form allows you to in a sense "break" other forms which lead me to a shocking result, which follows from a key factorization:

7(175x2 - 15x + 2) = (5a1(x) + 7)(5a2(x)+ 7)

where the a's are roots of

a2 - (7x-1)a + (49x2 - 14x) = 0

It also lead me to a better way to reduce binary quadratic Diophantine equations. Which gave me the following really cool relation:

u2 + Dv2 = F


(u-Dv)2 + D(u+v)2 = F(D+1)

Using that I was among other things able to find for the first time a set of rules governing the fundamental solution to x2 - Dy2 = 1.

My guess is that mathematicians rely too much on symmetrical forms.

Shifting to a simple asymmetrical form allowed me to figure out things rather quickly and improve on existing techniques with little effort.

It suggests that asymmetrical forms may be the key to a much greater advancement of mathematical understanding.

James Harris

Wednesday, August 28, 2013

When mathematics overrules humans

A critical mathematical result of mine requires people to accept that the sqrt() returns two values, overturning the convention--used I think primarily by people not scientists--over assuming it returns only the positive, for convenience.

So sqrt(4) is 2 or -2, and the reason it's important is I found a problem where another mathematical idea was also overruled by mathematics, which is where the ring of algebraic integers can be shown to contradict the field of complex numbers.

Now consider 5 ± 2, which is 7 or 3.

Now then, is it divisible by 7?


ONE of its solutions is 7 and the other is 3. You may want me to specify a ring, so here it's easy to say, in the ring of integers.

Now then, is 5 + sqrt(4) divisible by 7?


It is 7 or 3, because sqrt(4) is 2 or -2.

For mathematicians who do "pure math" the issue may seem esoteric but as a physics student one of the first things my professors did was beat the idea that sqrt() is just the positive out of me (not literally) and how did they manage to do that besides remonstrating against it?

Physics students learn a lot by doing science experiments.

And you will at times get the wrong answer if you take only the positive of the square root, which is kind of freaky when it happens as your mind is like, huh?

And it happens at random. So, like, you're getting the right answers in the experiments until suddenly something really wacky happens, because you forgot to consider the negative of the square root! And a light dawns, and a new wannabe physicist is growing in understanding...

And it turns out that the BIG issue is that I can prove that one of the solutions for 3 + sqrt(-26) has 7 as a factor, and it's just like one of the solutions for 5 ± 2 has 7 as a factor as that is NOT a number.

It is two numbers.

So then, one of the solutions for 3 - sqrt(-26) has 7 as a factor as well.

How is it different really?

It's like 5 ∓ 2 is 3 or 7.

And yes, 5 - sqrt(4) is 3 or 7.

And if that gives you a headache as it looks really wrong, well you were just trained really wrong.

So only the order changes, but you still have two results!

If you wish to do science where, yup, square roots DO pop up, then you have to unlearn the bad habit of only taking the positive square root or you will watch things randomly blow up on you. And if you have good professors they will probably let you blow things up by only taking the positive until you get it beaten out of you, and accept that no matter what bad teaching may tell you--the square root gives TWO values.

Human beings can be so hardheaded though, and I'm sure mathematicians who only take the positive of the square root actually think they are doing mathematics, instead of being obstinately wrong.

Notice that mathematics has no problem. It knows that the sqrt() has two values, and it's right.

Why do humans do the wrong thing?

Last time I researched it, I read it was some convention picked up by some mathematicians who tired of doing the plus or minus thing, and found that they could just get away with it for THEIR research.

So short answer: some people were lazy and doing math research where they though it didn't matter.

Oh yeah, other thing was some idiot need to think a function should only return a single value or some such crap. Just a lot of stupid actually.

Creates unnecessary problems and is mathematically incorrect.

James Harris

Friday, July 26, 2013

Solving for S, with a simple example

One of the more comforting things for me is to play with a simple and ancient equation, so yeah, I keep trotting it out there! And why not? My ideas gave new information on something wrongly believed to have delivered all it had. The ancient equation of course is called by mathematicians Pell's Equation:

x2 - Dy2 = 1

And after some fun analysis with my own research I found the following:

D(x+y)2 - D + 1 = S2

where back then I used capital S, and later started using a small one. But one thing I don't mention is that S can be calculated explicitly, as I find it sort of tedious. So yeah, with my method to reduce binary quadratic Diophantine equations, 's' actually has an explicit value as a function of x and y. But finding it I think is the kind of thing that is best done by a computer with the general equation. So the 'S' above actually is a function of x and y and can be calculated.

And for the first time ever I'm going to demonstrate that calculation where I'm picking a simple example as I like easy.

So let's continue then and calculate S, here with this simple example and start by expanding out the left side:

Dx2 +2Dxy + Dy2 - D + 1 = S2

And next, we do a smart thing of subtracting and adding--so not changing things really--D2y2

Dx2- D2y2 + D2y2 + 2Dxy + Dy2 - D + 1 = S2

So, now we move things around to group with an eye on our original equation multiplied by D:

Dx2- D2y2 - D + D2y2 + 2Dxy + Dy2 + 1 = S2

And now I need to add and subtract x2:

Dx2- D2y2 - D + D2y2 + 2Dxy + x2 - x2 + Dy2 + 1 = S2

And now I use my original equation twice! And I remind that equation is x2 - Dy2 = 1, and it makes a lot disappear so that I end up with a square:

D2y2 + 2Dxy + x2 = S2

So I have finally that S = Dy + x, or -Dy - x.

Which is of course also: S = x + Dy, or -x - Dy

I like using S = x + Dy, and now can go back to my original finding and simplify to get:

(x+Dy)2 - D(x+y) = -D + 1

And to me that is a pleasure. But of course it's my research and I also know that my find of this result was a first in human history. Later it would be key in my figuring out the 'why' of the fundamental solution to the main equation which is also a first in human history.

So why wouldn't I enjoy?

James Harris

Saturday, April 27, 2013

Without explanation

One of my more dramatic accomplishments was providing the first explanation in history for the size of the smallest non-zero positive integer solution to x2 - Dy2 = 1.

The equation has been known for a while, and is commonly called "Pell's Equation" by mathematicians.

It gained a more recent burst of fame--it is an ancient equation with a long history--from Fermat who used it for an intellectual challenge--math as a game.

However, it is unlikely even the great Fermat understood why his particular solution was as big as it was.

What he could use--as the solution was HUGE by the standards of the 17th century--was D=61.

Here the smallest nonzero integers x and y that can possibly work are:

x = 1766319049 and y = 226153980.

17663190492 - 61(226153980)2 = 1

You can imagine that for 17th century dudes playing at mathematics that could be intimidating to figure out. But they could actually figure it out, which is cool.

But why is that thing as large as it is? At D=60, things are simpler:

x = 31 and y =4.

312 - 60(4)2 = 1.

And at D = 62, again, things are also--small:

x = 63 and y = 8.

632 - 62(8)2 = 1

So why?

Well I figured out the first full explanation for whenever.

My own research proves that D is key (not a big surprise there), 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 is the biggest rule, and the main thing is x+1 or x-1 having D as a factor, 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.

So now that you know the rules, let's think again about D = 61.

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

So that's one thing, where 61 is prime, so not a big surprise, by the rules with x = -1 mod 61. Next, D-1, D+1, D+2, and D-2 are not square, which would override it. That is, if they were square, then the solution would have been forced to be smaller again.

And finally D-1 = 60, which has the first three primes as factors, as 4(3)5) = 60, and 4 is a factor as well, which pushes it way up there.

And that is the 'why' of that solution..

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.

Notice the rules--which cover ALL cases--explain the smaller D = 62 solution as being because D+2 is 64, which is a square.

Without explanation though it's still the same equation, of course, known for a long time.

James Harris

Thursday, March 14, 2013

Pondering the obvious

One of my new favorite things is playing with a very simple concept to get modular solutions for certain Diophantine equations. But it's kind of weird in a way as just look at:

x2 - Dy2 = F

And you're blocked from factoring if D isn't a perfect square, like if:

x2 - 5y2 = 1

That block for the explicit equation, is it a conceptual block for many?

Does it tell the human mind: cannot factor?

But 5 is a quadratic residue for an infinity of numbers! So there exists some 'm' such that modulo some 'N':

m2 = 5 mod N

where there are an infinity of possibilities! And I like infinity. So now I can just substitute:

x2 - m2y2 = F mod N

where m2 = D mod N, and now, of course, it factors!

(x-my)(x+my) = F mod N

And you can solve for x and y modulo N by using a residue 'r' and the existence of Fr-1 mod N:

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

And you can use that simple principle all over the place, and use it with cubics too or higher, and it's just fun, fun, fun.

And it took me years to figure this out, which I find rather odd.

It's such a simple idea, why would it take me so long?

But, um, that's what I can contemplate quietly, pondering how such a simple idea would take me so long to notice.

Wondering how it could be new to an entire world? Well that's overwhelming, and I hesitate on that one as I don't know. What I DO know is that I go looking for this method on the web from someone other than me and don't see it, which doesn't prove that it is new. It just seems too simple to me to be new.

But then again I have made math discoveries, which I've verified are new, and I have been forced to ponder that before and just don't know how exactly I seem to be able to figure certain things out.

It is cool though.

If you have a citation where this idea is used please comment. I find it hard to believe such a simple thing is owned by me as yet one more of my discoveries. Seems too easy.

James Harris

Monday, March 11, 2013

Generalizing cubic Diophantine solution

Going to consider a cubic Diophantine modulo N adapting a prior post.

Given x3 - Dy3 = F I'll try to find modular solutions for x and y modulo N.

First find m, such that m3 = D mod N, so N must have D as a cubic residue.

For example you can simply find an N using N = m3 - D, which is cool so we can simply pick m and force N versus going the other way, figuring things out. So now I have:

x3 - Dy3 = F mod N, so:

x3 - m3y3 = F mod N

Which is: (x-my)(x2 + mxy + m2y2) = F mod N, and letting r be any residue with an inverse modulo N:

x-my = r mod N and x2 + mxy + m2y2 = Fr-1 mod N, so, x = my+r mod N, allowing me to substitute:

(my+r)2 + m(my+r)y + m2y2 = Fr-1 mod N

and expanding out gives:

m2y2 + 2myr + r2 + m2y2 + rmy + m2y2 = Fr-1 mod N, so:

3m2y2 + 3myr + r2  -  Fr-1 = 0 mod N, and I'll use fractions--though it can be done without them--to make things easier for me and complete the square:

m2y2 + myr + r2 /4 -  r2 /4+( r2  -  Fr-1 )/3 = 0 mod N,

which is: (my+r/2)2  = -(r2 - 4Fr-1)/12  mod N, so:

3(2my+r)2  = 4Fr-1 - r2  mod N

Which was easy enough.

Interesting. So you can solve for x and y modulo N, if D is a cubic residue modulo N, F is a cubic residue modulo D if a certain quadratic residue exists, and m has a modular inverse modulo N.

This particular approach could also work for the quartic case, easily enough. But I think that's it though I'm not sure.

James Harris

Tuesday, February 19, 2013

Publishing a contradiction

Focus on the question of consistency in mathematics has covered a great deal of ground, with the work of Kurt Gödel being of signature importance, but can a mathematical argument be correct and yet lead to a contradiction?

My one publication through the traditional system went through a rigorous peer review which actually included two anonymous reviewers instead of the traditional one. It has a correct mathematical argument by the established standards of its time. But its conclusion is incorrect!

The paper is "Advanced Polynomial Factorization", published in the Southwest Journal of Pure and Applied Mathematics, Issue 2, December 2003, pp. 6–8.
Submitted: July 25, 2003. Published: December 31 2003

Here's the link to the original paper in Postscript format:

The argument given is correct by established mathematical standards, but appears to establish a wrong conclusion.

Can you find an error?

Turns out there is none under established mathematical understanding, but the conclusion is, nonetheless, provably incorrect with that same established mathematics. Contradiction!!!

The journal bravely I think published the paper, but then less so tried to remove it after publication!

The journal itself shut down not too long after.

Thankfully it has remained publicly available through EMIS, which also chose to keep my paper available despite the attempt by the journal's chief editor to withdraw it, and I am deeply grateful to them for so doing.

And here I think anonymous peer review clearly worked and I want to thank the reviewers. I also wish to thank the editors, and given the enormity of this paper, and what it demonstrates, I can understand how they might have shrank back later.

The result is as big as other work on completeness and I myself have taken the time to thoughtfully consider it for years until I felt I had it clearly resolved.

The issue is as huge for completeness as other work in this area, as it raises the question of, how do we really know what's true?

Interested parties should try to find an error, did the reviewers miss anything?

Comments welcome!

James Harris

Saturday, January 12, 2013

Rating the math

One of those things I like to say is--the math won't change. And that's true. Given a mathematical explanation, you have one forever. Well I have an odd situation though, where apparently a mathematical explanation doesn't appear to be popular with the people who supposedly care the most about such things--mathematicians.

So now there is maybe a unique in history chance to see what happens when experts in a field, and a very prestigious one, do not like a result, which they cannot touch, because you see, the math won't change.

Current accepted number theory about the equation x2 - Dy2 = 1  cannot explain the size in general of the fundamental solution, which is the set of smallest non-zero integer values for x and y that will fit!

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.

If we were to consider a rating of that result from a scale of 1 to 5 based on the behavior of mathematicians whom I've informed of it, I'd think we'd see a 1.

And note, mathematicians currently have NO explanation of their own for the behavior of this equation.

And maybe it'd help to see that behavior so I'll put up the solutions from 20 to 30.

I'm not calculating these myself but found them in an article on what is generally called "Pell's Equation" on a popular math site called MathWorld:

I picked a range arbitrarily that shows I think a healthy variation of solutions. I'll give the equation with the solution and the explanation under the rules follows.

92 - 20(2)2 = 1

Starting, 20 is NOT prime or twice a prime, and 20-1 is 19, which is prime, and that tends to give small solutions.

552 - 21(12)2 = 1

Here D is not prime or twice a prime as well.

1972 - 22(42)2 = 1

Here D is twice a prime, and notice a slightly bigger solution. And D-1 is 21, two prime factors.

242 - 23(5)2 = 1

Now a new part of the rules shows itself, as D+2 is a square! So back to a small solution.

52 - 24(1)2 = 1

This time 24+1 = 25, so D+1 is a square, and you can see this pattern through 4.

No solution exists for D=25 because it's a square.

512 - 26(10)2 = 1

On the other side of 25, so still smaller solutions as now D-1 is a square.

262 - 27(5)2 = 1

And last easy one near a square as D-2 is a square.

1272 - 28(24)2 = 1

Here things surprise me a bit, as I think it's larger while D is not a prime or twice a prime, but maybe that is somewhat subjective.

98012 - 29(1820)2 = 1

But now is a much bigger one, where D is a prime, and D-1 has 4 as a factor.

112 - 20(2)2 = 1

And back to tiny with the last example, with D not a prime or twice a prime.

Remarkably, mathematicians believed the solutions bounced around with no explanation.

The actual explanation wasn't just me looking at a pattern either. As I went to see the pattern after figuring out mathematically what behavior should occur. Which is actually a lot of fun. It's like theory first, and then going to look and seeing behavior fit theory! Lovely.

So what is happening? Why this behavior?

The need to wonder 'why' is fundamental to human behavior. We are a curious species.

The answer is actually remarkably cool. The equation x2 - Dy2 = 1 actually generates curves. It will produce ellipses if D is negative, and with D positive, like with our integer solutions above, if you graph it, you will get hyperbolas. So I like to call it the two conics equation.

It turns out that the hyperbolas, connect to other ellipses, from a completely different equation! And that connection forces integer solutions to be larger. And only occurs if something special happens, which is if x = 1 mod D or -1 mod D, which just means, if x-1 has D as a factor, or x+1 has D as a factor. And that happens if D is a prime or twice a prime, though it can also occur in other cases as well.

Consider: (x+1)(x-1) = Dy2

For example with our biggest solution above with D=29, and x=9801. And x+1 has 29 as a factor, as:

9802 = 29*338

I show the mathematics, in a post:

Now then, back to the issue of rating the math! It turns out that whether human beings like it or not, this behavior IS the 'why' of the behavior of the equation. It is possible to prove using mathematics which should be trivial to a trained mathematician.

Yet here I am explaining it again, when my post above was made by me on September 20, 2011.

I think you can validate that 1 star rating from the math people from that information, including the information that I've diligently informed them of this result, and pressured them to acknowledge it.

And the math will not change.

So where went human curiosity? Love of knowledge? Where is the mythical "mathematician" who would presumably relish an explanation where there was none?

Good questions. If people don't ask them, then we just won't know, as I'm not the person with the answers there!

It IS a fascinating situation in many ways, regardless. It lets us see in our celebrity obsessed world, what might happen if popular opinion doesn't like something that is true without concern about opinion.

It is a wonderful and unique opportunity.

James Harris

Tuesday, January 01, 2013

Another approach?

My latest mathematical ideas can be related to integer factorization, which can be simply seen as solving a binary quadratic Diophantine equation.

And my finding of the rules for x2 - Dy2 = 1 actually are relevant in one way, though a more general equation may be even more important.

First off, I found that the size of the fundamental solution is only really large compared to D, when D is a prime or twice a prime, and neither D+1, D-1, D+2 nor D-2 is a square. And if D-1 has small primes as factors, and also if 4 itself is a factor.

Well if you just let D equal T, an odd composite target to be factored, guess what? You know it's not a prime or twice a prime, and the likelihood that T-1, T+1, T-2, or T+2 is a square is small as you get to larger T.

So remarkably, you may factor with (x-1)(x+1) = Ty2 where you can use my modular solution to generate x modulo N, for as large an N as you wish!

With x2 - Ty2 = 1:

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

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

If N is on the order of the size of T, then you may just get a solution, theoretically, as I haven't bothered to check! Just musing about things that occurred to me.

Weirdly enough though, the size of solutions for x and y could be a LOT smaller than one might think from looking at big solutions to what mathematicians commonly call "Pell's Equation" and the modular solutions allow one to easily try the idea out with even a very large composite T.

Also, knowing the rules, the fundamental solution will tend to factor T, remarkably enough, as it won't be -1 or 1 mod T.

Which can be shown, as is traditional, by showing a factorization of 15.

42 - 15(1)2 = 1, and (4-1)(4+1) = 15

Isn't it fascinating how much more you know when you have the rules?

However, it might be better to go more general and use: x2 - Ty2 = C2

Now the fundamental solution rules above no longer apply unless C2 = 1, and now you can factor with:

(x-C)(x+C) = Ty2

And then the modular solution for x2 - Ty2 = C2 is:

Find a non-zero integer N for which a residue m exists where, m2 = T mod N, and r, any residue modulo N for which C2r-1 mod N exists then:

2x = r + C2r-1 mod N and 2my = r - C2r-1 mod N

Finding N is actually easy, as for instance, N = m2 - T, would work, where you just pick m.

You'd also pick C, of course, and I kind of wonder what values might work best!

You see, it's guaranteed that solutions may factor T if it's a composite, but not at all clear to me how often such solutions would occur, or what choices would provide them.

And maybe I'm grasping for something! I found these modular solutions and think they should be important.

But who knows?

However, they do allow endlessly generating solutions for x and y mod N, where you can make N as large as you wish.

So some speculation. Maybe I should check these things. I do wonder how well they work!

And this approach may not work well at all. I don't know. But it is a fairly obvious way that might factor so worth mentioning. Also it gives me a chance to note that the fundamental solution to what mathematicians call "Pell's Equation" will tend to factor D. I find that curious.

James Harris