## Wednesday, December 31, 2014

### Year in Review 2014: Sharing certainty

Nothing like thinking you found something really interesting and then finding out you've been suckered, misled or otherwise lead astray, and for the first time this year I focused a serious amount of effort with addressing reasonable concerns people might have when they come across some math of mine. And managed to cap the year off with really the ultimate in shutting up "haters" as they are so colloquially called which is an easily checkable original result for the ages:

x2 + (m-1)y2 = mn+1

Has non-zero integer solutions for x and y, for m = 3 or higher, and n = 0, or higher, and I have m raised to n+1 so that n is a count of iterations, if you want m raised to the nth power you just start n at 1 instead of starting at 0.

For example, with m = 5, and n = 6, x = 29 and y = 139 is a solution:

(29)2 + 4(139)2 = 78125 = 57

Which is the answer to a challenge I gave as I look at better ways of sharing.

Of course the 4 can be pulled into the square so you have a sum of squares:

(29)2 + (278)2 = 78125 = 57

What an odd thing to be able to say: I found that.

The necessity of answering reasonable concerns has helped me find the fun as well. As some people LOVE that "gotcha" thing where you find something you think is cool and they toss something back at you to try and make you look like an idiot or a sucker. Critics abound.

And I finally addressed one of the biggest insults tossed my way, which is the horror, horror, horror that I once worked on trying to solve Fermat's Last Theorem!!! And yeah, I'll admit I find it fascinating that certain math people feel so calm about that as an insult as if it is one of the ultimate terrible things that you can do! It's weird. And I haven't worked at it in over a decade, but that one seems to be so useful that people keep using it. So yeah, before you talk to anyone about my research you need to know that one, so when they throw it at you, you can say, oh yeah, know about that....

To me though the most important thing was sharing an example of a mathematical proof that demonstrates how you can know absolutely that it is correct. That to me is actually a lot more important than the social stuff, as it tells you how to be confident regardless of what people believe. As people can be supremely confident in completely wrong stuff.

Probably ended the year out strongest by the more spectacular, as giving an original basic research result that also can answer and expand upon Euler and Ramanujan is about as big as it gets. But back in March with my second post of the year I talked about my mainstream concerns. Which also explained the name change of this blog.

It may seem strange how often I make it clear I'm not a mathematician, but I think that is important. And each person can consider their own personal beliefs about what kind of person can make an important mathematical discovery. I suggest that the math doesn't care.

Our beliefs can be so important though, which is why I accepted the responsibility of sharing certainty this year. It's not enough for me to have all the fun. And it's not fair for me to leave others defenseless to people trying to mislead them about my work.

Sharing certainty really was the theme for the year, and that is also about sharing the joy of discovery, with defense against those people who will try to ruin it for you. And I think that's a good thing and a responsibility I should take seriously. This year I did.

So what about next year? Well, we'll see.

James Harris

## Monday, December 29, 2014

### Some open problems

Recent efforts have convinced me that giving people some challenging problems might be helpful both in advancing knowledge of my research as well as getting some questions about which I'm curious answered.

1. To what extent does what I call the Binary Quadratic Diophantine iterator explain the behavior of Mersenne numbers?

It is quite well-known that Mersenne primes are rather rare, and considerable effort is being expended in finding them. But can we use the BQD iterator to explain why they are so rare?

2. How well does what I call the Prime Residue Axiom predict the behavior of actual numbers?

The PRA examples I give are with twin primes, but if it is an axiom then it gives predictions for indefinite sized gaps. While such evidence might not be convincing for those doubting the axiomatic quality they would definitely be compelling. While a predictive failure would force me to answer or accept that the notion fails.

3. Does numeric integration of the partial differential equation that follows from my prime counting function agree with the predictions of other mathematics?

The prime counting function I found is distinctive in that it recurses, and counts primes by calling itself.

That means it leads to a difference equation which is still Diophantine. But that leads to a partial differential equation which is not. So yeah, I do have research which is not just about integers! This one is good to leave as the last as it relates to the Riemann Hypothesis which I tend to very deliberately avoid. To me RH is just too emotional an area for mathematicians so I don't think they're rational about it. And I'm not a mathematician so I don't really care about it anyway. So I leave everything open for others as I think it worthless for me to bother.

Gave links I think related to the questions in each though there are more blog posts around each. I have fiddled around with lots of things as the spirit moved me.

And those are some open research areas.

And that's good for a start! Just had an impulse to write a post like this one and see how it goes, so quickly came up with some things that occur to me.

I actually have quite a lot of open research areas. May as well talk some of them out. These are three that might intrigue others, I'm guessing.

If this way of doing things works well maybe I'll toss some others up.

James Harris

## Saturday, December 27, 2014

### Celebrating simple math

People naturally have preferences, and preferences can shift what one sees--or can see. Then certain things can simply fall between the cracks. Which is why I'm proud to celebrate simple math. Past mathematicians had the thrill of these types of results, but so can we.

And that pleasure can be had by anyone including those NOT mathematicians, as I'm not one.

Maybe that's why I can find things that well-trained modern mathematicians I think have been indoctrinated to think do not exist. There's some notion that all the good simple discoveries were already discovered long ago.

Like recently I had the pleasure of offering a challenge based on a simple solution, giving those who like to play with numbers a chance to learn something not known to greats like Euler and Gauss, or was it?

The other thing you get from a challenge presumably is people digging into the archives, and I'm curious! I want to know. Will not hurt my feelings if Gauss and Euler knew of it. As I still re-discovered in that case meaning I get that precious thrill that few ever get to experience.

And discovering an infinity result with THIS level of simplicity offers that rare thrill that few human beings will ever get in the history of our species.

And I'm proud to dedicate this result to my parents, even though they think this math stuff is nuts. And once it's verified to be unique to me--if it is--to my country. My story is so much, only in America. But that's getting ahead of things. Someone may have had this thing before. Will take time to see if it's an original result.

The other thing you get with an excitement over just playing with numbers is that enthusiastic need to show over and over again, so here's another simple list, which follows from the general result.

Let's let m = 17, and u = v = 1. Then, the BQD iterator is:

(u - 16v)2 + 16(u + v)2 = 17*F

Start is:

12 + 16*12 = 17

then it must also be true that

(-15)2 + 16(2)2 = 289 = 172

Next iteration: (-47)2 + 16(-13)2 = 4913 = 173

And third iteration: (161)2 + 16(-60)2 = 83521 = 174

Fourth iteration: (1121)2 + 16(101)2 = 1419857 = 175

Fifth iteration: (-495)2 + 16(1222)2 = 24137569 = 176

Sixth iteration: (-20047)2 + 16(727)2 = 410338673 =  177

---------------------------------------------------

And in general: x2 + 16y2 = 17n+1

Or you could just say: x2 + y2 = 17n

If you were going for the ultimate in pretty simplicity. Where x and y will always exist, where they are nonzero integers, where we know y will always be even, and n is a positive integer.

Interesting to look at this thing with somewhat bigger numbers. I do wonder if Gauss or Euler or anyone else for that matter besides myself figured this thing out before me. My suspicion is evidence strongly suggests they didn't, as I've used this result to explain and extend a result from Euler and from Ramanujan. It also gives a new way to analyze Mersenne numbers. If they had it, they'd have used it like I just did. Broad reach is often an indicator of how fundamental a result is.

And not claiming it's complicated of course. Celebrating simple here! But it may give a reminder of just how HUGE mathematics is. Mathematics is an infinite subject. We get lucky with pieces of that infinity.

Of course being trained not to find simple results may affect a math student, but has no meaning to someone like me, as I'm NOT a mathematician. I got physics training. I was actually surprised to read that mathematicians believed all simple fundamental results had already been found. I never received that training.

So the future of finding these types of results may be about non-mathematicians. Which I think is ok.

Maybe that's actually better, at least for someone like me! As if certain attitudes hadn't taken over the mathematical community, the result may not have been left around for me to find.

Want to help support basic math research of this kind? Then please talk about it. That's the best thing.

Have also a discussion group which should mention: My math group

James Harris

## Thursday, December 25, 2014

### Give the gift of being stumped

Discovering new math is very rare. Discovering new simple math is the rarest of all, so why not use it to stump that wannabe genius in YOUR social circles?

Here is a puzzle that no math person without knowing my research should be able to solve without using brute force:

What is next in the series?

1) 12 + 22 = 5

2) 32 + 42 = 25 = 52

3) 112 + 22 = 125 = 53

4) 72 + 242 = 625 = 54

5) 412 + 382 = 3125 = 55

6) 1172 + 442 = 15625 = 56

7)  (?)2 + (?)2 = 78125 = 57

---------------------------------------------------

Of course just pulling from a prior post where I left off the last answer I derived, so the answer IS posted! So it's silly to solve the problem by brute force which is also cheating. Brute force involves programming a computer to just try different squares until you find some that fit. That only requires knowing how to code, and involves no mathematical intuition at all.

But the bigger question in my mind is: if some highly intelligent person is given that list can she or he figure out a pattern?

These type of problems are actually given in IQ tests I've seen or as mathematical challenge questions in other areas.

This particular one involves new math, so should stump any math expert in the world--unless they know of my research that is!

Or maybe not. Part of me wonders if some very high IQ person can figure it out. Or some person with exceptional math intuition. But I doubt it.

This particular result escaped Archimedes, Euler, Ramanujan and Gauss as well as every other major mathematical mind in human history, though before Archimedes I don't think it's fair to include anyone. So, let's say, every mathematical mind in the last two thousand years missed it. I doubt any human alive can just look at that series and figure it out. I couldn't have done it that way myself!

Putting it forward in this way can give a sense of what kind of an accomplishment it is, but of course news travels fast! So if you want to stump someone in your life you better hurry before this "think" is well-known.

Update January 5, 2015: One person answered the challenge on my Facebook page. So it can be done.

Update June 7, 2017: Removed link as I decided to delete my Some Math Facebook page some time ago.

This post generated a HUGE amount of interest in comparison to prior post and I consider it to be a great success. It has me thinking more about trying to find challenges from mathematical results as a better way to promote them.

James Harris

## Wednesday, December 24, 2014

### Infinite difference of squares

Using the BQD Iterator it's possible to get an infinite number of difference of squares for certain integers raised to successively higher powers.

The appropriate BQD iterator is, given:

u2 - (T+1)v2 = F

then it must also be true that

(u + (T + 1)v)2 - (T + 1)(u + v)2 = -T*F

So if you start the iterator with u = v = 1, or u = v = -1, F = -T.

That means that in general there must always exist nonzero x and y, such that for an integer n equal to 0 or higher, and an integer T equal to 2 or higher:

x2 - (T+1)y2 = (-T)n+1

I have -T raised to n+1 so that n is a count of iterations, if you want T raised to the nth power you just start n at 1 instead of starting at 0, which I prefer.

And if T+1 is a square you get an infinite series of difference of squares. That's cool. Will try it out.

So, let T = 15, and u = v = 1. Then, iterator is:

(u + 16v)2 - 16(u + v)2 = -15*F

Start is:

12 - 16*12 = -15

then it must also be true that

(17)2 - 16(2)2 = 225 = (-15)2

Next iteration: (49)2 - 16(19)2 = -3375 = (-15)3

And of course you can keep going out to infinity.

Notice the factorization at the first iteration: (17 - 8)(17 + 8) = 9(25) = 225

But of course if T + 1 is a square then it is trivial to factor.

James Harris

## Sunday, December 21, 2014

### Squares and nth powers

Using the BQD Iterator it's possible to show a connection between squares and integers raised to an arbitrarily high power.

The appropriate BQD iterator is, given:

u2 + (m - 1)v2 = F

then it must also be true that

(u - (m - 1)v)2 + (m - 1)(u + v)2 = m*F

So if you start the iterator with u = v = 1, or u = v = -1, F = m.

That means that in general there must always exist nonzero x and y, such that for an integer n equal to 0 or higher, and an integer m equal to 3 or higher:

x2 + (m-1)y2 = mn+1

I have m raised to n+1 so that n is a count of iterations, if you want m raised to the nth power you just start n at 1 instead of starting at 0, which I prefer.

Wow. That's really cool. Let's try it.

Let's let m = 5, and u = v = 1. Then, iterator is:

(u - 4v)2 + 4(u + v)2 = 5*F

Start is:

12 + 4*12 = 5

then it must also be true that

(-3)2 + 4(2)2 = 25 = 52

Next iteration: (-11)2 + 4(-1)2 = 125 = 53

And third iteration: (-7)2 + 4(-12)2 = 625 = 54

Fourth iteration: (41)2 + 4(-19)2 = 3125 = 55

Fifth iteration: (117)2 + 4(22)2 = 15625 = 56

Sixth iteration: (29)2 + 4(139)2 = 78125 = 57

---------------------------------------------------

So it's easy to see how in general: x2 + 4y2 = 5n+1

And calculating x and y is easy with the iterator. I'll note that it's important to keep up with signs. But of course x or y can be positive or negative, which means you can switch signs for either at each iteration.

That's how you can figure out each x and y that will work. But the math can also simply force in a factor of 5 into each, which means 25 divides off, pushing you backwards.

Oh yeah, and with m-1 a square you end up with a sum of squares, which is what happened with my example above, which is the first one since m must be 3 or greater. So if m - 1 is a square, then mn can be written as a sum of exactly two squares for any n greater than 0. That's cool too. Wow, love these infinity results.

So it is proven that in general squares have a connection to integers raised to an arbitrarily high power.

Actually, to be more specific, it's proven that m squares where m-1 of them are the same, can always be found to sum to m raised to an arbitrarily high power.

Watching those numbers behave as predicted is a thrill unlike any other. That thrill is what you want to share.

That's pretty cool. I wonder, did anyone know that before my research? Will have to go do some web searching.

James Harris

## Tuesday, December 16, 2014

### Validating historical connections

Nothing like having your mathematical research connect with and expand upon research by great mathematicians from history, and I got a double as my Diophantine research connected with a result from Euler, and a related result from Ramanujan.

According to MathWorld, check out eqn. 22 on this page, Euler had a result which I found I could also reach with my own research:

x2 + 7y2 = 2n

where with his result, n is 3 or greater, and x and y are odd integers. He also found a really cool solution for x and y, while with my research you can generate solutions with my BQD Iterator.

The applicable Binary Quadratic Diophantine iterator here, or BQD Iterator for short is:

u2 + 7v2 = F

means that

[(u-7v)/2]2 + 7[(u+v)/2]2 = 2*F

where the iterator has been adjusted by dividing 4 off, which requires u and v odd integers. Letting u = v = 1, gives F = 8, and you get Euler's result, as the iterator goes out to infinity.

Here are some iterations:

12 + 7*(1)2 = 23

Next iteration: (-3)2 + 7 = 24

And next is: (-5)2 + 7(-1)2 = 25

And another iteration gives: (1)2 + 7(-3)2 = 26

One more gives: (11)2 + 7(-1)2 = 27

Notice that 7 is bare, in all but one case:

12 + 7 = 23,  (-3)2 + 7 = 24,  (-5)2 + 7 = 25,  (11)2 + 7 = 27

And also according to MathWorld, Ramanujan has a result called Ramanujan's square equation:

x2 + 7 = 2n

Which they say only has solutions for n = 3, 4, 5, 7 and 15.

That is listed on the first page I referenced on MathWorld as eqn. 21, but also has its own page.

And I've shown the n = 3, 4, 5 and 7 cases, but am not interested in iterating 8 more times to get the last one.

I can generalize from both of these results.

For example, in general you always have at least one iteration of the Ramanujan type:

x2 + (2c - 1) = 22c-2

Where x = (1 - (2c - 1))/2

And you can generate endless Euler type equations:

x2 + (2c - 1)y2 = 2n(c-2)+c

Where here n = 0 or higher. I like starting it at 0, versus starting at c, as then n gives you a count of iterations.

And easily find x and y with the adjusted BQD Iterator, with u = v = 1, or u = v = -1:

u2 + (2c - 1)v2 = F

then it must also be true that

[(u - (2c - 1)v)/2]2 + (2c - 1)[(u + v)/2]2 = 2c-2*F

It's fascinating to ponder how Ramanujan found his result and Euler found his as well.

I can generate their results easily with my research plus go further, but there's nothing like getting that connection to two of the great mathematical minds of human history. It was the greatest thing for me stumbling across this connection.

It's extremely validating.

It makes it SO REAL as I know I could go to either of them if they were still alive and show them what I have, and know they'd cheer the math. Sure would make my life simpler if they were still alive. Or if any of the greats were still alive.

James Harris

## Saturday, December 13, 2014

### Mersenne primes and BQD Iterator

Using the BQD Iterator it's possible to generate an arbitrary number of binary quadratic Diophantine equations, including ones related to Mersenne primes.

Consider that given:

u2 + (2c - 1)v2 = F

then it must also be true that

(u - (2c - 1)v)2 + (2c - 1)(u + v)2 = 2c*F

And here an adjustment can be made if u and v are odd, as then 4 will automatically be a factor that can be divided off, so the adjusted BQD Iterator is:

Given odd u and v with:

u2 + (2c - 1)v2 = F

then it must also be true that

[(u - (2c - 1)v)/2]2 + (2c - 1)[(u + v)/2]2 = 2c-2*F

To get to Mersenne primes the iterator is started with: u = v = 1, or u = v = -1.

Which gets you at the start F = 2c.

And an infinite number of iterations are available.

The first iteration, with the adjusted BQD iterator, and u = v = 1, or u = v = -1, is:

[(1 - (2c - 1))/2]2 + (2c - 1) = 22c-2

Next iteration:

[(1 - (2c - 1))/2 - (2c - 1)]2 + (2c - 1)[(1 - (2c - 1))/2 + 1)2 = 23c-4

With that 2c in there it's a lot harder to get a sense of it, so will try some examples. I know that 7 is a Mersenne prime, so will show c=3, with u = v = 1. Then the start is:

12 + 7*(1)2 = 23

Next iteration: (-3)2 + 7 = 24

And next is: (-5)2 + 7(-1)2 = 25

And another iteration gives: (1)2 + 7(-3)2 = 26

One more gives: (11)2 + 7(-1)2 = 27

This example is interesting as 7 is left bare for several cases which is something that was noticed by Ramanujan. Let's move on now to 15, which is interesting for not being prime. Curious to see if there is some indication by this approach.

12 + 15*(1)2 = 24

Next iteration: (-7)2 + 15 = 26

And next is: (-11)2 + 15(-3)2 = 28

But I'll note you can also switch signs at these iterations, but you can't always switch signs as if u+v = 0 then you get no further iterations, but here it's ok, so let's use 7 instead of -7.

Then the next is: (-4)2 + 15(4)2 = 28

That gives a factor of 16 which when divided off just drops back to the earlier result. Interesting.

Now each of those examples can be used for factorizations, where the first non-trivially factors 15, the second does not, and that last does.

So it's not like the mathematics cares here to hide whether or not 2c - 1 is prime. It's not clear to me though if it's also giving a reason for when it would have to be composite though I suspect one may be in there. Of course it'd be more fun to me if there is. Doesn't make it so.

That mathematical intuition thing fascinates me.

The iterator goes out to infinity. That's a much bigger deal for a prime. Having to fulfill the conditions for the iterator is an infinite burden which the math handles of course. But I suspect it's a LOT easier to handle that infinity with a composite than with a prime.

Which is an attempt to try and get a handle. It's interesting you can look at an area where so many people have looked before, and see it in a different way. But of course, mathematics is an infinite subject. Somehow you can figure things out--though it may take a while.

Oh well, that's enough for a start to an investigation. Wanted to step through some early thoughts after seeing that the BQD Iterator could be used with Mersenne primes. Wondering if it can explain some things about them. It may take years but I suspect there is an answer in there, somewhere.

It's interesting that Ramanujan helped me in this direction. I've had what I now call the Binary Quadratic iterator for over 6 years. It was just one more of those things I have laying around.

I had used it for things like explaining the fundamental solution to x2 - Dy2 = 1. But a few days ago I decided I needed to name the thing. And after naming it, I decided to try something.

I wonder why naming things can have such power on the imagination.

Oh well.

Whatever works, I like to say. Definitely have had fun.

And learned a few things. And that works for me.

James Harris

## Wednesday, December 10, 2014

### Considering further connection to Euler and Ramanujan results

A couple of days ago I found an intriguing connection between my research and results attributed to Euler and Ramanujan, as I found using my Binary Quadratic iterator:

x2 + 7y2 = 23n+3

has nonzero integer solutions for x and y, for n = 0, or any positive integer.

According to the MathWorld website, check out eqn. 22 on this page, Euler had a similar result with an "unpublished proof":

x2 + 7y2 = 2n

where with his result, n is 3 or greater, and x and y are odd integers.

But also according to MathWorld, Ramanujan has a result called Ramanujan's square equation:

x2 + 7 = 2n

Which only has solutions for n = 3, 5, 7 and 15. That is listed on the first page I referenced on MathWorld as eqn. 21, but also has its own page.

The Binary Quadratic iterator here, or BQD Iterator for short is:

u2 + 7v2 = F

means that

(u-7v)2 + 7(u+v)2 = 23*F

The great thing then is that we can actually look inside this thing, and see why it behaves as it does.

The Ramanujan one is actually a little more interesting to me at first because it requires that v = 1 or -1.

Here is the start that gives F = 23:

12 + 7*12 = 23

So you have u = v = 1, or u = v = -1, is possible I'll note as well, so of course with the iterator those will divide off!

So the next value from the iterator is:

(-6)2 + 7*(2)2 = (23)23

And of course 4 divides off to give: (-3)2 + 7 = (2)23

And the mystery is solved. Doing further iterations is tedious and boring to me, but I'm guessing if I keep going the 7 would be left bare for exactly as many results as will give Ramanujan's finding, while have u and v must be odd, will always mean their sum gives at least a factor of 4 that divides off.

That explains Euler's result as well, as with each iteration you're multiplying by 8, but dividing 4 off automatically, so net result is multiplying by 2. Wow.

That is kind of interesting, I wonder if Euler or Ramanujan knew the why of these results. I have a mathematical tool that lets me peer inside easily.

With great mathematicians you can never be sure. They may have had this tool and didn't see any reason to tell it, or they did and that's buried somewhere or lost. So I should say, I've seen no record that they had it.

But now we can extend beyond these results, like the next one is:

x2 + 15y2 = 24n+4

has nonzero integer solutions for x and y, for n = 0, or any positive integer.

And we can adjust it a bit immediately by dividing off 4 to:

x2 + 15y2 = 22n+4

has nonzero integer solutions for x and y, for n = 0, or any positive integer, as we know that 4 will divide off.

That gives us a more general result with 2c where c is a positive integer greater than 1:

x2 + (2c - 1)y2 = 2(c-2)n+c

when c is 2 or greater, has nonzero integer solutions for x and y, for n = 0, or any positive integer.

So we have a general path for Ramanujan type results, as there will always be at least one bare iteration which is:

x2 + (2c - 1) = 2(c-2)+c

Which is of course: x2 + (2c - 1) = 22c-2

James Harris

## Tuesday, December 09, 2014

### Doing some iterations with BQD Iterator

Decided to check something yesterday using what I call the BQD Iterator, and was thrilled at intriguing results. Here I'll focus a little more closely on one of them.

Yesterday I found the following:

x2 + 2y2 = 3n+2

must have integer non-zero solutions for x and y, for n = 0 and any positive integer n.

The iterator here is:

u2 + 2v2 = F

then it must also be true that

(u-2v)2 + 2(u+v)2 = 3F

One starting result for x and y, for n=0, is u = 1 and u = 2:

12 + 2*22 = 9

For n = 1, iterator gives next solution is x = -3, y = 3:

32 + 2*32 = 27

For n = 2, next is x = 3, y = 0. So that stops. Interesting.

Here's another option.

Starting for n=0, with x = 1, and y=-2:

12 + 2*(-2)2 = 9

For n = 1, iterator says next is x = 5, y = -1:

52 + 2*(-1)2 = 27

For n = 2, next is x = 7, y = 4:

72 + 2*42 = 81

For n = 3, next is x = 1, y = 11:

12 + 2*112 = 243

That's more interesting for a bit, but next iteration with positives will pull in factors of 3 again. Which can be avoided by using a negative value again.

I'll do that just to see, using -11 for y.

For n = 4, with that change iterator says next is x = 23, y = -10:

232 + 2*(-10)2 = 729

So you can just switch to a negative whenever to avoid having 3 as a factor of x or y.

Interesting but tedious.

So it can be a little tricky using the iterator.

James Harris

## Friday, November 28, 2014

It must be that if you have:

u2 + Dv2 = F

then it must also be true that

(u-Dv)2 + D(u+v)2 = F(D+1)

-------------------------------------------------------------

In general true binary quadratics, except for the cases D = 0 or D = -1, F = 0, are connected to an infinity of others, as all can be reduced to this form, so it's what I like to call an infinity result.

By true binary quadratic I mean one that does not have a form like this example:

x2 + 2xy + y2 = c4 + x + y

is: (x+y)2 - (x+y) - c4 = 0

So I don't consider that to be a true binary quadratic because it behaves like a single variable equation, with x+y as the single variable.

One of my favorite discoveries I've decided to call it a binary quadratic Diophantine iterator.

The beauty of the iterator is that an already reduced equation simply gives you another of the same basic form, as:

(u-Dv)2 + D(u+v)2 = F(D+1)

Is equivalent to the original as notice, with u' = u-Dv, v' = u+v, and F' = F(D+1), you have:

u'2 + Dv'2 = F'

You get the iterator when you use my method for generally reducing binary quadratic Diophantine equations on the already reduced form, and to me is one of the coolest things ever. Looks like I first gave it in its general form back October 2008. This thing conceivably could have been discovered 2000 years ago, but somehow it wasn't so I had the honor of finding it.

While it is derived it can be verified to be correct simply by using the equations given.

Given:  u2 + Dv2 = F

Verify: (u-Dv)2 + D(u+v)2 = F(D+1)

Verification:

Expand out: (u-Dv)2 + D(u+v)2 = F(D+1),

which gives:

u2 - 2Duv +  D2v2 + Du2 + 2Duv + Dv2 = FD + F

Notice -2Duv cancels out +2Duv, and then it's just a matter of grouping:

u2 + Dv2  - F + Du2 +D2v2 = FD

And: u2 + Dv - F = 0, so:  Du2 + D2v2 = FD, and dividing off D, gives:

u2 + Dv2 = F

Verification complete.

Example using iterator:

Let F = 1, u=x, v=y, and D = -2.

1. x2 - 2y2 = 1

2. (x+2y)2 - 2(x+y)2 = -1

3. (3x+4y)2 - 2(2x + 3y)2 = 1

4. (7x + 10y)2 - 2(5x + 7y)2 = -1

5. (17x + 24y)2 - 2(12x + 17y)2 = 1

and you can keep going out to infinity, but I'll stop with 4 iterations.

Notice now you can use the simple case of x=1 and y = 0:

1. 12 = 1

2. (1)2 - 2(1)2 = -1

3. (3)2 - 2(2)2 = 1

4. (7)2 - 2(5)2 = -1

5. (17)2 - 2(12)2 = 1

And you have answers to x2 - 2y2 = 1 and x2 - 2y2 = -1.

When I found it years ago back in 2008 I extended the symbolic version further. Copying four of the five I see in that post to here for convenience. Here I'm using u = x, v = y.

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

While the symbols go out to infinity, when using actual numbers you may be able to divide off factors from D+1 routinely. So for instance with D = 1, every other iteration 4 can be divided off if you so choose. So it still can be iterated out to infinity, but that then seems trivial to me, and when 4 is divided off the series simply oscillates.

James Harris

## Monday, November 24, 2014

### So what results

My recent post covering an attempt at a research perspective on considering the value of web search results is useful to talk about another more troubling issue which I think is a reasonable concern when you hear some person claiming to have important results that aren't being properly accepted. After all, such a belief can simply be a hallmark of delusion, and I think it of interest to talk about that issue and in my case, how deep that delusion would have to go.

Turns out my first result where I ponder its importance where if it is correct there should be validation from established mathematical institutions is from 1996. I finally wrote about it on this blog May 2008:

somemath.blogspot.com/2008/05/spherical-packing-problem.html

And re-reading what I posted I can see the ambivalence but also I didn't want to talk about my own doubts. While here it is to note that I have a result from my mid-twenties which if correct would have been one of the greatest improbable finds in all of human history. So I naturally doubt it.

One of the biggest things that really bugs me about the concept of the result is that the problem is so old and Sir Isaac Newton tackled it. I find it just so difficult to imagine that he would have missed such a simple thing.

But I had a paper, sent it to a math journal, which rejected it as too simple, and I lost the paper. That is, I had it typed up or something on actual paper, while now everything is online as of course the web revolution happened, so the original paper as far as I'm concerned is lost. And I don't like talking about this thing. It irritates me. I can't find anything wrong with my approach or its logic, and no one else has ever told me anything wrong with it--too simple is not an actual mathematical error--but if that mathematical argument is correct its existence annoys me greatly. It should never have been left for me to solve.

But you know the best and worst thing about having such a result? It's the doubt.

I've had over 18 years now of doubt with that thing. Doubt can be such a treasure. In mathematics, doubt can push you to understand why you think you know something.

Rather than endlessly debate its correctness with myself or others, I gave myself a simple challenge: find something else.

No major discoverer at the highest levels in the area of mathematics to my knowledge has only a single major result.

It's like this tired refrain I'd taunt myself with through the years--find something else--no matter what it was always there until I just let it go.

But it's amazing to realize that 26 year old me, saved my future.

Moving on to something else was the best thing to do in that situation. Don't sit and argue with yourself or others, find something else. And there is always something else to do with your time and attention.

So I went from spherical packing to Fermat's Last Theorem, looking to find something else, where for a long time I've refused to discuss that most infamous, frustrating, and demeaning effort, as it is a lightning-rod for insults. And I think a lot of that pursuit was misguided. Chasing a famous problem can come from the wrong motivations I think. Great thing was I quit caring about that kind of stuff, and started just having fun.

But it's interesting to me that the stigma of that thing is so great. It stalks you, but that's not that big of a deal now.

For others pondering what they may see as indications of delusion, I think there is a natural concern. Especially if you worry about encouraging a person with some kind of mental disability or illness. But I think people naturally tend to focus on what they see as emphasized, with a continuing focus on that thing as somehow proof of something: like, why can't this person just let it go?

For me that's the fun part, I can. My spherical packing approach irritates me. Pursuit of Fermat's Last Theorem was humiliating, and turned into a continuing stain so great I hesitate to even mention the thing. But I stopped working on it over a decade ago. Was done in 2002.

So what about other results then?

It's odd to realize that I can idly throw away what I don't like as what I could consider my first major result. Dismiss my work on the next thing I spent so much time and effort on, and not even touch the areas of current potential controversy.

But I did develop mathematical analysis techniques from tackling Fermat's Last Theorem which I used with my next move on, which lead to my one and only mainstream publication in a formal math journal.

Oh yeah, so I DO have a published result. Turns out that is a lifetime thing which can't be taken away from you, as I checked! I'm listed as a published mathematical author. Where? I forget. It was over a decade ago that I checked. Actually called them up and chatted on the phone. It's funny the things that seem important until they don't. I think it was MathSciNet. Those who care about such things might be able to find it.

So do I hold on to any results?

No. My view is that I chase ideas, and chase certainty. And I pondered mathematical proof maybe because I had the spherical packing thing: How do we know what's true or not?

Correct mathematical results need no defense.

If every person believed just that one thing then you'd remove all of the heated argument out of mathematics.

People might still argue but it'd only be in teasing out the actual logical details and verifying them.

So I don't hold on to results, nor do I defend them. I present them. And thanks to the web that's enough.

James Harris

### Web acceptance research perspective

Obviously mathematical research stands on its own but it is of interest how others react to it. My position has given me a unique perspective where it actually helps to try and step away from it objectively, which is something I've valued trying to do for years, and I think it worth it to talk about some conclusions.

The acceptance of the web for a particular content author can arguably be considered reflected in web search rankings. One concern is in choosing a metric simply because one is available, but I think there is substantial quantitative data that supports this metric. And in fact my own experiences give both a subjective for me reference point and potentially objective for others reference.

The benefit of this research perspective has to do with best routes to sharing research information. The 'social disruptive' nature of the web is much discussed and the potential ideal of the web is to remove gatekeepers, like math journals, or force them to re-define their role.

Specifically recently I've noted the emergence of my own mathematical research in the specialized area of number theory which considers binary quadratic Diophantine equations. That is, I have a high ranking across web search engines, with the search: reduce binary quadratic Diophantine equations

The search is important in that it is asking how to do something. That would naturally seem to reflect a demonstrated need which can be satisfied by the results of the search. To my knowledge I am the only person doing that particular search string though I have suggested it now to others.

But also importantly the actual mathematics helps solve a highly particular type of problem and absolute correctness is required, that is 100% correctness, as the mathematics has to be perfect. So it's a great example that the web can find such things.

However, I have other search results, like the name of this blog. At one point I found it necessary to change the name of this blog. My choice of the rather generic "Some Math" was not just because I thought it appropriately descriptive but also to aid in considering this kind of research as to whether or not the web would rank it highly. This social experiment was telling in that the term--some math--gave results at first not related to my content, but over time my content rose in search results. And I could easily compare across search engines.

Another contrary result has been with the search: definition of mathematical proof

That has been extremely informative as the ranking of my page where I define mathematical proof has shown a negative reaction to my discussion of it. For instance emphasizing its search ranking in the past actually lead to it dropping completely from web search across web search engines. I will note using its search position in emotional tones that are not relevant for this post, but considering what is known about web search it involves people linking to your content. So what causes people to link, or de-link is of interest.

With years of experience now in this area I think I have a very clear idea about how the web behaves with regard to certain types of content, also which content has a weaker link attribute, like my definition of mathematical proof, versus content which has a very high link attribute.

It's telling that when I changed the name of this blog to its current one, the link for the definition of mathematical proof promptly disappeared from web search, but the web switched to other research on Usenet math newsgroups with other results!

Talk about validation of a research path, when one source was removed as changing the name of this blog broke web links to it, the web first switched to an alternate source of the same information, which were posts I had made on Usenet, and then switched back, as revealed by the new link name rising in web search.

The web simply went and found the research in a dynamic process.

From a research distribution perspective it appears that math journals may be irrrelevant: people will find useful content regardless which is reflected in web search results which adjust dynamically even if where that information can be found is artificially manipulated.

Presumably if I found a way to destroy all sources that I provided, others might simply put the content up again, which is an experiment I don't see necessary. The web reflects a continuing need for information, but presumably any number of people who needed the content have copied it off the web.

Looking at this situation as objectively as I can I will admit to a subjective reality, and maybe as an independent researcher will note a certain amount of relief. An independent researcher like myself can put up content directly without gatekeepers. Web validation can come from search engine ranking, which is dynamic in nature meaning that it is a constant evaluation, which can even shift to chase the research, if one source is made unavailable, indicating a real need.

Less dynamic behavior can also be associated with content that presumably is deemed less important in the dynamic process like my definition of mathematical proof, which presumably is less of a direct need for people who might be interested in this information.

All search results considered for this evaluation were at #1 in more than one search engine at one time or another, and this position was confirmed to be global, as in, those searches would come up at #1, anywhere in the world. Though confirming that has been easier or harder at times. Primary reference has been Google, which thankfully also during the time of this research inquiry has been the dominant search engine worldwide, which it is at the time of this writing.

There was some limited checking by trying web searches in different geographic areas though I have used primarily states in the United States including Hawaii.

James Harris

## Friday, November 21, 2014

### Crowd approval

My focus is on knowing things absolutely. So I made a post where I talked about how you can check for absolute proof. There is a value though in seeing that what you've found is appreciated by others.

For a sense of some of my math research rank, anywhere in the world you can go to a web search engine and you can get an objective measure by a simple test which tells you about the interests of other people besides myself.

Web search: reduce binary quadratic diophantine equations

Turns out my research dominates on that subject as amazingly enough I have the privilege of having improved upon methods that Gauss previously best developed. And Gauss is a hero of mine so I think that's really cool. And the point then is to see that's not just some personal opinion.

Look at the other search results besides my content that come up to see the competition, as the web is highly efficient for helping you with such comparisons. I guess that's the future.

My way of reducing binary quadratic Diophantine equations is one of the best in the world. That's the kind of objective assessment the web can give you. Don't trust me, trust it. Or don't, but it's not like that changes things. That pull is so strong that it's like a tidal wave. The pressure is actually more intense than possibly faced by anything but results at the highest level.

So the good news is that my research has to survive intense pressure and scrutiny.

Getting the attention of the world requires your math work.

Interesting to state it, which can sound so far-fetched--which is why the web search is so potent. It removes my opinion from the mix. All I'm doing here is pointing out what is verifiable. It's not even hard to check.

Unless possibly you think there's a vast conspiracy with the major search engines of the world desperate to fool you, it's an objective test.

There is no such vast worldwide conspiracy. I checked.

And if you think maybe there is, do you really think you're that important? Why should they bother fooling you?

Yet, it's one of my minor results which is part of one of the more fascinating advances in human thought that took over a century to occur.

And to me that's the kind of thing that should be cherished by the mathematical community, but instead it's supported by the world as revealed by web search. Um, between those two, which should concern me the most? Yup, the world.

So there is this world support for me and my ideas. That it comes from new technology? Should that bother me? Actually that should give more confidence to others. But no official support of which I'm aware, which I admit is probably more about lack of information. If you think I'm just some individual railing against the mathematical establishment, then that perception can control your response, which I think is not good. Why can't mathematical proof matter more?

Reality can defy perception.

But the situation has been helpful as a part of me needed to see what people would live up to a certain standard, in terms of truth.

But I've accepted I can't hold up everything while I push that kind of standard. I just wish it had worked. My goal was to find people for whom absolute mathematical proof was all that mattered. Possibly I was misguided in that search.

For fans of my research, yeah, there are lots of positives. So yeah, it's clearly been recognized as revealed by web search. That's better than having some people claim you're right. Let the crowd give its approval.

Turns out binary quadratic Diophantine equations are really important and not just for "pure math" so people want the best, regardless of who discovered it.

That is, world is weirdly efficient.

James Harris

## Wednesday, November 19, 2014

### Finding solid ground

There is an extra to the extraordinary issue of thinking you've found a problem where the mathematical field has an important error in that the field prides itself on not only not having error, but being the one major discipline where such a thing is impossible. Worse, if you "prove" the problem you also have to deal with where do you go talk about it? To the very people who couldn't see it before? Some of whom may have relied on it to build their careers?

But the personal side of it was far bigger for me for a long time. I grew up solidly within the system, went to a top college on an academic scholarship. Got my degree in physics. Was thoroughly indoctrinated in the mainstream, and didn't expect to find a problem with it. It took me a long time to work through things, and at times I'd try to discuss, but I now realize that the "personal" part means it isn't helpful to others.

So what about someone who comes across my mathematical research and now has to struggle with its implications? I'm ready to finally help there.

And I think that is a responsibility I should accept. For those who just dismiss my ideas of course it's irrelevant. Such people will not be convinced by what I say here. But this post is not for them.

First off, is the question of--how do we know anything? And in mathematics you can hear a lot of talk about mathematical proof and axioms, but also, especially after my research came out, unfortunately, you can see pushback where mathematicians are claiming that "mathematical proof" is actually about the weight of the opinion of mathematicians! So I gave a post, stepping you through an absolute mathematical truth for demonstration purposes:

somemath.blogspot.com/2014/10/example-showing-truth-logic-and.html

And the BIG thing that should jump out is that I rely a lot on identities. Like x = x is an identity. And of course that's true! So like if I'm thinking about arguments someone might make against my research, it's like there's a point where you can jump off. If some person wishes to debate whether or not x=x is true then you really aren't going to be able to reach common ground, you know?

But for those who will accept it, then it's not very far to solid ground. Then I have arguments that rely on identities with conditions. Like, if x2 + y2 = z2 then under those conditions these things that follow from identities must be true. Not complicated.

When you're pondering "truth" or what some person claims is truth, then you have to find that thing you accept from which all else must follow and I like that I can say: it's identity. If you believe in mathematical identities then you can get the rest.

It's important to have that linchpin.

So what about acceptance? Why can't you go to some mathematical conference on say, my work on the problems with using algebraic integers? Or why can't you read up on my prime counting function from others? Or the partial differential equation that follows from it from others?

There I have to hedge that maybe you can find others discussing such things, but to my knowledge the only other people I know of who talked about them publicly were people years ago arguing with me against their importance online.

It's up to other people where they choose to engage and what they choose to discuss publicly.

So the short answer is, I don't control what other people choose to engage upon publicly.

They do.

So the choice tells you more about them than me.

From a personal perspective though, it can be a lot about who you are. So it's like, if you conclude that something is true, like that identity is key, and everything else follows, but then get stuck on--but what about these experts in the field who aren't talking about this thing?--then you're dealing with two different issues.

And I've been there. Spent a lot of time there. That's one of those personal things I realized wasn't doing anyone any good by discussing publicly, like on this blog.

My research has drawn the attention one would expect, which means that on the social side I can put forward ideas that promptly and rapidly zip around the planet. That does take a while to really feel ok. Maybe it will never really feel ok to me, but that's not a good reason for me not to help other people.

After all, coming across these ideas, you can have a real concern about how to know if they are correct or not.

And that I can understand and feel I have a responsibility to give that help.

James Harris

## Monday, October 20, 2014

### General binary quadratic Diophantine solution algorithm?

Even if I'm not sure it works, it's worth it for me to put forward the ideas for what might be a general solution algorithm for binary quadratic Diophantine equations just in case it does work.

Coming up with a general solution algorithm would complete my research in this area, as I already have a general method for reducing. For me the idea of a complete theory is enticing.

A complete theory of binary quadratic Diophantine equations would be a boon to research.

The reduced form for binary quadratic Diophantine equations x2 - Dy2 = F can be easily solved modularly to give:

y2 = (x2 - F)/D mod N

where x is some residue modulo N. If x and y have explicit solutions there must be some x mod N, for which (x2 - F)/D will be a quadratic residue. And it is possible to probe more deeply with a slightly more complex solution to get more information about what is happening internally, and also reveal an additional constraint.

It can be shown that x2 - Dy2 = F can also be solved modularly to give:

y2 = (r2 - 2mr)/D mod N, x = r + (Dy2)/(2r) mod N

Where N can be any integer for which F is a quadratic residue, where: m2 = F mod N, and r is a residue modulo N, for which y exists.

This result is found by the modular factorization:

x2 - F = Dy2  = (x - m)(x + m) mod N

by considering r, a residue of N, where: x+m = r, and x-m = (Dy2)/r

Proof:

Given x2 - Dy2 = F, consider N such that F is a quadratic residue modulo N, so there must exist m such that m2 = F mod N.

Then we have trivially: x2 - F = x2 - m2 = (x - m)(x + m) mod N

Now consider r modulo N, such that:

x2 - F = r[(Dy2)/r] = (x + m)(x - m) mod N

The existence of r is a result of F being a quadratic residue of N, as only then must there exist at least two residues modulo N that give Dy2 mod N as their product, if solutions exist.

Let: x+m = r mod N, and x-m = (Dy2)/r mod N

Solving gives:

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

And using the second I can solve for y2:

y2 = (r2 - 2mr)/D mod N

So I have finally:

y2 = (r2 - 2mr)/D mod N, x = r + (Dy2)/(2r) mod N

Proof complete.

Notice that r = 2m mod N, will always give x = m mod N and y = 0 mod N, so it is not giving unique information, which is also true for r = -2m mod N. However, if explicit solutions do exist and there are no other solutions it means y = 0 mod N. If it's not already known that explicit solutions must exist, no other solutions except the default one may mean no explicit solutions exist.

This result shows that N being made up of primes for which F is a quadratic residue is mathematically significant, as it increases the complexity, with the existence of r.

The existence of r and the requirement that (r2 - 2mr)/D be a quadratic residue of N gives constraints which may highlight explicit solutions because they have a 100%  probability of giving an r, if they exist, which suggests a rather basic modular solution algorithm.

Modular algorithm:

Collect a list of primes p for which F is a quadratic residue. That should be the case for at least 50% of primes or for all primes if F is a square. The needed size of that list will depend on the size of x and y necessary to factor. The primorial gives roughly the size of solution that can be handled by any particular number of primes in the list.

Given two primes p1 and p2 on that list, Let N equal the product of the two primes, and calculate y and x, which may have multiple solutions. There will be two values for m and you need to use both! While there may be two distinct solutions for y for any particular r that works, as you're solving a quadratic residue to get it, you only need one, as modulo each prime they only vary by sign.

Discard values where r = 2m or -2m mod N, as those always exist and give no useful information, unless you know there must exist explicit solutions and no other values give solutions. In that case then you know y = 0 mod N is forced. For other cases where y2 = 0 mod N verify that calculated x and y = 0 mod N actually work with x2 - Dy2 = F mod N. Discard any cases where they do not.

Solutions can vary only by sign, where you need a consistent way of handling sign variations.

For example: 5 mod 10 and 10 mod 5 only vary by sign as 10 mod 5 is -5 mod 15. You don't necessarily want to look at those as different solutions since both x and y are squared in the root equation. You might always just take the smaller, so check to see if you have a negation.

There must be solutions if the explicit equation has solutions. If you find no solutions, except the default case, for any prime p for which F is a quadratic residue then there may be no solutions to the explicit equation. But you don't know for sure with just a single case.

Save possible x and y modular values, increment the count for a solution residue by prime.

For example, with N = p1p2, you have x mod p1p2, but would save x mod p1 and x mod p2, so you save by prime.

Save linked prime and its residue for a particular prime if the other prime is the next smallest prime, assuming your list goes from smallest prime to greatest. That way each prime is linked to only one other.

Repeat with each combination of your list of primes by 2, and keep a hit count for each factor modulo a particular prime.

After you have completed looping through all prime combinations by two, and stored values by prime, incrementing the counts of particular residues by prime as they are verified, you should have resonance spikes around values that come from the explicit solutions.

That is, the values with the highest counts should be values from the explicit solutions.

If there are no spikes then you have a growing probability that no explicit solution exists. As the alternative is that every explicit solution for y has each of your primes as a factor.

If you do have resonance spikes, solve for the explicit x and y using the solutions at each prime, which is of course easy, as for instance you could use the Chinese Remainder Theorem. If you have multiple factors you can use linked primes to tell which  go with each other.

I'll use a small example to show some numbers without demonstrating the full algorithm as described above.

Let D = 4, F = -119. I need it negative as I need F mod 4 = 1 for an explicit solution. I note that 3 and 5 are primes that will work, so will take them as a pair with N = 15.

As m2 = F mod N = -119 mod 15 = 1 mod 15, m = 1 mod 15 is a solution. And m = 11 mod 15 is another. Will first use m = 1 mod 15.

So: y2 = 4(r2 - 2r) mod 15

And: x = r + (4y2)/(2r) mod 15

The quadratic residues of 15 are: 1, 4, 6, 9, 10

We find that r = 2, gives 0, and works, but no other values work except:

r = 6 and r = 11, where both give y2 = 6 mod 15, which has y = 6 mod 15 as a solution, and is a non-trivial case.

Solving for x with the first: 2x = 6 + (4(6))/6 mod 15 = 6+4 mod 15 = 10 mod 15, so x = 5 mod 15.

With the second you get 2x = 11 + (4(6))/11 mod 15 = 11 + 9 = 20 mod 15, so x = 10 mod 15, which is of course -5 mod 15.

And x = 5 mod 15 and y = 6 mod 15 work.

And x = 5, and y = 6 work explicitly, which is why it's a small example for basic demonstration purposes.

Now using m = 11 mod 15.

So: y2 = 4(r2 - 22r) mod 15 = = 4(r2 - 7r) mod 15

And: x = r + (4y2)/(2r) mod 15

And r = 7 works, to give x = 11 mod 15 and y = 0 mod 15. That gives 11(11) = 121 which is larger than 119, so beyond interest here.

So why might this thing work where other modular approaches have failed?

Well that limiting quadratic residues is a BIG DEAL.

There is a 100% certainty of a quadratic residue for an r for an explicit solution, but for others the likelihood should be less! Which means you can simply look at counts by r, with resonance spikes around the explicit solutions.

That I could bring in the concept of resonance was really exciting.

Actually running the algorithm could look like watching a spectrometer as peaks built around the explicit solutions while other values had lesser heights. Kind of wonder what it would look like exactly. Would there be lesser mounds or just sharp spikes surrounded by a few odd counts randomly distributed? If it works...

So why really? Well now with x + m = r and x - m = (Dy2)/r, you're balancing the variable x against the constant m, which is the kind of thing that has come up in my research before, so it was eery here to see it valuable again. That gives very little wiggle room. The math has to figure out all kinds of things to make it work! Which means it's hard for me to comprehend the complexity. But it means some values for r are not available, as both x+m and x-m can't behave as needed.

Of course if it DOES resolve by prime then building the full factor is easy. There are a couple of techniques including the Chinese Remainder Theorem, but I like one I found on my own and then discovered was not new.

For instance 7 = 1 mod 3 and and 7 =  2 mod 5.

Using the formula gives it right back: 1 + 3(2) mod 15 = 7 mod 15

So 1 mod 3 and 2 mod 5 give you 7.

The ability to describe a number with its residues by prime is kind of cool. So yeah, if you have the prime residues then you have the number, where you need fewer primes than the closest value of the primorial to your number. And the primorial rises faster than the factorial!

James Harris

## Friday, October 17, 2014

### Modular integer factorization algorithm

The general reduced form of the binary quadratic Diophantine equation x2 - Dy2 = F can be solved modularly to give:

y2 = (r2 - 2m'r)/D mod N'

2x = r + (Dy2)/r mod N'

Where N' is any integer for which F is a quadratic residue, where: m'2 = F mod N'

This result is found by the modular factorization:

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

by considering r, a residue of N', where: x+m' = r, and x-m' = (Dy2)/r

That (r2 - 2m'r)/D must be a quadratic residue of N gives a constraint which can be used for a modular integer factorization algorithm.

Modular algorithm:

Let D = 4. Then x2 - 4y2 = F, will have solutions for any odd F, as long as F = 1 mod 4. So F is positive or negative as every odd integer is either 1 or -1 mod 4.

And of course we have a factorization with: (x-2y)(x+2y) = F.

Then collect a list of primes p for which F is a quadratic residue. That should be the case for roughly 50% of primes. The needed size of that list will depend on the size of x and y necessary to factor. The primorial gives roughly the size of the factor that can be handled by any particular number of primes in the list.

Given two primes p1 and p2 on that list, Let N' equal the product of the two primes, and calculate y and x, which may have multiple solutions. There will be two values for m' and you need to use both! While y will tend to have two distinct solutions for each r that works, they differ by each prime only by sign, so you only need to use one.

Discard values where r = 2m' or -2m' mod N', as those always exist and give no useful information, and cover the trivial factorizations, unless there are no other solutions, as then y = 0 mod N is forced. For other cases where y2 = 0 mod N' verify that x and y calculated actually work with x2 - 4y2 = F mod N'. Discard any cases where they do not.

Use your x and y to find the factors mod N'. The two factors are (x-2y) and (x+2y).

For example, f1 = x - 2y mod N', and f2 = x+2y mod N' is an option.

There must be solutions since the explicit equation has solutions. But this approach automatically leaves off the trivial factorization. If no other solutions besides the trivial factorization, represented by r = 2m' or -2m mod N', exist then F must be a prime number.

Save possible factors by prime, so for instance if you have f1 mod p1p2 you'd take f1 mod p1 and store it, and then f1 mod p2 and store it, and increment the count for each factor residue by prime.

Save linked prime for each if it is less than the larger prime, assuming you go from smaller primes to greater in your list or vice versa otherwise. That way you save only one linked prime per prime.

Repeat with each combination of your list of primes by 2, and keep a hit count for each factor modulo a particular prime. You can optionally store x and y as well, but shouldn't need them.

Values that are the same across both primes have the highest probability of coming from the explicit solutions as they must always exist.

After you have completed looping through all prime combinations by two, and stored values by prime incrementing the counts as they are verified, you should have resonance spikes around values that come from the explicit solutions.

That is, the values with the highest counts should be values from the explicit solutions.

Now solve for the explicit factors using the solutions at each prime, which is of course easy, as for instance you could use the Chinese Remainder Theorem. If you have multiple factors you can use linked primes to tell which factors go with each other.

I'll use a small example to show some numbers without demonstrating the full algorithm as described above.

Let F = -119. I need it negative as I need F mod 4 = 1. I note that 3 and 5 are primes that will work, so will take them as a pair with N' = 15.

As m'2 = F mod N' = -119 mod 15 = 1 mod 15, m' = 1 mod 15 is one solution, while m' = 11 mod 15 is the other. Using m' = 1 mod 15 first.

So: y2 = 4(r2 - 2r) mod 15

And: 2x = r + (4y2)/r mod 15

The quadratic residues of 15 are: 1, 4, 6, 9, 10

We find that r = 2, gives 0, and works, but no other values work except:

r = 6 and r = 11, both give y2 = 6 mod 15, which has y = 6 mod 15 as a solution, and is a non-trivial case.

Solving for x with the first: 2x = 6 + (4(6))/6 mod 15 = 6+4 mod 15 = 10 mod 15, so x = 5 mod 15.

And r = 11, gives x = 10 mod 15, which of course is -5 mod 15.

So x = 5 mod 15 and y = 6 mod 15 are solutions.

And x = 5 and y = 6 work explicitly, which is why it's a small example for basic demonstration purposes. In general of course the factor can be much larger than the primes being used.

x2 - 4y2 = -119 = (x+2y)(x-2y) = (5 + 2*6)(5 - 2*6) = (17)(-7)

The other case is m' = 11 mod 15.

So: y2 = 4(r2 - 22r) mod 15 = 4(r2 - 7r) mod 15

And: 2x = r + (4y2)/r mod 15

And r = 7 mod 15 works, to give x = 11 mod 15, and y = 0 mod 15. This case is removed though because 11(11) = 121, which is larger than 119.

If you were using the algorithm you would increment the count for 2 mod 3, and for 2 mod 5, for one factor, and increment 2 mod 3 for the second and 3 mod 5 for the second.

So for factor one, 0 mod 3, and 1 mod 3 would have zero hits, while 2 mod 3 would have one hit, and 2 mod 5 would have one hit, while all other residues and 0 mod 5 would have zero hits. And for factor two, 2 mod 3 would be the only one with a hit modulo 3, while 3 mod 5 would be the only one modulo 5.

The linked prime for 2 mod 3 would be 5 with the solution 2 mod 5.

You only need to link one way, so conversely could link 2 mod 5 to 3 with the solution 2 mod 3. And probably only need one link, so might choose the next closest prime, either larger or smaller.

Linking primes allows you to distinguish multiple factorizations. Information would be unnecessary if there are only two prime factors, as in the example.

James Harris

## Wednesday, October 08, 2014

### Example showing truth, logic and absolute proof

Thought I'd talk out a simple example showing absolute proof with a basic mathematical argument, from beginning with a truth, to logical steps, to a conclusion which must be true. So I'll give a lot of detail.

This example will prove that if x2 + y2 = z2 then (v2 - 1)z2 - 2xy = 0(mod x+y+vz), where v can be any value.

1. x+y+vz = x+y+vz

A lof of my research begins with an identity, which of course is true. For me a truth is something that does not change. Which means it must be true no matter what.

Proceed using logical steps, where some might prefer to say, valid mathematical steps.

2. x+y+vz = 0(mod x+y+vz)

(If "mod" is unfamiliar to you or you have questions about my usage of it, please check out this post.)

That simply states that x+y+vz has itself as a factor, and is actually a truth from which one could start, but I like to begin with the explicit identity, when I'm explaining in a lot of detail.

3. x+y = -vz (mod x+y+v)

Subtracted vz mod x+y+vz from both sides.

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

Here I've just continued with basic algebra, squaring both sides. If you're worried you can do it all explicitly. Now it's time to introduce what I call a conditional.

5. Let x2 + y2 = z2 and subtract from 4.

It is, unlike the identity with which we started, not always true. So now things remain true under certain conditions.

6. 2xy = (v2 - 1)z2 (mod x+y+vz)

Simplifying a bit, more for presentation.

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

Proof complete.

This type of result is true with the condition that x2 + y2 = z2, so it's like, if that is true then

(v2 - 1)z2 - 2xy = 0(mod x+y+vz) must be true.

That conditional result is what I call a conditional residue.

This simple example shows how you begin with a truth, and connect truth to truth, as well as use a conditional to find what must be true when that condition is met.

Notice v is a perfectly independent variable.

As long as x, y and z fit the conditional, for instance, x = 3, y = 4, z = 5, then v can take any value.

For example: (v2 - 1)25 - 2(3)(4) = 0(mod 3+4+5v)

Which is just: 25v2 - 49 = 0(mod 5v + 7), which can be seen to be true by inspection.

For a second example, let v=1, then you have -2xy = 0(mod x+y+z) is true if x2 + y2 = z2.

With our values: -2(3)(4) = 0( mod 3+4+5), which is -24 = 0 mod 12

So now you know that if x2 + y2 = zthen 2xy has x+y+z as a factor.

As v is completely free there are an infinity of such relations available. Notice you can also probe behavior of x, y and z by what v you pick, so it can be an analysis tool.

The conditional residue contains all information about the conditional itself, which allows you to probe behavior of the conditional through it.

The purpose of this post is to step through showing a truth, logical steps, which are also valid mathematical steps, to get to a conclusion which must be true.

Once you understand the basics, you can handle far more complex examples and understand why you have absolute proof without any possibility of doubt.

Each of the steps above, is both logical, and a valid mathematical step. I call those steps, linkages.

For example, the linkage from x+y+vz = 0(mod x+y+vz) to x+y = -vz(mod x+y+vz), is the logical step of subtracting vz mod x+y+vz from both sides. Notice that is a valid mathematical step.

One of those things I rarely do with my own arguments is state when a proof is complete, but do so at times when explaining in detail.

You may notice that the argument was perfect at each step.

Correct mathematical arguments have the property of being correct at each step.

James Harris

## Friday, September 26, 2014

### Question of absolutes

One of the more common refrains with someone claiming important mathematics is: I have mathematical proof!

And a great draw of mathematics is the stated ability to be absolutely correct, though there is a problem, how do you know when something is actually proof?

That question drove me to a simple answer: begin with a truth, and proceed by logical steps, then whatever you have at each step must be true.

That is, I concluded that truth follows from truth. Which revolutionized my thinking, but I had a problem: how did I know that I could really trust logic?

My problem was with the oxymoron "logical paradox", and so I went to look the so-called ones over, and concluded that there were form issues and a failure to appreciate exceptions. For instance, consider the following set:

Consider a set that includes all and only sets that exclude themselves, except itself.

And I found I could remove "paradox" from every so-called example questioning the foundations of logic, as it was a necessity! By linking logic to mathematical proof, I had to resolve any issue bringing the consistency of logic into question. That effort is shown in several posts on this blog including this one:

somemath.blogspot.com/2005/06/three-valued-logic.html

Ultimately I concluded that mathematics is provably a subset of logic, and came up with a rather simple definition of truth itself. That may sound pretentious but it was a necessity of my proof definition. I had to know what "truth" meant in it. By claiming "logical steps" were necessary I had to consider logic and conclude that mathematics depended on logic and not the other way around. From those foundations I found I could move forward, and importantly knew what a mathematical proof was, and how to check a claim of one.

So I went around using it! And found my definition worked perfectly. I could even scan through complex mathematical arguments where much of the math was unfamiliar to me, by keeping up with linkages--individual steps in the argument. Checking linkages can be easier than checking every detail of the mathematical argument. Any break in any one of the linkages mean the conclusion is not proven by the argument given.

It's actually really cool. And the idea that mathematical proofs simply link truth to truth is one I pondered over for quite some time, and it's kind of weird! But it also helped me turn to identities with what I call tautological spaces.

Of course one person can claim mathematical proof, and you can have a perfect way to check one, but what if people disagree anyway? Is there nothing left but endless argument? Nope.

Correct mathematical arguments are not refutable, right?

So far beyond what you may think is possible, the mathematics will work! While incorrect mathematical arguments are limited within perception, the idea that they work operates within a limited zone, with a refutation out there somewhere.

Mathematics is an infinite subject.

Like with my mathematical research, it builds forward I like to say, where there is more that follows from it than I will ever know! Like with algebra. There is no end to it.

In contrast, flawed mathematical arguments are error constrained--even the appearance of success must operate in a special zone.

For instance, I innovated by factoring polynomials into non-polynomial factorizations, which turned out to be the thing which broke mathematical ideas which seemed ok as long as, yup, you didn't factor polynomials into anything other than polynomials!

And it IS remarkable when you see where people have stayed within the lines in such a way that they can't see something that jumps out at you if you just try something different.

One of the troubling things I've seen at times from the mathematical community at large is an emphasis on personal opinion, where things are believed to be true because a bunch of mathematicians say they are true, which is a running away from certainty.

Which is why I emphasize checks which do not require expertise, like asking: what does your research predict about what actual numbers will do?

The innovations I introduced allow you to do more. And the mathematics that follows from the truth will always exceed our finite human imaginations. It is the promise that has been fulfilled with advances like algebra and calculus--there is always an infinite more out there than we now know, and is the promise that makes mathematics such a power in all our lives.

James Harris

## Tuesday, September 16, 2014

### Power of prediction put to the test

If number theory means knowing what numbers will do then it has predictive power, right? Well there is an ancient equation which puts predictive power to the test.

Years ago I found myself at a critical point with the equation x2 - Dy2 = 1.

Some of my mathematical theory told me that there should be a large fundamental solution, the smallest positive integer solution with y not equal to 0, if D is prime and D-1 had 4 as a factor and small primes as factors, as long as neither D-1, D+1, D-2 nor D+2 is a square.

The most famous case is D=61, with a very large fundamental solution, which I already knew behaved as my research predicted:

61-1 = 60 = 4*3*5

And mathematical science is the art of prediction about the behavior of numbers. So in front of me was a simple test. What about 4*3*5*7 + 1?

I could just go to the next prime and see what happened. If my mathematical research was correct then the numbers had no choice. They would have to behave as predicted. But what if they didn't?

Well, if they didn't, I was wrong.

Turns out that 4*3*5*7 + 1 is 421 and I found the fundamental solution for D=421 on the web, with a quick search, why? Because it's a HUGE solution, so it was listed out there, by people who probably had no clue why it was so huge. Actually I'm sure they had no clue, any more than ancient people understood why they have weight before Sir Isaac Newton. But you see, they didn't have the mathematical science I had.

I felt relief.

If you're curious you can keep checking like check D = 4*3*3*5 + 1 = 181. Did some more web searching and found this useful table and yup, it's on it!

That's with just some of the rules of the mathematical predictive framework.

I put all the rules in a blog post:

somemath.blogspot.com/2011/09/cool-rules-renaming-pells-equation.html

My research tells you predictably how the fundamental solution will behave, so is mathematical science research. Science has a predictive power which people may fail to realize, but it gives predictive certainty.

Mathematical science gives predictive certainty about the behavior of numbers.

So mathematics is a science when it gives you predictive ability as that is a key component of science in general.

Now we can contrast with mathematical scholarship.

Mathematical scholarship has an emphasis on history and other facts of interest to scholars, so for instance you might learn of a guy named John Pell, and find that irrelevant to understanding the ancient equation. I don't care about such things. But my scientific perspective looks for predictive power.

Notice mathematical scholars can give you a way to solve the equation with integers, which does work. But ask them to explain why it works. (I think I can but the probable answer I have in mind is not very exciting to me.) Or also, ask them to tell you when solutions will be larger or not, which is the predictive power mathematical science offers.

This example is a great way to see how traditional mathematicians differ from scientists. Do a web search on Pell's Equation size, read through some published papers, and ask yourself, is that what mathematics is to you?

From those papers, it is clear that for a math scholar it isn't necessary for a mathematical technique to actually tell you much that is useful about how actual numbers should behave! Those papers reveal to me more of an exercise in style.

The scientific itch can drive discovery: what if mathematicians had been irritated by the lack of ability to do something as basic as figure out when the fundamental solution would be large or not?

Well then, the answer might not have been left out there for me to discover.

Predictive power can appear to tell you why, when it actually tells you what will happen. Knowing what will happen with electricity in circuits and through a filament can give you light at the flip of a switch.

James Harris

## Saturday, September 06, 2014

### Why predict with numbers?

If your math is wrong, one may assume it does not work, as it does not, but how do you know, if you don't DO anything with it with numbers?

Surprisingly you can find mathematical papers which never demonstrate with actual numbers, where in number theory that amazes me to no end.

What good is a math paper in number theory where at the end you cannot see with some number what was supposedly proven?

How is that even satisfying to anyone?

I find that when I look over math papers now (yes I do at times) I skim to the end, and see if they ever show anything with actual numbers.

Seeing mathematicians at their best as mathematical scientists puts the pressure on them to be correct, and numbers are brutally efficient at weeding out charlatans: the math of charlatans does not work!!!

The issue of certainty is one I'm aiming to put front and center across the academic world. That will make funding easy, as policymakers have the tools to tell when researchers are doing real work or not, regardless of the complexity of that research.

I covered that issue on one of my other blogs:

beyondmund.blogspot.com/2014/08/functional-certainty.html

Simply forcing mathematicians to demonstrate what you can predict with their research about numbers and then see if that prediction is verified--where in mathematics 100% is usually required--removes human error in a functional way.

That does not replace mathematical proof. And since I defined mathematical proof I am very well aware of how it works.

However, mathematicians have the skills to evaluate proof. But even a college administrator with no math skills whatsoever can evaluate a predictive test.

If you said one thing would happen and it didn't, then your math is wrong.

James Harris

## Sunday, August 31, 2014

### For mathematical scientists

There have been problematic aspects to my discoveries where my ability to easily prove that an accepted mathematical framework allowed one to appear to prove two different and conflicting things should have been picked up and widely discussed by others, leading to changes to address the problem. If that happened I missed it. And since it has been over a decade since my 2003 paper I can assume something went wrong.

My reaction was to figure out exactly what, as I had a functional need which is reminiscent of when I found I needed a definition of mathematical proof which could tell me when something was a proof or not.

There is something weird about finding something wrong on such a scale. It shook me to my core. So I began to search earnestly for a firm foundation again. Needed it badly. There were such strange times when I questioned everything for a bit. My skepticism went total.

Looking at behavior that didn't fit my ideas of a mathematician, I had to figure out what I thought one was. Also I needed to understand how the problem had not impacted science, as well as why I didn't seem to be able to fix it by going to scientists. And then I wondered what relation there was between math and science and whether mathematicians were scientists.

Struggling with accepted definitions I realized I needed a functional definition of science in order to figure it out, and along with it came a definition of scientific theory:

science (noun): the art of prediction using methodologies and tools to expand zones of certainty by discovery of a predictive framework.

scientific theory (noun): predictive framework found by using science.

Those are functional definitions that allow me to determine when science is being done.

For instance I was able to determine that many mathematicians are actually mathematical scholars, and do not behave as scientists. Their emphasis is on gathering knowledge, not advancing predictive frameworks related to numbers.

My analysis also indicated that the people who would help me would need to be mathematical scientists, people who use scientific methods to advance mathematical theory.

That allowed me to finally know the most important audience for this blog, and that audience is mathematical scientists.

Now I can easily explain the behavior of mathematicians, including those I consider to be mathematical scientists and those who are not.

Turns out it's not complicated. Modern mathematics as a discipline for the most part was established in the 1800's. At that time European ideas were dominant and included the now completely out-dated notion of the gentleman scholar, the idea that certain men of distinguished breeding could figure things out because they had superior intellects, and working for a living was a mean activity which they should not face.

Well the error I discovered came into the discipline right around its inception which isn't surprising to me. Some of these hero-worship decisions were fascinating like the decision to hand Euler's zeta function to Riemann. The notion that certain human beings are imbued with extra special superpowers is rampant across fan decisions in the math field of the 20th century. Those ideas were completely discredited in other areas and the cult of celebrity around them was mostly extinguished. But in modern mathematics these heroes of the discipline were deemed infallible, and mathematics was thought to just be a building on the foundations they laid.

But they laid an egg.

Thankfully a lot of important mathematics got done regardless, but most of that was outside of pure mathematics where the gentleman scholars held most sway. Today most of what they did is junk.

Didn't take me long to handle all of those.

The amount of social structure built around these bad ways of thinking is immense and is resistant to change.

But never fear, it will crumble into dust, eventually.

I can use scientific methods easily. Presumably I might use scientific methods against the social problem of mathematical scholars preferring to continue in error in mathematics. My one concern though was in doing so before I fully understood the problem. Science is the most powerful intellectual tool ever found by humanity.

Just because you can do something doesn't necessarily mean you should.

Science gives certainty, but more importantly, predictive certainty. It's not about just believing something, but about knowing what you can make happen. Like flipping a light switch. For so much of human history such a thing would have been extraordinary, but today it's very predictable. Science has a seductive power.

My analysis is complete and my conclusion is I should rely on natural social processes rather than force anything. That analysis indicates that force would do more harm than good.

Analysis indicates a timeframe for change which is acceptable to me. Humanity will be ok.

In the meantime, mathematical scientists who are the future of mathematics can do their own research regardless.

And have all the real fun.

James Harris