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