Monday, November 23, 2015

Pressure to be perfect

To me what makes mathematical discovery really hard is the pressure to get to a point of no error. That pressure to be perfect is so daunting but so absolute! As with a single error an entire framework of thought can collapse.

Pursuing perfection defines mathematical discovery unlike any other human endeavor, except I guess logic, because there are so few areas where a perfect argument can be known!

For other areas there is an approximation, or finding of things that work where it may not be clear why.

But in mathematics for a mathematical proof every single connecting element must be known, and every single one must be absolutely perfect. And every single piece of the mathematical argument must be absolutely correct.

Or you have nothing.

James Harris

Friday, November 20, 2015

Push beyond with simple relations

Playing with simple mathematical relations you can push the instinctive limits of human thought in key mathematical areas.

For example, if you dare, in the complex plane consider:

k*P(x) = k*(g1(x) + 1)(g2(x) + 2)

where P(x) is a quadratic with integer coefficients, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x. And k is any nonzero non-unit integer.

Yes, I know the k can be removed, but now see what I do as instead I use it to allow me to force symmetry on the right.

k*P(x) = (f1(x) + k)(f2(x) + k)

where: f1(0) = 0, f2(0) = -k+2

And now the k is wrapped up! How do you remove now?

There ARE simple solutions, for example, try: g1(x) = x and g2(x) = x

That forces P(x) = x2 + 3x + 2

Which means, switchable by indices of course:

f1(x) = kx and f2(x) = x - (k-2)

Which is the simplest case of it. Polynomial cases are easy. But can you go beyond?

If you DO try and get stuck I have answers. And guess what? The absolute truth requires only algebra. And understanding that truth requires that you can step through a mathematical argument and care fiercely about wanting the truth. Because the typical human mind may balk at trying to go beyond polynomials.

I figured out a way to get a handle on the equations with a function H(x), where yeah the H is about "handle", and I capitalize because I like the look, though actually it can be an infinity of functions in the complex plane:

f1(x) + f2(x) = H(x)

So now I can substitute out for one of the f's, and choose easily:

f12(x) - H(x)f1(x) - kH(x) - k2 + k*P(x) = 0

Which covers an infinite class of equations now. Easy.

Now we can see the obvious, but also see how you step off into beyond. And notice that all I did was do something counter to habit. That is easier said than done.

So many talk about stepping outside the box, but the point is that thinking a certain way is easier than not.

That I have a quadratic isn't a surprise of course, but now can use the quadratic formula which has a square root! And that has HUGE consequences.

And you can solve for f1(x) using the quadratic formula:

f1(x) = (H(x) +/- sqrt[(H(x) + 2k)2 - 4k*P(x)])/2

Because H(x) is actually a function with x as a variable, turns out that to get that square root to go away, you need a quadratic. And that means that then the f's are just simple polynomials themselves, linear.

There is no way around it.

Also if H(x) is itself an algebraic integer function, that is, give algebraic integers for algebraic integer x, then know that the f's are as well.

Weirdly enough, H(x) can cover ALL possible solutions for the g's and f's, but is limited with my proof by: H(0) = -k + 2

That just amazes me. It in essence tells the math all it needs to know about what I want, as I picked the initial conditions. And wanted 1 and 2 in there for simplicity.

Here's something with actual numbers:

7*162 = (5(3+sqrt(-26)) + 7)(5(3-sqrt(-26)) + 7)

Ask yourself, what are the values for the g's here?

And I've talked about this subject quite a bit including a recent mathematical proof. While I've also noted that the first post on this blog back in 2005 relates to this subject as I've been considering mathematics in this area now for over a decade.

LOTS of controversy in this subject area as well, including a dead mathematical journal no less.

These kinds of things are fun, weird, and fascinating as in human history they don't happen that often.

People living through such things? Rarely have a clue how huge it all is. Which probably has its benefits I guess. Maybe I shouldn't tell you then, eh? But I'm not worried, only a special group of you will believe me early anyway, I'm sure.

At least at first, while eventually of course will just be settled mathematical history.

That weird feeling though that some may have with the result and its consequences is all about limitations of the normal. We think a certain way! It can be SO HARD to shift even with absolute proof.

Am guessing the human brain wasn't built for such mathematics as I've discovered, as even for me has taken some time to get comfortable with it. But the great thing is with effort our minds can comprehend the math anyway.

Makes me question how any people could think that human beings create mathematics rather than discover it, which is why an ego check like this result may have such a hard way to go! After all, for some self-important person that may be a cherished belief.

So maybe for some people it's a welcome to humility if they accept it.

Welcome to mathematical reality where you're small. The Math is infinite.

And guess what? The Math does not need you. It does not need you to understand.

The Math does not care.

James Harris

Tuesday, November 17, 2015

Non polynomial factorization short argument

It can be shown that there must exist additional numbers besides algebraic integers, which are also integer-like numbers.

Proof:

In the complex plane given:

P(x) = (g1(x) + 1)(g2(x) + 2)

where P(x) is a primitive quadratic with integer coefficients, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.

Introduce k, where k is a nonzero integer, and not 1 or -1, and new functions f1(x), and f2(x), where:

g2(x) = f2(x) + k-2 and g1(x) = f1(x)/k,

multiply both sides by k, and substitute for the g's, which gives me the now symmetrical form:

k*P(x) =  (f1(x) + k)(f2(x) + k)

And introduce H(x), where I like using the capital letter here for visual reasons, but it is not to signify H(x) must be a polynomial, where:

f1(x) + f2(x) = H(x)

So I can find:

f12(x) - H(x)f1(x) - kH(x) - k2 + k*P(x) = 0

And you can solve for f1(x) using the quadratic formula:

f1(x) = (H(x) +/- sqrt[(H(x) + 2k)2 - 4k*P(x)])/2

And sqrt[(H(x) + 2k)2 - 4k*P(x)] will only resolve if it is to a linear function, which proves that the f's are only polynomials when H(x) is, which means the g's are then as well.

And H(x) is a handle for every possible factorization with the g's. While it has a key constraint, H(0) = -k + 2.

By smoothly transitioning H(x) through all possible equations with the given constraint can encompass all possible equations for the g's with the given requirements at x=0, and in no case can they both be equations that produce algebraic integers with algebraic integer x, except the polynomial case, when k is not zero or a unit.

Not really part of the main argument, but notice if we relax the restriction on k and let it be a unit, like try k=1, then necessarily the g's give algebraic integers. And using k =1 or -1 and using all possible H(x) in that ring covers ALL algebraic integer solutions for the g's.

But with the condition that k not be zero or a unit, there can be values of x for which you get algebraic integers, for instance x=0 will always work. But such exceptions don't change the typical case.

But regardless of whether k is a unit or not, the f's must be in the ring of algebraic integers if H(x) is. So it is shown that in those cases ANY k with the requirements given that it be a nonzero integer, not 1 or -1, will give algebraic integer f's.

Therefore there must exist numbers which I will call objects o, such that k*o must be an algebraic integer, when o is not an algebraic integer, for any k with the requirements.

Proof complete.

But what about fields? Does this impact them?

And it is easy to show why fields are not impacted, as consider o2/o1, which might not seem to be in the field of algebraic numbers, but you just have to multiply top and bottom by some nonzero integer not 1 or -1, like:

2o2/2o1

and you have a ratio of algebraic integers, showing that it is indeed an algebraic number.

These numbers are such that k*o will give an algebraic integer, regardless of k as long as not 0, 1 or -1, with it an integer. So for a particular o not an algebraic integer, 3o and 5o will both be algebraic integers.

With these new numbers needed a ring to contain them as well as algebraic integers.

And after pondering them for some time I came up with a more robust classification scheme with what I decided to call the object ring.

That link goes to the first post on this blog. Posted back in 2005.

These numbers were the biggest reason for the existence of this blog.

But I hadn't discovered this simple approach back then. Wonder if I had, would it have made a difference? Will it make one now?

These results change so much in mathematics. Resistance has been so fierce for over a decade now.

James Harris

Monday, November 16, 2015

Beyond polynomial mental barrier

Thinking about ways of looking at problems intrigues me, where I recently posted about something rather simple to try, which has remarkable consequences:

k*P(x) = k*(g1(x) + 1)(g2(x) + 2)

where P(x) is a quadratic with integer coefficients, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x. And k is any nonzero non-unit integer.

And:

k*P(x) =  (f1(x) + k)(f2(x) + k)

These two expressions force you in certain directions while including the polynomial case which I think is the one the human brain by design prefers. So then:

g1(x) = x and g2(x) = x

That forces P(x) = x2 + 3x + 2

Which means, switchable by indices of course:

f1(x) =  kx and f2(x) = x - (k-2)

Which is the simplest case of it. Polynomial cases are easy. But I suggest to you there is a mental barrier to accepting that the g's are not limited to what is familiar to a human mind! But if you push beyond polynomials you break what most people think they know about mathematics and numbers.

Which is why I like this example.

Test yourself--find another set of f's for a different set of g's.

Polynomials are boring. There are more of those of course, an infinity more but that pattern is easy.

If you're stuck, welcome to the limitations of your brain. Some may be incapable of seeing anything other than polynomials here. You may have an inherent limitation in the wiring of your brain. And if so that's not my fault! Don't get mad at me because of it.

Notice I didn't specify a ring. See how that matters. Try different rings, or even the complex field! The f's and g's will still exist there of course.

Push beyond your mental barriers--if you can.

But some of you don't need to be so enticed, as you've been looking for a door into something beyond, maybe your entire life. For those people, you're welcome to the fun!

James Harris

Sunday, October 18, 2015

Weird Diophantine products

Found myself contemplating Diophantine solutions for:

(x2 + ay2)(u2 + bv2) = p2 + cq2

And realized with my BQD Iterator can find easily enough, when c = a+b+ab:

(x2 + ay2)(u2 + bv2) = p2 + (a + b + ab)q2

Oh yeah c = a+b+ab is actually forced for it to work! Which I forgot so need to remind myself here. That keeps them all in sync. Which will always have solutions with any given 'a' and 'b' if you start iterators at 1 or -1, for x, y, u and v.

Here's an example where they can't stay in sync, let a = -2, b = -3, -2 - 3 + (-2)(-3) = 1:

(x2 - 2y2)(u2 - 3v2) = p2 + q2

Which doesn't iterate. The right side doesn't.

Ok, continue up a bit, let a = -3, b = -5, -3 - 5 + (-3)(-5) = 7:

(x2 - 3y2)(u2 - 5v2) = p2 + 7q2

That does iterate. Or could do a = 3, b = -5, 3 - 5 + 3(-5) = -17:

(x2 + 3y2)(u2 - 5v2) = p2 - 17q2

And that does as well.

Still continuing up by primes, let a = -5, b = -7, -5 - 7 + (-5)(-7) = 23:

(x2 - 5y2)(u2 - 7v2) = p2 + 23q2

That does iterate. Or could do a = 5, b = -7, 5 - 7 + 5(-7) = -37:

(x2 + 5y2)(u2 - 7v2) = p2 - 37q2

And that does as well. You know at first I'm wondering if prime 'c' is just a coincidence but it gives the math more flexibility but that also is something would need investigation. Will note here as speculative but maybe worth considering further someday?

Oh, should note why more flexibility with prime 'c'. The left side is a quadratic residue, for instance with the last modulo 37. And primes have more quadratic residues available.

Got me curious. Guess easy enough to jump up just a bit without going to very BIG numbers.

Still continuing up by primes, let a = -97, b = -101, -97 - 101 + (-97)(-101) = 9599 = 29(331):

(x2 - 97y2)(u2 - 101v2) = p2 + 9599q2

And switching signs a = 97, b = -101, 97 - 101 + 97(-101) = -9801 = -992:

(x2 + 97y2)(u2 - 101v2) = p2 - 992q= (p - 99q)(p + 99q)

Glad I went up a bit. So yeah, found non-prime cases for 'c' easily enough. And found situation where is a square.

So these are Diophantine factorizations. Kind of things that really only make sense with the BQD Iterator as in every case where iterates are an infinite set of solutions in existence.

Maybe should iterate some to look at some solutions, but just don't feel like it right now. Would just be too tedious. I do know that if I start with variables set to 1 or -1 do guarantee solutions. So yeah, easy to work out actual answers but don't feel like it. Kind of arena where I think it'd be more fun to use a computer.

James Harris

Saturday, September 26, 2015

Abstracted one generator for approximate square root of 3

Found myself intrigued by putting up just enough information to do the calculations for square root of 3 with one of an infinity of possible relations:

sqrt(3) approximately equals xn+1/yn+1, where:

xn+1 = 97xn + 168yn

yn+1 = 56xn + 97yn

and x0 = 1, and y0 = 0;

Does an n exist for: x = 1351, and y = 780?

With those if it does then: xn+1 = 262087, and yn+1 = 151316

And 262087/151316 equals approximately: 1.73205080758

Correct for sqrt(3), same number of digits: 1.73205080756

And differs from sqrt(3) by approximately: 1.26e-11

That pc calculator is so handy.

Oh yeah, negative DOES work. Just been kind of sloppy, and just talking the positive, so you can start with: x0 = -1, and y0 = 0

The math does NOT care. And of course you have two solutions! Positive and negative for sqrt(3). So yes, try x and y with different negations and see what happens! (Still works.)

Wondering if should fix prior posts, occuring to me don't really care as is freaking obvious anyway. Moving on to other more important things.

Still just doing these for fun.

James Harris

Sunday, September 06, 2015

More approximating square root of three

Did some math for fun where I used my BQD Iterator to approximate the square root of three, and found myself still thinking about it. So will extend things a bit. And will formalize a little so things will look a bit different from that post.

Ended with sqrt(3) approximately equals xn+1/yn+1, where:

xn+1 = 7xn + 12yn

yn+1 = 4xn + 7yn

and x0 = 1, and y0 = 0;

And I stopped with x = 1351, and y = 780, which I got using:

(7x + 12y)2 - 3(4x + 7y)2 = 1

And going to use the BQD Iterator to advance that a couple more, and then use that equation, to get a better approximation.

So next is:

(19x + 33y)2 - 3(11x + 19y)2 = -2

And next iteration is:

(52x + 90y)2 - 3(30x + 52y)2 = 4

And can divide off that 4 to finally get:

(26x + 45y)2 - 3(15x + 26y)2 = 1

Which is my new set and a quick check with x = 1, and y = 0, shows it works.

Now lets advance our x and y, so plugging in, x = 1351, and y = 780, I get:

(26(1351) + 45(780))2 - 3(15(1351) + 26(780))2 = 1

Which gives me my new x and y:

(70226)2 - 3(40545)2 = 1

And sqrt(3) approximately equals 70226/40545 which is about: 1.7320508077

Which is really close.

Now the new equations for approximating sqrt(3) are:

sqrt(3) approximately equals xn+1/yn+1, where:

xn+1 = 26xn + 45yn

yn+1 = 15xn + 26yn

and x0 = 1, and y0 = 0;

And why do it? Just for fun. Will admit it started to interest me that you have these recursive equations. So mainly did this post just to write it out that way.

There is some symmetry with the coefficients. And 3 itself is in there as a factor. Interesting.

You know you can just endlessly play with numbers just for fun. Of course it's trivial to get a calculation for the square root of 3, from the pc, but it's not like I need one.

Just playing around.

James Harris

Sunday, August 30, 2015

Actual value of recognition

To some extent the way attention works for something you find is easy--other people learn of it and use it.

The more people who learn of it and use it, the more influence some thing you found can have.

The true value of recognition by someone of your work I think is in referral, which can happen a couple of ways as to how someone refers things to someone else. Usually it's by mentioning it to someone, or using something and having someone ask about it. That latter is a BIG driver. Human curiosity can demand to know--how did you do that?

The web facilitates referral.

Am confident that most of my influence from my mathematical discovery happens because someone needs the mathematics.

Which is why I often focus on some math of mine that fulfills a real world need, as then I can explain why it gets the functional recognition that drives it around the world. And the web gives me objective measures to show that happened.

The web simplifies distribution of information. It is an information distribution mechanism.

James Harris

Sunday, August 23, 2015

My trust in math

Maybe I should get serious and find math software that I can recommend to people for when I talk about how computers are my friends as they can easily verify some my math.

Mathematical discovery has the awesome property that a correct result is immutable. That's a fancy word for unchanging. So you can chill out, relax, sit back and enjoy the ride if you have one. But for other people looking on it can be stressful--how can they know? Sure you can say, just work through the proof, but realistically how many people really want to do that work before they have confidence you are correct?

So I've been working on what I call the social problem, which includes ways to help, like pointing out that computers verify certain really cool things I found perfectly and easily.

Checking by computer is quick and easy if you know your way around math software.

That social stuff can mess with people's minds and I believe quickly checkable information helps, so I've talked more in such areas recently, like my time on the Usenet newsgroup sci.math and other things that for the most part are really about perception.

How might it look to someone just coming across these things who doesn't have over a decade of experience playing with the math?

To me numbers are just FUN. So I at times I find myself stuck thinking: why can't I just show them the fun? Don't they get it?

Like, one of my favorite math expressions now:

(462 + 482 + 722)(1722 + 258+ 430+ 6022 +17622) =

615+ 30752 + 141452 + 159902  + 1884972   =  114*74*210

That relies on some math I found to get it. As I can do lots with squares and sums of squares to get squares. Like I posted recently to emphasize a way to get multiple sums of squares for the same value, which might be useful to math hobbyists.

What other people do with the discoveries, especially if they find new math of their own is what will shift the social reality.

What changes things is what others find as that can make it really real to a person.

Like, long ago, I just dreamed of having my own mathematical discoveries and wondering what it would feel like.

It's like--the math is your best friend EVER.

The math will never let you down. The math will never betray you. The math will never change on you.

If you find out something was wrong, it was people who screwed up, not the math. That's a weird thing to contemplate which I love so much. Guarantee of human error when it doesn't work!

Put your trust in math, if you can, and then you can have a certainty for which others can only dream.

But you can't manufacture that want. For some people like me it is an aching need.

I had to have that certainty, and knew that mathematics could provide, so I was going to go get it.

The truly motivated will figure it out and take the time required. But to do the work you have to believe in the math. As otherwise why would you do what it takes?

James Harris

Wednesday, August 19, 2015

How computers help me

Our world shifted a lot very rapidly and my story is very much a web story, where computers around the globe endlessly help me, and without them, well you wouldn't be reading these words.

And it's not just about computers and the web allowing me to communicate my mathematical findings, but about them making checking trivial.

So for instance I've decided to focus lately on a better way to reduce binary quadratic Diophantine equations, which doesn't even care about the discriminant.

Well, turns out you can just plug in my way into math software--which I've never done--and it will just check it for you, and for instance solve for a variable I call 's'.

Here's me doing the calculation with a simple example. You can see how it could be difficult with the general case.

If someone plugs my method for reducing into the appropriate math software it gives s, as a function of x and y, which I know because years ago, back in 2008 I think, some people did it.

I've posted about my years on Usenet, where most was on the sci.math newsgroup which some people seem to think is significant from search results I've seen. For me it was a way to get things checked, and I've concluded that math people there thought I looked stupid because of some of the things I was trying. And for them, showing errors in what I was doing was proving my stupidity! So they'd fall all over themselves critiquing my ideas, which often I'd simply brainstorm. So I could generate the ideas meaning credit was always mine alone, while they'd simply critique! Often insultingly, which I don't condone, but it's not like I could control it either. That was their choice. To me it would be easy enough to point out an error without also having to insult a person.

Because it is an example where computers make checking easy, it means there's little effort in some math student verifying. Weirdly enough, if you know the history in this area I can take it up a notch and note that I did in fact improve on Gauss. The computer does ALL the real work. All that student needs to do is know how to use it, and just type in the equations and watch it whirl.

And Gauss is a hero of mine, so that's a big deal to me. And I do not say it lightly.

Actually I'm sure I overrated his importance to the modern math field as for a while I thought that would be it! No way something so huge could be ignored. But mathematicians today are more focused on Riemann I think who came after Gauss.

With such a claim it is a relief for me that any person with a computer and math software who knows how to use it can verify. Computers back me up.

While I wouldn't mind established mathematicians validating my research, this way is more fun. And is less work for me! Besides we're in a fascinating new situation with the arrival of the web, and wouldn't it have been so much more boring without a nice dramatic story like mine?

Why does history seem to require such?

Reality is I'd prefer computers to validate me than mathematicians. Wouldn't you?

Human beings can just be wrong.

And computers have made this story possible.

Puzzling these things out helps me I think and maybe others.

Computers let me reach the world, validate my research, and are tireless in that defense.

I'd rather be validated by computers.

James Harris

Sunday, August 16, 2015

Approximating square root of three

Sometimes I figure things out just for the fun of it.

When I get bored I can play with some math of mine. And noticed this thing years ago, but don't think I've put it out there.

With my BQD Iterator let F = 1, u=x, v=y, and D = -3. Then:

1. x2 - 3y2 = 1

2. (x+3y)2 - 3(x+y)2 = -2

3. (4x+6y)2 - 3(2x + 4y)2 = 4

Divide off 4, and you have: (2x+3y)2 - 3(x + 2y)2 = 1

4. (5x + 9y)2 - 3(3x + 5y)2 = -2

5. (14x + 24y)2 - 3(8x + 14y)2 = 4

Divide off 4 again, and you get: (7x + 12y)2 - 3(4x + 7y)2 = 1

And x = 1, y = 0 works, so I have:

1. 12 - 3(0)2 = 1

2. (1)2 - 3(1)2 = -2

3. (2)2 - 3(1)2 = 1

4. (5)2 - 3(3)2 = -2

5. (7)2 - 3(4)2 = 1

and you can keep going out to infinity.

And you're getting approximations to square root of 3, with the ratio of x/y, like 7/4 = 1.75.

But wow that is SLOW.

Oh yeah, you can use the algebraic ones to move things along as well. So let's plug the last result of x = 7, y =4, into the last algebraic one: (7x + 12y)2 - 3(4x + 7y)2 = 1

(7*7 + 12*4)2 - 3(4*7 + 7*4)2 = 1

(97)2 - 3(56)2 = 1

And is approximately 1.732 to three significant figures. And yes, you can now plug x = 97 and y = 56 back into it, if you wish to continue. Or you could move further out with an algebraic solution. So yeah there are an infinity of such equations to generate an approximate square root of 3. Guess that's kind of cool.

I'm curious! So I'll plug them back into what I have. Not quite curious enough to extend to the next algebraic one:

(7*97 + 12*56)2 - 3(4*97 + 7*56)2 = 1

(1351)2 - 3(780)2 = 1

So I now have 1351/780 which is approximately 1.73205, which I think is more satisfying.

You can also look at 13512/7802 is approximately 3.000001.

That was fun! And it was something quick to do.

I LOVE to play with actual numbers. My discoveries mean I can always get my number theory fix whenever I need it. Often I'll figure something out just to feel better.

Works marvelously for some reason.

James Harris

Thursday, August 13, 2015

Considering modular with a cubic Diophantine

Spend a lot of time with quadratics but also did extension of some ideas to the cubic equation:

x3 - Dy3 = F

And found that I could solve modulo N, where N is a cubic residue of D that does not share any prime factors with F, m is the cubed residue, so: m3 = D mod N, and r is some residue modulo N, where Fr-1 exists.

Then:

3(2my+r)2  = 4Fr-1 - r2  mod N

and

x = my + r mod N

Here's a link to derivation of the result. The gist is just factoring x3 - m3y3 = F mod N, in order to get those modular solutions.

Usage steps:

First step then is to find some N for which D is a cubic residue, which does not share prime factors with F, then find m, which is the cubed residue, then for some residue r, you can find y if it exists, and then x.

Some numeric examples:

To explore and confirm equations are correct will not specify F, but will see what I can do knowing just D, so for example if you have D = 11, a close cube is 27, so N = 16 will work if F is odd, and I have then m = 3.

So I have:

3(6y+r)2  = 4Fr-1 - r2  mod 16,

And 3(11) = 1 mod 16, so: (6y+r)2  = 12Fr-1 - 11r2  mod 16

And the only quadratic residues of 16, are: 1, 4 and 9

Can pick any odd r and have its modular inverse exist, so let r =3 to start. Then:

(6y+3)2  = 4F - 3  mod 16, which means:

F = 1, 5 mod 16 for 1, 4 is blocked, and 3, or 7 mod 16 for 9.

Which means F = 1, 3, 5, or 7 mod 16, which is all of the odd residues. Well that didn't tell me much then, but at least equations did work.

Noticed that 33 - 11 = 16, but that was excluded with N = 16, to keep F and N from sharing prime factors. But should be able to find that one with a different N.

So will try an odd N. Another close cube is 64, and 64 - 11 = 53, so N = 53, and m = 4.

3(8y+r)2  = 4Fr-1 - r2  mod 53, and 3(18) = 1 mod 53, so:

(8y+r)2  = 19Fr-1 - 18r2  mod 53.

And can figure out r, if x = 3, y =1, as then x = 4y + r mod 53.

So r = -1 mod 53 = 52 mod 53 = r-1 mod 53.

1, 4, 9, 12, 13, 16, 17, 25, 29, 36, 40, 48, 49

(8y+r)2  = 19(16)(-1) - 18  mod 53 = -322 mod 53 = -4 mod 53 = 49 mod 53

Which works with 8y - 1 = 7 mod 53, so y = 1 mod 53.

Lessons learned from examples:

Glad I just calculated an r that would work, as tried from the bottom with r = 2 mod 53, which didn't work at all. But I already knew that not every r would work.

And finding N and m is trivially easy. That's good. You can just find cubes near your D, which gives you m, and then N, with N = m3 - D.

Here the solution for y modulo N involving a quadratic residue means it might be possible to check for existence of solutions.

Proving explicit x and y do not exist would definitely involve checking each residue r available for a given N that fits the rules.

If you try each r modulo N, and find that no y modulo N exists, then of course there are no integer solutions for x and y.

Most of my focus with modular solutions has been on quadratics, but it's worth reminding some of the techniques can go beyond quadratics.

James Harris

Wednesday, August 12, 2015

Finding the discussion

Remarkably, over two decades ago I was some guy who decided as a hobby it would be fun to work on old, hard math problems. Needed to talk things out, and found the Usenet newsgroup sci.math was a place that worked. Math ideas I presented would immediately get critiqued by a large number of people, who would also often hurl insults as well, so it was an odd thing. Chasing math ideas is a LOT easier when others help you find errors.

Of course reality can be tough as you see yourself fail a lot. And you can't have an ego about it, as often you look really silly.

My guess is that one of the deep terrors of many mathematicians is looking stupid!!!

If true, then to mathematicians being insulted for looking stupid may be one of the greatest horrors in life, tapping into their deepest fears. While insulting someone for what is seen as stupidity may feel for them to be deeply empowering.

But if my guess is correct that meant that those Usenet people in attacking my ideas may have felt I was the one who looked stupid, so they'd point out errors to show this perceived stupidity, and then on top of that hurl insults with the help! Which explains their motivation. So wacky.

Can you imagine? As you come up with math ideas, people from all over the world falling all over themselves to evaluate them? Day after day for years?

In contrast this blog has an extraordinary peacefulness to it. Though I was briefly intrigued by a burst of discussion recently on one of my posts.

While I pursued several problems during my Usenet posting phase my time focused on Fermat's Last Theorem garnered the greatest attention. For years various attacks on the problem failed, very publicly, and often spectacularly.

But it was so much fun.

That intense discussion hasn't come back, like so far am struggling on Facebook where I last noted that this blog had visits from 441 cities in 65 countries in the last year. That is, I checked in Google Analytics for the previous 365 days. But the thing is, here it's now just math. It's like take it or leave it. No entertainment. And that IS the difference. So I know how it was done, but I like it better to just be some math.

As for the discussions at their height, if you're curious, much of it is still out there. Search!

Did my best back then to be entertaining, as you DO have to work to keep that kind of attention. I think I did ok. That was work! My hobby really turned into a job of sorts given how much time I put into it.

So I think the quiet here is well earned.

James Harris

Tuesday, August 11, 2015

Infinity and binary quadratic Diophantine equations

Possibility is amazing. How do we know what we don't know? But we can have clues from what things people find that are very surprising. Like one of my discoveries years ago was proof that binary quadratic Diophantine equations are in general connected to an infinity of binary quadratic Diophantine equations where my discovery relies on using an identity.

So I found years ago a way to reduce binary quadratic Diophantine equations, which in general look like:

c1x2 + c2xy + c3y2 = c4 + c5x + c6y

You could reduce to a simpler form, where with my method you do NOT care about the discriminant. Turns out the discriminant is unnecessary when reducing these equations, and 'discriminant' is a technical word familiar to people taught classical methods to reduce.

It being unnecessary to me is really cool! And is one reason my way is simply better as I've discussed recently, because I bring it up routinely.

To put things into some perspective, number theory students who have learned about the discriminant for reducing, can imagine a future where students are not even taught it, in that context, as it's unnecessary. Which it is.

I've tried to usually link to one post for my method to generally reduce, where I use a variable 's', which is a function of x and y, where it was ego that I don't give it there.

Turns out that: s = (c2 - 2c1)x - (c2 - 2c3)y - (c6 - c5)

Which was determined by math software as it's just too freaking hard to figure it out by hand, which I started trying to do a couple of times and never finished, so I wouldn't give the value. Others used the math software I should add. I've decided I was just being silly.

The reduction method gives you a reduced form, but if you use it against a reduced form, it gives back a reduced form, as you can reduce no further, which looks like:

u2 + Dv2 = F

when reduced you get

(u-Dv)2 + D(u+v)2 = F(D+1)

Which I've had for years but recently decided maybe I could jazz up interest by naming it, so call it now a Binary Quadratic Diophantine iterator, or BQD Iterator for short.

Oh yeah, so now we've connected almost all binary quadratic Diophantine equations to an infinity of other ones. Turns out there are some exceptions, but they are trivial cases.

Check out one of my fun things that I did with the result years before I named it:

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.

Which I like to trot out. To me it's just AWESOME.

I really don't know if that series is in an established number theory text, but it does use my BQD Iterator, which appears to be mine alone, so I wonder how it might. Of course someone could have got lucky, just playing around. But then they'd have no clue why that worked!

So yeah, my way of reducing binary quadratic Diophantine equations just throws away the discriminant. And works on any such which is not trivial. That is, you can find ones where my method doesn't work, but those are easily solvable, and equivalent to a unary case anyway. Like the example I usually give:

x2 + 2xy + y2 = c4 + x + y

is: (x+y)2 - (x+y) - c4 = 0

Having the BEST way to reduce these things is guaranteed attention. Turns out they have practical uses, so there is no doubt that people will find this method and use it.

And since the method was discovered with an identity there is no doubt about correctness.

It is perfect. Absolutely perfect.

That an identity can be used in this way is fascinating to me. Finding the identity requires more advanced techniques which to me is just more modular algebra symbology.

If I had done nothing else in my life just this discovery alone would give me my place in mathematical history. So it's like, I'm good. And of course got LOTS of other things, many of which are much, MUCH bigger. So I'm like, chill.

If you believe in mathematics then it's a simple situation.

So I found something cool, but who knows what someone else might find? We don't know what we don't know, you know?

Mathematics is an infinite subject. To me it is a no-brainer that it has an endless ability to surprise us, keeping the spirit of adventure alive. We can never fully encompass the infinity that is Mathematics.

The web links to my pages but years ago I changed the name of this blog, and the web promptly linked to pages on Usenet where I'd discussed it. Until the blog regained position in web search under its new name of Some Math, and the web links switched back!

Our world is hungry for new useful knowledge. It will do what it takes to have and keep it.

That fascinates me so discussed it more than once, where here is a highly analytical post.

But even without them I'm sure plenty of people all over the world have copied the method down, so it's a permanent part of the world's body of knowledge.

And with it comes the stunning reality that in general binary quadratic Diophantine equations are connected to an infinity of other ones, like some massively huge freaking number theoretic family.

You can't stop the connection. It rules.

James Harris

Friday, August 07, 2015

Finding primes using non-square residues

Back in 2007, was probably just trying to come up with something, and put forward an idea for finding primes by using non-square residues. The idea is simple enough.

Given a known prime p, use one of its non-quadratic residues r, with:

p2 - r

and factor the result, to get a new larger prime in it. If a new large one is in it.

At that time I used 29 and 17.

The quadratic residues of 29 are:

1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25, 28

So yeah, 17 is not one of them.

Start:

292 - 17 = 824 = 8*103

Next iteration:

1032 - 17 = 32*331

Second iteration:

3312 - 17 = 8*13693

Third iteration:

136932 - 17 = ( 23 )( 19 )( 43 )( 28687 )

Fourth iteration:

286872 - 17 = ( 24 )( 1429 )( 35993 )

Fifth iteration:

359932 - 17 = ( 25 )( 179 )( 226169 )

And I mused at that time about why it was going so slow. I did do two more iterations and you can see my musings back then in a post.

But it's interesting, I'm not sure why I thought that would work. And I'm also not so sure on why it should work. Though I'm guessing there is a simple reason. Putting it here to kind of collect it.

Maybe not the most practical thing for finding primes from what I'm seeing, but kind of curious. I think you can have cases where no larger prime comes up, but not sure.

Gives me something to do if I'm bored trying to understand something I used to know, years ago.

It does fade away. Sometimes I struggle to understand my own postings, with enough distance I am not quite certain how I figured something out.

But back then it was so easy I often zoomed through the results, not always leaving enough detail even for myself later on. I try not to do that now.

Have gained perspective.

James Harris

Much better than expected

There is that wonderful thing that time can do which is settle things. And as things have settled I've become far more appreciative of how well things have gone for me, which requires focus on the mathematical community.

Turns out it's very vibrant, and quite healthy, driving a TREMENDOUS amount of progress globally, and in the past I was worried about that community as I lacked perspective, and didn't realize that I'm a member of it.

Reality is that most mathematics used today is in science and technology, but also in finance, and vast other areas. And if you look at drivers for progress in mathematics itself, the real heat is in such areas.

It may not be admitted but the dying areas are "pure". And while there is a lot of funding for those areas which may dominate at universities, my experiences which I've pointed out here more recently explain why it is dying. Quite simply some have chosen to see mathematics as a social enterprise where the opinion of mathematicians is what's right.

Linking to an article from 2003 which I'm glad is still up. I've wondered why it was never pulled as it puts forward a notion that mathematics is too complex for absolute proof, so you need a committee of expert mathematicians, to decide.

My reaction was to define proof, where I focused on beginning with a truth and proceeding by logical steps so that you know it is absolute. And I recently put up an example of an absolute proof to refresh.

That attitude that the opinions of mathematicians are real in mathematics is so ludicrous it could never hold sway outside of "pure mathematics" which is useless, as in the real world buildings would literally crumble, as mathematics is not about opinion.

However I DO have a responsibility for the health of pure mathematics. And reality is that much of what I've accomplished is pure. But there is no rush. The world isn't pressuring me to do anything spectacular in that regard, which is a relief.

The ability of some to block mathematical proof, like with my stunning story of a paper revealing my ability to put forward a mathematically correct paper by established standards which appears to establish something not true is all about the power of uselessness.

The fight for the soul of mathematics could not be an easy one. But my relief was in knowing that the mathematical community is hale and hearty, powering our world forward. Which gives me time to just grind out the rest.

At times I've debated with myself if I should just force the situation, but use of force is not in my nature. While I might simply break the resistance of the pure mathematicians in an instant, the fallout could be problematic for decades.

Besides, the world does not need that to happen or it would pressure me to make it happen.

And so far the world seems ok with things going as they have.

Time can be a wonderful way for all concerned to gain perspective, especially me.

James Harris

Tuesday, August 04, 2015

Quadratic residue pairs and urge to know why

One of my favorite results explains the count of things called quadratic residue pairs. And while the way to count for primes has been known, I found a way to derive the result previously deduced. And it took some modular arithmetic and an alternate form to a well-known equation.

For example the quadratic residues of 31 are:

1, 2, 4, 5, 7, 8, 9, 10, 14, 16, 18, 19, 20, 25, 28

And there are 7 pairs: {1,2}, {4,5}, {7,8}, {8,9}, {9,10}, {18,19}, {19, 20}

Why would people care? Curiosity. And it's logical.

For a prime p, the quadratic residue pair count is [(p-1)/4], if p mod 4 = -1, or

(p-5)/4 if p mod 4 = 1.

Here I'm using [] for the floor() function, where for example [10/6] = 1, so just means throw away the remainder.

So with 31, since 31 mod 4 = -1, you have [(31 - 1)/4] = 7

That has been known for some time, but was figured out in some boring way that doesn't interest me as I found my own way!

I use (n-m)2 - Dm2 = 1, which is just an alternate form of x2 - Dy2 = 1. And consider it modulo D-1 to get:

2m = n-1(n2 - 1) mod D-1

(n - m)2 = m2 + 1 mod D-1

That second expression lets you see the quadratic residue pairs being forced! For people who are into numbers it's like this profound thing. Um, I guess it is. Should I try to speak for everyone interested in numbers? Well, for me it was just stunning. It still is. I just like to stare at it.

Notice there is no mention of primeness there as D-1 doesn't have to be prime. So it turns out there may not be solutions for m and n modulo D-1, because there are situations where there are no quadratic residues pairs. The math has to handle those situations of course.

I just LOVE that thing, so came up with a name for it.

Call it a quadratic residue engine.

Coming up with names is an awesome perk of discovery, but it's not intuitive! You need to think long and hard on it to get a style. It really bounces things along a bit. People like it more when you name these things.

Notice it doesn't care if D-1 is prime either, but that makes it easier to solve for the count, while you can use these equations to count in general, which I discussed as a possibility and is an example of something I can do later if get curious enough.

Go over far more including full derivation of count for primes in a post:

What makes this result more interesting is that it is the first time the count is derived versus being deduced. Go out there and find the result from others to see how that was done. Think it was some kind of pigeonhole thing or something.

But I just figure it out directly. To me that's a LOT more satisfying as you can even in this instance literally see the mechanism forcing the existence of quadratic residue pairs and then work out the logic for how this mathematical machinery lets you count them.

Not surprisingly to me, it's one of my most popular results based on search dominance.

It tells you the 'why' and satisfies the urge to know with numbers.

James Harris

Monday, August 03, 2015

My math paper and recent discussion here

Finally got a bit of discussion on this blog! And was on my post with some important facts and observations, where I think the two people commenting expressed things I should address in a post.

It has been well over a decade since I had a paper published in a formally peer reviewed mathematical journal, and maybe shouldn't be surprised if people end up speculating around the details. The paper was pulled from the electronic journal directly, where I guess they just tried to delete out the content. The table of contents were changed to say the paper was withdrawn. I did not withdraw the paper. The editors tried.

But EMIS kept the paper up, as well as the rest of the journal, as it managed one more edition and then ended operations. I like to say, it keeled over and died.

Am going to ask some questions on behalf of others based on what the commenters said. I'm guessing, so if you have other questions feel free to give them in the comments!

Ok, so some may wonder, did the editors know I was not a mathematician?

Yes. I told them at the outset when I submitted the paper. It took nine months for them to publish and was in communication with the editors routinely to see how it was going, and also wondering if they would just tell me no at some point. Was surprised and gratified when they notified me they would publish.

Is it possible the editors were not aware of the implications of the paper?

Can't say for sure, but I doubt it. The point of the paper was to show an important result, which is related to a coverage problem, but also to reveal that the currently established methods of rigor could be breached. As the paper by the rules appears to prove a result true in the ring of algebraic integers, which can be shown to NOT be true in that ring. It may be subtle to non-mathematicians but these were mathematical experts who probably understood it better than I did!

Did Usenet play a significant role as some posters seem to think?

Maybe but not the one they think they did, I'm sure. The reality is that Usenet is a fringe area, where people spend a lot of time making bold claims and well, insulting each other. My own view is that the editors expected a reaction from the global mathematical community of their peers, which did not occur. Then the Usenet posters emailing them would have been insult to injury as they say.

What if the editors had backed the paper do I think much would have changed?

In retrospect, I don't. My own guess is that the situation could be much as it is to this day! Only thing is I'd have some established mathematicians along for the ride. Maybe we would get together every once in a while and have some beers grousing over the situation. And possibly those mathematicians might have destroyed their careers, if they didn't anyway. But for me it's not a big deal in the same way as I'm NOT a mathematician.

Is a big deal in ways, of course. But not a career impacting big deal.

I say the situation is very political. And in fact I think mathematicians moved away from rigor especially with journals to human judgment and the view that mathematicians ultimately decide what is mathematically correct or not.

And I responded eventually by putting up a definition of mathematical proof on this blog, which removed any doubt about mathematical correctness being an absolute and NOT about shades of gray or opinion of mathematical experts.

My current view is that a political situation in this area will play itself out over time, am gratified that EMIS kept my paper up. Feel a bit sorry for mathematicians pulled into this drama. And am not terribly surprised that my paper wasn't simply accepted by the mathematical establishment with its implications.

Of course I'm a really big deal as a result of it I'm sure. I feel confident every major mathematician on the planet knows who I am.

To accept it mathematicians would have to shift from some of their most deeply held views and accept that like other disciplines, like my favorite which is physics of course, they can be forced to adjust.

Oh yeah, and feedback DOES matter! It helps me a lot to see what people may find puzzling or where they need clarification on things. Otherwise I'm just putting up what I think might be useful to people who have an interest in these things.

James Harris

Thursday, July 30, 2015

Sums of squares for same value

Handling sums of squares can be mostly automated using some of my mathematical research. For instance consider:

52 + 202 = 17*52

82 + 192 = 17*52

132 + 162 = 17*52

These were found using what I decided to call a Binary Quadratic Diophantine iterator, or BQD Iterator for short, which lets you find in general solutions for x and y with:

x2 + (m-1)y2 = F*mn

When F = x02 + (m-1)y02, my research proves there are non-zero integer solutions for x and y, where n is a count of iterations.

To get my examples used x0 = 1, and y0 = 2, and picked m = 5 so that I had 4 which I could pull into the square, to get a sum of two squares.

At each iteration you get a split point, where you can go positive or negative, which means as you iterate you may generate extra solutions for the same sum.

For my example I had a duplicate in the second iteration which is why there are only 3 distinct solutions instead of 4.

And the 5 being squared in this example is not an accident, as that's the second iteration. The first iteration has 17*5.

So yeah I could have kept going, and would have had a maximum of 6 cases of two squares summing to equal 17*53, but less if there were duplicates.

I have an earlier post which shows how the solutions were found.

And I have additional research using the BQD Iterator showing how you get a desired number of sums of squares.  Which could mean for math hobbyists, maybe could be useful in generating magic squares of squares?

James Harris

Wednesday, July 29, 2015

Things can sound different depending on how you present them, and it occurs to me that I should note my surprising success following academic rules for presenting research results.

So yeah, I got published. And always add that the editors then got weird on me by trying to pull the paper--against established rules. And that didn't matter. The original paper is held online by EMIS and I'm a published mathematical author.

Theoretically I could simply put forward certain positions as formally peer reviewed, but defer to the reality of the weirdness from the editors, where one admitted the chief editor doubted his peer reviewers! They used two, unlike most journals from what I've heard.

So yeah, my published research passed a higher standard than most math papers ever see.

But clearly the editors maybe had less faith in the peer review system than I did.

So of course I support academic rules.

When followed they can only help me.

That shouldn't be a surprise. If I believe my own research then naturally I'm a supporter of the global system. I need it.

If I'm correct, then people following the rules will support that conclusion, which is what happened, when rules were followed.

My greatest interest is in the health of the global intellectual community not just as a concerned human being but also as a practical matter. And journals are an important part of our academic world, which is key to our intellectual community.

There are winners and losers in these kinds of things. And my place in mathematical history was set years ago. Not that I didn't keep building on it.

But you see, I have faith in the system.

In contrast to the well ordered world of academia consider the mess that is all over the web, like I had a fun time years ago for a while talking out some ideas on a Usenet newsgroup called sci.math where you had to keep people entertained. I got creative.

Some of them were nasty though, and quite a few apparently had little if any real mathematical knowledge. Wading through the useless postings could get tedious.

But some actually did know math and there were some actual professors including one of the most notable a professor from the prestigious Hamilton College that I posted about, who helped a lot. You have to grab those opportunities. Better than paying tuition! And I got a lot of useful info there as I'm not a mathematician. However the egos were out of control and some had an irritating habit of routinely proclaiming they had destroyed you, which it turned out involved trying to smear you across the web. Some of those people firmly believed in insults as a tool of power, in their bizarre imaginations. So absurd.

Of course I knew they didn't matter. When time came to get things done, notice I got a paper published.

Turns out I have only one mathematical discovery that really needed formal peer reviewed publication, otherwise I wouldn't have needed to bother with it at all. But I didn't know that before. Regardless I like the idea of at least one published result and do have a lot of respect for the system.

You have to know what you're doing, you know?

And don't take yourself too seriously! Life's to be enjoyed. Gotta have fun! And follow your own schedule I say. Don't let others pressure you onto theirs.

People can think what they want. Doesn't make it true.

James Harris

Saturday, July 25, 2015

Simply more effective

Check out:

x2 + 2xy + 3y2 = 4 + 5x + 6y

Are there any integer solutions? Yup.

Mathematical techniques where Gauss had a big role with them give one way to reduce it to a simpler form, but I found another.

That reduction to the simpler form using my own research gives:

(-4(x+y) + 10)2 + 2s2 = 166

It's easy to see how I made my example as I just counted up from 1 to 6, and now you can see the reduced form, and can if you wish click the link to see how it was done.

There's a hard to explain satisfaction in mathematics when you can find a simpler way, especially when your hero had the earlier techniques, as Gauss had a big role with early methods, while I've had the luck of being able to innovate and find there was something further along.

And it's fun to discuss, of course, so will do that here.

A reason to reduce to a simpler form is to aid in finding solutions, so let's keep going.

Subtracting 2s2 from both sides the equation above is:

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

And got lucky as no, didn't know years ago when I first thought to use it as demonstration but it DOES have an easy answer, and you can see that 83 - 81 = 2, and get a 4 on the right side. God knows I like lucky!

So you can see that s=9 works, giving -4(x+y) + 10 = 2 or -2, so x+y = 2, or x+y = 3.

And can also easily determine these are the only integer solutions that will work, as s = 9 is as high as you can go without going negative, and only an odd s can work, and none of the other lesser odd values do. Way cool, eh? Just from reducing to another form.

You can solve for x and y also by substituting with the original equation. However you do it, you get two answers.

And x = 4, y = -2, will work or x = 5, y = -2.

Will show with the first which gives:

42 + 2(4)(-2) + 3(-2)2 = 4 + 5(4) + 6(-2)

You can actually fully solve for s as a function of x and y, which was never high on my to-do list. Others have done it in general using math software, and may as well go ahead and give it versus being silly:

s = (c2 - 2c1)x - (c2 - 2c3)y - (c6 - c5)

With this example c2 = 2, c1 = 1, c3 = 3, c6 = 6, and c5 = 5, as I did that on purpose.

So s =  4y -1, and -9 works, to give y = -2. Am slipping in here as it's one of  the few results I have where others figured it out first, but they just plugged everything into math software, so it's not like they did any real work. Like to keep this blog exclusively my findings.

There's an odd feeling when you see the numbers behave as your research says they should. Will admit find it comforting, but also feel a bit odd. I found it. Wonder what Gauss would say to me if he were still alive.

If you think my claim of a better approach ludicrous then reduce the given equation using other techniques.

To see more you can check out my post on concepts in binary quadratic Diophantine equations.

That method for reducing to a simple form does something interesting when confronted with an equation already in a simpler form:

u2 + Dv2 = F

as it gives you back an equation that is also in that form:

(u-Dv)2 + D(u+v)2 = F(D+1)

And I've used that a lot and talk about it a lot, where you may not know it comes from using a method to reduce binary quadratic Diophantine equations that improves upon techniques pioneered by Gauss. And Gauss is a major hero of mine in the mathematical field.

One social problem I've seen is when people who lack even the most basic math knowledge cast judgement because they have no clue how these things work, but can react to a statement on a feeling. Math does not care about your feelings.

Actually some things I do really relate a lot back to him, for instance, I read that he kept up with news by relying on lots of newspapers. And it is remarkable to consider this man of a different time, using the tools available, having all these physical newspapers mailed to him from all over. Clearly it was important to him.

While today I can rely for lots of news on Twitter, as well as other web sources, where yeah, part of the reason for making more effort there is trying to emulate Gauss, and I have it so much easier than he had it.

Originally I was frustrated with why a better and more simple way to do something, like reducing binary quadratic Diophantine equations wouldn't be cheered loudly, as it took over. And you know? The mathematical community DID apparently pick it up quickly. And is using it.

And why wouldn't they? Can you think of a reason? I can't.

Can you think of why some people though, might prefer less effective, more complex techniques? I can.

Simplifying mathematics is a joy. And apparently has endless power. I could go on and on, with things I've done with just this one thing. Oh yeah, I have, on this blog.

James Harris

Friday, July 24, 2015

Feel better remaining cautious

Took me a while to start really considering things from the perspective of someone else, who might look over my mathematical ideas and come to certain conclusions and then I keep talking for over a decade like so much is up in the air.

Reality is I've always preferred an extraordinary amount of caution, while also at times getting into arguments with some people in the past, and more recently expressing frustration at times when what I expected didn't happen.

But what about for those who have worked it all out? Who know the math? Know the story?

Going forward I've decided to accept the things proven as proven without endlessly giving out warnings about who accepts what or not, though it feels better to remain cautious.

And even more importantly realize that the "mathematical community" is made up of people around the globe using valid mathematics. In the past I'd use that phrase in a way that didn't fit the reality. Wonder if I should go back and clean that up. Was how I used the phrase really that bad though?

For someone who knows the story then, I can say that our world consumes and uses a vast amount of mathematics, without which modern civilization would not be here.

A tiny sliver of some of the mathematical ideas out there are negatively impacted by some of my research. So much of it all involves esoteric areas. And the mathematical community is healthier than ever.

Working mathematics is growing rapidly, and my ideas help. Where they help, people are using them.

I could waste my time worrying about worthless things, or focus on the positives while remaining cautious. I like remaining cautious and sticking with the positives.

These are the kind of posts where I debate with myself, about what the point is. But yeah, for some people these things are just settled. While to me there is a bit of a sense of something not done until the established experts acknowledge.

But that's me stuck in the past. Was born into a different time and really a different world as a person born in the 20th century. This one is so much more fun though.

These things matter to a lot of people all over the world and I need to respect that reality. I can't just go with, oh, I was taught things have to happen a certain way so forget you. I need to accept that it's a great thing. Our world is moving forward. I need to move with it.

James Harris

Thursday, July 23, 2015

Some important facts and observations

Recently I have begun noting that the mathematical community can be presumed to have certain characteristics like use of correct mathematical results. That puts me firmly within that community, while I also note I'm not a mathematician, as my educational specialty was actually in the sciences. I do consider myself to be a mathematical discoverer.

What I think is my most controversial result at this time apparently pulls in others things which I think can be broadly considered to be a social problem, which is the label for this post and others like it. That social problem includes issues around my find of a coverage problem where the ring of algebraic integers can be shown to leave out certain integer-like numbers.

Explaining the coverage problem can be done with a simple factorization:

P(x) = (g1(x) + 1)(g2(x) + 2)

where P(x) is a quadratic with integer coefficients, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.

At the trivial level, it's very easy to get started, as of course:

P(x) = (x+1)(x+2) fits the requirements.

But if you look at non-polynomial g's then it's easy to find examples where algebraic integers are excluded, where you do not need a field.

The result here backs up a previously published result, where I should note an attempt by editors to remove the paper after publication, doing so from an electronic journal, while the original remains available where I've discussed the situation many times, where here I think is a decent reference post.

That paper served two purposes, both to highlight the coverage problem, and to give me the distinction in human history of being I'm sure the only person to be published with a formally peer reviewed and correct result, under established mathematical views, which nonetheless leads to a contradiction.

I doubt any other human will get that chance, so seized it.

I guess in some ways it's a dubious accomplishment, but I think I had a duty to grab it anyway, as no other human being will probably even have the opportunity.

Supposedly that was impossible, but this weird coverage thing made it possible. So yes, the paper IS correct by accepted standards of rigor, and it does lead to an apparent mathematical contradiction. Resolving that apparent contradiction is easy though. So it actually invalidates "established mathematical views". Human beings got some things wrong, but the math is still perfect.

In general with all of my mathematical results have taken the long view, where early on with this result and others I leaned heavily on simply reporting to mathematicians, making sure it was in their field of interest as best I could, and leaving it up to them to do the right thing. Which included journal submissions.

However, historically except for one case already mentioned, all submissions to journals have been rejected by their editors. Which was fascinating, curious, and intriguing as to the various reasons given. It is of little interest to expand upon further here.

At some point if I chose I could again send papers to journals. It just doesn't seem necessary as I can make the information publicly available myself, like on this blog.

Given that I do have a formally peer reviewed result, with the one published paper, even if the editors tried to withdraw, am officially a mathematical author, and also have the right to present as an established result. There has been no formal refutation of the paper which fulfills the requirements of such, which are that same be itself formally peer reviewed and published. There were attempts in the past at attacking these ideas, which are not worth elevating in this discussion as did not fulfill minimum requirements for mention at my level.

Any attempts, however, to refute these ideas are of continuing interest to me.

Members of the mathematical community can then by the rules consider them valid. And I can present them as valid, while I also like to caution that to my knowledge the results are not accepted by mainstream mathematicians, which is actually irrelevant.

But in terms of the social problem I find the situation intriguing, and have noted I have little motivation to end the situation.

Members of the mathematical community are presumed to have very high standards, rely on mathematical proof, and act on the basis of their own conscience and judgment.

Members of the mathematical community will of course be held by the absolute standard of the discipline that correctness rules, which is true through its long history, with no excuses.

This blog will continue as a reference for much of my mathematical research. And I like it more as a reference rather than dealing with these type issues which I see as more social and political. However, I continue to reserve the right to discuss what I see fit here, as it is my blog.

Other research results are far less controversial than the one dealing with the coverage problem, but much of the above still applies. Members of the mathematical community who may wish more from me are simply not going to get it at this time.

My assessment of the global mathematical community is that it is extremely healthy at this time, with working mathematics helping to drive a tremendous amount of global progress. I expect that will continue with little reason to believe anything of importance will interfere.

James Harris

Wednesday, July 22, 2015

Reality check with Ramanujan

Last year I was gratified to see I could expand upon something noticed by Ramanujan.

He noticed that 2n - 7 = x2 had integer solutions for 5 values of n, and you can read up on that at MathWorld. And Euler had considered something more general. While I could generate those solutions with my own research.

Using my research I can show how to find integer solutions for x and y, when:

x2 + 7y2 = 2n

where with Euler's 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.

So you just plug the number in and watch it go. I actually did that with a prior post last year when I discussed these things as I found them.

For me it's a maybe minor result which is so awesomely important because it relates to Ramanujan and Euler. Which is the kind of thing that is remarkable for someone. I wonder how many people in history will ever be able to explain something beyond what was known by those two when they were working on the same problem?

Is there anyone else in mathematics?

And there you go. It's a reality check with Ramanujan. And I guess Euler but that would make too long of a blog post title.

I have no clue how many people have managed such a thing, but as one of them I'll admit that it was a confidence booster. And it was me heading off in that direction after working at generating solutions for equations with Mersenne numbers, and I just picked one to see, and found that Euler and Ramanujan had been intrigued by similar things.

To me that's just awesome.

Got a bit of a reaction last year as apparently that find zipped around the globe. Didn't notice? If so, that's not my problem. My duty was to present publicly, which I did, on this blog.

The mathematical community cheers mathematics, its discovery, and the profundity of a discipline that spans all of human civilization.

Sometimes you can get a reality check that points out the obvious. But for the discipline as a whole, its history speaks for itself.

James Harris

Monday, July 20, 2015

Weird power of the individual

Years ago I realized I needed to get comfortable with drawing attention from, well, from possibly every country on Earth. It is an odd thing as an individual to step through a door knowing there is no return, but that is the potential of mathematics.

One of my favorite results highlights an oddity to knowledge which I guess is inescapable:

It turns out that if you have:

u2 + Dv2 = F

then it must be true that

(u-Dv)2 + D(u+v)2 = F(D+1)

Rather simple algebra, and I found it, but why wasn't that found thousands of years ago? Why was it there for me to discover?

We don't know what we don't yet know! It is inescapable.

Yet the information has always been there. It just took someone to find it. But why me?

But it is just one of so many now I find it hard to keep count. For some reason discovery in mathematics doesn't stop you at one but seems to require an individual have many huge results. I do wonder why.

While it is a derived result it's verifiable by just simplifying the second expression by expanding out and using the first. I've stared at it so many times through the years wondering why it was for me to find. That can mess with your sense of sanity.

Recently I was jarred yet again when I found it even allowed me to explain things that had been considered by Ramanujan and Euler. And that was just one thing. If either were alive today, he'd probably chuckle in appreciation at finally learning the 'why'. I've been using it since 2008, here and there to figure things out.

Information travels fast in our world, and did before, while now the web means it moves faster. For years I've watched bursts of attention with fascination, knowing I have the best seat in the world. Others can just guess.

Mathematical discoveries are like perfect engines of attention: incorruptible, indestructible with no emotion, nor any care about human things like race, sex, nationality or anything else human. They don't care if people wish them to be true. And useful ones roar across the planet, whether you see that happen or not.

It really does feel like being pulled by something that is inhuman. It will not stop either.

And I get to be at the center of it all, contemplating the oddity of knowledge, how it does not seem to care about human social needs or structures and does not fit into the plans of man.

The world starts changing you immediately. It's like being pulled into an intense school like no other as yes, I can accidentally impact all kinds of things, if I'm not careful. You learn rules of politicians, consider economic impact, and worry about destabilizing things.

The attention gives you the influence. These words will be read in many countries. That gives them an impact just from that reality by itself.

But the best thing for me is to realize the weird power in the potential of the individual human, beyond what we can explain or predict. And I know what so many people around the planet may not realize: how much potential they have.

My focus is not just on world systems, but also on the individuals that make up our many human societies. God only knows what they will discover. We get to wait in anxious anticipation as the discovery engine continues to roar around our world. It's so exciting to contemplate.

The additional attention starts immediately. Information moves fast in our modern world. Very fast.

James Harris

Tuesday, July 14, 2015

For those who are fascinated by them the absolute perfection possible with numbers is a thrill.

To others it might seem strange but for me there is something electric:

(462 + 482 + 722)(1722 + 258+ 430+ 6022 +17622) =

615+ 30752 + 141452 + 159902  + 1884972   =  114*74*210

My research let me easily find that perfectly balanced result. Just like to stare at it.

Yet the techniques for finding it traceback to a simple cool little result:

u2 + Dv2 = F

requires that

(u-Dv)2 + D(u+v)2 = F(D+1)

Which is one of my favorite discoveries.

For me there is a thrill in the absolute logic, knowing something that is perfection, and in the beauty of numbers following rules with absolute precision. Nice to find incorruptible things, you know?

Connections. And it's interesting seeing them separated for this post, and wondering how might it look to me without knowing how one leads to the other?

I've accepted you cannot just assume that thrill about numbers and logic exists in someone.

Some love it.

James Harris

Monday, July 13, 2015

When math surprises

Years ago I thought it simple enough if you actually found some important math--contact leading mathematicians who specialized in that area, notify them of it, and sit back and watch it get picked up, if it were valid. If it weren't then of course, no.

Not a mathematician myself, when that scenario didn't play out, I was at a loss. Check and re-check, see that it is correct and then got a bit angry, until I wondered to myself if I were antagonistic to math society itself. But of course that's not possible if my results are valid! Why not?

It's kind of weird, but by definition valid mathematics is important to the mathematical community. That is, the global mathematical community is actually focused on valid mathematical results! Obviously it doesn't willingly waste tons of time focused on false results, eh? Of course not.

But there can still be a challenge if you find people who may for whatever reason believe they are members of that community which may not be as simple an assessment as you might think where I have some ideas worth highlighting here and now.

For instance I was published in a troubling story, where recently I noted how much support has been required for that published paper to remain available from an official source despite the attempt to remove it after publication by editors of a math journal which soon after decided to simply roll over and die. I love that story. It's so wild.

Kind of story that just has to give me a chuckle. I readily admit.

For those who may think that math must be wrong, or way too complex, I've spent more time going through the argument, and actually got I think a slightly simpler angle recently, focusing on the factorization:

P(x) = (g1(x) + 1)(g2(x) + 2)

where P(x) is a quadratic with integer coefficients, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.

The conclusions reached from that expression validate my earlier paper, and I can comfortably say there really is no doubt, mathematically. But it's a surprising result! And for some who currently think they are mathematicians, may seem to be a troubling one.

Of course I have other results as well, where am gratified that my work with counting prime numbers seems to have a recent surge in popularity. I not only innovated slightly on ideas that go back to Legendre, I found a way to get a difference equation in a prime counting function that works by calling itself. What was my innovation? I focused on a function that counts composites by each prime excluding counts by smaller primes.

Weird how such a simple thing could have such profound implications! Leading to a far simpler prime counting function which uses a function I decided to call P(x,y), where using my approach I can also get the most succinct, fast prime number counter available. You can get faster with more involved algorithms, but it's a powerful and simple introduction to counting primes, with very simple ideas.

I love that thing.

But it's also surprising mathematics, maybe giving a route to answering some of the great problems in mathematics that some people who the world thinks are mathematicians do not want.

These are not issues that should concern me, however. My joy is in the discovery itself.

You see, I'm a member of the mathematical community--though I don't consider myself to be a mathematician--who greatly prizes valid mathematical results, which I assure you, is the dominant quality of the mathematical community through thousands of years of human history.

More so than any other field I think that mathematics has a problem with people who cannot measure up to its standards for truth.

But thankfully there will always be those like me, who would prefer to stand with the greats in one of the greatest intellectual disciplines of all time. What good is it, if not correct? Through thousands of years it is those who echo the truth who define the discipline. A mere hundred or so with some delusion or other pales in comparison.

Those who prize truth in mathematics are its community.

Our record speaks for itself.

James Harris

Saturday, May 30, 2015

Can you explain this mathematical absolute?

Here is a fun little result, which is a mathematical absolute, with all positive integers:

x + y + j + 1 = n2 or 2n2

if x = 1 mod D, x2 - Dy2 = 1, and j = (x+Dy-1)/D

For example, 82 - 7*32 = 1, so x = 8, and 8 mod 7 = 1.

j = (8 + 7*3 - 1)/7 = (7 + 7*3)/7 = 4

8 + 3 + 4 + 1 = 16 = 42

That's a simple example to make this post easy to write for me, but the result is true over infinity.

So whenever x = 1 mod D, with x2 - Dy2 = 1, then these rules are forced, absolutely.

Seem easy or trivial?

It's another key piece of the explanation for an ancient math mystery.

James Harris

Sunday, May 10, 2015

Part of the key to an ancient math mystery

Known since antiquity: x2 - Dy2 = 1

The smallest integer solution by absolute value for x and y, not zero is called the fundamental solution. Sometimes it can be quite large, but why?

For example, for D = 61, the fundamental solution is quite large:

17663190492 - 61(226153980)2 = 1

Part of the key to the answer--hidden in simple math for millennia:

(D-1)j2 + (j - 1)2 = (x+y)2,

where j = (x+Dy+1)/D, so our x, gives: j = 255110030.

60*2551100302 + 2551100292 = 19924730292

A connection like no other, an ellipse to a hyperbola--an answer? Part of it.

Here's another example, for D = 313, which I found in a Wikipedia article:

321881208291348492 - 313(1819380158564160)2 = 1

so:

j = 1922217605302610

312*19222176053026102 + 19222176053026092 = 340075009876990092

That one is too big for my pc calculator, though I could program something to check it, I'll use another trick.

I know that when x = -1 mod D that x+y+j-1 must be a perfect square or twice one.

That's an absolute by the way.

And yup, x+y+j-1 = 2*1340330532 so it works! Oh yeah, same thing works above, which is another clue.

If x = -1 mod D and x2 - Dy2 = 1, with all nonzero positive integers, then these equations apply, across infinity. So oh yeah, it's an infinity result. I get a kick out of those things, try now to note when I have one.

Oh yeah, remembered I could use Wolfram Alpha and everything checked out ok there.

Who knew? Ellipses and hyperbolas, like a freaky number family?

Of course I have the full answer. It's ok. To me it's just nice to know, but I like having answers to mysteries.

James Harris

Tuesday, April 21, 2015

How social can work with math

Mathematics is one of those rare areas of human endeavor where a result is absolutely right or wrong.

Absolutes are a special kind. And nothing any human being can do can shift that absolute reality one way or the other. Either the math works or it does not.

But people can SAY anything, which is where social effects can be fascinating.

If you argue against an absolute truth, you can not only destroy the trust of others in your opinion, but also your trust in your own opinion, within yourself.

Regardless, the math doesn't care.

Social helps though, I think. Talking out ideas can be extremely useful. But you have to keep it in perspective.

Or in other words, you can waste your time arguing with people about math or you can enjoy your time finding it.

And if the mathematics works, trust human curiosity and need, but don't worry about it.

Want to see an example of an absolute mathematical truth?

Here's one:

If x2 + y2 = z2 then (v2 - 1)z2 - 2xy = 0(mod x+y+vz), where v can be any value.

And I have up a post stepping through in enough detail to make certain, well, of absolute certainty: Example showing truth logic and absolute proof

There is a plodding quality to valid mathematics where you can just check step-by-step, and every piece must be perfect. I find that comforting. But to some the effort may seem difficult, but I think it is refreshing. Passion for what you do, gives you the energy to do what it takes.

Maybe social can help reinforce views that give permission to escape the work, if people tell each other what they wish to hear.

But the math doesn't care how hard you think it is, or not.

The proper point of view with mathematics is: absolute proof.

Accept nothing less.

James Harris