Tuesday, January 27, 2015

Looking at sum of squares product

Find myself intrigued a bit at looking at how sums of squares can multiply to give another sum of squares, like with:

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

            615+ 30752 + 141452 + 159902  + 1884972   =  774*210

Just like to stare at it for some reason.

Monday, January 26, 2015

Distinct square products

One of my earlier results is a connection between products of sums of squares:

The Diophantine equation:

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

will always have an infinity of nonzero integer solutions which you can find using BQD Iterators.

Thought it might be fun to consider some recent sums of distinct squares I've given to play with the result.

Let a = 4 + 9  = 13, and b = 4 + 9 + 25 + 49 = 87, so I can find where I'm pulling sums of squares from prior posts:

(922 + 962 + 1442)(70482 + 6882 + 1032+ 1720+ 24082) = p2 + 1231q2 

Matched by iterations which is necessary and used the third iteration, so p and q should be easy to calculate.

The BQD iterator is:

Given nonzero integers u and v with

u2 + 1231v2 = F

then it must also be true that

(u - 1231v)2 + 1231(u + v)2 = 1232*F

Where u = v = 1 to start will give F = 1232.

12 + 1231*12 = 1232

First iteration: (-1230)2 + 1231*(2)2 = 12322

Second iteration: (-3692)2 + 1231*(-1228)2 = 12323

Third iteration: (1507976)2 + 1231*(-4920)2 = 12324

Which means that p = 1507976, and q = 4920 are solutions, taking the positive values.

(922 + 962 + 1442)(70482 + 6882 + 1032+ 1720+ 24082) = 15079762 + 1231*49202 

Of course now I wonder what is the smallest number of distinct squares necessary to sum to 1231. Please feel free to reply with an answer in the comments. Doubt I'll work at figuring that one out.

Got bored, found 262 + 232 + 52 + 12 = 1231

Guess I should show that then:

(922 + 962 + 1442)(70482 + 6882 + 1032+ 1720+ 24082) = 

15079762 + 1279202  + 1131602  + 246002 + 49202 

Doesn't mean much to me. And don't know if that's the smallest number of distinct squares that will work.

My mind will just start playing with numbers whether I like it or not, almost will just go on automatic. That doesn't sound normal to me. But I'm ways away from anything close to normal I guess.

James Harris

Thursday, January 22, 2015

Something you can check

Here's an easy result, two sums of five distinct squares that give a square:

42 + 6+ 10+ 14862 = 882

862 + 129+ 215+ 301+ 8812 = 9682

They are examples I'm using, pulled from the same series that generates an infinity of them.

Web searching I DID find three squares summed to give a square as there is a known identity, but haven't found a square the sum of five distinct squares or more. So suspect these techniques are new.

Innovation is a critical sign. You can't fake it.

Finding new things in number theory is very exciting for people who love numbers. But it's also incredibly hard, so I like to search thoroughly, and get as much help as possible before trumpeting too much.

My web searching might miss something! So very curious to see if anyone can find prior research showing how to sum an arbitrary number of distinct squares to get a square.

Or you can just try to generate such a series yourself: test your math ability.

If you can't easily find your own sum of 5 distinct squares that is itself a square, why not?

Maybe more fun to try that before looking at how I do it.

So yeah, I posted about it, after first posting the easy algebra for getting an arbitrary number of squares that give a square.

The solution is surprisingly easy and can generate an infinity of such solutions or of any arbitrary number of squares. You could sum a hundred squares or even a thousand to get a square if you wished, though I'd think it better to let a computer do it!

It's SO easy to do though that I don't think of it as fascinating for difficulty, but only for whether it was known or not.

It's something you can check.

Do you love numbers? I do. So I'm curious about them.

Looking for help from other curious people who love numbers.

James Harris

Wednesday, January 21, 2015

Square sum of five squares

Figured out a general method for summing squares to an nth power, and found myself looking at summing three squares to get a square just to test it out. However, doing some web search found that is known and solved. Did a web search just now and didn't see summing five squares to get a square so think it worth showing just to see. The approach can sum any arbitrary number of squares to get a square or higher power though.

My most general result with sums of squares is that in general there must always exist nonzero x and y, such that for an integer n equal to 1 or higher, and an integer m equal to 3 or higher:

x2 + (m-1)y2 = mn

Where n starts at 1. Often I like to start it at zero so it's a count of iterations but is prettier starting at 1.

For a sum of c+1 squares: m = s12 +...+sc2 + 1

x2 + (s12 +...+sc2)y2 = mn

The BQD Iterator is:

Given nonzero integers u and v with

u2 + (s12 +...+ sc2)v2 = F

then it must also be true that

(u - (s12 +..+sc2)v)2 + (s12 +...+sc2)(u + v)2 = (s12 +...+ sc2 + 1)*F

So for 5 squares, I'll need 4 s's and I'll use primes: 2, 3, 5, and 7

Then m = 4 + 9 + 25 + 49 + 1 = 88

12 + 87*12 = 88

First iteration: (-86)2 + 87*(2)2 = 882

Which is: (-86)2 + 4*(2)2 + 9*(2)+ 25*(2)+ 49*(2)2= 882

Which is: 862 + 42 + 6+ 10+ 142= 882

Second iteration: (-260)2 + 87*(-84)2 = 883

Which is: (-260)2 + 4*(-84)2 + 9*(-84)2+ 25*(-84)2+ 49*(-84)= 883

Which is: 2602 + 1682  + 2522 +  4202+ 588= 883

Third iteration: (7048)2 + 87*(-344)2 = 884

Which is: (7048)2 + 4*(-344)2 + 9*(-344)+ 25*(-344)+ 49*(-344)2= 884

Which is: 70482 + 6882 + 1032+ 1720+ 2408= 884

Thank you modern computer systems where the calculator lets me do all of the above easily as I type out this post. Would be easy to program but not sure yet that there's a point. But then could go anywhere like do a square as the sum of 20 squares or whatever. Ok, now will do more with the look of my two results that sum to give a square.

First: 42 + 6+ 10+ 14+ 862  = 882

Second: 6882 + 1032+ 1720+ 2408 + 70482  = 884

Where I can divide off 64 to get:

862 + 129+ 215+ 301+ 8812  = 64*114

Which is: 862 + 129+ 215+ 301+ 8812  = 9682

So I found, two sums of fives squares that give a square:

42 + 6+ 10+ 14862 = 882

862 + 129+ 215+ 301+ 8812 = 9682

Some things are worth noting. First I'm setting some factors of the squares, so the other thing is I can pick some factors. And for this example chose to use prime numbers.

But those can conceivably divide off at some iteration, or be divided off, as you can choose to divide off common factors or not. Oh yeah, signs can be switched at each iteration to get alternate values. I tend to just go with the flow in terms of signs. Makes it easier for me just typing up demonstration posts.

It is interesting that 86 popped up twice. There could be all kinds of patterns that emerge, but to go searching for them generating a lot more would be best so a computer program would be better.

Guess it could the kind of project some enterprising number theorist might do as who knows what could be learned when you can play at this scale so easily.

But you don't have to be a specialist, as the mathematics is so elementary. Really cool! You only need to know basic algebra and have the desire to explore.

James Harris

Tuesday, January 20, 2015

Unfortunate number theory coincidence?

Number theory deals with integers, like 1, 2 and 3, or -1, -2, and -3, and of course 0. But for over a century it has also been believed to also have some relationship to complex radicals, like sqrt(-D),  like when considering:

x2 - Dy2  = 1

For years I've pondered something which I think might explain it as a number theory coincidence. 

Consider this number theory tool requiring only integers.

Given: x2 - Dy2 = F

(x+Dy)2 - D(x+y)2 = F(-D+1)

I decided to call it a binary quadratic Diophantine iterator and have an entire post as a reference to it, but you may wonder what does that have to do with sqrt(-D)?

Consider: (x - sqrt(-D)y)(x - sqrt(-D)y) = x2 + Dy2 - sqrt(-D)(2xy)

It's also iterative, and works a lot like mine if x = y = 1. That square root of a negative number lets you get the positive where x2 + Dy2 tracks with my x+Dy, and 2xy tracks with my x+y.

With x = y = 1, that's 1+D for both with the former and 2 for both for the latter.

Of course a big difference is that -D+1 that gets stuck on the end with mine.

So far though I've been able to do a lot with no radicals needed. And that's just what I'm figuring out on my own.

I've thought about the similarity noted above for years. Figured I may as well make a post about it, finally.

James Harris

Saturday, January 17, 2015

Numbering sum of squares

My most general result with sums of squares is 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. But you can start n at one, which I'll show, with a sum of exactly two squares, which is true in general if m-1 is a square.

x2 + y2 = mn

will always have nonzero integer solutions for x and y, as long as m - 1 is a square and n is greater than 0. Here n starts at 1, because I think it looks prettier and you have a sum of exactly two squares.

And the general principle is then obvious: m = s12 +...+sc2 + 1

x2 + (s12 +...+sc2)y2 = mn

Where c is the number of sums of squares desired minus one. For example for two squares, c = 1.

In general the BQD Iterator is:

Given nonzero integers u and v with

u2 + (s12 +...+ sc2)v2 = F

then it must also be true that

(u - (s12 +..+sc2)v)2 + (s12 +...+sc2)(u + v)2 = (s12 +...+ sc2 + 1)*F

And starting with u = v = 1 or -1, is necessary so that the initial F equals m.

Notice with this result if you square the square which is the result of this sum of squares then that square is also a sum of squares which is you can generate with just two more iterations.

And I'll do a numerical example with three squares, so c= 2 then: m = s12 + s22 + 1

x2 + (s12 + s22)y2 = mn

And the BQD Iterator is:

Given nonzero integers u and v with:

u2 + (s12 + s22)v2 = F

then it must also be true that

(u - (s12 +s22)v)2 + (s12 +s22)(u + v)2 = (s12 + s22 + 1)*F

And I'll keep it simple with s1 = 2, and s2 = 3, so m = 4 + 9 + 1 = 14.

And I'll use u = v = 1, so F = 14, to start.

Then first iteration: (-12)2 + 13(2)2 = (-12)2 + 4(2)2 + 9(2)2 =142

Second iteration:  (-38)2 + 13(-10)2 = (-38)2 + 4(-10)2 + 9(-10)2 =143

And you can keep going indefinitely. But I'll stop here, and write that last prettier:

382 + 202 + 302 =143

And I have a sum of three squares with a simple method for finding them. You can vary signs at each iteration to find other solutions.

Oh yeah, I tend to go higher but you also always have a result with all squares from the first iteration, or of course with any odd iteration as for instance 144 is also a square, and writing the first iteration with my numerical example out prettier:

122 + 42 + 62 =142

Interesting. You can divide 4 off, and order from least to greatest:

22 + 32 + 62 = 72

Guess I shouldn't just ignore just a sum of squares. Though to me that's the Old School, which is well worked out already.

Am curious enough to see that third iteration.

Third iteration:  (92)2 + 13(-48)2 = (92)2 + 4(-48)2 + 9(-48)2 =144

And prettifying, as well as looking at as all squares:

922 + 962 + 1442 = 1962 = (142)2

Oh yeah, can divide off 16 here, and will as I'm curious:

232 + 242 + 362 = 492 = (72)2

Now that has a curious feature:

232 + 242 + (62)2 = (72)2

No radicals needed. Integers rule.

That is MY research result! Yeah!!! Woo hoo! You can in essence just dial up whatever number of squares you wish! Just for the fun of it. Probably would be better to automate it with a computer program.

This result lets you put any positive integer you want on that last exponent.

Doing web searches as usual and found there are results with sums of three squares, including identities which include my examples. But that's just confirmation of the approach.

The advantage I see is the generality and simplicity of this route.

So I can just pick any arbitrary sum of squares if I wished and generate solutions. Could do a hundred if I wanted, but would definitely program a computer to do such a result!

James Harris

Friday, January 16, 2015

Considering my public responsibility

It occurs to me that I have some responsibilities to the public, and this post is meant to address those of which I'm aware in as plain language as possible, which means there will be no math in this post.

The intention of this post is to inform, and is an expression of my opinion. And I am NOT a mathematician and am not presenting myself as a mathematical expert. I think that's important to emphasize. Nothing in this post should be construed as legal guidance nor an expression of a legal opinion.

I consider myself to just be some guy who found some math.

And I think it's important but that's my opinion.

To me the best case scenario would be you find something mathematical you think is important and you send it to the experts, who are mathematicians. If it is important then I firmly believe that it's their job to act on it.

However, I am not in a position to make anything happen in that regard. So yes, I can send it, or try to send it, but can't make anything happen as a result as that is up to other people.

It's their decision.

And I can say that I have sent various findings of mine to mathematicians, and to my knowledge they have not acted on that information as I expected.

But that behavior in my opinion has not impacted me negatively. In fact, I probably benefited from it in multiple ways, including extended time to continue research with little competition in areas that I opened up.

So an important thing to do here is note my conflict of interest, with what may be seen to be the public good.

And it's not clear to me how I might have resolved this situation in the past.

It's not completely clear how to resolve it now.

It is important that I inform about my conflict of interest.

I believe I had little reason to be concerned with recognition by mathematicians, except as a responsibility to the public, and I do believe I've made good faith efforts in that regard and continue to make such efforts.

Mathematics might be profitable in and of itself with savvy efforts in that direction, but that is not a path that interests me greatly. Recognition from mathematicians is irrelevant to me, and I like to emphasize I'm not a mathematician so it has no career impact. Potential mathematical prizes pail in comparison to what I might make in other directions. And recognition could have brought potential pressure to spend time with mathematicians, which I have no intention of doing.

Actual mathematicians are just potential competition. It's better for me if they keep their distance.

By "mathematician" in general I mean a person who has a degree in mathematics. That does not necessarily apply to someone who is interested in mathematics. I am specifically talking about those academically trained in a mathematical speciality.

By noting competition I am pointing out the obvious, as it makes sense that people with specialized training and interest in mathematics could have competitive advantage over me. Potentially I might have found some collaborative relationship with someone so trained but that did not happen. For me that turned out well.

Of far greater benefit to me has been the ability to pile up results. Over time that is extremely more valuable in many ways. Today I have far greater potential stature in human history than I might have had with early recognition, especially if that had occurred a decade ago and I'd faced competition for my results.

It is my desire that this situation is resolved to the benefit of humanity as quickly as possible. And I am increasing my efforts to make that happen.

Decided to make this information public because I think it is contrary to what many might think. And others might behave differently believing that I was somehow worsened by the situation or greatly at a disadvantage by it, when in actuality it has greatly helped me.

James Harris

Sunday, January 11, 2015

Preserving information

There is one problem with ideas on blogs which is the potential impermanence. Which is why it's worth reminding that all information here is freely available for copying and sharing for the good of humanity.

Some ideas I know are permanently recorded elsewhere like my only published paper.

Never hurts to plan ahead I like to say.

James Harris

Saturday, January 10, 2015

Nearly twenty years

Back in 1995 I realized that moving into the world of work things would be shifting in terms of motivation to continue learning in certain areas, and wondered how I might sustain it. Mathematics seemed to be a great arena for pure intellectual pursuits.

I'd been studying modern problem solving since I was a kid, and realized that classical math problems considered hard offered a unique opportunity, as I could work on them indefinitely with no reason to think I'd succeed.

That might seem strange to deliberately look for problems you don't expect to solve, but that guaranteed I could work on them indefinitely. Sometimes I've likened it to a dog gnawing on a bone--the point is the effort.

And I also knew I had to limit myself, so decided to focus on what mathematicians call elementary methods, while I emphasize I am NOT a mathematician. And I have no interest in becoming one either. I just wanted the kind of thrill I'd had through a lot of my life as a supposedly smart kid. And at 25 years old I knew I didn't want to do graduate studies, as I was done with school, and didn't want to get pulled into additional formal academic study in any field.

That was back sometime in April 1995 and nearly 20 years later I'm glad my younger self sent me down this path, but little did he know I'd someday have a result like this one:

x2 + y2 = mn

will always have nonzero integer solutions for x and y, as long as m - 1 is a square and n is greater than 0.

For example: 4952 + 48882 = 176

That result looks like something from antiquity. But I found it.

Though the search is still on to see if anyone else discovered it first. Which has its own benefits, as I solved a related problem because I was searching the web looking.

And it is not like my discovery is hard once found, as calculating x and y is easy, and you can see for yourself in a post where I talk about celebrating simple math, and muse about why the result was available to me to discover.

And I strongly suspect it is my discovery alone as I used the underlying mathematics to explain a result from Euler and a related result from Ramanujan, as well as extend from them, and they are HUGE in mathematics. There's almost no scenario where if this thing were already known that wouldn't have been done before.

Like, who wouldn't want to be able to say they not only could explain a result from Euler and Ramanujan but extend it as well beyond what those two great minds started?

And it's not the only thing I've found but presents well and even it is backed by over a decade of research.

Treasure Island, San Francisco, CA, USA  April 20, 2010

Not being a mathematician probably helped me ironically enough. While being motivated to just grind at problems primarily as mental exercise did as well, and using elementary methods kept me from getting bogged down learning techniques versus inventing my own.

It's weird: sometimes constraints no matter how arbitrary can make all the difference as they guide you in a particular direction.

Resistant to using complex methods by my own choice, I was pushed to range more widely for simple ones. Not concerned really about success--though I do like it--I could keep at it for the sheer enjoyment, even in the face of 'haters' as I think motivation is very important. And some of the nastiest people out there simply have the wrong motivations.

Reality is I've had fun beyond my wildest expectations and kept my mind occupied, though lately this hobby is becoming a bigger deal as I consider what responsibilities I have to others.

I guess that's why some things are shifting as I approach 20 years of this particular intellectual pursuit and consider, now what?

And I'll readily admit I'm struggling now as I try to figure out what responsibility to others there is. Worse I'd actually like to succeed so it's not like I can just shrug off failure like I could with the math. Right now the social is mostly a mystery to me. I'm not sure what's going on, or why, or what I should do.

The math research doesn't care though, so I have no responsibility to it. A result like the one I show in this post quite simply does not need me.

A hundred thousand years from now if some being we can't imagine is reading these words that entity could appreciate my mathematical result, as it would work just as perfectly: pristine, absolute, incorruptible.

There is an odd beauty to these simple mathematical expressions and a sense of timelessness.

I think of mathematical results as like beautiful gems buried all over the place, as there is an infinity of them, and they do not care who discovers them any more than physical gems.

It's a great thing.

Playing around with simple math is a great way to exercise the mind.

I highly recommend it.

James Harris

Wednesday, January 07, 2015

Implications of latest BQD Iterator result

Was searching the web trying to find any established mathematics related to recent research interests and came across a posted problem, which I realized I could probably answer. And I did. For more detail go to this post:

And I realized the problem is actually another interesting one in its own right, so here is another BQD Iterator result.

Given is the Diophantine equation:

(x2 + ay2)(u2 + bv2) = p2 + cq2

You can find an infinity of solutions as long as c = ab + a + b, using BQD Iterators.

Like with a = 2, b = 3, which gives c = 11:

(x2 + 2y2)(u2 + 3v2) = p2 + 11q2

And to me that's really interesting in terms of compactness.

To see two of the the three iterators needed go to the post, but I'll note that you can get c easily by letting all other variables equal 1, as then you have:

(1 + a)(1 + b) = 1 + c

That using that value will work over infinity is the advance this research brings,and calculating x, y, u, v, p and q is easy with three BQD Iterators. For those who know my research, it's like, oh yeah, easy.

But notice that two products of squares can compress into another, so I'm thinking to myself that's worth noting. There is an asymmetry there I wouldn't have just assumed. Part of me for some reason isn't totally surprised, so I guess I'm building a mathematical intuition about these things.

There is a product of exponents raised to 2 on one side of the equals with a sum of them raised to 2 on the other, but that product is with independent variables. Still, I don't see myself considering this type of problem on my own, which shows the importance of looking at what others are doing.

Asymmetry rears up again as a norm. Interesting.

Clearly there is a special mathematics to these kind of sums of squares, which is rather compact. Worth noting in its own post.

Curious readers can try to solve the Diophantine equations with what they think they know to compare and contrast. Can you figure out c? How do you get answers? If you can, how easily can you do it? If you struggle, why?

To me mathematics should be not just what you think you know, but what you can learn that works best.

James Harris

Monday, January 05, 2015

Considering product of sum of squares

Found this post on mathoverflow which I figured would be fun as another easy result to be had with my BQD Iterator.

(I answered the question there and can no longer find it, so must assume my answer was deleted.)

Editing May 28, 2016: What follows does work to give solutions but have my doubts it gives a complete solution which is important. I am certain I DID find a general way to get infinite sets of solutions. The word "complete" does matter here though. Am studying this thing, but will note my confidence is shaken. Trouble with coming back later, and wondering. Oh yeah, someone questioned me on a more recent post where I refer back to this one, which got me to puzzling. Why do I think I have a proof here of a complete solution?

The poster asked for a complete solution to the Diophantine equation:

(x2 + ay2)(u2 + bv2) = p2 + cq2

And I already know that I can find integers for the following.

(x2 + ay2) = f1(a+1)n, (u2 + bv2) = f2(b+1)n

So I need: p2 + cq2 = f1f2(a+1)(b+1) to have solutions.

That forces (a+1)(b+1) = c+1, so c = (a+1)(b+1) - 1 = ab + a+b

Oh yeah, since c = ab + a+ b, that means if 'a' and 'b' are coprime then 'c' is as well. Easy. Fun!

But also it is required that: p02 + cq02 = f1f2

Which means p0 and q0 must exist such that:

p02 + (ab + a+b)q02 = f1f2

Oh yeah, I know that: f1 = x02 + ay02

And f2 = u02 + bv02

So: p02 + (ab + a+b)q02 = (x02 + ay02)(u02 + bv02)

Interesting. So for an infinite set of solutions that Diophantine equation must have solutions. If it does, then rest is easy, and it will have solutions for a given 'a' and 'b' if you set all other variables to 1, or -1, and then you can simply iterate to get more solutions. And get every possible by varying signs at each iteration.

Here are two of the BQD iterators, given:

x2 + ay2 = f1

then it must also be true that

(x - av)2 + a(x + y)2 = (a+1)*f1

And you start the iterator with f1 = x02 + ay02, which is easy enough.

Next one is:

u2 + bv2 = f2

then it must also be true that

(u - bv)2 + b(u + v)2 = (b+1)*f2

And you start the iterator with f2 = u02 + bv02. You can figure out the third iterator easily from the pattern shown.

Let's try some numbers! See if I did my math right.

Luckily see an easy path with a = 2, b = 3, which gives c = 11, and get 1 as a possible solution for all the others.


(x2 + 2y2)(u2 + 3v2) = p2 + 11q2

And the base passes ok, where all variables equal 1, so let's look at the first iteration.

Then x = -1, y = 2, u = -2, v = 2, p = -10, q = 2.

((-1)2 + 2(2)2)((-2)2 + 3(2)2)= (-10)2 + 11(2)2

Which is: (9)(16) = 144


Next iteration gives: x = -5, y = 1, u = -8, v = 0, so need to switch signs, using u = 2 from the prior iteration, u = -4, v = 4,  p = -32, q = -8

((-5)2 + 2(1)2)((-4)2 + 3(4)2)= (-32)2 + 11(-8)2

Which is: (27)(64) = 1728

Actually looks like you can always find solutions with a given 'a' and 'b' as I did above, by setting other variables to 1, and it is a complete solution as you can get all possible solutions by varying signs as you iterate.

It was an interesting experiment to see what would happen if I answered. Don't regret trying.

That wasn't completely not fun. So there's this kind of complicated looking Diophantine equation which is the control equation. I like control equations. Kind of a funky solution but sort of interesting.

James Harris

Friday, January 02, 2015

Foundations of a result

Recently I noticed a sum of squares result with integers, which is:

x2 + y2 = mn

will always have nonzero integer solutions for x and y, as long as m - 1 is a square and n is greater than 0.

For example: 1612 + 2402 = 174

Where I'm using a more elegant form than the more functional form I tend to show in posts when I'm talking about making it work, like how you calculate x and y.

Here though I want to talk about the mathematical foundations of the result, so it's worth it to show it at its most elegant as a basic fundamental result of number theory, which it is!

But from where does it follow?

Turns out there is a bit of history and the result builds on research of mine over more than a decade, which ironically resulted from my working at a problem so infamous in math circles I find I hesitate to mention it.

However, for a result this basic to have remained hidden for centuries, actually maybe a couple of thousand years, it follows there must be some new math involved. And there is, as I came up with an analytical approach I call tautological spaces, which was just extending modular algebra symbology.

That such an option was still available is extraordinary to me. But experienced math people can maybe see the difference with something like:

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

Having a modulus which is all symbols is a usual characteristic of this path of research. And you can see that result proven in a post where I step through a mathematical proof where its absolute truth is easy to show, where there is no room for logical doubt.

Using those techniques on a more complex equation I found a simpler way to generally reduce binary quadratic Diophantine equations. Which when used on the already reduced form of a general binary quadratic, gave what I now call the Binary Quadratic Diophantine iterator.

And I cover some of those foundations in mathematical detail in a prior post:

This mathematical approach of using more symbols in the modulus with modular algebra is the foundation, so links back really to Gauss and his work laying the foundations of modular arithmetic, so I like to say it's just a natural extension from there.

It's incredibly powerful as an approach, which tends to give what I call infinity results, as modular algebra kind of can't help but cover infinity.

Having scratched the surface and looked for easy answers, I got a simpler more general way to reduce binary quadratic Diophantine equations, new ways to calculate the number of quadratic residue pairs, and a way to explain the fundamental solution of x2 - Dy2 = 1, among other things, like my result above!

And that was just looking for the easy things.

You have to choose discovery.

James Harris