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.