Tuesday, October 16, 2012

Very useful little math result

One of my favorite math discoveries is a surprisingly simple result which I've used for years where a lot of recent posts are I think a culmination of years of following down paths that flowed from this one thing.

Oh yeah, it's worth mentioning again that I am NOT a mathematician. And I have no intentions of ever being one, while I have training in physics with a 4 year degree, but then with that little training I'm not a physicist either. But it's intriguing to me that while I have a lot of math in my background, it was mostly in the field of complex numbers. And with physics you end up doing a lot of approximations.

And I don't really like approximating.

I love exactness, and to me playing with integers is relaxing where at the end I can look at exact answers as you're looking at integers.

Like getting back to this extremely useful result.

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)

And it's easy to get to actual integers, as let F = 1, u=x, v=y, and D = -2.

Then just plug those values in and let if fly:

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. The solidity of exactness is so comforting. Perfection.

You can actually verify the result by just multiplying out and simplifying:

(u-Dv)2 + D(u+v)2 = F(D+1)

Which is kind of fun, and if you do it, you'll find it just reduces back to the original:

u2 + Dv2 = F

The cool thing is that I figured it out using my general method to reduce binary quadratic Diophantine equations on that original.

Where that was about curiosity as the reduction method allows you to reduce any binary quadratic Diophantine equation to that simpler form, so I wondered to myself, so what happens with the simpler form if I use the same process on it?

It just kind of gives a variation on itself, as the math really had nowhere else to go.

The next one in the series is easy enough: (3(17)+4(12))2 - 2(2(17) + 3(12))2 = 1

Which is: (99)2 - 2(70)2 = 1

And that gives an approximation the square root of 2, as 99/70 is about: 1.4142

And if you're bored you can just keep going! Next one is: x=577, y=408, and 577/408 is approximately 1.41421.

But I don't like approximations and just throw that in there because it seems I should.

Now I've only touched the surface as you can also figure out a connection between discrete ellipses and hyperbolas. But keep going, as you can also use it to figure out the reason behind the size of the fundamental solution for something mathematicians like to call Pell's Equation.

But keep going, and the research path can lead you to a general modular solution for binary quadratic Diophantine equations in the reduced form.

For me it was years of chasing down those paths which was very satisfying, with exact answers in so many places. Nothing like the certainty of integers, which is why for my hobby I much prefer these types of intellectual exercises to physics problems, where approximate might be the best answer.

James Harris
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