c

_{1}x

^{2}+ c

_{2}xy + c

_{3}y

^{2}= c

_{4}+ c

_{5}x + c

_{6}y

where x and y are the unknowns to be figured out. A simple example of such an equation is:

x

^{2}+ 2xy + 3y

^{2}= 4 + 5x + 6y

where I've simply used, c

_{1}= 1, c

_{2}= 2, c

_{3}= 3, c

_{4}= 4, c

_{5}= 5, c

_{6}= 6.

With my simple example above I can reduce to a simpler form using my own research to get:

(-4(x+y) + 10)

^{2}+ 2s

^{2}= 166

Now you can solve for x+y and s, and it's easier to find that s = 9 works, and x+y = 2 or 3, and x = 4, y = -2 is a solution. To see me work through in more detail, click here.

Now you may say, so what? Well, turns out we can immediately use this thing to get a simple result previously unknown before I discovered it, which is one of the greatest math finds of all time, and we're going to use it for some simple math trivia and approximate the square root of 2 with it.

And I start with a simple equation:

u

^{2}+ Dv

^{2}= F.

And with my general method to reduce binary quadratic Diophantine equation we can find that:

(u-Dv)

^{2}+ D(u+v)

^{2}= F(D+1)

And I'm now going to let D = -2, F = 1, since we're going after the square root of 2, and to make the equation look like a more familiar one I'm going to shift variables with: u=x, v=y, so my original is now:

x

^{2}- 2y

^{2}= 1

And now I can crank through my result to get:

(x+2y)

^{2}- 2(x+y)

^{2}= -1

And it's iterative! So I can do it again and again:

(3x+4y)

^{2}- 2(2x + 3y)

^{2}= 1

Next is:

(7x + 10y)

^{2}- 2(5x + 7y)

^{2}= -1

And iterating one more time:

(17x + 24y)

^{2}- 2(12x + 17y)

^{2}= 1

And the more astute of you may have noticed that x = 1, y = 0 is a solution to the original equation, so guess what? We've solved the original equation as well with JUST my research result. Using that on the last:

(17)

^{2}- 2(12)

^{2}= 1

And we can still iterate, but let's do it now with just the numbers.

So now for another iteration: x = 17, y = 12,

so: (3(17)+4(12))

^{2}- 2(2(17) + 3(12))

^{2}= -1

Which is: (99)

^{2}- 2(70)

^{2}= -1

And that gives the slightly more impressive approximation of 99/70 is about: 1.4142

And if you're bored you can just keep going! Where now x=99 and y=70. And it works out to infinity with ever more precise approximations to sqrt(2).

Next one is: x=577, y=408, and 577/408 is approximately 1.41421.

Why do these solutions approximate the positive square root of 2?

Because x

^{2}- 2y

^{2}= 1, is:

(x

^{2}- 1)/y

^{2}= 2, and you can just take the square root of both sides now:

sqrt(x

^{2}- 1)/y = sqrt(2), so the trick then is that

*approximately*x/y = sqrt(2).

And x

^{2}- 2y

^{2}= 1 was used over a thousand years ago, and one of its uses was, yup, approximating square roots, and I wonder if I'd have been cheered if I showed some of the ancients my simple result above?

I've used my result with much bigger things though.

James Harris