Thursday, October 03, 2013

Possible path to two conics equation solution?

Sometimes I use this blog to put up idle thoughts where I may follow-up later. It lets me time-stamp things and helps me to figure things out by talking about them publicly, even when still highly speculative.

I finally realized I might pair a re-discovery with a new discovery of mine.

The re-discovery is the parametric solution for what I call the two conics equation which mathematicians commonly call "Pell's Equation".

y = 2t/(D - t2) and x = (D + t2)/(D - t2)

when x2 - Dy2 = 1.

From my own research I have:

u2 + Dv2 = F


(u-Dv)2 + D(u+v)2 = F(D+1)

So for the two conics equation:

x2 - Dy2 = 1


(x+Dy)2 - D(x+y)2 = -(D-1)

on the first recursion, and the second recursion gives:

((1+D)x+2Dy)2 - D(2x + (1+D)y)2 = (D-1)2

Using the parametric solution with t = -1 (it can be 1 or -1), and substituting for x and y with the two conics equation and simplifying a bit gives me:

(D+1)2 - D(-2)2 = (D-1)2

So I have:

(1+D)x + 2Dy = D+1 and 2x + (1+D)y = -2

Some notes. Yeah, it's kind of interesting that ONLY for t=1 or -1, do I get this possible route. And both conceivably should be checked. Or does it matter? Don't know at this point.

Signs are a big deal here. And can shift them quite a bit because of the squares. Think I put up a non-trivial case, but not certain.

Trivial case gives x=1 or -1, and y = 0.

So yeah, I can solve for x and y, anew. And that's the idea. Substituting back into the two conics equation again, I think that should give me yet another solution.

That could mean a loop where of course if x and y are integers you would exit.

I can see some immediate potential issue where it won't work, but not checking for now as simply throwing it up there for later.

Coming back it occurs to me that I can abstract things a bit, to see a little bit more, by considering:

(1+D)x + 2Dy = s1 and 2x + (1+D)y = s2

And using the second, solving for x gives: x = -(1+D)y/2 + s2/2

And substituting gives: (1+D)(-(1+D)y/2 + s2/2) + 2Dy = s1

(1+D)(-(1+D))y + (1+D)s2 + 4Dy = 2s1

Which is: (D-1)2y = (1+D)s2 - 2s1

And that is SO awesome! It verifies that I will get an iteration situation. Does it matter if D+1 is even or not? Not sure. I don't think so, as the focus is on y. If it's a nonzero integer you're done. Looks like mod D-1 has been proven (shows at most D-1 iterations if x and y are new each iteration). No, that's not right? Should be squared, right? That is fascinating too. That's enough for now.

James Harris