###
Connecting Diophantine hyperbola to ellipse

The equation x^{2} - Dy^{2} = 1 has a direct Diophantine connection to an ellipse for a special case.

The ellipse is: (D-1)j^{2} + (j +1)^{2} = (x+y)^{2} where j = ((x+Dy)-1)/D, when x = 1 mod D.

Or: (D-1)j^{2} + (j -1)^{2} = (x+y)^{2} where j = ((x+Dy)+1)/D, when x = -1 mod D.

So the special case is x = 1 mod D or -1 mod D, which must be true if D is a prime or twice a prime, but the connection will exist whenever x meets the conditions.

Proof:

Using my method to reduce binary quadratic Diophantine equations on x^{2} - Dy^{2} = 1, gives:

(x+Dy)^{2} - D(x+y)^{2} = -(D-1)

(If you multiply out and simplify you simply return to x^{2} - Dy^{2} = 1.)

Shifting things around a bit, I have:

(x+Dy)^{2} + (D-1) = D(x+y)^{2}

And I can group to divide off D:

[(x+Dy)^{2} - 1]/D + 1 = (x+y)^{2}

Now we'll shift to the special case of x = 1 mod D or x = - 1 mod D.

Introducing integer j, I have x+Dy = jD +/- 1, and only one of those cases will be true. That usage is to keep from writing x+Dy = jD + 1 or x+Dy = jD - 1, over and over again.

And making that substitution gives:

[(jD +/- 1)^{2} - 1]/D + 1 = (x+y)^{2}

And squaring and simplifying a bit gives:

Dj^{2} +/- 2j + 1 = (x+y)^{2}

So now I just subtract and add j^{2} so that I can complete the square to get:

(D-1)j^{2} + (j +/- 1)^{2} = (x+y)^{2}

And you have your ellipse.

Proof complete.

Here is an example of the result with the famous case of D=61:

1766319049^{2} - 61(226153980)^{2} = 1

x = 1766319049, y = 226153980, and x = 1766319049 = -1 mod 61, so the second equations apply.

(D-1)j^{2} + (j - 1)^{2} = (x+y)^{2}, j = (x+Dy+1)/D

Which gives j = 255110030.

60*255110030^{2} + 255110029^{2} = 1992473029

^{2}

If you'd like to continue further, additional research shows you can now generate yet another hyperbola from the ellipse.

James Harris
Post a Comment