Here I write that result as, given
x2 + Dy2 = F
it is forced that
z2 + D(x+y)2 = F(D+1)
and you can verify trivially that z=x-Dy, by substituting out F, from the first equation into the second:
z2 + D(x+y)2 = (x2 + Dy2)(D+1)
so, expanding and simplifying:
z2 = (D+1)x2 + D2y2 + Dy2 - Dx2 - 2Dxy - Dy2
z2 = x2 - 2Dxy + D2y2 = (x-Dy)2.
And it may seem like a trivial result, but now you have a chain where the first 3 equations are:
x2 + Dy2 = F
(x-Dy)2 + D(x+y)2 = F(D+1)
(x-Dy - D(x+y))2 + D(x-Dy + x+y)2 = F(D+1)2
((1-D)x-2Dy)2 + D(2x + (1-D)y)2 = F(D+1)2.
For each following equation you use the general form that with:
u2 + Dv2 = C
the next equation is given by
(u-Dv)2 + D(u+v)2 = C(D+1)
so the next in the series is
(((1-D)x-2Dy)-D(2x + (1-D)y))2 + D(((1-D)x-2Dy)+(2x + (1-D)y))2 = (F(D+1)2)(D+1).
Which simplifies to
((1-3D)x + (D2 - 3D)y)2 + D((3-D)x + (1-3D)y)2 = F(D+1)3.
And that may not seem remarkable unless you realize that for every solution to the first equation, you have a solution to that last with (D+1)x, and (D+1)y, so what if you go backwards and ask, can
(1-D)x-2Dy = +/-(D+1)x or +/-(D+1)y
2x + (1-D)y = +/-(D+1)y or +/-(D+1)x?
But if so, then you have two solutions for x in terms of y, so BOTH must give the same answer or there is a contradiction.
For instance if
(1-D)x-2Dy = (D+1)x, then x=-y, is a solution, and then
2x + (1-D)y = (D+1)y, gives that x = Dy, and only D=-1 can work.
But then I have x2 - y2 = 0 = F, so no solutions.
So, it's now possible to see that all integer solutions to equations of the form x2 + Dy2 = F, are part of this immensely huge number theoretic structure, which is infinitely sized, where depending on the value of D, and the value of F, at various points in the chain you will have situations when you have multiples of x and y.
What's remarkable about that super structure is it allows an explanation for all previously noted behavior and indicates that solutions have to do with linear equations and nothing else.
Intriguingly there is other research noting a lack of a need for non-rational numbers with these equations:
Pell's equation without irrational numbers
Authors: N. J. Wildberger
(Submitted on 16 Jun 2008)
Abstract: We solve Pell's equation in a simple way without continued fractions or irrational numbers, and relate the algorithm to the Stern Brocot tree.
Comments: 9 pages, 3 figures
Subjects: Number Theory (math.NT)
Cite as: arXiv:0806.2490v1 [math.NT]
So there is an infinitely sized and static number theoretic structure which is just an infinite expansion where the first four equations are:
1. x2 + Dy2 = F
2. (x-Dy)2 + D(x+y)2 = F(D+1)
3. ((1-D)x-2Dy)2 + D(2x + (1-D)y)2 = F(D+1)2
4. ((1-3D)x + (D2 - 3D)y)2 + D((3-D)x + (1-3D)y)2 = F(D+1)3
and that goes out to infinity. To get successive terms in the series you use the algebraic result that given:
u2 + Dv2 = C
it must be true that
(u-Dv)2 + D(u+v)2 = C(D+1).
And where whenever the exponent of (D+1) is even, you can have a case where you just have a multiple of x and y, so you can solve for D, which defines possible values for F in terms of x or y.
So the previously seen behavior with solutions to these equations can be explained as finding points in the expansion where you have a multiple with a given D, which will allow your F.