Binary Quadratic Diophantine Iterator

It must be that if you have:

u² + Dv² = F

then it must also be true that

(u-Dv)² + D(u+v)² = F(D+1)

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In general true binary quadratics, except for the cases D = 0 or D = -1, F = 0, are connected to an infinity of others, as all can be reduced to this form, so it's what I like to call an infinity result.

By true binary quadratic I mean one that does not have a form like this example:

x² + 2xy + y² = c₄ + x + y

is: (x+y)² - (x+y) - c₄ = 0

So I don't consider that to be a true binary quadratic because it behaves like a single variable equation, with x+y as the single variable.

One of my favorite discoveries I've decided to call it a binary quadratic Diophantine iterator.

The beauty of the iterator is that an already reduced equation simply gives you another of the same basic form, as:

(u-Dv)² + D(u+v)² = F(D+1)

Is equivalent to the original as notice, with u' = u-Dv, v' = u+v, and F' = F(D+1), you have:

u'² + Dv'² = F'

You get the iterator when you use my method for generally reducing binary quadratic Diophantine equations on the already reduced form, and to me is one of the coolest things ever. Looks like I first gave it in its general form back October 2008. This thing conceivably could have been discovered 2000 years ago, but somehow it wasn't so I had the honor of finding it.

While it is derived it can be verified to be correct simply by using the equations given.

Given:  u² + Dv² = F

Verify: (u-Dv)² + D(u+v)² = F(D+1)

Verification:

Expand out: (u-Dv)² + D(u+v)² = F(D+1),

which gives:

u² - 2Duv +  D²v² + Du² + 2Duv + Dv² = FD + F

Notice -2Duv cancels out +2Duv, and then it's just a matter of grouping:

u² + Dv²  - F + Du² +D²v² = FD

And: u² + Dv²  - F = 0, so:  Du² + D²v² = FD, and dividing off D, gives:

u² + Dv² = F

Verification complete.

Example using iterator:

Let F = 1, u=x, v=y, and D = -2.

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 to x² - 2y² = 1 and x² - 2y² = -1.

When I found it years ago back in 2008 I extended the symbolic version further. Copying four of the five I see in that post to here for convenience. Here I'm using u = x, v = y.

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

While the symbols go out to infinity, when using actual numbers you may be able to divide off factors from D+1 routinely. So for instance with D = 1, every other iteration 4 can be divided off if you so choose. So it still can be iterated out to infinity, but that then seems trivial to me, and when 4 is divided off the series simply oscillates.


James Harris