Infinite difference of squares

Using the BQD Iterator it's possible to get an infinite number of difference of squares for certain integers raised to successively higher powers.

The appropriate BQD iterator is, given:

u² - (T+1)v² = F

then it must also be true that

(u + (T + 1)v)² - (T + 1)(u + v)² = -T*F

So if you start the iterator with u = v = 1, or u = v = -1, F = -T.

That means that in general there must always exist nonzero x and y, such that for an integer n equal to 0 or higher, and an integer T equal to 2 or higher:

x² - (T+1)y² = (-T)ⁿ⁺¹

I have -T raised to n+1 so that n is a count of iterations, if you want T raised to the nth power you just start n at 1 instead of starting at 0, which I prefer.

And if T+1 is a square you get an infinite series of difference of squares. That's cool. Will try it out.

So, let T = 15, and u = v = 1. Then, iterator is:

(u + 16v)² - 16(u + v)² = -15*F

Start is:

1² - 16*1² = -15

then it must also be true that

(17)² - 16(2)² = 225 = (-15)²

Next iteration: (49)² - 16(19)² = -3375 = (-15)³

And of course you can keep going out to infinity.

Notice the factorization at the first iteration: (17 - 8)(17 + 8) = 9(25) = 225 

But of course if T + 1 is a square then it is trivial to factor.

James Harris