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:

The appropriate BQD iterator is, given:

u

^{2}- (T+1)v

^{2}= 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:

**x**

^{2}- (T+1)y^{2}= (-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:

1

^{2}- 16*1

^{2}= -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