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:
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
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