I realized something today interesting or not, known or unknown, but that tickled my fancy.

The equation we have:

nn + 2d(n - 1) + f - 1

can also be seen as an expanded form of the difference between two arbitrary squares. We can rewrite it as follows:

nn + 2d(n - 1) + f - 1

nn + 2d(n - 1) + (2*d + 1 - e) - 1

nn + 2d*n - e

n(2d + n) - e

Which (when you ignore the e) is the result of the difference between two squares.

I don't think this will be mindblowing, or anything, but it gave me some insight.

>>5990

Just spitballin' here. We don't know (x+n) yet, but we know that there is a number in (e, <some n>) that is equal to (x+n) + e.

>>5994

I was looking into this, just as a curiosity and it makes sense.

(x+n)*2 + e will exist with a=n, b=2d + n.

It will have the same e as the original record and a few other interesting properties. For the record, the a=n, b=2d + n is simply because of (x+n)2 + e = n(2d + n).

The interesting part is that the n in the record for a=n, b=2*d + n is equal to a from the original record.

I guess this will only hold for the records where (x+n)**2 = n(2d + n) - e.