>>5718

Yeah it wasn't as foolproof but yeah I'll post some output.

(b-a)/2 = (x+n) and (b+a)/2 = (d+n)

>>5722

>>5723

My suspicion is x+n is either:

gcd(n,d+n)

gcd(f,x+n)

sometimes a multiple of the gcd.

>>5724

So I'm thinking when f = d, we have a = 6k+1.

If f is a square, (x+n) is the square root of f.

If the gcd(f,x+n) = f, then (x+n) is fj where j is a factor of the original (x+n) = (c-1)/2

There is probably other stuff but I'm sleepy