I'm not gonna explain a long math magic behind this, but here is much faster version, that can give correct anstwer for every 8-digit n under 7s

` def bound(n): return int(n*(1-math.sqrt(2)/2))

def calc_b(nn, n, a): if nn % (n-a) == 0: return (nn-a*n)//(n-a)

def pit(n): if n % 2 == 1: return -1

nn = n*(n >> 1)  # nn = (n^2)/2
for a in range(bound(n), 0, -1):
    if b := calc_b(nn, n, a):
        return a*b*(n-a-b)
return -1

`