Harristotelian Logic

Consider

x^2 - ax + n = 0

and

y^2 - by + n = 0

with a, b and n natural numbers and now subtract the first from the second to get

y^2 - x^2 = by - ax

and group a bit, and divide by y - x to get

y + x = (b-a)*y/(y-x) + a

so

(y + x - a)*(y-x) = (b-a)*y

which is the generalized factorization result.

And, of course, you can solve for x and y to get

x = (a + sqrt(a^2 - 4n))/2 and y = (b + sqrt(b^2 - 4n))/2

so you can use this result to further factor y.