JCSBimp's Hexagonal Orthogonal

When m is the product of two odd primes p and q, with p < q, and m is also equal to s^2 – r where s is the integer ceiling of the square root of m,

the polynomial x^2 + n*x – (r + s*n) for the positive integer n that allows it to factor, and thus lead to a factorization of m by setting x equal to s, produces a discriminant in this case, n^2 + 4*(r + s*n), with a square root that turns out to be equal to q – p.

Though this discovery surprised me, I suspect it is neither profound nor particularly helpful in determining the value of n that will produce the desired result. Back to the search for something that will be helpful.