Let m be an odd discrete semiprime. (I.e., m is the product of two different odd prime numbers.)
Let c be the square root of the least perfect square greater than m, and let r be the positive difference between them, so that m = c2-r.
Consider the quadratic polynomial y = x2+2cx+r. Determine the smallest positive value of x for which y is a perfect square. [This is the step that is currently difficult for large m unless we can formulaically simplify it.]
For that value of x, (c+x)2-m is also a perfect square, leading to a Fermat two-squares factorization of m.
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