JCSBimp's Hexagonal Orthogonal

[To the regular reader: I do apologize for a bit of repetition of definitions, but I do this mainly for someone just jumping in to my posts for the first time, so that they’ll get up to speed quickly.]

Let m be a discrete (square-free) semiprime, c its Ceiling Root and r the difference c²-m, so that m=c²-r.

c, as defined in previous blog posts, is the least positive integer that is odd or even, as (m+1)/2 is odd or even, such that c²>m.

Let c₀, c₁, c₂ be the integer sequence starting with c and increasing by two, so that c_n=c+2n. Construct an integer sequence of remainders r_n by subtracting m from each c_n². For my studies, I only need to increase n until it reaches the value for which c_n=(m+1)/2, the greater square corresponding to the “trivial” factorization, 1 times m.

This series of remainders r_n are the values of the quadratic polynomial x²+2cx+r evaluated at x=2n. Similar to what I’ve established elsewhere, the least non-negative value of n for which n²+2cn+r is a perfect square gives a remainder that allows for factoriztion of m as (c_n-sqrt(r_n))(c_n+sqrt(r_n)).

The real trick, of course, is predicting when this will happen, using the values of c and r which we have. The quest continues!