*[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 c^{2}-m, so that m=c^{2}-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 c^{2}>m.

Let c_{0}, c_{1}, c_{2} 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_{n}^{2}. 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 x^{2}+2cx+r evaluated at x=2n. Similar to what I’ve established elsewhere, the least non-negative value of n for which n^{2}+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!

## Leave a Reply Cancel reply