Hello AA! Here's a Pattern I'm working on to factor the odd (x+n) square, using (x+n)^2 -1 to get 4 rectangles or 8 triangles.

For c6107, interesting that our u1 and u2 values are 41 and 42, since our f derived polite numbers = 3+4=7

The total x+n dimensions are 83 * 83 = 6889.

4 rectangles of 41 * 42 +1 = 6889

It’s the (n-1) pattern playing out at 2 levels simultaneously.

Maybe a fluke, just brainstorming over here.

c6107 f= 134

f div 8 = 16

Sqrt(16) = 4

Sqrt(16)- 1 = 3

4+3=7. These are our two polite triangle number bases added together.

7 * unknown + 1 = correct (x+n)^2

For this example it turns out to be 984.

The factor will scale with c, since it's derived from f.

This should be blazing fast at computer calc speed! Since the two polite triangle bases come from f, it should scale upwards with increasing c values.

All we have to do is check each iteration to see if it's a perfect square.

This is the two equations merging together or running alongside each other. (i think)

Equation 1 is: SQRT(c + (x+n)^2) - d = n

Equation 2 is: f factor * iteration + 1 = a perfect square? If not, don't run it through equation 1.

This was why VQC told us we would be estimating n0 when we used the (f-1) div 8 method. We’re looking for a way to use f to construct (x+n)^2

This is one of the ways to do it. There are multiple correct methods.

I’m just working on the one that I understand the most.

Can we try it out on the RSA100 Numbers we have all the solutions for?

Essentially I'm postulating that (x+n)^2 is a multiple of f or one of its roots, so a big difference from just (x+n).

The factor derived from f * iteration forms / fills the area of the square.

Odd (x+n) formula is f * unknown + 1 = (x+n)^2

Even (x+n) formula is f * unknown = (x+n)^2 (possible)