VA !!Nf9AmQNR7I ID: 174e7d Dec. 2, 2018, 10:29 a.m. No.8423   🗄️.is 🔗kun   >>8424

>>8422

Farey Tree Anon, I studied that Plimpton 322 article. Hilarious that all this probably started as a project cost analysis for some engineer or business manager ~6000 years ago. Like "Shit guys we gotta figure out the dimensions of this ramp some we know how much limestone to order from the quarry for this 1000 cubit length ramp. Go figure this shit out, we're on the King's payroll anyways, and we don't want too many slaves dying hauling extra rock from the quarry if we're gonna make our deadlines. Tell that fucking nerdy kid who sits around thinking all day to figure it out. He's a fucking weirdo, oh well, he's great at drawing triangles all day."

Thus was born the First Autist, lol.

VA !!Nf9AmQNR7I ID: 174e7d Dec. 2, 2018, 9:15 p.m. No.8425   🗄️.is 🔗kun   >>8426

>>8424

>>8420

Here’s another example, this time for even x+n

Even (x+n)^2 = 144

f=2d+1-e = 24

Sqrt(24) = 4 r 8

4 * 36 =144

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

VA !!Nf9AmQNR7I ID: 174e7d Dec. 2, 2018, 11:01 p.m. No.8426   🗄️.is 🔗kun   >>8429

>>8425

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

Correction: 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)