Good to see you VQC! Will get to work on the new hints.
The key is that the area of the triangles is the same no matter what method you use. I think the sharp triangles very accurate, and the block method makes visualization of the base much easier, ie (n-1) and (f-2)/8 = 5. VQC said “this” to my example, but then posted his example in block style >>4344
Both are valid, blocks are easier for understanding and calculating the base, which is soon belong to us ;)
I support this theory >>4285 for the following reasons:
Theory: The reason why one triangle side is less than the other is this:
3^2+4^2=5^2
You need one side to be less so the pythagorean theorem works out. Thoughts?
Come on Anons, need your input. My time is limited by IRL shit. Jump in fags!
Sorry for being short, Anons. I know we ALL have IRL shit to attend to. I apologize for being this: >>4431
Mistake:
>12^2+15^2= 144+225 = 369 Sorry Topol!
Should be:
12^2 + 16^2 = 20^2, 144+256 = 400, still base of 5.
Base = GCD????????
O Captain, My Captain!
https:/ /youtu.be/j64SctPKmqk
Hey AA! Just getting off work. Ok, here’s my current understanding, other Anons please correct me if I’m wrong on anything.
(n-1) is the small capstone of each of 8 triangles.
n0 = (f-2)/8 gives you a triangle with a base divisible by 5. Bigger than the capstone triangle, smaller than the entire triangle. I also think you could get this base by doing (n-1)*5?? Not sure on this idea. VQC is hinting that multiples of this n0 triangle can be used to fill up the remaining space inside each triangle, allowing us to solve for x+n or n.
Then the bottom base portion is ((x+n)-1)/2 as you know. So n0 is somehow going to help us solve x+n, possibly in combo with the factor trees.? Working to understand over here.
>These diagrams are fantastic
Blessings To You, MM.
When a lot of us were tired from working so long, you got us fired up again. Get some rest, then come back and bust ass again.
Although I guess it's irrelevant for RSA numbers. Just trying to find proper sized examples to work on so I can hopefully see patterns. What about (x+n)=15? f=30 so f-2=28. what next?
28/40 = n0= 7/10?
28/8 = n0 = 3 + (28 mod 8) = 3 + remainder 4? If this, then what to do with the remainder?
Here's the example for c=259, (x+n)=15. Attaching PMA's output for clarity.
We start with c = 259
d = 16
e = 3
f = 30
n0 =(f-2)/8 = 3 remainder 4
Note: (f-2) mod 8 = 4 (this calcs the remainder)
Now we iterate n0 triangle base using multiples of n0 = 3
n0*1 = 3 (no match)
n0*2 = 6 (no match)
n0*3 = 9 (no match)
n0*4 = 12 (no match)
n0*5 = 15 (Match!)
So TuXPN (Triangle Base x plus n) = 15
Note: Method needed to verify when we reach the correct (x+n) value.
Now we use the quadratic formula.
n = SQRT(c + (TuXPN)^2) - d
n = SQRT(259 + (15)^2) - 16 = 6
Now we can also solve for x.
x = SQRT((d+n)^2 - c) - n
x = SQRT((16+6)^2 - 259) = 9
Now we can solve for a and b.
d - x = a = 16 - 9 = 7
c/a = b = 259 / 7 = 37
Complete element. {3:6:16:9:7:37}
Thoughts, Anons?
Typo, lads.
>x = SQRT((16+6)^2 - 259) = 9
Should be:
x = SQRT((16+6)^2 - 259) - 6 = 9
CA and PMA tearing it up! I've studied all your guys' output, and it looks good. I'm working to understand the underlying ideas, I've got about 90% of it down.
>The theory is that each pair of numbers points you to a different prime (maybe not even a prime idk) as you recurse up the tree and it should end at the right factor.
>Also my idea is a way to generate prime numbers.
What are the algebra formulas for web.one and web.two? Or how to calculate web1 and web2? I understand this idea:
>for any a,b, where a or b is even (so that ab is even), we can make a continued fraction [(ab)/2; (a, ab)] which evaluates to the square root of an integer value. The integer value would again be dd+e which is (aabb)/4 + b.
Can CA or PMA give a quick explanation of the formulas and how they work? Almost there in my understanding.
Alright, I've found an appropriate shitposting niche! If PMA is commenting on it, it's gold.