>>5251

Hello PMA! Studied your output closely. So the (x+n) and u values are the same for both (1,c) and (prime). This the essence of your capstone idea, correct?

Also, I understand and follow your idea about dealing with the remainders, especially the large remainders in the (1,c) calcs.

For this improved method, we are now essentially running an iterative search that also accounts for the remainder at each iteration and searches for a value where rm (x+n) = 0 and rm 2d(n-1) = 0. Have I understood your idea correctly?

>>5287

>Coincidence? I think not.

Very cool!

>>5288

Welcome, Anon! Glad to have you on board.

>>5283

Howdy Topol! Thanks for the snacks.

>>5285

Hello PMA! Thanks for the new ideas and output. I've studied them closely and understand. So at this point you're working to speed up the iteration process by looking for quicker ways to deal with the remainders, correct? Remember these crumbs? Handling remainders is key, base selection is arbitrary. We're on the right track, lads!

>The objective is to find a base larger than n and smaller than x+n at this stage.

>>4338

>>4339

>>4340

>>5274

>>5263

Hello AA! Thanks for keeping shills off the board, and thanks for your work on the binary ideas. I've checked out your output and there are definitely patterns there. I have no clue yet how to decode them, but thanks for chasing down new ideas.

VQC was saying the connection between (1,c) and (prime) records is easier to see in binary. We got very close to connecting them a few breads back. PMA's output had us one step away from a connection. Takes a look at the numbers I've put a box around in binary??

>>5331

Hey Topol! I'm all signed up. Don't like giving my email (of many) but fuck it. NSA isn't stopping our Rothschild Annihilation Mission.

Hello Lads! Just checking in over here to see what's going on. I've been turning over all the ideas in my brain and studying our recent posts, along with VQC crumbs. It's pretty Mellow over here.

>>5335

Hey AA! Eh, Discord is cool for chatting, but I prefer working here.

>>5355

Welcome back Senpai!! What a nice surprise to wake up and find your new posts. What’s the best way to check iterations to know when you’re getting close to a (x+n) match? Is there a way, or do we just iterate until a match shows no remainder?

>>5364

Hey boss, PMA has been kicking ass while you’ve been away. If there’s a VQC trophy you should award it to him. We’re all following the big ideas, thanks to PMA pumping out diagrams, ideas, and code.

>>5433

>>5434

>>5435

>>5437

Yeah MM! Thanks for the cool output! I’ll study it now.

>>5439

Damn. Sorry MM, had no clue, just trying to help out. All the other breads are archived there too. What are your thoughts?

>>5439

Eh. NSA knows all. “And by all I mean all.”

Google is about to suck some DJT NSA dick. If what Q is saying is true, the good guys are about to mop up 400 years of Evil Bastards and foil their ill-begotten plans. Google can suck our dicks.

>>5442

>>5445

I guess what I mean is that they will be forced to serve patriots instead of the Cabal. God, we dodged a bullet. We were about to go full totalitarian regime.

>>5447

Try all of human history. What is the source of Evil? Primary question of human existence.

>>5562

>>5563

>>5564

>>5573

Good to see you VQC! Whoa, death threats are not good. (((They're scared.))) Q team should send you a security detail! If you can't post anything else at the moment, we'll work on your Recap and exercise.

Hey Lads, cell (2,1) has these exact ascending (x+n) examples to work through. They're also in(4,1) (6,1) etc as var e increases. I can see that as e ascends by 2's, d ascends by 1. Grid snapshot attached. Cool patterns everywhere. Let's delve into these (x+n) patterns.

(2,1,1) x+n = 1

(2,1,2) x+n = 3

(2,1,3) x+n = 5

(2,1,4) x+n = 7

etc.

VQC's questions:

>For each odd square what are the permutations of possible values of f,d and (n-1)?

>What are the patterns?

>What do these look like?

I also attached the graphic VQC was referencing, and my working spreadsheet for the exercise. Working up triangle diagrams now. I've got famalam stuff today, and I'll stop in as often as possible.

Question: For the example I've posted, is (n-1) just 0 for every example? Or is (n-1) a reference to something else I'm missing or not understanding yet?

Here's a quick workup of the increasing (x+n) squares. Cool find, the correct length of (x+n) is always formed by u + (u+1),

or (x+n) = 2u+1

or (x+n) = correct n0 + correct n0 +1

>>5576

Whoa. Another cool reminder of why we can solve for all of row one using SQRT(f). Check it out. (x+n) = SQRT(f) = SQRT(2d+1-e) for row one.

Oops, here's the screenshot.

>>5582

This is a great explanation, AA.

>The main points of comparison are f, d and (n-1), as VQC said. Within the infinite sets where n=1, f will always equal (x+n)^2 and (n-1) will equal 0. This will be true for all odd values of (x+n) where n=1 (which is an infinite set). In the case of nn+2d(n-1)+f-1, the value of d is irrelevant, because 2d(1-1) is 0. Also, nn-1 will be 0. That means, for the infinite set for which (x+n) equals an odd number and n=1 (which will begin after a certain increment of a and b), (x+n)(x+n) = f. That means we can very easily calculate x and n, since, if f=(x+n)(x+n) and n=1, x will be sqrt(f)-1. This is true for an infinite set of numbers. Either I’m somehow wrong (I don’t see how I could be) or this has been staring us in the face the whole time.

>>5623

Sure, Anon! More diagrams to follow after work today. Attaching VQC's diagram to help explain.

1. We start with c.

2. Then we get d, e, and f using known formulas.

3. Then we use f for two tasks related to constructing and searching for the correct (x+n) square.

4. First, we use f to estimate n0. This "Starting n" will allow us to iterate through possible values of n to look for a match.

5. Second, we use f to begin filling the (x+n) square. We don't yet know how big the square is. The formula for the Area of the (x+n) square is nn+2d(n-1)+f-1, so we plug in our full f value.

6. Now we begin to iterate n0. As we increase n0, we get new values for nn + 2d(n-1). f -1 remains constant in the formula. This is the bottom part of my spreadsheet.

7. We know that (x+n) is a perfect square, so each time we iterate, we check to see if we have a perfect square for the Area of (x+n)(x+n). We do this by taking the square root of our new (x+n)^2 value we've created. If it has a remainder, it's not a match. If it has no remainder, it's our match!

8. We now have the correct (x+n) and the correct n. We can also solve for x easily by taking SQRT(correct (x+n)^2) - correct n0 = x

9. We can now solve for a and b.

10. BOOM. Done. Simple and clean.

11. It works for all the small examples I've tested, but we need to test on many more examples.

12. The key idea I was missing is that f is also used to CONSTRUCT the (x+n) square. In fact the size of f fits perfectly with a multiple of nn+2d(n-1) -1.

13. VQC gave us all the pieces and formulas, but there was some assembly required. He kept hinting to build out all the (x+n) squares and play with the pieces to see how things fit together. Try it out for yourselves, Anons. When we iterate n0, the part that's increasing is nn+2d(n-1). f -1 remains constant.

14. I think we're about ready to move on to part 3b.

15. Can some of you excellent program Anons give this a full testing out? Let's run this bitch on the RSA 100 numbers.

Good discussion going on on Topol's second Bake. Here it is anons.

>>5562

Recap and exercise for odd (x+n) squares.

>>5592

I'm going to work on this, in addition to the (x+n) exercises. This part is cool:

>The remainder of f when divided by 8 is a key to what values of d and (n-1) can be used. Playing with these patterns will organically lead you exactly where we will be going.