>>887

I'm trying to understand now. Whats the latest equation to test?

sqrt( c - (aaxbb).x ) == (axb).a?

>>888

I fear you may be proving d-x=a?

>>924

This looks pretty good.

Can you run through another example? I think I'm following, but i'm not sure.

>>942

>in my example above, I don't think it's coincidence that the c^2 record is at (1,10368,73) and it's corresponding a=na record is at (0,1,73).

I agree. Very interesting.

From (0,1) we should easily be able to find a, right? Since if p is a factor of a at t, p will also be a factor at p+t, 2p+t, 3p+t, and p+1-t, 2p+1-t, 3p+1-t, etc.

>>942

sqrt(a/2) in the (0,1, t for ccna) row is always equal to the 1xc row x+n.

>>951

I agree, bitcoin price reaction is unrelated.

I expect we're a month out from a true crash due to cracking ecc. If we're lucky.

I've been monitoring bitcoin for a while, there's always a dip at every major milestone, normally 25%, but this time I think people are getting scrared, its dropping closer to 33%.

>>1026

I'm still here and working too.

I'm going back through Chris's posts and trying to re-read all the information to see if we're missing anything.

I'll post what I find.

>>1028

One thing I'm obsessed with is the fact that x&n are 2 squares, as are (d+n) and (x+n).

There's some pattern either, x+n's or x's for 1xc and axb are always factors of each other.

There's more too it then that, but I think maybe the triangles made by the axb and the 1xc x+n squares may be similar.

>>1030

To continue that, VQC said:

>I thought from near the start in 2011 that it would feel good if "x" marked the spot.

I think factoring x, or 2x might be useful, so we recursively repeat that until row 1 or column 0.

>>1031

x^2 + e = 2an

and (d+n)^2 -(x+n)^2 I think are the algebraic keys to this theory.

x1 = x at 1 * c

xa = x at a * b

n1 = n at 1 * c

na = n at a * b

n1 - (x1^2 - xa^2) = aa * m

Where m is a odd number.

This implies that if we knew the multiple, we'd know a.

Examples (sorry about the lack of cell references):

1261 - (50^2 - 8^2)/2 = 43 * 1

13 - (4^2 - 0)/2 = 5 * 1

69 - (11^2 - 5^5)/2 = 7 * 3

1237 - (49^2 - 19^2)/2 = 31 * 7

369 - (27^2 - 15^2)/2 = 13 * 9

815 - (40^2 - 12^2)/2 = 29 * 3

>>1033

I'm working on the algebraic side now, since we know e=e, c=c and d=d at both rows, I'm setting the equations as equal to each other and solving.

>>1037

This looks interesting, I'll take a look.

>>1036

Another quote by VQC:

>(e,n,d,x,a,b) in a grid (e,n)

>That is one way to organise the grid.

>Selecting two of the six variables (dimensions).

>There are SEVEN dimensions because each (e,n,d,x,a,b) element represent c.

>What are ways or grids or cubes are there to represent each elements?

>You could use another pair apart from (e,n)

>What would that produce?

>What other ways or shapes are there to view the six (seven) dimensions?

>What images result?

>Does this help find patterns?

>Does this help you c(see)?

Here's something else for you, assuming our 2 rows, 1xc and axb, labeling all my variables n1 or na for the n values corresponding to the cells etc.

2n1 = -(aa)^2 + 2aa(na + d) - (2d-1)

For those who dont want to click around for all VQC's posts, all the important stuff in 1 place:

https:/ /pastebin.com/eHJTEwWY

Can I get some eyes on something real quick please?

We made an assumption that the equations with t applied to more than n=1, but I'm not convinced they do.

If you assume that in every cell t always starts from 1, then, using the grid:

Choose your 2 primes, and find the cell 1xc.

Then if your n here is odd, the following will hopefully work (I haven't found a way to do it when n is even).

Move to the negative side by following F.

1xc for F will always be the first element in the cell.

Find the element in the cell when a=c, look at the t (assuming that always t starts from 1).

So far for me, t has always been a factor.

Am I crazy? Maybe? I dunno?

Help please guys!?

So far, whenever I

>>1051

Forget the last line, I dun goofed.

>>1053

I think you understand correctly. But so far, I've only been using the grid to check, because if so our formulas for t are wrong.

I'm coding something now to test.

>>1053

actually, correction f_t will be a factor of c I'm finding.

This would explain the p being a factor of a at t then it also is one at p+t, 2p+t, 3p+t, etc, and p+1-t, 2p+1-t, 3p+1-t

>>1056

I'm sorry, this is wrong.

Seems to be just luck when its correct or not.

Very sorry all.

>>1058

Yeah, agreed. I'll use dot notation from now on.

Yep I need a hint.

Hey Chris, there are a number of different paths that I can see us exploring that would take us further. Would you suggest any of the following, or some other?

• transposing the grid

• exploring c*c cells more

• exploring the x and n of the primes

• exploring the f cell more

• exploring the c* a prime relationship

• exploring the "p at t being a factor, hence p+t being a factor" etc

And if you get time, one more question. Looking at the grid does t start at 1 and increase by 1 each element in a cell?

Or is t related directly to x via the relationships already exposed.

We seem to be split on 2 definitions for t.

Thank you!

>>1064

Or one more:

• pythagorean triples

Thank you!

>>1066

Of course we're cool!

Sorry! Didn't mean to be rude, my head was just down in the numbers today.

>>1066

Also, what happened to Topol and CollegeAnon?

We're losing numbers!

I'm looking into Pythagorean triples and they appear everywhere in the grid, I wish I knew more math.

I am learning more math in this process though!

>>1069

I'm sad Topol got booted. I like his posts, makes it feel alive here, and his pics are always great.

>>1166

Thanks for this post Topol, well said, and I appreciate the kind mention.

And headpats go a long way, no doubt!

I've been working in moderate silence the last couple of days because I haven't had any new insights, and I have only had intermittent time to spend on the problem.

Mostly reading not much writing.

A few attachments…

1) Andrew Wiles paper on Fermat's Last, and its relation to Elliptic curves.

2) A picture of 1*cc for c=15, a=3, b=5. Explained below.

3) Output in code for similar semi-primes (the 2 numbers per row are described below).

So I think PMA might have already identified this relationship. And I makes perfect sense, and I'm not sure what value it adds.

But I've been playing with geometrically viewing the 1*cc.

For both the 1cc row, and the aabb row, (when e=0), both n's are always halfway between 2 perfect squares.

Now of course, like I said, this is "obvious", since d+n is always halfway between a & b.

So in the e=0 column, for both 1cc and aabb, sqrt(n/2) is a perfect square, as is sqrt(n*2).

And sqrt(n*2) will always be equal b-a.

Often the sqrt(n2) for the aabb row is a factor of the 1cc row. You can see that in the image with colored squares. Both x and n at the 1cc row are divisible by 7.

So if you take x in the 1cc row, divide it by 7, square that then halve it, you'll get n at aabb.

This is only mostly true though!

There are many rows where this is not the case.

Examples include:

313, 317, 319, 717, etc

But most of the time this is true.

Here's a pastebin of the full results:

https:/ /pastebin.com/a9KwHCFC

The 2 numbers at the end of each line are sqrt(n*2) and sqrt(n/2).

The row beginning with is the values above for the 1cc row / aabb row.

Any thoughts guys?

PMA, VA, MA, am I just restating what is already known?

>>1169

Thanks PMA, I just re-read.

I don't fully get it. Do you think you could explain the part from:

>(0,1) perfect squares to analyze

on?

And what exactly do you mean by substituting a and b?

Sorry for the questions, I want to understand!

I feel like we may be close again…

>>1171

Ok, I think I gotcha. I'm going to need to sit with this a bit.

I'm going to run my own tests to re-prove your tests, and see how they correlate to my path. I think we might be coming full circle to the same point!

Do you think you could test with one of these pairs next?

313, 317, 319, 717

>>1171

>move it's a to the next record b (figure out the d)

So I'm with you until here. You move b to a. you know e=0, and you need 1 more variable to find a specific cell, and you're saying you can calculate d?

I'm stuck PMA.

>>1179

Ok, I'm with you.

Look at your example. 2 things to notice (if you haven't already):

1) taking your a from (0,1,8), how many times in the 0 column will it appear as b? It is always going to be 1/2 of a perfect square. And as such it'll appear in the 0 column at every n where n is a perfect square (in your example n was 9). Because this value (98) appears many time, the pattern of the value that it appears next to (the a to its b) is all possible values of 1 half perfect squares, that have an e=0.

2) There's a row directly below that one that, (0,9,14), where a = 18, and b = 128, and n will equal the n of (0,9,13).n.

I'm not sure what this means, I'm having trouble visualizing it now.

Were you already aware of these patterns?

>>1181

Can you show me with this example?

{59:5:61:19:42:90}

From the grid, the cell values are:

{59:5:61:19:42:90}

{59:5:71:21:50:102}

{59:5:119:29:90:158}

{59:5:133:31:102:174}

{59:5:197:39:158:246}

{59:5:215:41:174:266}

>>1193

Ok, I'm with you now.

So does it always work for walking in steps of 5? or n?

I also still don't get how this helps us get to aa*bb, but i'm trying!

>>1196

Merry Christmas to you VA. And PMA, and Topol, and MA, and Hobo, and AA, and CA (if he's still alive), and Baker and all the other anons!

Merry Christmas VQC, I hope the plan plays out smoothly over the coming days, and I hope we can help you.

I enjoy being here with you all too!

>>1198

Good job PMA.

I am worried about the t– though…

I like the process, but so far we're just decrementing x by 2 each time, which is essentially a search for "a".

But I agree that this might unlock the ability to algorithmically move through it.

For this example the n = 61 from 1xc the t == 61 from 0,1 at the end…

Is that a coincidence?

>>1205

Do you know whats happening?

I'm concerned for Chris.

>>1204

+1 for math lsd pics

and +1 for topol summoning generally

I hate to be that guy, but I fear we're just translating the problem to different variables, but we're unable to do no better than iterate t.

I've been looking at it a look, and I think this may be the case.

Sorry to be a downer.

>>1276

I'm working on a different path regarding that actually.

You can kinda swap some variables around to calculate for any integer i, the cell in (0,n) with a=ii and b=cc.

If you iterate i, growth in a is quadratic, but growth in n, x, and d are kinda linear (they increase in a arithmetic series).

And at the point that i is a factor of c, x+n / n will be a whole number.

I think this is saying the same thing, we're just increasing the square size, but maintaining the same ratio.

The interesting point to this direction is the sorta linear growth in x,n and d.

I'm not sure if we're just restating the problem again though, stuck with iterating i.

>>1278

Feel free to explain it and i'll take a look next.

>>1277

Something very interesting about this approach is the relationship between the x+n/n ratio for the 1xc row, and the aa x cc and bb x cc row. It always appears to be very close to the first, or in between both.

>>1280

>x+n/n

correction

x+n/x

>>1283

This is assuming a fixed e?

While I'm taking a look, do you mind explaining your reasoning?

>>1283

Starting at 72.

36 = 72 * 1/2

24 = 36 * 2/3

18 = 24 * 3/4

etc.

>>1283

>{23:18:15:11:04:62|

Also, this line looks very unique to me…

Its the only one ending in |

;)

>>1289

Correct, d = a + x, and x is constant.

n is the reverse set, obviously, e is fixed, b can be calculated.

There should even be a way to jump any amount of records algebraically.

>>1282

Hey topol, any cool math visualizations?

>>1293

>>1294

Math is so beautiful.

I am in the process of making my own mini tesla coil too!

>>1380

Thank you man, I can't wait!

You're incredible Chris!

Hey guys, its been a while between posts, but I've been following along with the progress.

I've been focused on my own theories mostly, and none have been fruitful.

>>1731

Hey PMA, the size of the small square has been where my attention has been lately.

I've also been spending some time going through reading for ECC.

I want to point out something that I've realized too…

n = (x^2 + e) / 2(d - x)

This is just a simple rewrite of x^2+e=2na.

But what I like about this representation, is that d and e are constants and so we have a relationship between n and x for any c.

I'm trying to use this equation to figure out the possible sizes of the small square.

Anyway, how is everyone?

Also welcome back CollegeAnon! How were your exams?

>>1735

I was having a really hard time with understanding what VQC meant when he said:

>At the correct element in the grid at (e,1) where the value of a at that element equals the na we want, if you subtract 2d+1 from na then you are in the negative half of the grid in terms of e and the value at the same element in the first row will be (n-1)a

I believe the "na" that we want refers to the na in the 2na = x^2 + e.

This is why I think the equation is important.

I'm trying to work out what the (n-1)a cell means now.