>>3263

You know how to increase d by one. Use f for e.

That is simply subtracting 2d +1 from e.

As d increased by one the remainder became negative because it's larger than c now.

A and b stayed the same.

X also goes up by 1 and n is reduced by 1. You can do this over and over. Remainder keeps getting more negative as d is increasingly larger than c. Each increase in d makes the remainder 2d+1 smaller than the previous d.

The formula allows you to move multiple Ds. When you have done that n times then e = -(x+n)^2

Then n=0 and x= x+n and that is the solution.

Just replaced my router cause it sucked. Now I have a new ID

>>3260

>>3264

>>3266

>>3167

Look at the records in 3167 all have same a and b. They are different ways of expressing the same information as d grows larger the remainder the difference between d^2 and c becomes increasingly negative. Every move up 1 by d increases x by 1 and reduces n by 1. The goal is to get rid of n.

>>3266

Its really easy to understand if you draw it on graph paper. After you draw a few you realize that you are basically putting dots representing squares on the diagonal. d=t

>>3271

e = c -d^2

As d increases remainder becomes increasing negative.

>>3270

Your program is wrong. x = d - a it isn't negative

made slight typo corrected

>>3269

Test them!

I have given the rules for generating them over and over.

For any record you {E, N, D, X, A, B}

You can generate an equivalent record

{E-2*D-1, N-1, D+1, X+1, A, B}

If you do it again the the new record will be

{E-(2d+1)-(2(d+1)+1), N-2, D+2, X+2, A, B}

>>3275

C=d^2 + e

C also equals (d+1)^2 + e - (2d+1)

f = e - (2d+1)

new record is {f, n-1, d+1, x+1, a, b}

Now do it again

new f = f - (2*(d + 1) + 1)

new record = {f - (2*(d + 1) + 1), n-2, d+2, x+2, a, b}

Draw them on graph paper

>>3280

Again c = d^2 plus e

Is it also possible for c to equal a bigger d^2 with a negative remainder? Is it possible that there are an infinite number of larger ds with increasingly larger negative remainders? All describing the SAME a and b

How about this pick a relatively small record like a =13 b = 37

{40, 4, 21, 8, 13, 37}

Now make e = -(x+n)^2

n=0

d=d+n long side

x = x+n short side

a=a b=b

{-144, 0, 25, 12, 13, 37} your solution record.

>>3282

>>3258

{40, 4, 21, 8, 13, 37}

{-144, 0, 25, 12, 13, 37}

Use formula to change e from bottom of 3258

new e = e -2dm-mm

original e =40 d =21 n=4

so new e =40 -2214-4*4=40-168-16=-144

e equals - (D+N)^2 or -(long side)^2

x=x+n or short side. Anyone rcognize the difference between two squares with n=0

Solution

>>3284

d=t

d=1 is a square 1^2

d=2 is a square 2^2

when you go to a negative remainder then d is getting bigger which means t is getting bigger

d=t

>>3283

You can go in the opposite direction but its the wrong way you want n to disappear. But you're right there are an infinite ways to describe a*b=c.

Also the formula for moving multiple Ds down is not the negative of moving up it is

e+2dm-mm

>>3284

>{40:220:21:20:1:481} (40, 220, 11)

I don't see where this record comes from.

The path I took goes directly from start record based on moving n steps into negative e space. d=t

There is a record for every t therefore d that can describe the original record.

>>3293

yes

>>3293

Unfortunately we still don't have a way to figure n. But the grid is making more sense

>>3290

You aren't missing anything just better understanding the grid. We still don't know n.

>>3297

Great find I don't remember that crumb.

Yes you can describe a*b=c with an infinite number of Ds and each D has a different f.

It may or may not help finding solution. I'm confident that as I learn more about the grid eventually the solution will appear.

Anyone looked at the (f,1) records? They aren't just mirrors and they don't have negatives except for e. I have a formula but it uses nested if statements. Got a DM today referring to the (f,1) records. I've been playng with them for a few days.

f = 2d+1-e

When e = -f, a[t] = d[t]-d

Look for pattern of a[t] and n, and d[t]-d and (n-1)

>>3299

I've been talking about negative e's. If you draw a record where e=-(x+n)^2 you will understand what I'm talking about. d=d+n longside. x=x+n shortside. a and b remain constant. To draw both the x and y axis contain equal values (squares) for a x d n b. E doesn't appear directly but is derived from other values.

>>3305

The invalid e should be -(x+n)^2 which was 81 or -6561 that will work. I haven't looked at n,0 for (1,c) that a good idea thanks.

>>3309

>>3305

I haven't looked at relations for between (n,0) for c,1 and a,b records it looks like the smaller e may be factor of larger (ABS). Question to check is difference between factor when a and b are close versus far apart. Big n small n. If there is a difference that's a big clue. There is a difference between the a of the (f,1) records when a and b are far apart. the opposite of what I would have guessed. So we finally have a way of reducing the time it takes to solve RSA

>>3312

All ears? eyes?

I don't feel any of this is in vain. Its just a huge puzzle that I keep seeing around the corner then when I get there I find another corner. Eventually there won't be anymore corners

Just thought i would put up a few (f,1) records

E N D X A B

-48 16 31 20 11 83

-48 1 16 8 8 26

-17 8 34 17 17 67

-17 1 9 5 4 16

-23 77 58 45 13 257

-23 1 6 5 1 13

-84 2 59 16 43 79

-84 1 18 10 8 30

I think the relationship i saw yesterday with a in the f,1 records may have been a figment.

>>3317

For movements in N d=t. Depends what variable is changing what t is

>>3317

As d increase you are subtracting 2d +1 from e then 2d+3 then 2d+5. I looked at patterns 500 deep. All of a sudden there is a perfect square. The grid is amazing at keeping its secrets.

>>3320

Excellent post. I think you may have to take a root of e to make it scalar. E isn't a distance its a 2 dimensional volume. So taking its root would be a distance.

>>3323

e=2na-xx both terms are two dimensional.

Also e= a*b -dd both terms again two dimensional.

>>3320

The idea of flipping e and n is excellent