I've made a brute-force version of this algorithm

>>6910

It doesn't use the root of d (since obviously we don't know how you're meant to use that to do this), but it's useful for considering the context in which this function fits within the entire algorithm. It doesn't do (1,c) for even cs since n would have .5 at the end with an odd a+b (so those cells aren't in the grid).

import java.util.;public class brute{ public static void main(String[] args){ int e = Integer.parseInt(args[0]); int f = Integer.parseInt(args[1]); int d = (e-f-1)/2; int c = (dd)+e; int twod = d2; int roottwod = (int)Math.floor((int)Math.sqrt(twod)); System.out.println("c="+c+", d="+d+", e="+e+", f="+f+", 2d="+twod+", sqrt(2d)="+roottwod); for(int i=0; i<((((1+c)/2)-d)+1); i++){ int aplusb = (i+d)2; int x = ((int)Math.floor((int)Math.sqrt(((d+i)*(d+i))-c)))-i; int testa = d-x; int testb = aplusb-testa; if(testa*testb==c){ System.out.println(i + " is a valid n, with a=" + (int)testa + " and b=" + (int)testb); } } }}

So you input e and f values (f has to be negative) and it outputs every valid n value for that pairing

>>7481

This thing about c being divisible by either a or b and BigN-n definitely still seems to stand. I've verified it with several other test examples.

>>7480

But with this thing, it seems like by pure coincidence, both of my test examples happen to be the only ones that actually make this work.

Here's a thing based on the latest talk about the sum of two squares.

Since c=(d+n)(d+n)-(x+n)(x+n), the difference of two squares, you can also find another c where c'=(d+n)(d+n)+(x+n)(x+n), the sum of two squares. The difference between these c values appears to be the smaller square times two.

559=13x43

559=(28x28)-(15x15)

(28x28)+(15x15)=1009

1009-559=(15x15)x2

203=7x29

203=(18x18)-(11x11)

(18x18)+(11x11)=445

445-203=(11x11)x2

>>7485

Oh hey, nice timing, I'm pretty sure everyone else is asleep

>>7485

>Any number of the form n(n-1) or multiple of it, has distinguishing properties and can be pick out of a crowd, especially and more easily if the multiple of n(n-1) is a square.

>What if you multiplied each corresponding element in (-f,1) and (e,1)?

So if you take an and a(n-1) and multiply them together, you get a^2 and n(n-1)

c559 13*43, n=5

an=65, a(n-1)=52

65*52=3380

And apparently it will be easy to factor that number somehow.

>What if you multiplied each corresponding element in (-f,1) and (e,1)?

>Would that series be totally predictable?

Not that I can tell straight away but probably, if you're saying it

>Could you quickly lookup or figure out any values that are a square multiples by two consecutive numbers like n(n-1)?

So I guess this means (an)(a(n-1)) pops up somewhere, and there's an easy way to factor it. I feel like this will require a look at some of the old crumbs.

>If you do this, you will quickly spot the value of t or x that is a.a.n.(n-1) at x.

This seems like another one of those clues that we'll misinterpret several times before actually understanding it.

>you will quickly spot the value of t or x that is a.a.n.(n-1) at x.

>the value of t or x that is a.n.n.(n-1) at x

>the value of x

>at x

Huh?

>>7486

Also don't worry, I immediately realized how blatantly obvious the math in this post is

>>7485

Triangle numbers are (n(n+1))/2

The triangle number before the nth triangle number is then therefore (n(n-1))/2, since n+1 becomes n and n becomes n-1.

So (an)(a(n-1)) is a^2 + 2T(n-1). If that means anything.

>>7492

This is a list of all (an)(a(n-1)) for the early numbers where a and b are odd and aren't 1.

18

50

54

98

108

150

162

180

242

270

294

300

338

378

450

486

500

504

578

588

648

722

Interesting to note is that some of these numbers are an odd square times 2. 18=92, 50=252, 98=492, 162=812. I'm not sure what the deal with the other numbers is.