Also another thing I remember VQC said a while ago was that c^2 was significant, and I don't think we've used it yet. We've found the 'key' could c^2 be related to the lock?
This does work I don't think it solves the equation though, but it is a way to use negative x's to calculate a previously unknown record. The following stuff I posted is bunk though it doesn't work all the time I wasn't sure if that made people overlook this original note which was legit.
Another thing:
Here are the A[t] values in (e,1) for c=13*41 and their hits for A mod a, b, aa, bb. Notice that in the A/a column, 4,10 come up, which are entries in (4,1) and the record is (4,1,6,2,4,10). Then you get 50,68, which are entries in (36,1) and the record is (36,1,58,8,50,68). Then the next two are 148,178 which are entries in (100,1) and the record is (100,1,162,14,148,178). This is a weird pattern.
so:
(4,10) -(4, 1, 6, 2, 4, 10)
(50,68) -(36, 1, 58, 8, 50, 68)
(148,178) -(100, 1, 162, 14, 148, 178)
for (-f,1) we get A/a
(3,7) -(5, 1, 4, 1, 3, 7)
(51,63) -(77, 1, 56, 5, 51, 63)
(151,171) -(221, 1, 160, 9, 151, 171)
Whats cool is these are always in n=1.
Also if you take instead of (4,10), (50,68), (148,178) you do (10,50), (68,148), etc, they are all in the same n row again, but this time its different (in this case n=8).
Here is some triangle stuff. I'm going to try and narrow this down because VQC said you can make assumptions and cross out big chunks and narrow it down. VQC also said c^2 was importand so here goes.
So for any c value, we can have a "goal d" that you want. This is the d where n=0. This is a cool record because you only need d and x to define it. Also, since (d+n)^2 - (x+n)^2 = d^2 - x^2 [these are not the traditional d and x]:
(-x^2, 0, d, x, d-x, d+x) (c = c)
Then there is another record that is the "square d", which also can be similarly defined (note they are capital letters).
(-X^2, 0, D, X, D-X, D+X) (c=c^2)
Also there are some neat relationships with these numbers:
(D-c)(D+c) = X^2
D + X = b^2
D - X = a^2
D = (a^2 + b^2)/2
D = d^2 + x^2
2X = b^2 - a^2 = (b-a)(b+a)
X = 2dx
b^2 - a^2 = 4dx
b = a + 2x
This is cool because now we have D^2 - X^2 = c^2, which can be looked at as a right triangle like c^2 + X^2 = D^2. Now we know there are some rules about Pythagorean triples.
(https://en.wikipedia.org/wiki/Pythagorean_triple#Elementary_properties_of_primitive_Pythagorean_triples)
So one thing is we know that that either c, X, or D is divisible by 5. If c is divisible by 5 it is over, so we know that D or X is divisible by 5. Also we know that c or X is divisible by 3 and c or X is divisible by 4. The number that always divides cXD is 60. Also 60=345 and the babylonians wrote in base 60. So I decided to mod everything by 60, because thats the biggest thing we can mod and still keep these patterns.
There are only really 4 cases [screenshot of the code to find these]
c%60 = {1, 11, 19, 29, 31, 41, 49, 59} :
X=0 D=49 (mod 60)
or
X=0 D=1 (mod 60)
c%60 = {7, 13, 17, 23, 37, 43, 47, 53}
X=36, D=25 (mod 60)
or
X=24, D=25 (mod 60)
These are all the possibilities mod 60.
So if we have our c value, we can know definitively 1 of 2 things.
i) if c = 30k +/- {1,11}, then X=0 (mod 60)
ii) if c = 30k +/- {7,13}, then D=25 (mod 60)
Suppose we have situation (i), then we know that X=0 (mod 60)
X = 0 (mod 60)
2X = 0 (mod 60)
b^2-a^2 = 0 (mod 60)
(b-a)(b+a) = 0 (mod 60)
(2d + 2n)(2x + 2n) = 0 (mod 60)
[remember we're using our row where n=0, but this would hold true for any n I suppose..]
(2d)(2x) = 0 (mod 60)
4dx = 0 (mod 60)
dx = {0, 15, 30, 45} (mod 60)
dx must be odd so
dx = {15,45} (mod 60)
Suppose we have situation (ii), then we know that D=25 (mod 60)
D=25 (mod 60)
(2 more situations)
a) X=24 (mod 60)
2X = 48 (mod 60)
b^2-a^2 = 48 (mod 60)
(b-a)(b+a) = 48 (mod 60)
(2d)(2x) = 48 (mod 60)
4dx = 48 (mod 60)
dx = {12,27,42,57} (mod 60)
We also know that d or x MUST be odd because d-x = a which is odd, so
dx = {27,57} (mod 60)
b) X=36
2X = 12 (mod 60)
b^2-a^2 = 12 (mod 60)
(b+a)(b-a) = 12 (mod 60)
4dx = 12 (mod 60)
dx = {3, 18, 33, 48} (mod 60)
>again we know it must be odd so
dx = {3, 33} (mod 60)
Could this narrow it down a little closer ????