Been visiting briefly not doing much work but these posts have reinvigorated me completely!!
I LOVE ALL OF YOU
> c % d is equal to what else % d?
c % d = ((d+n)^2 - (x+n)^2) % d
c % d = ( dd + 2dn + nn - xx - 2xn - nn ) % d
c % d = ( -xx - 2xn ) % d
c % d = ( -(d-a)(d-a) - 2(d-a)n ) % d
c % d = ( -dd + 2da - aa - 2dn + 2an ) % d
c % d = ( 2an - aa ) % d
c % d = ( xx + e - aa ) % d
c % d = ( xx - aa + e ) % d
(dd + e) % d = ( xx - aa + e ) % d
dd % d = ( xx - aa ) % d
0 = (xx - aa) % d
If this is the solution then this would mean we have to find a number that is a difference of two squares that is a multiple of d.
For c=145 it would be
(1, 5, 12, 7, 5, 29)
49 - 25 = 24
f here is 2d+1-e or 2*12 + 1 - 1 = 24
For c=65 its
(1, 1, 8, 3, 5, 13)
9 - 25 = -16 (divisible by d=8)
f here is 2*8 + 1 - 1 = 16
For c=403
(3, 2, 20, 7, 13, 31)
49 - 169 = -120 (divisible by d=20)
Potential Alg Here
So I guess if we wanted to factor 145, we could try to factor 12. To do this we would make it dd + e which makes it 33 + 3. Obviously d=e indicates that 3 is a factor, so the we can find that 3 and 4 are the factors. Since these are not the same parity, it can't be a difference of squares so it can't be 12. Then we'd jump to 24 and try to factor this. If we do 24, we see its 44 + 8 =4 is factor. 4 * 6 = 24. This would make the diff of squares be (6+4)/2 and (6-4)/2 => 5^2 - 1^2 => 24. For this would indicate that x or a is 5 or 1. Well wouldn't you know 5 factors 145! (it doesn't give us 77 - 55 though) OR we could have done 2 * 12 => (12 + 2)/2 and (12-2)/2 which gives us 7 and 5 the REAL values.
Factoring d? Is this the reason for the factor trees we made earlier with d and e?
Another example for 11 * 31 = 341
d = 18
18*18=324
341 = 18 * 18 + 17
x SHOULD be 18 - 11 = 7
Lets look at d = 18
4 * 4 + 2 = 18 (2 is factor)
2 * 9 = 18 (different parity so it doesn't work) (only multiples of d that are congruent to 0 mod 4 so 18 wouldn't work)
9 isn't prime so split it up (factor it) into 3*3 so 18 = 2 * 3 * 3
another factorization is 3 * 6
3 * 6 = 18 so 3, 6 different parity so move on
Go to 36
[optional: We've already factored 18 so we know factors]
36 = 6 * 6 + 0
Square wouldn't work
Go to 54 (not 0 mod 4)
Go to 72
72 = 8 * 8 + 8 (divisible by 8)
72 = 8 * 9 (diff parity so shift factors)
72 = 4 * 18
(18+4)/2 and (18-4)/2
1111 - 77
Here we have our a and our x
GUIS IS THIS IT???
So those sums of squares are x^2 + (x-1)^2, in (1,1). if we look at (4,1) we get the sums of squares of x^2 + (x-2)^2 which are 4, 10, 20, 34. Then if you look at (9,1) you see the squares which are sums of squares which are x^2 + (x-3)^2, 9, 17, 29, 45. Then if you go to (16,1) you get x^2+ (x-4)^2, which are 8, 10, 16, 26, 40.
In short,
if (e,n) = (s^2, 1)
has A, B values equal to m^2 + (m-s)^2 for any m
This feels obvious now that I'm looking at it, so forgive me if you've already know this.
So our classic example:
c=145 is 12^2 + 1^2
12^2 + (12 - 11)^2
so it would be in cell (121, 1) as an a value. Look at pic related and there it is. The we could simply slink back a few x values (or forward) and do the gcd alg and we'll find a factor. This is because (I ran a script to basically prove this) for every record in (e,1), maybe everywhere but I haven't tested it yet, if you have an a value, then if you increase x you will find another value that shares a factor with a.
Maybe we can find this for any any c. We would want any e that is a square (e positive, negative already works because d^2 - m^2 is the solution) so
Suppose you split the amount of possibilties in two each division, and you know that it must be between 2 and d, then you would need
1/(d-1)
We would want 1/(2^y) <= 1/(d-1)
Invert
2^y >= d-1
y >= log_2 (d-1)
So the amount of iteratiosn would be the length in bits of (d-1)
From time to time yes. Not as vigorously as before. I am posting just in case thing brings any sort of morale boost to any of you. I'm still trying things too. I still believe it's possible we just haven't seen it the right way.