VQC lives in Britain. He won't be up til after midnight.
I'm getting close to having e right on (e,1). It gets a and b right but it doesn't get N right. Using a transform from (0,N) to (0,1). You still have to search for the first appearance without N working.
Do you have a derivation for the f formula. Chris gave us this formula with both negative and positive values. One of the two looks wrong to me. Seems one should be plus one and the other should be minus 1. I haven't been able to derive it. He says add f to make c a square.
Need to go get a few hours. Happy new year anons!
Tested on multiple c = ab !!
Easily calculated e such that:
a of (e,1,1) is factored by goal a goal b and N.
This is what Chris was talking about as a solution!
a repeats as factor every a*t +1
b repeats as factor every b*t +1
N repeats as factor every N*t + 1
I'm having trouble figuring out how to get to the aNt +1 record or the bNt+1 since we still don't know a or b! Any suggestions???
I used the formula Chris suggested. Subtract 2d +1 from a to get a new negative e. This works with one problem.We get a b and N as factors of a(new.e( e-a + 2d+1),1,1 ) I had to take all factors of 2 out of N to get negative e to work.
Figured it out.I have two separate e such that
goal.a goal.b and N are factors of…
a (e, 1, 1) and
a (-e(from formula), 1, 1) that Chris called N-1
Chris has given a number of hints about a special record in (e,1).
That record is e equals 2Nab.
a, b, c, N are all factors of (2Nab, 1, 1) and all through the cell at 1+ at, 1+bt, 1+Nt, 1+sqrt(N/2)*t etc.
They are all factors of the (-2Nab, 1, t) cell also.
This record is what a lot of Chris hints are referring to. We just need one more step!
Factors of a(2Nab, 1, t).
The records get hairy but I'll give a couple examples
(19, 2, 10, 3,7,17) yields
(1656956, 1, 828478, 9, 828478, 828480)
with 828478 divisible by 7, 17 and N 6962
{10, 1, 17, 4, 13 23} yields
{26552396, 1, 13276198, 0, 13276198, 13276200}
with a =13276198 divisible by 13, 23 and N 44402 and all up and down the cell.
N is big because x equals 1. It represents the only factor we know for cc besides (0,0,c,0,c,c)
yup
>>53423 RSA 2 is where Chris said f is what you add to c to make a square
you're
Remember all the discussion of moving up and down t to find records where target a is a factor of a(e,1, t).
So for (e=2c, 1, t)
a is factor every t= a*m+1
b is factor t = b*m + 1
c is a factor t = c*m +1
for integer m
Here's the interesting part….
For (e = 2c-1, 1, t)…
a is factor every t= a*m+1 and
a is a factor every t =a*m
b is factor t = b*m + 1 and
b is a factor t = b*m
c is a factor t = c*m +1 and
c is a factor t = c*m
for integer m
It works on the negative side too.
This seems highly useful but I haven't made any progress yet!
thanks
I have an idea but I'm having trouble creating records for larger (e,n) does anyone have a good algorithm.
In testing the even odd pairing with the same factors is e = 2c with factors a, b, c and
e = 2c-1. Not e + 1. For both even and odd records s is factor at (s + 1 - t), (2s +1 - t), but also at s,2s,3s The important part is that if s odd is a factor of 2e then s is a factor of all those described a of (2c, 1, t) and (2c-1,1, t)
Using e 2c or (2c -1) is a piece of the puzzle.
Big picture thoughts.
Start with ab=c record
Fully describe it.
(e, n, t) for c = { e, n, d, x, a, b}
we know c e and d
Can we answer the problem here.
It seems not.
We do our first transform.
We square the record.
No information lost.
Our new record is { 0,N, cc, cc-1, 1, cc}
Have we gained anything? Well if we know original n we can solve the problem. Now we know an N record. Its huge so it doesn't tell us much. we do know that it codes for the difference between perfect squares. If you examine these n's it turns out they are all twice a perfect square.
That pattern shows up a lot so its worth noting.
We still can't solve the problem.
Still need more information.
It turns out that the (e,1) row offers a lot of information. a and b of (e, 1) cell elements are factor a tree. The starting values of the tree show up the most frequently as they are mixed with other factors which also repeat.
If we choose 2c for e in ( 2c, 1, 1)
Then the record is ( 2c, 1, c, 0, c, b(e1)). This tree will flower more factors of c than any other factors. There is more information on e,1 but including odd and negative trees but still not enough to solve our problem.
So we need to go to (e,n)!
Going to (e, n) is tough. I believe there are going to be only 5 records in (e,n) matching the five (0, n) difference of perfect squares records of cc.
When we can fill in (e,n) computationally, not iteratively we solve the problem. First one to figure out that out cracks RSA. On your marks!
Good night anons!
Think how elegant Chris solution is. He defines coordinates based on (e,n). To crack RSA all you need is to fill in the (e, N) elements
Yes that's what big N is.
After I found the (e=2c) contained a factor tree full of a and b I tried to solve the problem by moving up and down the tree. I compared odd to even side by side. I didn't get anywhere. Then I decided to look at (e,n) and couldn't find them.
Here is one of the very simplest (e,n) cells that I believe solve our problem. See how quick you can find a solution.
For a b equals
5 11
7 17
the (e,N) records for those to simple pairs are
( 110, 1458 )
( 289, 6962 )
I was struggling all day with this banging my head. Then I got high and realized that of course I couldn't find any. Its the solution set to
cc and I wrote my post. Now I'm sure I can find those records. We already know a and b. These are the very simplest set of cells representing solution for cc. Imagine what solving it for RSA will be like.
We have to solve it without knowing a and b and algorithmically.