On Aug 22, 7:39*pm, Joshua Cranmer <(E-Mail Removed)> wrote:

> JSH wrote:

> > The Internet still uses public key encryption.

>

> Public key encryption does not equal RSA or other factoring. As I've

> said before, there's elliptic curve; there are other even more secure

> algorithms.

>

> > If P=NP, then a polynomial time solution is possible for factoring

> > meaning that public key encryption is no longer viable as a security

> > system.

>

> No, it just means you have to keep bumping up key sizes every few years.

>

> > I don't have to check it.

>

> You do if you want to claim that it's correct, which you do a lot of.

>

> > I'm not doing surrogate factoring further. *It's too dangerous.

>

> No, it's not. To suggest otherwise is to demonstrate your ignorance of

> computer security.
Ok. Here's the research result at which I stopped which is a way to

solve for quadratic residues mod p, but the "p" which is for an odd

prime, can be replaced with an "N" for natural number which can be a

composite to be factored.

Given a quadratic residue q modulo p where p is an odd prime, where

k^2 = q mod p

it is a method to find k, which comes from reversing some of the

surrogate factoring equations.

As usual with my research you get additional variables as now you need

T, where

T = 2q mod p

and while you'll want the smallest T--because it has to be factored--

you must pick an odd T, where T - 2q must be non-zero.

Next you have to factor T, as with integer factors f_1 and f_2 of T,

where f_1*f_2 = T:

k is given by

k = 3^{-1}(f_1 + f_2) mod p.

And my analysis indicates that there should be a 50% probability that

you will get the correct k with each set of factors (weirdly too

simple, I know, but hey, just wrong?). Checking is done by just

squaring your k modulo p to see if you get back the correct quadratic

residue q.

Example: Let q=2, p=17 so T = 2(2) mod 17 = 4 mod 17.

Here T=21 does not work, but T = 55 does, and the answer then from

3k = 2(

mod 17, is k = 11 mod 17, as 112 = 2 mod 17 as required.

To have an absolute case when you must get a solution for k, T mod 3 =

2 is required, and one of the factors f_1 or f_2 when both are

positive and non-unit must be greater than p (which again has to do

with why 15 doesn't work!), then k is given exactly by

k = (f_1 + f_2)/3

with 100% certainty.

If T mod 3 = 1, because p mod 3 = 1 and q is divisible by 3, then an

alternate set of equations can be used as then

T = 10q mod p

and

k = 19^{-1}(3(f_1 + f_2)) mod p

and an exact solution occurs if with positive factors f_1 or f_2 is

greater than p, both are non-unit, and z is divisible by 19 as then

k = 3(f_1 + f_2)/19.

That is an incredibly small bit of research in terms of physical size,

but if it's right, then public key encryption is dead, as replace the

p with N, where N is your target composite to factor, then calculate a

quadratic residue modulo N and then use the equations above to solve

for that same residue and you may get back your original or its pair.

i.e. If you start with 'a' you may get back 'b', where

a^2 = b^2 mod N

and factor N from (a-b) or (a+b), and THAT is what paused me and later

sent me looking for an algorithm for TSP.

But you say everything is fine, and no worries! So Joshua, we'll go

on your opinion here for the moment.

James Harris