I'm hoping to shorten the arguing phase with the remarkably simple
solution to the factoring problem I've outlined in a previous post by
pointing out a few things:

Given a target composite T, and

z^2 = y^2 + nT

where n is an integer chosen for z to have 3 as a factor--so if T mod
3 = 1 then n=5 will work--then z exists as

z = (f_1 + f_2)/2

when f_1*f_2 = nT, and of course you want integer factors.

The method I have then allows you to find z using a variable I call k,
where

z = 3k/2

and k^2 = 2^{-1} (nT) modulo p

where p is an odd prime less than k as k is an even positive integer.

May seem complicated all in a rush like that but it's incredibly
simple.

Wow. Years of research and then the answer turns out to be incredibly
simple, but that's how it can be with mathematics.

What I like about this method is that there isn't really any room for
people to argue with me about whether or not it solves the factoring
problem (I've posted it in another thread), and it has to be
incredibly fast, as easily shown by mathematical proof.

So no dumb arguments about me having to demonstrate as I haven't
gotten around to testing it yet, at all, as I've just relied on
mathematical proof and it's so simple.

It boggles the mind how simple it is.

Needless to say this research result gives FULL VALIDATION to my
research across the board and will usher in a new era in mathematics
as a discipline around the world.

That brings forward the correct prime counting function, the correct
information about randomness and primes, so that the Goldbach
Conjecture is handled and the Twin Primes conjecture is handled, and
the Riemann Hypothesis can probably be quickly handled.

If you have a target composite T, and use

z^2 = y^2 + nT

where n is used to force z to be divisible by 3, then there will be an
integer k such that

z = 3k/2.

But now find an integer x and odd prime p such that

x^2 = y^2 mod p

where also

2x = k.

So given k as defined before, you find an x that equals one half of
it, and consider a prime p where

(k/2)^2 = y2 mod p.

Then it's trivial to show that

k^2 = 2^{-1}(nT) mod p

and, if f_1*f_2 = nT, then

f_1 = k mod p

and

f_2 = 2k mod p.

I read this requirement in the FIPS 140-2 Derived test requirements:

AS05.20: (Levels 3 and 4) The tamper response and zeroization
circuitry shall remain operational when plaintext secret and private
cryptographic keys or CSPs are contained within the cryptographic
module.

Does this imply that if your plaintext secret/private key/CSP's are
stored in nonvolatile memory, that you must use battery-backed logic
that can perform zeroization in the event of power loss in order to be
FIPS 140-2 level 3 or 4 compliant? Is there any literature to support
this claim? Thanks.

Besides parents' rights, a great emphasis is also laid on the rights
of other relatives. In Islamic terms, 'Silah-Rahimi' is used to denote
good treatment towards the relatives'.

In the Qur'an, where the Muslims are enjoined to show kindness to
parents, they are also required to treat the other relatives with love
and sympathy and to pay due regard to their rights as well.

Allah has declared, "I am Allah, I am Ar-Rahman (The Merciful), I have
created the bond of kinship and named it Rahim,-which I have derived
from the root of my name of Rahman. Thus, whoever, shall join it tie.
Rahim, I shall join him, and whoever will break it, I shall break
him."

The Almighty has designed the system of birth in such a way, that
whoever is born, is tied to the bonds of kinship-and these bonds carry
certain claims and rights. Thus, whoever fulfils these claims, by
being kind to his relatives and treating them well, Allah will "join
him" i.e. He will make him His own and bestow His favor and mercy on
him. And whosoever will violate these claims, Allah will "break him"
i.e. He will have nothing to do with him.

Fulfilling The Rights Of Relatives

Hi
I have a setup where two systems needs to transfer some data securely
over an insecure connection.
Normally I would use a station-to-station(STS) setup with AES
encryption of the data. In this case, however,
one of the systems does not have the ability to store a private key
over time (used in STS signing to identify the system towards the
other system).
Does some protocol exist that does not need one system to store a
private key but still gives the ability to have the secret key K
generated at runtime (as STS does it)?

kim

I was wondering if someone tell me how I go about calculating
collision rates for a "key" that is 2 characters (lowercase alpha
only) long?

I realise it's 26x26 but if I generate two random numbers between 1-26
is that all that's involved? So each time I generate this key, I have
a 1/676 chance of a collision?

Thanks for the help

hamada gmail.com> wrote:

"Thou shalt love thy neighbour like thyself."

I win, you lose. On spelling as well as on principle.

Richard

I was wondering what the group here thinks is the set of
characteristics a good file compressor should have before one does an
ecnryption pass. I feel not only should it compress well but that it
shoulf be bijective and that it should not suffer from Uniciciy
distance problem like arithmetic or ppm.
What are your views?

David A. Scott
--
My Crypto code
http://bijective.dogma.net/crypto/scott19u.zip
http://www.jim.com/jamesd/Kong/scott19u.zip old version
My Compression code http://bijective.dogma.net/
**TO EMAIL ME drop the roman "five" **
Disclaimer:I am in no way responsible for any of the statements
made in the above text. For all I know I might be drugged.
As a famous person once said "any cryptograhic
system is only as strong as its weakest link"

Hi,

I am facing an issue for the authentication with RSA.
Scenario :

I have 2 application
1] Internet banking application
2] Armor (used for authentication and authorization)

The Armor is the agent host communicating with RSA.

Flow of the user request :

Internet banking application user ---> Armor ---> RSA

Internet banking application is in SUN cluster and Armor is in a
seperate server without any cluster.

We have generated the sdconf.rec file for the agent host "Armor"
Now when the user tries to login from the Internet banking application
the error at the RSA server comes as " Access denied ,Invalid
passcode".
The securid file does not get created.