[118978] in Cypherpunks
Square roots of nonsingular binary matrices
daemon@ATHENA.MIT.EDU (bill payne)
Mon Oct 11 23:02:53 1999
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Date: Mon, 11 Oct 1999 20:42:55 -0600
From: bill payne <billp@nmol.com>
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Reply-To: bill payne <billp@nmol.com>
Monday 10/11/99 7:32 PM
Alexandr
http://www.semionoff.com/cellular/hacking/phreaking/
Our communications on breaking crypto caused me to think.
What's the square root of the companion matrix
010
001
110
?
It’s
011
111
101
I think.
Let’s try it.
011 011 0 1 0
111 111 = 0 0 1
101 101 1 1 0
Mission accomplished.
With a bit of help from Lagrange.
Companion matrices form a cyclic group.
Math humor
Q What’s purple and commutes?
A An Abelian grape.
100
If the companion matrix is denoted M, then M, M^2, M^3, ... M^7 = 010
001
And therefore, [Lagrange]
The order of a subgroup S of a finite group G is a divisior of
the order of G.
Fermat’s theorem
If a is an integer and p a prime, then a^p [is congruent to] a
mod p
is merely a corollary of Lagrange’s theorem.
Allahu ahkbar ... and, of course, KEEP UP-WIND.
THEY are PISSED!
http://209.185.240.250/cgi-bin/linkrd?_lang=&lah=0344efe07e62d8f7dc1669ccadbdb9f3&lat=939693182&hm___action=http%3a%2f%2fhome%2ebip%2enet%2flaszlob%2fcryptoag%2fcrypto_ag%2ehtm
And the FORTH is with them.
http://209.185.240.250/cgi-bin/linkrd?_lang=&lah=0344efe07e62d8f7dc1669ccadbdb9f3&lat=939693182&hm___action=http%3a%2f%2fhome%2ebip%2enet%2flaszlob%2fcryptoag%2fcrypto_ag%2ehtm
http://www.apcatalog.com/cgi-bin/AP?ISBN=0125475705&LOCATION=US&FORM=FORM2
biru
