From: hughes@ah.com (Eric Hughes)
Date: Tue, 17 May 94 22:42:51 PDT
To: klbarrus@owlnet.rice.edu
Subject: Rabin
In-Reply-To: <9405170239.AA23367@flammulated.owlnet.rice.edu>
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Karl posted a good answer about square roots modulo a Blum integer.
I'd like to explain some of the context for this math.

Recall that a multiplicative group modulo n=pq is the product of two
multiplicative groups modulo p and modulo q.  That is,

	Z^*/nZ =~= Z^*/pZ  x  Z^*/qZ

(The superscript asterisks denote multiplication.)  So an element of
Z/nZ can be represented by an ordered pair of residues mod p and mod
q.  This same situation explains why there is another decryption
exponent in RSA, a previous thread.

Anyway, if p is prime, then every square mod p has two square roots.
When p = 3 (mod 4), these square roots are easy to find.  See the
article in the current MAA Monthly for a discussion of the other case.
If <m,n> is a square in Z/nZ, then each component m and n must also be
a square.  Thus if <m,n>=<a^2,b^2>, there are four possible square
roots <a,b>, <a,-b>, <-a,b>, and <-a,-b>.  These are additive inverses
in one pairing and conjugates in the other.

For completeness, it should be noted that the set of all squares of a
group is a subgroup.  The commutative case is easy; the
non-commutative case is much harder.  It is a good exercise to
calculate some square groups, to see how they generally behave, for
example, properties about their sizes.

Karl's explanations of using the Chinese remainder theorem to get the
canonical representations is fine, as is his observation about the
error in Schneier's text, although n-x = x (mod n), so the "n -" part
is unnecessary.

Eric