FINDING N-TH ROOTS IN NILPOTENT GROUPS AND ...

FINDING N-TH ROOTS IN NILPOTENT GROUPS... 589
a
n
Note that when the modulus is n = p t1
1 pt2 2 · · · ptm m , we use the Jacobi symbol,
, defined by
a
t1 a
=
n p1
t2 tm a a
· · · ,
p2 pm
where the symbols on the right-hand side are Legendre symbols.
Proposition 8.4. Let p be a prime such that p ≡ 3 (mod 4). The
solutions of x2 ≡ a (mod p) are x ≡ ±a p+1
4 (mod p).
Proposition 8.5. Let n = pq, where p and q are distinct odd primes. Let
0 < a < n and (a,n) = 1. Then x 2 ≡ a (mod n) has exactly four solutions
modulo n.
Example 8.6. Let n = 103·107 = 11021. Suppose we know that x 2 ≡ 860
(mod 11021) has solution. Then we need to solve
and
x 2 ≡ 860 ≡ 36 (mod 103)
x 2 ≡ 860 ≡ 4 (mod 107).
Since 103 ≡ 3 (mod 4) and 107 ≡ 3 (mod 4), we know that the solutions are
and
x ≡ ±36 26 ≡ ±97 ≡ ±6 (mod 103)
x ≡ ±4 27 ≡ ±105 ≡ ±2 (mod 107),
respectively. By the Chinese Remainder Theorem,
x ≡ ±109 (mod 11021) or x ≡ ±212 (mod 11021).
8.2. Rabin Public Key Encryption
As usual, we assume that Alice and Bob are sending messages through an
insecure channel. Alice will choose two large primes, p and q, which will be
private. Her public key is n = pq. Bob sends his message m by computing
c ≡ m 2 (mod n) and sending c to Alice. Alice recovers the message by using
the methods discussed above, but she has four possible values for m.
In order for Eve, the eavesdropper, to recover the message, she will need to
know the factors p and q. Factoring n and computing square roots modulo n
are computationally equivalent. Therefore, security of this encryption scheme
lies on the assumption that factoring n is computationally intractable.