Dietzfelbinger M. Primality testing in polynomial time ... - tiera.ru

88 6. The Solovay-Strassen Test
relatively prime. It is important to note that even in this case � �
a
n does not
indicate � � �whether � � � a is a quadratic residue modulo n or not. (For example,
2 2 2
15 = 3 · 5 =(−1)(−1) = 1 while 2 is a quadratic nonresidue modulo
15.) Our definition will have the useful consequence that the Jacobi symbol
is multiplicative in the upper and in the lower position, that square factors
in the upper and in the lower position can be ignored, and that � �
−1
n can
easily be evaluated.
Lemma 6.2.2. If n, m range over odd integers ≥ 3, anda, b range over
integers, then
(a) � � � � � �
a·b a b
n = n · n ;
(b) � a·b 2 � � �
a
n = n ,ifgcd(b, n) =1;
(c) � � � � � �
a a a
n·m = n · m ;
(d) � a
n·m2 � � �
a = n ,ifgcd(a, m) =1;
(e) � � � � � � � �
a+cn a
a a mod n
n = n , for all integers c (in particular, n = n );
(f) � 2 2k � � � � 2k+1 � � � � �
·a a 2 ·a 2 a
n = n and n = n · n ,fork≥1; (g) � �
−1
(n−1)/2
n =(−1) ;
(h) � � � �
0
1
n =0and n =1.
Proof. Parts (a) and (b) follow by applying Lemma 6.1.6(a) and (b) to the
factors in Definition 6.2.1. — Part (c) is an immediate consequence of Definition
6.2.1. — Part (d), in turn, follows by noting that � �
a
m ∈{1, −1} and
applying (c) twice. — For part (e), we use that for every prime factor p of
n we have that � � � �
a+cn a
p = p , by Lemma 6.1.6(c). — For part (f), we note
that � � � �
4 2 2
n = n = 1, and then apply (b) repeatedly to eliminate factors of
4� in� the upper � � position. � � — For part (g), we note the following. By definition,
−1 −1 −1
n = ··· ,wheren = p1 ···pr is the prime number decomposi-
p1 pr
tion of n. We apply Lemma 6.1.6 to see that this product equals (−1) s ,where
s is the number of indices i so that pi ≡ 3 (mod 4). On the other hand, we
have
n ≡ (p1 mod 4) ···(pr mod 4) ≡ (−1) s
(mod 4).
This means that 1
2 (n − 1) is odd if and only if s is odd, and hence (−1)s =
(−1) (n−1)/2 . — Part (h) is trivial. ⊓⊔
6.3 The Law of Quadratic Reciprocity
In this section, we identify a method to efficiently calculate � �
a
n . Most helpful
for this is an amazing relationship that connects the values � � � �
a n
n and a ,for
a, n ≥ 3 odd.
To gather some intuition, we put up a table of some of the values � �
a
n ;
see Table 6.1. We cover odd a ≥ 3, but also include the special cases a = −1
(congruent modulo n to the odd number 2n − 1) and a = 2 (congruent to
n +2).