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

142 A. Appendix
Induction step: Assume r + s ≥ 3, and the claim is true for all n ′ , m ′ that
together have fewer than r + s prime factors. By symmetry, we may assume
that n is not a prime number. We write n = kℓ for numbers k, ℓ ≥ 3. By the
induction hypothesis, we have
� � � �
m k
· =(−1)
k m
k−1
2 · m−1
2 and
� � � �
m ℓ
· =(−1)
ℓ m
ℓ−1
2 · m−1
2 .
(A.3.15)
Using multiplicativity in both upper and lower positions (Lemma 6.2.2(a)
and (c)) we get by multiplying both equations:
� � � �
m n
· =(−1)
n m
k−1
2 · m−1 ℓ−1
2 + 2 · m−1 �
2 = (−1) k−1
2
Now (k−1)(ℓ−1)
2
n − 1
2
is an even number, hence
= (k − 1)(ℓ − 1) + k + ℓ − 2
2
≡
k − 1
2
+ ℓ − 1
2
+ ℓ−1
2
(mod 2).
� m−1
2
.
(A.3.16)
Plugging this into (A.3.16) yields the inductive assertion. Thus the theorem
is proved. ⊓⊔
Proof of Proposition 6.3.2. We show that for n ≥ 3anoddintegerwe
have
� �
2
n
=(−1) n2 −1
8 .
Again, we consider the prime factorization n = p1 ···pr and prove the
claim by induction on r. Ifr = 1, i.e., n is a prime number, the assertion
is just Corollary A.3.2. For the induction step, assume r ≥ 2, and that the
claim is true for n ′ with fewer than r prime factors. Write n = kℓ for numbers
k, ℓ ≥ 3. By multiplicativity and the induction hypothesis, we have
� �
2
n
=
� 2
k
Now (k2 −1)(ℓ 2 −1)
8
n 2 − 1
8
�
·
� 2
ℓ
�
=(−1) k2−1 8
is divisible by 8, hence
= (k2 − 1)(ℓ 2 − 1) + k 2 + ℓ 2 − 2
8
+ ℓ2 −1
8 . (A.3.17)
≡ k2 − 1
8 + ℓ2 − 1
8
(mod 2).
(A.3.18)
If we plug this into (A.3.17), we obtain the inductive assertion. Thus the
proposition is proved. ⊓⊔