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

3.4 The Chinese Remainder Theorem 35
Proof. (a) is trivial. — For (b) and (c) we make heavy use of Proposition
3.1.13:
(b) Assume gcd(a, m) =gcd(b, m) =1.Wewrite
1=ax + my and 1 = bu + mv,
for integers x, y, u, v. Then
(ab) · (xu) =(ax)(bu) =(1− my) · (1 − mv) =1− m(y + v − myv),
hence gcd(ab, m) = 1. By Proposition 3.1.10, this implies gcd(ab mod m, m) =
1.
(c) “⇒”: Assume gcd(a, m) = 1. Then there are integers x and y with
1=ax + my.
This implies ax − 1=−my ≡ 0(modm). Let b = x mod m. Thenab ≡
ax ≡ 1(modm), which implies that a ·m b =1.
“⇐”: Assume ab mod m = 1. This means that ab−1 =mx for some x, which
implies gcd(a, m) =1. ⊓⊔
Note that by the formula in the proof of (c) “⇒” it is implied that for
a ∈ Z ∗ m some b with a ·m b = 1 can be found efficiently by applying the
Extended Euclidean Algorithm 3.2.4 to a and m.
Example 3.3.9. (a) In Z∗ 7 = {1, 2, 3, 4, 5, 6} we have 3 ·7 3=2and3·75=1. By inspection, ϕ(7) = 6.
(b) In Z∗ 15 = {1, 2, 4, 7, 8, 11, 13, 14} the products 4·151, 4·152, 4·154,...,4·1514
of 4 with elements of Z ∗ 15
are 4, 8, 1, 13, 2, 14, 7, 11. By inspection, ϕ(15) = 8.
(c) In Z ∗ 27 = {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26} we have
8 ·27 17 = 1. By inspection, ϕ(27) = 18.
In Sect. 3.5 we will see how to calculate ϕ(n) fornumbersn whose prime
decomposition is known.
3.4 The Chinese Remainder Theorem
We start with an example. Consider 24 = 3 · 8, a product of two numbers
that are relatively prime. We set up a table of the remainders a mod 3 and
a mod 8, for 0 ≤ a