By VICTOR SHOUP
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Additional resources for A COMPUTATIONAL INTRODUCTION TO NUMBER THEORY AND ALGEBRA (VERSION 1)
6, we can compute successive powers of a modulo n to ﬁnd its multiplicative order modulo n. 1i 2i 3i 4i 5i 6i i mod 7 mod 7 mod 7 mod 7 mod 7 mod 7 1 1 2 3 4 5 6 2 1 4 2 2 4 1 3 1 1 6 1 6 6 4 1 2 4 4 2 1 5 1 4 5 2 3 6 6 1 1 1 1 1 1 So we conclude that modulo 7: 1 has order 1; 6 has order 2; 2 and 4 have order 3; and 3 and 5 have order 6. 15 (Euler’s Theorem). For any positive integer n, and any integer a relatively prime to n, we have aφ(n) ≡ 1 (mod n). In particular, the multiplicative order of a modulo n divides φ(n).
Pe − 1 are 0 · p, 1 · p, . . , (pe−1 − 1) · p, of which there are precisely pe−1 . Thus, φ(pe ) = pe − pe−1 = pe−1 (p − 1). 14. If n = pe11 · · · perr is the factorization of n into primes, then r pei i −1 (pi φ(n) = i=1 r − 1) = n (1 − 1/pi ). 14. Show that φ(nm) = gcd(n, m) · φ(lcm(n, m)). 5 Fermat’s little theorem Let n be a positive integer, and let a ∈ Z with gcd(a, n) = 1. Consider the sequence of powers of α := [a]n ∈ Z∗n : n = α0 , α1 , α2 , . . TEAM LinG 26 Congruences Since each such power is an element of Z∗n , and since Z∗n is a ﬁnite set, this sequence of powers must start to repeat at some point; that is, there must be a positive integer k such that αk = αi for some i = 0, .
For example, suppose f (d) = 1/d in the above theorem, and let n = · · · perr be the prime factorization of n. Then we obtain: pe11 µ(d)/d = (1 − 1/p1 ) · · · (1 − 1/pr ). 8) d|n TEAM LinG 30 Congruences As another example, suppose f = J. Then we obtain r (1 − 1), µ(d) = (µ J)(n) = i=1 d|n which is 1 if n = 1, and is zero if n > 1. Thus, we have µ J = I. 18 (M¨ obius inversion formula). Let f and F be arithmetic functions. Then we have F = J f if and only if f = µ F . Proof. If F = J f , then µ F = µ (J f ) = (µ J) f = I f = f, and conversely, if f = µ F , then J f =J (µ F ) = (J µ) F = I F = F.