By VICTOR SHOUP

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**Extra info for A COMPUTATIONAL INTRODUCTION TO NUMBER THEORY AND ALGEBRA (VERSION 1)**

**Example text**

6 Arithmetic functions and M¨ obius inversion 29 as well, which one can see by checking that (f (g h))(n) = f (d1 )g(d2 )h(d3 ) = ((f g) h)(n), n=d1 d2 d3 the sum being over all triples (d1 , d2 , d3 ) of positive integers with d1 d2 d3 = n. We now introduce three special arithmetic functions: I, J, and µ. The function I(n) is deﬁned to be 1 when n = 1 and 0 when n > 1. The function J(n) is deﬁned to be 1 for all n. The M¨ obius function µ is deﬁned for positive integers n as follows: µ(n) := 0 if n is divisible by a square other than 1; r (−1) if n is the product of r ≥ 0 distinct primes.

In the third row, the only numbers hit are the multiples of 3, which follows from the theorem and the fact that gcd(3, 15) = 3. Also note that the pattern in this row repeats every ﬁve columns; that is also implied by the theorem; that is, 3z ≡ 3z (mod 15) if and only if z ≡ z (mod 5). In the fourth row, we again see a permutation of the ﬁrst row, which follows from the theorem and the fact that gcd(4, 15) = 1. In the ﬁfth row, the only numbers hit are the multiples of 5, which follows from the theorem and the fact that gcd(5, 15) = 5.

That proves the ﬁrst statement. For the second statement, let d = gcd(a, n). Simply from the deﬁnition of congruences, one sees that in general, az ≡ az (mod n) holds if and only if (a/d)z ≡ (a/d)z (mod n/d). 9), the ﬁrst statement of the theorem implies that (a/d)z ≡ (a/d)z (mod n/d) holds if and only if z ≡ z (mod n/d). That proves the second statement. 5 implies that multiplicative inverses modulo n are uniquely determined modulo n; indeed, if a is relatively prime to n, and if aa ≡ 1 ≡ aa (mod n), then we may cancel a from the left- and right-hand sides of this congruence, obtaining a ≡ a (mod n).

### A COMPUTATIONAL INTRODUCTION TO NUMBER THEORY AND ALGEBRA (VERSION 1) by VICTOR SHOUP

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