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Exercise A.2.16. Suppose gcd(a, n) = 1. Show that there exists b such that [a] [b]n = [1]n. Exercise A.2.17. (1) Show that gcd

Exercise A.2.16. Suppose ( operatorname{gcd}(a, n)=1 ). Show that there exists ( b ) such that ( [a]_{n} cdot[b]_{n}=[


 

Exercise A.2.16. Suppose gcd(a, n) = 1. Show that there exists b such that [a] [b]n = [1]n. Exercise A.2.17. (1) Show that gcd (a,n) = 1 if and only if there exist m and k such that ma + kn 1. (2) Use part (1) to prove that if there is b such that [a]n [b]n[1], then gcd (a, n) = 1.

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