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 b such that [a], · [b]n = [1]n, then gcd(a, n) = 1.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.2: Properties Of Division
Problem 51E
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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]n, then gcd (a,n) = 1.
Transcribed Image Text: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]n, then gcd (a,n) = 1.
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