Prove that if gcd(a, m)=1 and gcd(a - 1, m) =1, then 1+a+a²+...+ a² + a(m)−1 = 0 (mod m).

Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter7: Real And Complex Numbers
Section7.3: De Moivre’s Theorem And Roots Of Complex Numbers
Problem 21E
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Prove that if gcd(a, m) =1 and gcd(a - 1, m) =1, then
1+a+a²+...+ a
+ a(m)−1 = 0 (modm).
(Hint: What do you get when you multiply the congruence by a - 1 ?)
Transcribed Image Text:Prove that if gcd(a, m) =1 and gcd(a - 1, m) =1, then 1+a+a²+...+ a + a(m)−1 = 0 (modm). (Hint: What do you get when you multiply the congruence by a - 1 ?)
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