Prove that if gcd(a, m)=1 and gcd(a - 1, m) =1, then 1+a+a²+...+ a² + a(m)−1 = 0 (mod m).
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 ?)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2fd56cd7-16d6-42a3-873f-51e9748dabe0%2F4219b299-8fa7-4182-876b-2828cffb31fe%2Fsz98hpd_processed.jpeg&w=3840&q=75)
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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