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For this question, recall that a Carmichael number is a composite number n such that an-11 mod n whenever ged(a, n) = 1. Let m = P1P2. pr. for distinct primes pi, such that (p:- 1) divides m - 1 for all i. (p₁ (a) If ged(a,m) = 1 show that api-1 = 1 mod pi, for each i. (b) Using (a), show that am-1 = 1 mod p; for each i. (c) Using (b) and the Chinese Remainder Theorem, show that am-1 = 1 mod m.

For this question, recall that a Carmichael number is a composite number n such that an-11 mod n whenever ged(a, n) = 1. Let m = P1P2. pr. for distinct primes pi, such that (p:- 1) divides m - 1 for all i. (p₁ (a) If ged(a,m) = 1 show that api-1 = 1 mod pi, for each i. (b) Using (a), show that am-1 = 1 mod p; for each i. (c) Using (b) and the Chinese Remainder Theorem, show that am-1 = 1 mod m.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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4. For this question, recall that a Carmichael number is a composite number n such that an-11 mod n whenever gcd(a, n) = 1. Let m = P1P2. pr. for distinct primes pi, such that (p; - 1) divides m - 1 for all i. (a) If ged(a,m) = 1 show that api-1 = 1 mod p₁, for each i. (b) Using (a), show that am-1 = 1 mod p; for each i. (c) Using (b) and the Chinese Remainder Theorem, show that am-1 = 1 mod m. (d) Using the above, prove that if there are infinitely many triples of prime numbers of the form (6k + 1, 12k + 1, 18k + 1) for some k ≥ 1, then there are infinitely many Carmichael numbers.
Transcribed Image Text:4. For this question, recall that a Carmichael number is a composite number n such that an-11 mod n whenever gcd(a, n) = 1. Let m = P1P2. pr. for distinct primes pi, such that (p; - 1) divides m - 1 for all i. (a) If ged(a,m) = 1 show that api-1 = 1 mod p₁, for each i. (b) Using (a), show that am-1 = 1 mod p; for each i. (c) Using (b) and the Chinese Remainder Theorem, show that am-1 = 1 mod m. (d) Using the above, prove that if there are infinitely many triples of prime numbers of the form (6k + 1, 12k + 1, 18k + 1) for some k ≥ 1, then there are infinitely many Carmichael numbers.
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