We stated that the number 561 is a Carmichael number, but we never checked that a561 ≡ a (mod 561) for every value of a. (a) The number 561 factors as 3 · 11 · 17. First use Fermat’s little theorem to prove that a561 ≡ a (mod 3), a561 ≡ a (mod 11), and a561 ≡ a (mod 17) for every value of a. Then explain why these three congruences imply that a561 ≡ a (mod 561) for every value of a. (b) Mimic the idea used in (a) to prove that each of the following numbers is a Carmichael number. (To assist you, we have factored each number into primes.) (i) 1729 = 7 · 13 · 19 (ii) 10585 = 5 · 29 · 73 (iii) 75361 = 11 · 13 · 17 · 31
We stated that the number 561 is a Carmichael number, but we never checked that a561 ≡ a (mod 561) for every value of a. (a) The number 561 factors as 3 · 11 · 17. First use Fermat’s little theorem to prove that a561 ≡ a (mod 3), a561 ≡ a (mod 11), and a561 ≡ a (mod 17) for every value of a. Then explain why these three congruences imply that a561 ≡ a (mod 561) for every value of a. (b) Mimic the idea used in (a) to prove that each of the following numbers is a Carmichael number. (To assist you, we have factored each number into primes.) (i) 1729 = 7 · 13 · 19 (ii) 10585 = 5 · 29 · 73 (iii) 75361 = 11 · 13 · 17 · 31
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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We stated that the number 561 is a Carmichael number, but we never checked that a561 ≡ a (mod 561) for every value of a.
(a) The number 561 factors as 3 · 11 · 17. First use Fermat’s little theorem to prove that
a561 ≡ a (mod 3), a561 ≡ a (mod 11), and a561 ≡ a (mod 17)
for every value of a. Then explain why these three congruences imply that a561 ≡ a (mod 561) for every value of a.
(b) Mimic the idea used in (a) to prove that each of the following numbers is a
Carmichael number. (To assist you, we have factored each number into primes.)
(i) 1729 = 7 · 13 · 19
(ii) 10585 = 5 · 29 · 73
(iii) 75361 = 11 · 13 · 17 · 31
(iv) 1024651 = 19 · 199 · 271
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