Math 328K (Daniel Allcock)
Homework 12, due Friday Nov. 22, 2016
The following is a useful formulation of a special case of the Chinese Remainder Theorem. It is well-adapted for use in proving that things are true modulo
composite moduli. (Hint: it is useful in several problems on this homework set.)
Problem 1. Suppose x and a are integers and M is a product pairwise coprime
integers m1 , . . . , mk . Then x ≡ a mod M if and only if x ≡ a mod mi for all
i = 1, . . . , k.
Problem 2. If p and q are distinct primes then pq−1 + q p−1 ≡ 1 modulo pq.
Problem 3. Show that a6 ≡ 1 mod 168 whenever gcd(a, 42) = 1. (This is stronger
than what Euler’s theorem promises; what does Euler’s theorem say in this case?)
Problem 4. Factor n = 2821 and use this factorization to show that for any integer
a coprime to n, we have an−1 ≡ 1 mod n. (This property means that n defeats
the Fermat test for compositeness. It is composite, but no a detects this except for
those that have a factor in common with n. So trying to prove its compositeness
with Fermat’s little theorem would be equivalent to just randomly guessing numbers
a and taking their gcd’s with n, hoping to find a gcd bigger than 1. Numbers like
this are called Carmichael numbers; see example 6.12 in the book if you get stuck.)
Problem 5. Compute the Euler totient function φ(n) for n = 13, . . . , 20, directly
from its definition.
Problem 6. Find φ(10!).
Problem 7. Show there is no positive integer n with φ(n) = 14.