Math 328K (Daniel Allcock)
Homework 10, due Friday April 8, 2016
Problem 1. Use FLT to simplify 3999,999,999 modulo 7.
Problem 2. Write n for 161038 and show that 2n ≡ 2 modulo n. The point is that if
you misunderstood Fermat’s little theorem, then you might erroneously think that
n is prime. Which it obviously is not. (It is legitimate or is it obnoxious to say
“obviously” here?)
Problem 3. Use Fermat’s little theorem to show that 1387 is not prime. (This
number is small enough that brute-force factorization would work. But you must
use FLT.)
Problem 4. Show that if p is an odd prime then 2(p − 3)! ≡ −1 mod p.
Problem 5. Show that if p is prime then 2p
p ≡ 2 mod p.
The point of the next problem is to prepare you for the Chinese Remainder
Theorem, to come next week. You don’t need to look up that theorem to do the
problem—this is just a warmup.
Problem 6. Consider the following statements about an integer z, which turn out
to be equivalent.
Statement A: z ≡ 53 mod 84
Statement B: z satisfies all three conditions z ≡ 2 mod 3, z ≡ 5 mod 4 and
z ≡ −3 mod 7.
(a) (the easy part) show that A implies B by observing that 53 is congruent to
2, 5 or −3 modulo 3, 4 or 7.
(b) For i = 1, 2, 3 we will write ai for the right sides in statement B, and mi for
the moduli. Define Mi as the product of the m’s other than mi . So for example
M1 = m2 m3 . Work all three M ’s out explicitly.
(c) For each i = 1, 2, 3, check that Mi is coprime to mi , so it has a multiplicative
inverse modulo mi . Call this multiplicative inverse yi , and compute all three y’s
explicitly.
(d) Now, using the data from statement B you will write down a solution to
statement A. Namely: define z = a1 M1 y1 + a2 M2 y2 + a3 M3 y3 . Check that z ≡ 53
mod 84.
(e) Finally, show that z satisfies all three congruences in statement B. Actually
(d) and (a) together imply this already, but I want you to do it a better way. Using
Mi yi ≡ 1 mod mi and Mi ≡ 0 mod mj (for i 6= j), show without your calculator
that z from (d) satisfies the congruences in statement B.