MATH3320 HW11 (Solution)
Outline of methods and proofs with remarks
p.153, Q1 We will repeatedly make use of Theorems (5.10) and (5.18) and then
Theorems (5.11), (5.12) and (5.13).
(70/97) = (2/97)(5/97)(7/97) = 1 1 (6/7) = (2/7)(3/7) = 1.
(263/331) = (33

MATH3320 HW10 (Solution)
Outline of methods and proofs with remarks
(1) x2 1 (mod 561) if and only if x 1 or 1 (mod 3), (mod 11)
and (mod 17) respectively. Therefore, the Chinese Remainder Theorem tells us that there are exactly 8 solutions modulo 561, wh

MATH3320 HW9 (Solution)
Outline of methods and proofs with remarks
(1) Since s18 0 (mod M19 ) and s30 0 (mod M31 ), we conclude that
M19 and M31 are primes. Next, s10 1736 0 (mod M11 ) and s22
313912 0 (mod M23 ) so we conclude that M11 and M23 are not p

MATH3320 HW8 (Solution)
Outline of methods and proofs with remarks
p.64, Q36 Chinese Remainder Theorem.
Q39 This question is tricky. We will see how to use the Chinese Remainder
Theorem to solve it. Let m = r pki . So we have
i=1 i
#cfw_0 x < m : m|x2 x =

MATH3320 HW7 (Solution)
Outline of methods and proofs with remarks
Q4 27 28 36 1 2 10 6 (mod 13).
you dont have to calculate the number 1 2 10 = 10! before doing
modulo 13 because the product is large. You can just carry out each
calculation modulo 13: 1

MATH3320 HW6 (Solution)
Q1 The fundamental solution is the smallest positive solution, which is
(10, 3).
General form is (xn , yn ) by Theorem 10.10 where xn + yn 11 =
(10 + 3 11)n .
Q3 The 4th positive solution is (32257, 12192).
Q4
(a) The least positi

MATH3320 HW3 (Solution)
Outline of methods and remarks
Q1 Since all the factors of 15 are 1, 3, 5 and 15, we have to solve for each
one for a and b where (x, y, z) = (a2 b2 , 2ab, a2 +b2 ) for possible values
of x, y and z.
For 1, there is no good values

MATH3320 HW2 (Solution)
Outline of methods and remarks
Q30 Use the identity
a3 + b3 = (a + b)(a2 ab + b2 ).
Q32 Do not use brute force. Re-write the question as:
33a + 39b + 47(100 a b) = 3998.
and solve for a, b N with 35 b 40.
Q33 First, solve 455 = 10a

MATH3320 HW1 (Solution)
Outline of methods and remarks
B solution exists if and only if the gcd divides the number on the right
hand side.
D For those who are unfamiliar with doing induction, I am going to write
out all the steps with extra details here.

MATH 3320: ASSIGNMENT #1
Due Monday September 14th. Try as many of the following questions as you can. Note that
grades will be based not only on correctness but also clarity of exposition. For the computational
questions you can use a calculator to perfo