Answer to MT 2
the solution is for reference only
1. We only need to nd d such that cd 1 mod (n), then we can use the
exponentiation method to calculate the message with the help of the table.
Since (n) = 2c + 662, c = 2 662 + 1, we clearly have 2(n) = 4c
Math3080: Number Theory
HW 2, Due Sep 25(Wed) 6pm
Charles Li
1. Let a, b be two non-zero integers.
(a) (Burton Sect 2.4 Q9) Show that gcd(a, b)| lcm(a, b).
(b) (Burton Sect 2.4 Q10a) gcd(a, b) = lcm(a, b) if and only if
a = b.
(c) (Burton Sect 2.4 Q10c).
Math3080: Number Theory
HW 1, Due Sep 18(Wed) 6pm
Charles Li
1. (a) Show that for any integer n,
n(n+1)(n+2)
6
is an integer.
(b) (Burton, Sect 2.3 Q17) Show that for any positive integer n,
an integer.
(3n)!
(3!)n
is
2. (Burton, Sect 2.3 Q14(a) Show that
2013 Math3080 Number Theory
Lecture 4: The Fundamental theorem of Arithmetic
Charles Li
1
The fundamental theorem of arithmetic
Denition 1 A positive integer p > 1 is said to be a prime number if p has no positive factors other than 1 and p.
Recall Euclid
2013 Math3080 Number Theory
Lecture 1 : Introduction
Charles Li
Mathematics is the queen of the sciences and number theory is the queen of mathematics by Gauss (the prince of
mathematics)
1
Class info
Instructor: Charles Li
Oce: LSB 218 (next to the gen
Math 3080 Number Theory Lecture 3 : LCM and linear
diophantine equations
Charles Li
1
The least common multiples
Reference: Burton Section 2.4
Denition 1 Given two non-zero integers a and b. An integer c is
said to be a common multiple of a and b if a|c a
2013 Math 3080 Notes
Innitude of primes
Charles Li
Burton, section 3.2, 3.3
1
Euclids proof of the innitude of primes
Let pk be the k -th prime. So p1 = 2, p2 = 3, p3 = 5, p4 = 7, etc.
Let nk = p1 pk + 1.
Lemma 1 For k i 1, then pi nk .
Proof. nk = pi (p1
Math3080: Number Theory Week I review
Charles Li
Division algorithm, proof, other versions (e.g. |r| b/2), applications (e.g. x2 divided by 3 is either 0 or 1).
Denition of a|b, related properties and the proofs.
Denition of common divisors and the gre
2013 Math3080
Lecture 2 : Divisibility
Charles Li
Reference: Burton, Chapter 2.
Unless otherwise stated, all the variables in this lecture are integers.
1
The division algorithm
Reference: Burton, Section 2.2
Theorem 1 (Division Algorithm) Given integers
Math3080: Number Theory Tutorial I
Charles Li
The purpose of these extra exercises are to see how well you understand lecture 2.
1. Let a1 , . . . , ar be integers, non all zero. Dene common divisors and the greatest common divisors for this set. Denote t
Math3080: Number Theory Tutorial 2
Charles Li
1. Let a, b be relatively prime positive integers. Show that the
Diophantine equation ax by = c has innitely many positive
integer solutions.
2. (a) Use the previous tutorial exercise, discuss when the Diophan
Math3080 Number Theory Notes
Modular Arithmetic
Charles Li, 2013
1
Modular Arithmetic
Reference: Burton, Section 4.2
+, , , and modular arithmetic are ve basic operations in Math.
Denition 1 Let m be an integer. Two integers a and b are said
to be congrue
Solution to HW1
Section 2.2 Problem 3(c)
Use the division algorithm to establish the following:
(c)The fourth power of any integer is either of the form 5k or 5k + 1.
Proof. For an integer n, by the division algorithm we have
n = 5m + r, |r| = 0, 1, 2, 3,
Solution to HW2
Section 2.3 Problem 20(a)
Proof. There exist x, y, u, v Z such that 1 = ax + by = au + cv . Then
1 = (ax + by )(au + cv )
= a(axu + cxv + byu) + bc(vy )
Therefore, gcd(a, bc) = 1.
Section 2.3 Problem 20(c)
Proof. Since gcd(a, b) = 1, for a
Solution to HW3
Section 3.1 Q15
Proof. Its self-evident that if all the exponents of the primes in the canonical
form of a are even, then a is a square. Now suppose a is a square, then a = b2
j (i)
for some integer b. Write b in the canonical form i pi ,
Solution to HW4
the solution is for reference only
Section 4.2 Q1(c)
Proof. Given a, b, n are divisible by d, then a = rd, b = sd and n = td for
some integers r, s, t.
Since a b (mod n), there exists an integer p such that a b = pn.
Then, rd sd = ptd and
Solution to HW5
Section 4.4 Q1(c)
Proof. The equation is equivalent to the Diophantine equation 6x +21y = 15.
Since gcd(6, 21) = 3 and 3|15, we see that the equation has 3 mutually
incongruent solutions modulo 21. Since 3 = 21 + (3) 6, multiplying by
5 on
Solution to HW6
the solution is for reference only
Section 5.2 Q15
Proof. (a) Since p is prime, by Fermats theorem we have Mp 2 1 1
mod p. Thus there exists some k Z such that Mp 1 = kp. Since 2p
1|2kp 1 by factorization, we have 2Mp 1 1 mod Mp , which i
Solution to HW7
the solution is for reference only
Section 7.3 Q4
Proof. Since gcd(a, n)=1,
(a 1)(a(n)1 + a(n)2 + + a + 1) a(n) 1 0 (mod n).
Furthermore, since gcd(a 1, n)=1, we have a(n)1 + a(n)2 + + a + 1 0
(mod n).
Section 7.3 Q5
Proof. We have m(n) 0
Solution to HW8
the solution is for reference only
Section 8.1 Q3
Proof. We have 2n 1 (mod 2n 1).
Then for all k with 1 k n 1, we have 1 = 20 21 22 2n1 <
2n 1.
Therefore, 2k = 1 (mod 2n 1) and so 2 has order n modulo 2n 1.
By theorem 8.1, n|(2n 1) and so
Solution to HW9
the solution is for reference only
Section 8.2 Q4
Proof. (a) As 3 is a primitive root of 43, 3 has order 42.
42
= 6. Therefore, h = 7 or 35.
Since 3h has order 6, we have
gcd(h, 42)
In fact, we have 37 37 (mod 43) and 335 7 (mod 43), so 7