Chapter 21
Addition and Multiplication in
Zm
In this chapter we show how to dene addition and multiplication of residue
classes modulo m. With respect to these binary operations Zm is a ring as
dened in Appendix A.
Denition 21.1. For [a], [b] Zm we dene
[
Chapter 25
The Base b Representation of n
Denition 25.1. Let b 2 and n > 0. We write
(1)
n = [ak , ak1 , . . . , a1 , a0 ]b
if and only if for some k 0
n = ak bk + ak1 bk1 + + a1 b + a0
where ai cfw_0, 1, . . . , b 1 for i = 0, 1, . . . , k. [ak , ak1 , .
Chapter 26
Computation of aN mod m
Lets rst consider the question: What is the smallest number of multiplications required to compute aN where N is any positive integer?
Suppose we want to calculate 28 . One way is to perform the following 7
multiplicatio
Chapter 15
Congruences
Denition 15.1. Let m 0. We write a b (mod m) if m | a b, and
we say that a is congruent to b modulo m. Here m is said to be the modulus
of the congruence. The notation a b (mod m) means that it is false that
a b (mod m).
Examples 15
MAS 4214/001 Elementary Number Theory, Fall 2015 CRN; 80518
21
Answers, Set 2 (There are ,Llfioiuts and 10 points equal 100%)
Question 1' (1 pt) (Exercise 5.4) (a) Prove that if a | b or a I c, then a I bc.
(1) Give an example to show that the converse
To prove that the converse is false, we need an example. Take I : 3 and a1 = 6,0,2 2 10413 = 15. Then
gcd(6,10,15) = gcd(21 15) := 1,
but gc(l(6, 10) x 2 1. [1
Remarks: In fact, the god of each pair of integers from 6, 10, 15 is greater than 1.
Question 5
Chapter 22
The Groups Um
Denition 22.1. Let m > 0. A residue class [a] Zm is called a unit if
there is another residue class [b] Zm such that [a][b] = [1]. In this case [a]
and [b] are said to be inverses of each other in Zm .
Theorem 22.1. Let m > 0. A r
(b) Prove that for m n, gcd(Gn, 4m) 2 1. (Hint. Use the result: gcd(bq +1',b) : gcd(r, 12)
Solution. (:1) We shall be using the equation $2 - 1 = (:c ~ 1)(a; + 1) repeatedly. For n 2 1, we have
n
Gnz = b2 1
(1? 1)(b2" + 1)
(122 new
11
II
II
(1)21 1)G1-G
cfw_.sws. .
Question 6. (1 pt) (True or false) Suppose that a, b are integers, not both zero. Let d = gcd(a,b). Assume
that there are integers s and t so that
18 = as + bt.
Prove or disprove the following statements:
(a) d | 18.
(b) d = 18.
(c) If e is
Chapter 19
Residue Classes
Denition 19.1. Let m > 0 be given. For each integer a we dene
(1)
[a] = cfw_x : x a
(mod m).
In other words, [a] is the set of all integers that are congruent to a modulo
m. We call [a] the residue class of a modulo m. Some peop
Question 3. (1 pt) (Exercise 7.1.)
(a) Prove that d I a implies that (l I a.
(b) Prove that d I a iff d I a. '5
(c) Prove that d I (1 iff d I IaI.
190171177071. (3.) Assume that d I (L. Then there is q so that a = dq. Thus,
-a : d(1)a
and since q is an in
giving that
ac=~3w+5k, y=2w3k, kEZ.
Since in = 56 + 15h7 the general solution is
9;:16845h-i5k, y=112+30h3k, 2:8211, h,keZ.
Please note that the solutions can appear in other formats.
(1) (1.5 pts) We rst solve the rst equation in the system. By inspectio
MAS 4214/001 Elementary Number Theory, Fall 2015 CRN: 80518
Answers, Set 4 (There are 11 points and 10 points equal 100%)
Question 1. (1 pt) Consider my solution to Exercise 13.4 in Assignment 3 (to be on the class web page
soon) for proving that for in
Chapter 23
Two Theorems of Euler and
Fermat
Fermats Big Theorem or, as it is also called, Fermats Last Theorem states
that xn + y n = z n has no solutions in positive integers x, y, z when n > 2.
This was proved by Andrew Wiles in 1995 over 350 years afte
(3) True, because of the cancellation property with gcd(12, 15) = 3. The converse is also true.
(h) True, since a; E 73 (mod 75) is the same as saying that a; and 73 have the same remainder when divided
by 75. Now 0 S 73 < 75, implying that 73 is indeed t
A ~4va._
Wwww*v_v7w 7
By FLT, we have
(11 E 1 (mod 5), (11;2 E 1 (mod 13) am E 1 (mod 17).
Since 771 -~ 1 = 1104 is divisible by 4, 1:2, and 16, we have for some integers ([1,(]2,cfw_13 that
(rm1 = (12%) E 1 (mod 5),
am1 = (any: E 1 (mod 13),
owl1 = (Gwy
cfw_(7) Find all solfiuveises modulo m. = 1417 a (3) (49).
Solution. (:1) cfw_:>" part.) Suppose that, x is a selfinverse, moudlo 1), Then 1'2 E 1 (mod ,0"), giving that
p l cfw_:17 1)(~1: + 1).
By Euclid's Lcumm, we have p I (:1: 1) or p l (a: + 1.) Howe
MAS4214: Practice problems
November 1, 2016
Question 1 We want to find all self-inverses modulo 3720151123 .
Let p be a prime. Prove that if p | (x 1) and p | (x + 1) then p = 2.
Prove that if 37 | (x 1)(x + 1) then 37 cannot divide both x 1 and x + 1.
Chapter 27
The RSA Scheme
In this chapter we discuss the basis of the so-called RSA scheme. This is
the most important example of a public key cryptographic scheme. The RSA
scheme is due to R. Rivest, A. Shamir and L. Adelman 1 and was discovered
by them
Chapter 18
More Properties of
Congruences
Theorem 18.1. Let m 2. If a and m are relatively prime, there exists a
unique integer a such that aa 1 (mod m) and 0 < a < m.
We call a the inverse of a modulo m. Note that we do not denote a by
a1 since this migh
Chapter 20
Zm and Complete Residue
Systems
Throughout this section we assume a xed modulus m > 0.
Denition 20.1. We dene
Zm = cfw_[a] | a Z,
that is, Zm is the set of all residue classes modulo m. We call Zm the ring
of integers modulo m. In the next chap
MAS4214: review session for Test 1
September 1, 2016
Question 1 Show by induction that for all n 2, 4n > 6n.
Answer
Let us prove by induction that for all n 2, 4n > 6n.
We define P (n) by P (n) = cfw_4n > 6n.
42 = 16 and 6 2 = 12, therefore 42 > 6 2 an
IVIAS 4214/ 001 Elementary Number Theory, Fall 2015 CRN: 80518
Answers, Set 5 (There are 12 points and 10 points equal 100%)
destion 1. (2 pts) This problem is about solving an: E 1) (mod m), where a. = 2123 and m = 11632.
(2) Use the extended Euclidean
destion 4- (1 pt) Solve the following systems of congruences:
(a)
7.7: = 2 (mod 9)
7.7: E 2 (mod 15).
(3)
5w E 3 (mod 9)
41' E 1 (mod 15).
Saint-ion. (it) Since lcni(9, 15) = 45, the two congruences can be combined to form
7:1: E 2 (mod 45),
and since 1
MAS4214: review session for Test 3
November 3, 2016
Question 1 Suppose that 15x 3 mod 21, what is the congruence class of x modulo 7 ?
Question 2 Is there an integer x such that 26x 11 mod 65 ? (Justify your answer).
Question 3 Find an inverse of 3 modulo
IVIAS 4214/001 Elementary Number Theory, Fall 2015 CRN: 80518
Answers, Set 6 (There are 12 points and 10 points equal 100%)
destion 1. (2+1 pts) Similar to (but a little harder than) HVVS, Q6, and Exercise 23.3.
(:1) Use the CRT to solve for m, where a,b,
MAS4214: Test 1
September 6, 2016
Question 1 Show by induction that for all n 6, 2n > 10n [3pt].
Answer
Let us prove by induction that for all n 6, 2n > 10n.
We define P (n) by P (n) = cfw_2n > 10n.
26 = 64 and 6 10 = 60, therefore 26 > 10 6 and P (6)
MAS4214: Practice problems
November 15, 2016
k
Question 1 Implement a function that computes x2 mod N using the repeated squaring
approach.
Question 2 Implement a function that computes xm mod N using the square-and-multiply
approach.
Question 3 Implement
lVIAS 4214/001 Elementary Number Theory, Fall 2015 GEN! 80518
Answers, Set 3 (There are 11 points and 10 points equal 100%)
Question 1. (2 pts) (Similar to Exercises 12.1, 12.2, 12.3 (and 10.1, 10.2):
(3.) Use the extended Euclidean algorithm to nd the
The above gives
28w E 1 (mod 29), and (30)(29)(28)x E 1 (mod 31).
From these two congruences, we see that
a: E 1 (mod 29), and ' E 5 (mod 31).
Using the CRT, we obtain that
a; a (31)(15)(1) + (29)(15)(5) = 1710 E 88 (mod 899).
[:1
Therefore, the remainder