# 2x 1 7. A linear congruence in modular arithmetic is an equation of the form. Why is the solution a set of integers rather than a unique integer?

Save this PDF as:

Size: px
Start display at page:

Download "2x 1 7. A linear congruence in modular arithmetic is an equation of the form. Why is the solution a set of integers rather than a unique integer?"

## Transcription

1 Chapter 3: Theory of Modular Arithmetic 25 SECTION C Solving Linear Congruences By the end of this section you will be able to solve congruence equations determine the number of solutions find the multiplicative inverse C Solving Linear Congruences In algebra we have linear equations in one unknown x given by Solving this equation gives x 3. 2x 7 A linear congruence in modular arithmetic is an equation of the form ax b mod n The solution to this linear congruence is the set of integers x which satisfy this: ax b mod n Why is the solution a set of integers rather than a unique integer? Recall ax b mod n integer k. means that ax b is a multiple of n or ax b kn for some As we have often stated, in mathematics we are interested in a unique solution if possible. How can we find a unique solution to the linear congruence? If two solutions x x and x x satisfy the linear congruence 0 ax b mod n and they are congruent modulo n, that is x x 0 mod n same solution and count them as one solution. For example let us consider the linear congruence We can trial a table of integers for x: 2 x mod 5 (*), then we say these are the x x mod Therefore x 3 and x 8 Table Shows the junctions of 2x mod 5 satisfy

2 Chapter 3: Theory of Modular Arithmetic 26 2 x mod 5 because 23 6 mod 5 and mod 5. Hence integers 3 and 8 satisfy the linear congruence (*). They are the same solution. Why? Because 8 3 mod 5. We count this as one solution not two. [It is the same station in modulo 5 clock.] Since we are interested in solutions modulo 5 we only need to consider integers amongst the list; x 0,, 2, 3 and 4 [least non-negative residues modulo 5] Because all the other solutions will be one of these in modulo 5 which is illustrated below: x 0,, 2, 3, 4 mod 5 these junctions. Figure 7 Modulo 5 clock covers all the stops. Any other integer will stop at one of A more systematic way of solving the above linear congruence is given next. Example 0 Determine the integers x of the following linear congruence: 2 x mod 5 By definition of congruence 2x mod 5 implies that 2x is a multiple of 5: 2x 5y where y is an integer Rearranging this we have 2x 5y which is a Diophantine equation and we solved these type of equations in chapter. Making x the subject of 2x 5y gives 5y x 2 Remember x must be an integer. So what values of y can we use?

3 Chapter 3: Theory of Modular Arithmetic 27 Only the odd integers because if we choose even then we get even plus which does not give a whole number after dividing by 2. Substituting y, 3, 5,,, 3, 5, gives y y x Table 2 x values for particular y values Therefore x, 2, 7, 2, 3, 8, 3, Clearly there are an infinite number of integers which satisfy the given congruence 2 x mod 5 We count all these solutions as one or the same solution because they are all congruent to each other modulo 5: mod 5 See the above Fig. 7 and you will notice that all these numbers 3, 8, 3,, 2, 7 stop at the same junction, 3 mod 5. We normally write this as just one solution which is the least non-negative residue modulo 5: x 3 mod 5 We say the solution of 2x mod 5 is 3 mod 5 x. Example Solve the linear congruence: 2 x mod 6 The solution (if it exists) of this linear congruence 2x mod 6 x 0,, 2, 3, 4 and 5 must be in the list: Because we are working with modulo 6 in this case. Evaluating these we have x x mod Table 3 Junctions of 2x mod 6. [Note that the numbers 0, 2 and 4 are repeated]. By examining this table we find that there are no x values which satisfy

4 Chapter 3: Theory of Modular Arithmetic 28 2 x mod 6. The set of integers 2x will not stop at junction modulo 6. This means that there is no solution to the given linear congruence 2x mod 6. (Try doing this by solving the Diophantine equation 2x 6y.) If we have x b mod n In solving 2x mod 5 then we only have to consider values of x 0,, 2, 3,, n. we tried values of x up to 9 (see Table at the beginning of this section), but we only need to try x 0,, 2, 3, 4. C2 Number of solutions of a Linear Congruence The above Example demonstrates that there are some linear congruences which have no solution. How do we know which congruences have a solution? The next proposition gives the criteria for a solution. Proposition (3.5). The linear congruence ax b mod n has a solution g b where g gcd a, n. Note that in Example 0 we had 2x mod 5. The the linear congruence 2x mod 5 has a solution. g gcd 2, 5 and so On the other hand in Example we had 2x mod 6 the g gcd 2, 6 2 and 2 does not divide so there are no solutions to this linear congruence. (For this example you would have noticed from the previous table that 2x mod 6 only stops at 0, 2 and 4 modulo 6 because these numbers are multiples of g gcd2, 6 2.) How do we prove this proposition (3.5)? By Proposition (.6) of chapter : Proof. ax by c has integer solutions g c We have ax b mod n where gcd, a b g which means that there is an integer k such that ax b kn ax kn ax k n b implies

5 Chapter 3: Theory of Modular Arithmetic 29 Let g gcd a, n. Then by Proposition (.6) we conclude that the Diophantine equation ax k n b has a solution g b which is our required result. Example 2 Which of the following linear congruences have solutions? (a) 7x 8 mod 4 (b) 2x 8 mod 6 (c) 5x 2 mod 9 (d) 36x 54 mod 90 (a) We are given 7x 8 mod 4. The greatest common divisor of 7 and 4, that is gcd 7, congruence 4, is 7 but 7 does not divide 8 so by the above Proposition (3.5) the linear 7 x 8 mod 4 (b) For 2x 8 mod 6 the congruence 2 x 8 mod 6 (c) For 5x 2 mod 9 we have linear congruence has no solution gcd 2, 6 6 but 6 8 so the given linear has no solution 5 x 2 (d) We are given 36x 54 mod 90. The the given linear congruence 36x 54 mod 90 gcd 5, 9 3 and 3 divides 2 so the given has solutions (we are not asked to find them) gcd 36, 90 8 and 8 divides 54 so has solutions. Next we show that the linear congruence ax b mod n has exactly g gcd a, n incongruent solutions. What does incongruent mean? Not congruent. For example 6 x 2 mod 4 has solutions x mod 4 and x 3 mod 4 but

6 Chapter 3: Theory of Modular Arithmetic 30 3 mod 4 [ 3 mod 4 is not congruent to We say x mod 4 and x 3 mod 4 linear congruence 6x 2 mod 4. These congruences x mod 4 and x 3 mod 4 Example 3 are two different stops on the modulo 4 clock. Solve the linear congruence 6x 3 mod 9. mod 4 ] are two incongruent solutions of the given We use the above Proposition (3.5) to test whether the given congruence has a solution: mod has a solution g b where g gcd a, n ax b n We first find the greatest common divisor of 6 and 9: g gcd 6, 9 3 Since 3 3 the given linear congruence 6x 3 mod 9. has solutions. In this case we are working with modulo 9 so we only need to consider x 0,, 2, 3, 4, 5, 6, 7 and 8 Evaluating these gives x x mod By using this table we see our solutions are Table 4 Shows junctions of 6x mod 9 x 2, 5, 8 mod 9 The linear congruence 6x 3 mod 9 has three solutions 2, 5, 8 mod 9 By observing Table 4 in Example 3 we have that the congruences 6 x 0, 3, 6 mod 9 x. have solutions because all these numbers 0, 3, 6 are multiples of 3 which is the gcd of 6 and 9. The set of integers represented by 6x mod 9 0, 3, 6 on the modulo 9 clock. only stops at junctions However the following congruence equations 6x, 2, 4 and 5 mod 9 will have no solutions because g 3 does not divide into any of these numbers, 2, 4 and 5.

7 Chapter 3: Theory of Modular Arithmetic 3 Proposition (3.6). The linear congruence ax b mod n Has exactly g incongruent solutions modulo n provided g b where g gcd a, n How do we prove this result?. We use Proposition (.7) of chapter which gives the solutions of the Diophantine equation: If x 0, y 0 are particular solutions of the Diophantine equation and g c where gcd a, b g by Proof. ax by c then all the other solutions of this equation are given b x x t 0 g a and y y t 0 g We do the proof in two parts. First we list the solutions and then we show there are exactly g incongruent solutions. The given congruence ax b mod n ax b kn implies that there is an integer k such that ax kn ax n k b which implies Let x 0 be a particular solution to this equation then by applying the above Proposition (.7) to ax n k b gives the other solutions as n x x t 0 g where t is an arbitrary integer and g gcd a, n Substituting t 0, t, t 2, and t g we have n n n n x x, x, x 2, x 3,, x g g g g g (*) Need to show that each of these are not congruent modulo n. How? Use proof by contradiction. Suppose any two numbers in the list (*) are congruent modulo n: n n x t x t mod n g g Where 0 t t g. By using Definition (3.): 2 a b mod n a b kn

8 Chapter 3: Theory of Modular Arithmetic 32 g g On n n x t 0 2 x t 0 mod n implies there is an integer k such that n n x t 0 2 x t k n 0 g g n t t 2 k n Simplifying and factorizing g t t k g implies t t k g 2 2 Cancelling n's We have t 2 t k g t g g (because k ) and earlier we had t 2 g. This is a contradiction because we have t 2 g and t 2 g. Hence none of the congruences in the above list (*) are congruent to each other modulo n. This means that the list of numbers: are incongruent modulo n. n n n n x x, x, x 2, x 3,, x g g g g g n Any other solution x x t 0 g Why? is congruent to one of these in the list modulo n. We can show this by using the Division Algorithm on integers t and g. The Division Algorithm (.7) of chapter : Let a, b be given. Then there are unique integers q and r such that a bq r 0 r b Applying this on the above integers t and g means there are integers q and r such that: t gq r 0 r g ( ) n Substituting this t gq r into x x t 0 g gives n n x x 0 t x 0 gq r g g x nq r n 0 g n x 0 0 r Because g nq0 mod n n x r 0 mod n g

9 Chapter 3: Theory of Modular Arithmetic 33 From ( ) we have 0 r g which means r 0,, 2, 3,, g so this n x x t 0 g is in the above list (*) which is: n n n n x x, x, x 2, x 3,, x g g g g g Hence we have exactly g incongruent solutions to ax b mod n. C3 Solving Linear Congruence Equations The list (*) produced in the proof of the above Proposition (3.6) is used to find the g solutions of ax b mod n. Hence the following integers: g g g g n n n n (3.7) x x, x, 2, 3,, 0 0 x x x g mod n are the g solutions of ax b mod n provided g b where g gcd a, n These numbers can be written in compact form as: g (3.8) x n x t 0 mod n for t 0,, 2,, g. Example 4 Find all the solutions of the linear congruence equation: 7 x 35 mod 70 First we determine the greatest common divisor, gcd, of 7 and 70: gcd 7, 70 7 What next? We need to check that 7 divides into 35. Since 7 Why? Because by the above Proposition (3.6): 35 we have 7 incongruent solutions. mod has exactly g solutions provided g b where g gcd a, n ax b n We can find the first solution by trial and error. Is there an obvious solution? Yes, it is x 5 mod 70 because

10 Chapter 3: Theory of Modular Arithmetic 34 How do we find the other six solutions? Using the list of numbers given in (3.8): g (3.8) x n x t 0 mod n With x 5 0 mod 70, n 70 and g 7 numbers 5 0 mod 70 for t 0,, 2,, g we have n x, t 0,, 2,, 7 and 0 g n g Substituting these 7 into this (3.8) gives x 5, 5 0, 5 2 0, 5 3 0,, , 5, 25, 35, 45, 55, 65 mod 70 Simplifying You can check that each of these is a solution by substituting these into 7 x 35 mod 70 Example 5 Find all the solutions of the linear congruence equation: First gcd7, x 34 mod 70 the solutions of 7x 34 mod 70? but 7 does not divide into 34. What does this mean in relation to There are no solutions because by Proposition (3.5) which says: mod has a solution g b where g gcd a, n ax b n Hence there are no solutions to 7x 34 mod 70.. Example 6 Find all the incongruent solutions of the linear congruence: 5 x 34 mod 7 The greatest common divisor of 5 and 7 is, that is gcd5, How many solutions do we have of the given linear congruence? and divides into One solution which means this linear congruence has a unique solution. How can we find this?

11 Chapter 3: Theory of Modular Arithmetic 35 We are given 5x 34 mod 7. We can simplify this to make the arithmetic easier; note that 34 6 mod 7 therefore 5x 34 6 mod 7. It is simpler to solve 5x 6 mod 7 rather than 5x 34 mod 7 Also note that 5 2 mod 7 and 6 mod 7 can solve the equivalent easier equation:. Using these results means that we 2x mod 7 2x mod 7 Multiplying by By observation we know x 4 mod 7 is a solution because mod 4 Checking that this solution is correct: mod 7 Therefore 5x 34 mod 7 has the unique solution 4 mod 7 x., To solve the linear equation 6x 5 0 It is easier to divide through by 3 and solve 2x 5 0. Can we divide through by a common factor for congruences? We need to be careful because we are dealing with integers. The next example demonstrates this. Example 7 Find all the incongruent solutions of the linear congruence: 6 x 5 mod 2 The gcd6, 2 3 and 3 5 so there are 3 incongruent solutions modulo 2. If you only have paper and pen then modulo 2 is too tedious to work with. Can we convert this to a smaller modulus and work with that? Yes. By Proposition (3.0) of the previous section: ac bc mod n implies a b mod n g where g gcd c, n This n g gives us a smaller modulus. We are given 6x 5 mod 2 gcd 6, 2 3 and. We can write the given linear congruence 6x 5 mod 2 as

12 Chapter 3: Theory of Modular Arithmetic 36 32x 35 mod 7 3 The above Proposition (3.0) allows us to cancel the 3 s which in this case gives: 2 x 5 mod 7 By inspection we have the solution x 6 mod 7 From this x 6 mod 7 we have What are values of k? x 6 7k where k is an integer to this congruence. Since we only have 3 solutions so k 0, and 2. By substituting these values k 0, and 2 into x 6 7k we have x 6, 3, 20 mod 2 These are the 3 incongruent solutions modulo 2. You can check these by substituting them into the given linear congruence 6x 5 mod 2. It is generally easier to divide through by the gcd a, n to find possible solutions of ax b mod n work with. because then we are dealing with a smaller modulus which is easier to C4 Unique s How many solutions does the general linear congruence have if gcd a, n? ax b mod n Just one, a unique solution because g is the number of solutions of ax b mod n provided g divides b. We can write this as a general result. Corollary (3.9). If gcd a, n then the linear congruence ax b mod n has a unique solution modulo n. Proof. Applying Proposition (3.6) with g :

13 Chapter 3: Theory of Modular Arithmetic 37 ax b mod n has exactly g solutions provided g b where g gcd a, n We are given g gcd a, n and b so we have a unique solution to ax b mod n. Example 8 Solve the linear congruence: Since gcd6, 3 6 x mod 3 6 x mod 3 so we have a unique solution modulo 3. The congruence means that 3k 6x 3k implies x where k is an integer 6 35 We choose k so that x is an integer. Let k 5 then x. Hence 6 In ordinary algebra when we have 6x x mod 3 x. This 6 x is the inverse of 6. 6 Similarly the unique solution of the above linear congruence 6x mod 3 x mod 3 We call this x mod 3 the (multiplicative) inverse of 6 modulo 3. C5 Multiplicative Inverse is Definition (3.20). If ax mod n then the unique solution of this congruence is called the multiplicative inverse of a modulo n and is denoted by a mod n In Example 0 we had 2x mod 5 x 3 mod 5 Therefore we write this in compact notation as 2 3 mod 5..

14 Chapter 3: Theory of Modular Arithmetic 38 In the exercises we will show that a mod n has an inverse a and n are relatively prime. Example 9 Determine the inverse of 3 mod 4 To find the inverse means we need to solve 3x mod 4. By inspection x 5 mod 4 Because mod 4 Inverse of 3 modulo 4 is 5 modulo 4 or in notation form 3 5 mod 4.. Example 20 Determine the inverse of 3 modulo 5. In this case we need to solve 3x mod 5. Note that the not divide into so there are no solutions to this congruence 3x mod 5. Therefore 3 modulo 5 has no inverse. gcd 3, 5 3 but 3 does It is critical that a mod n has an inverse if and only if a and n are relatively prime. This implies that only the relative prime integers to n have inverses. Which integers have inverses modulo 0?, 3, 7 and 9 The integers 2, 4, 5, 6, 8 and 0 will not have inverses modulo 0 because they are not relatively prime with 0. SUMMARY (3.5) ax b mod n has solutions g b where g gcd a, n. The multiplicative inverse of a modulo n is the unique solution x mod n of ax mod n and is denoted by a mod n.

### MATH 433 Applied Algebra Lecture 4: Modular arithmetic (continued). Linear congruences.

MATH 433 Applied Algebra Lecture 4: Modular arithmetic (continued). Linear congruences. Congruences Let n be a postive integer. The integers a and b are called congruent modulo n if they have the same

### Elementary Number Theory. Franz Luef

Elementary Number Theory Congruences Modular Arithmetic Congruence The notion of congruence allows one to treat remainders in a systematic manner. For each positive integer greater than 1 there is an arithmetic

### 11 Division Mod n, Linear Integer Equations, Random Numbers, The Fundamental Theorem of Arithmetic

11 Division Mod n, Linear Integer Equations, Random Numbers, The Fundamental Theorem of Arithmetic Bezout s Lemma Let's look at the values of 4x + 6y when x and y are integers. If x is -6 and y is 4 we

### This is a recursive algorithm. The procedure is guaranteed to terminate, since the second argument decreases each time.

8 Modular Arithmetic We introduce an operator mod. Let d be a positive integer. For c a nonnegative integer, the value c mod d is the remainder when c is divided by d. For example, c mod d = 0 if and only

### 3.7 Non-linear Diophantine Equations

37 Non-linear Diophantine Equations As an example of the use of congruences we can use them to show when some Diophantine equations do not have integer solutions This is quite a negative application -

### Modular Arithmetic Instructor: Marizza Bailey Name:

Modular Arithmetic Instructor: Marizza Bailey Name: 1. Introduction to Modular Arithmetic If someone asks you what day it is 145 days from now, what would you answer? Would you count 145 days, or find

### Commutative Rings and Fields

Commutative Rings and Fields 1-22-2017 Different algebraic systems are used in linear algebra. The most important are commutative rings with identity and fields. Definition. A ring is a set R with two

### Chapter 5: The Integers

c Dr Oksana Shatalov, Fall 2014 1 Chapter 5: The Integers 5.1: Axioms and Basic Properties Operations on the set of integers, Z: addition and multiplication with the following properties: A1. Addition

### Math Circle Beginners Group February 28, 2016 Euclid and Prime Numbers Solutions

Math Circle Beginners Group February 28, 2016 Euclid and Prime Numbers Solutions Warm-up Problems 1. What is a prime number? Give an example of an even prime number and an odd prime number. A prime number

### A field F is a set of numbers that includes the two numbers 0 and 1 and satisfies the properties:

Byte multiplication 1 Field arithmetic A field F is a set of numbers that includes the two numbers 0 and 1 and satisfies the properties: F is an abelian group under addition, meaning - F is closed under

### Ch 4.2 Divisibility Properties

Ch 4.2 Divisibility Properties - Prime numbers and composite numbers - Procedure for determining whether or not a positive integer is a prime - GCF: procedure for finding gcf (Euclidean Algorithm) - Definition:

### LEGENDRE S THEOREM, LEGRANGE S DESCENT

LEGENDRE S THEOREM, LEGRANGE S DESCENT SUPPLEMENT FOR MATH 370: NUMBER THEORY Abstract. Legendre gave simple necessary and sufficient conditions for the solvablility of the diophantine equation ax 2 +

### Solving the general quadratic congruence. y 2 Δ (mod p),

Quadratic Congruences Solving the general quadratic congruence ax 2 +bx + c 0 (mod p) for an odd prime p (with (a, p) = 1) is equivalent to solving the simpler congruence y 2 Δ (mod p), where Δ = b 2 4ac

### Rings of Residues. S. F. Ellermeyer. September 18, ; [1] m

Rings of Residues S F Ellermeyer September 18, 2006 If m is a positive integer, then we obtain the partition C = f[0] m ; [1] m ; : : : ; [m 1] m g of Z into m congruence classes (This is discussed in

### Wednesday, February 21. Today we will begin Course Notes Chapter 5 (Number Theory).

Wednesday, February 21 Today we will begin Course Notes Chapter 5 (Number Theory). 1 Return to Chapter 5 In discussing Methods of Proof (Chapter 3, Section 2) we introduced the divisibility relation from

### MATH 145 Algebra, Solutions to Assignment 4

MATH 145 Algebra, Solutions to Assignment 4 1: a) Find the inverse of 178 in Z 365. Solution: We find s and t so that 178s + 365t = 1, and then 178 1 = s. The Euclidean Algorithm gives 365 = 178 + 9 178

### A SURVEY OF PRIMALITY TESTS

A SURVEY OF PRIMALITY TESTS STEFAN LANCE Abstract. In this paper, we show how modular arithmetic and Euler s totient function are applied to elementary number theory. In particular, we use only arithmetic

### Primes and Modular Arithmetic! CSCI 2824, Fall 2014!!

Primes and Modular Arithmetic! CSCI 2824, Fall 2014!!! Scheme version of the algorithm! for finding the GCD (define (gcd a b)! (if!(= b 0)!!!!a!!!!(gcd b (remainder a b))))!! gcd (812, 17) = gcd(17, 13)

### 4.4 Solving Congruences using Inverses

4.4 Solving Congruences using Inverses Solving linear congruences is analogous to solving linear equations in calculus. Our first goal is to solve the linear congruence ax b pmod mq for x. Unfortunately

### The next sequence of lectures in on the topic of Arithmetic Algorithms. We shall build up to an understanding of the RSA public-key cryptosystem.

CS 70 Discrete Mathematics for CS Fall 2003 Wagner Lecture 10 The next sequence of lectures in on the topic of Arithmetic Algorithms. We shall build up to an understanding of the RSA public-key cryptosystem.

### MATH 2200 Final Review

MATH 00 Final Review Thomas Goller December 7, 01 1 Exam Format The final exam will consist of 8-10 proofs It will take place on Tuesday, December 11, from 10:30 AM - 1:30 PM, in the usual room Topics

### Homework 7 solutions M328K by Mark Lindberg/Marie-Amelie Lawn

Homework 7 solutions M328K by Mark Lindberg/Marie-Amelie Lawn Problem 1: 4.4 # 2:x 3 + 8x 2 x 1 0 (mod 1331). a) x 3 + 8x 2 x 1 0 (mod 11). This does not break down, so trial and error gives: x = 0 : f(0)

### Algebra Introduction to Polynomials

Introduction to Polynomials What is a Polynomial? A polynomial is an expression that can be written as a term or a sum of terms, each of which is the product of a scalar (the coefficient) and a series

### Number Theory and Group Theoryfor Public-Key Cryptography

Number Theory and Group Theory for Public-Key Cryptography TDA352, DIT250 Wissam Aoudi Chalmers University of Technology November 21, 2017 Wissam Aoudi Number Theory and Group Theoryfor Public-Key Cryptography

### Primes. Rational, Gaussian, Industrial Strength, etc. Robert Campbell 11/29/2010 1

Primes Rational, Gaussian, Industrial Strength, etc Robert Campbell 11/29/2010 1 Primes and Theory Number Theory to Abstract Algebra History Euclid to Wiles Computation pencil to supercomputer Practical

### Direct Proof MAT231. Fall Transition to Higher Mathematics. MAT231 (Transition to Higher Math) Direct Proof Fall / 24

Direct Proof MAT231 Transition to Higher Mathematics Fall 2014 MAT231 (Transition to Higher Math) Direct Proof Fall 2014 1 / 24 Outline 1 Overview of Proof 2 Theorems 3 Definitions 4 Direct Proof 5 Using

### Numbers. 2.1 Integers. P(n) = n(n 4 5n 2 + 4) = n(n 2 1)(n 2 4) = (n 2)(n 1)n(n + 1)(n + 2); 120 =

2 Numbers 2.1 Integers You remember the definition of a prime number. On p. 7, we defined a prime number and formulated the Fundamental Theorem of Arithmetic. Numerous beautiful results can be presented

### Number theory. Myrto Arapinis School of Informatics University of Edinburgh. October 9, /29

Number theory Myrto Arapinis School of Informatics University of Edinburgh October 9, 2014 1/29 Division Definition If a and b are integers with a 6= 0, then a divides b if there exists an integer c such

### Gaussian integers. 1 = a 2 + b 2 = c 2 + d 2.

Gaussian integers 1 Units in Z[i] An element x = a + bi Z[i], a, b Z is a unit if there exists y = c + di Z[i] such that xy = 1. This implies 1 = x 2 y 2 = (a 2 + b 2 )(c 2 + d 2 ) But a 2, b 2, c 2, d

### Fermat s Little Theorem. Fermat s little theorem is a statement about primes that nearly characterizes them.

Fermat s Little Theorem Fermat s little theorem is a statement about primes that nearly characterizes them. Theorem: Let p be prime and a be an integer that is not a multiple of p. Then a p 1 1 (mod p).

### LARGE PRIME NUMBERS (32, 42; 4) (32, 24; 2) (32, 20; 1) ( 105, 20; 0).

LARGE PRIME NUMBERS 1. Fast Modular Exponentiation Given positive integers a, e, and n, the following algorithm quickly computes the reduced power a e % n. (Here x % n denotes the element of {0,, n 1}

### A Readable Introduction to Real Mathematics

Solutions to selected problems in the book A Readable Introduction to Real Mathematics D. Rosenthal, D. Rosenthal, P. Rosenthal Chapter 7: The Euclidean Algorithm and Applications 1. Find the greatest

### Solutions to Problem Set 3 - Fall 2008 Due Tuesday, Sep. 30 at 1:00

Solutions to 18.781 Problem Set 3 - Fall 2008 Due Tuesday, Sep. 30 at 1:00 1. (Niven 2.3.3) Solve the congruences x 1 (mod 4), x 0 (mod 3), x 5 (mod 7). First we note that 4, 3, and 7 are pairwise relatively

### THE GAUSSIAN INTEGERS

THE GAUSSIAN INTEGERS KEITH CONRAD Since the work of Gauss, number theorists have been interested in analogues of Z where concepts from arithmetic can also be developed. The example we will look at in

### 2.2 Inverses and GCDs

34 CHAPTER 2. CRYPTOGRAPHY AND NUMBER THEORY 2.2 Inverses and GCDs 2.2.1 Inverses mod p In the last section we explored the multiplication in Z n. We saw in the special case with n =12 and a = 4 that if

### Math 4400/6400 Homework #8 solutions. 1. Let P be an odd integer (not necessarily prime). Show that modulo 2,

MATH 4400 roblems. Math 4400/6400 Homework # solutions 1. Let P be an odd integer not necessarily rime. Show that modulo, { P 1 0 if P 1, 7 mod, 1 if P 3, mod. Proof. Suose that P 1 mod. Then we can write

### However another possibility is

19. Special Domains Let R be an integral domain. Recall that an element a 0, of R is said to be prime, if the corresponding principal ideal p is prime and a is not a unit. Definition 19.1. Let a and b

### ORDERS OF ELEMENTS IN A GROUP

ORDERS OF ELEMENTS IN A GROUP KEITH CONRAD 1. Introduction Let G be a group and g G. We say g has finite order if g n = e for some positive integer n. For example, 1 and i have finite order in C, since

### Proofs. Chapter 2 P P Q Q

Chapter Proofs In this chapter we develop three methods for proving a statement. To start let s suppose the statement is of the form P Q or if P, then Q. Direct: This method typically starts with P. Then,

### Our Number Theory Textbook

Our Number Theory Textbook Math 311: Fall 2015 December, 2015 Contents 1 Divisibility 2 M. Gonsalves, L. Lewis.......................... 2 1.1 Introduction.............................. 2 1.2 Glossary

### Mathematics of Cryptography

Modulo arithmetic Fermat's Little Theorem If p is prime and 0 < a < p, then a p 1 = 1 mod p Ex: 3 (5 1) = 81 = 1 mod 5 36 (29 1) = 37711171281396032013366321198900157303750656 = 1 mod 29 (see http://gauss.ececs.uc.edu/courses/c472/java/fermat/fermat.html)

Proof by Contradiction MAT231 Transition to Higher Mathematics Fall 2014 MAT231 (Transition to Higher Math) Proof by Contradiction Fall 2014 1 / 12 Outline 1 Proving Statements with Contradiction 2 Proving

### Finding Prime Factors

Section 3.2 PRE-ACTIVITY PREPARATION Finding Prime Factors Note: While this section on fi nding prime factors does not include fraction notation, it does address an intermediate and necessary concept to

### Mathematical Foundations of Cryptography

Mathematical Foundations of Cryptography Cryptography is based on mathematics In this chapter we study finite fields, the basis of the Advanced Encryption Standard (AES) and elliptical curve cryptography

### Experience in Factoring Large Integers Using Quadratic Sieve

Experience in Factoring Large Integers Using Quadratic Sieve D. J. Guan Department of Computer Science, National Sun Yat-Sen University, Kaohsiung, Taiwan 80424 guan@cse.nsysu.edu.tw April 19, 2005 Abstract

### THE TRIANGULAR THEOREM OF THE PRIMES : BINARY QUADRATIC FORMS AND PRIMITIVE PYTHAGOREAN TRIPLES

THE TRIANGULAR THEOREM OF THE PRIMES : BINARY QUADRATIC FORMS AND PRIMITIVE PYTHAGOREAN TRIPLES Abstract. This article reports the occurrence of binary quadratic forms in primitive Pythagorean triangles

### Definitions, Theorems and Exercises. Abstract Algebra Math 332. Ethan D. Bloch

Definitions, Theorems and Exercises Abstract Algebra Math 332 Ethan D. Bloch December 26, 2013 ii Contents 1 Binary Operations 3 1.1 Binary Operations............................... 4 1.2 Isomorphic Binary

### Discrete Mathematics and Probability Theory Summer 2017 Course Notes Note 6

CS 70 Discrete Mathematics and Probability Theory Summer 2017 Course Notes Note 6 Modular Arithmetic In several settings, such as error-correcting codes and cryptography, we sometimes wish to work over

### Cullen Numbers in Binary Recurrent Sequences

Cullen Numbers in Binary Recurrent Sequences Florian Luca 1 and Pantelimon Stănică 2 1 IMATE-UNAM, Ap. Postal 61-3 (Xangari), CP 58 089 Morelia, Michoacán, Mexico; e-mail: fluca@matmor.unam.mx 2 Auburn

### Sums of Squares. Bianca Homberg and Minna Liu

Sums of Squares Bianca Homberg and Minna Liu June 24, 2010 Abstract For our exploration topic, we researched the sums of squares. Certain properties of numbers that can be written as the sum of two squares

### Rings and modular arithmetic

Chapter 8 Rings and modular arithmetic So far, we have been working with just one operation at a time. But standard number systems, such as Z, have two operations + and which interact. It is useful to

### The Impossibility of Certain Types of Carmichael Numbers

The Impossibility of Certain Types of Carmichael Numbers Thomas Wright Abstract This paper proves that if a Carmichael number is composed of primes p i, then the LCM of the p i 1 s can never be of the

### cse547, math547 DISCRETE MATHEMATICS Professor Anita Wasilewska

cse547, math547 DISCRETE MATHEMATICS Professor Anita Wasilewska LECTURE 12 CHAPTER 4 NUMBER THEORY PART1: Divisibility PART 2: Primes PART 1: DIVISIBILITY Basic Definitions Definition Given m,n Z, we say

### OBTAINING SQUARES FROM THE PRODUCTS OF NON-SQUARE INTEGERS

OBTAINING SQUARES FROM THE PRODUCTS OF NON-SQUARE INTEGERS The difference between two neighboring squares n 2 and (n+1) 2 is equal to 2n+1 for any integer n=1,2,3,. Thus the numbers generated by n 2 -A

### Primes in arithmetic progressions

(September 26, 205) Primes in arithmetic progressions Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ [This document is http://www.math.umn.edu/ garrett/m/mfms/notes 205-6/06 Dirichlet.pdf].

### Finite Fields: An introduction through exercises Jonathan Buss Spring 2014

Finite Fields: An introduction through exercises Jonathan Buss Spring 2014 A typical course in abstract algebra starts with groups, and then moves on to rings, vector spaces, fields, etc. This sequence

### Mersenne and Fermat Numbers

NUMBER THEORY CHARLES LEYTEM Mersenne and Fermat Numbers CONTENTS 1. The Little Fermat theorem 2 2. Mersenne numbers 2 3. Fermat numbers 4 4. An IMO roblem 5 1 2 CHARLES LEYTEM 1. THE LITTLE FERMAT THEOREM

### Rings If R is a commutative ring, a zero divisor is a nonzero element x such that xy = 0 for some nonzero element y R.

Rings 10-26-2008 A ring is an abelian group R with binary operation + ( addition ), together with a second binary operation ( multiplication ). Multiplication must be associative, and must distribute over

### Math 324, Fall 2011 Assignment 6 Solutions

Math 324, Fall 2011 Assignment 6 Solutions Exercise 1. (a) Find all positive integers n such that φ(n) = 12. (b) Show that there is no positive integer n such that φ(n) = 14. (c) Let be a positive integer.

### Factorisation CHAPTER Introduction

FACTORISATION 217 Factorisation CHAPTER 14 14.1 Introduction 14.1.1 Factors of natural numbers You will remember what you learnt about factors in Class VI. Let us take a natural number, say 30, and write

### Fast Fraction-Integer Method for Computing Multiplicative Inverse

Fast Fraction-Integer Method for Computing Multiplicative Inverse Hani M AL-Matari 1 and Sattar J Aboud 2 and Nidal F Shilbayeh 1 1 Middle East University for Graduate Studies, Faculty of IT, Jordan-Amman

Table of Contents Chapter 1 What is Number Theory? 1 Chapter Pythagorean Triples 5 Chapter 3 Pythagorean Triples and the Unit Circle 11 Chapter 4 Sums of Higher Powers and Fermat s Last Theorem 16 Chapter

### * 8 Groups, with Appendix containing Rings and Fields.

* 8 Groups, with Appendix containing Rings and Fields Binary Operations Definition We say that is a binary operation on a set S if, and only if, a, b, a b S Implicit in this definition is the idea that

### Math Review. for the Quantitative Reasoning measure of the GRE General Test

Math Review for the Quantitative Reasoning measure of the GRE General Test www.ets.org Overview This Math Review will familiarize you with the mathematical skills and concepts that are important for solving

### . In particular if a b then N(

Gaussian Integers II Let us summarise what we now about Gaussian integers so far: If a, b Z[ i], then N( ab) N( a) N( b). In particular if a b then N( a ) N( b). Let z Z[i]. If N( z ) is an integer prime,

### and LCM (a, b, c) LCM ( a, b) LCM ( b, c) LCM ( a, c)

CHAPTER 1 Points to Remember : REAL NUMBERS 1. Euclid s division lemma : Given positive integers a and b, there exists whole numbers q and r satisfying a = bq + r, 0 r < b.. Euclid s division algorithm

### 3 The language of proof

3 The language of proof After working through this section, you should be able to: (a) understand what is asserted by various types of mathematical statements, in particular implications and equivalences;

### Arithmetic and Algebra

Arithmetic and Algebra Daniel Butnaru daniel.butnaru@uni-konstanz.de 15. Dezember 2006 Daniel Butnaru daniel.butnaru@uni-konstanz.de Arithmetic and Algebra 1/39 Outline 1 Introduction 2 Big Number Arithmetic

### Solutions to Assignment 3

Solutions to Assignment 3 Question 1. [Exercises 3.1 # 2] Let R = {0 e b c} with addition multiplication defined by the following tables. Assume associativity distributivity show that R is a ring with

### The Chinese Remainder Theorem

Chapter 4 The Chinese Remainder Theorem The Monkey-Sailor-Coconut Problem Three sailors pick up a number of coconuts, place them in a pile and retire for the night. During the night, the first sailor wanting

### Pade Approximations and the Transcendence

Pade Approximations and the Transcendence of π Ernie Croot March 9, 27 1 Introduction Lindemann proved the following theorem, which implies that π is transcendental: Theorem 1 Suppose that α 1,..., α k

### Dixon s Factorization method

Dixon s Factorization method Nikithkumarreddy yellu December 2015 1 Contents 1 Introduction 3 2 History 3 3 Method 4 3.1 Factor-base.............................. 4 3.2 B-smooth...............................

### Finite Fields. Mike Reiter

1 Finite Fields Mike Reiter reiter@cs.unc.edu Based on Chapter 4 of: W. Stallings. Cryptography and Network Security, Principles and Practices. 3 rd Edition, 2003. Groups 2 A group G, is a set G of elements

### CS 360, Winter Morphology of Proof: An introduction to rigorous proof techniques

CS 30, Winter 2011 Morphology of Proof: An introduction to rigorous proof techniques 1 Methodology of Proof An example Deep down, all theorems are of the form If A then B, though they may be expressed

### Number Theory in Problem Solving. Konrad Pilch

Number Theory in Problem Solving Konrad Pilch April 7, 2016 1 Divisibility Number Theory concerns itself mostly with the study of the natural numbers (N) and the integers (Z). As a consequence, it deals

### Modular Arithmetic Investigation

The basics Modular Arithmetic Investigation Definition: Two numbers are considered congruent, modulo n if they are exactly a multiple of n apart. Equivalently, if the two numbers are the same distance

### x y z 2x y 2y z 2z x n

Integer Solutions, Rational solutions of the equations 4 4 4 x y z x y y z z x n 4 4 and x y z xy xz y z n; And Crux Mathematicorum Contest Corner problem CC4 Konstantine Zelator P.O. Box 480 Pittsburgh,

### 5.1. Primes, Composites, and Tests for Divisibility

CHAPTER 5 Number Theory 5.1. Primes, Composites, and Tests for Divisibility Definition. A counting number with exactly two di erent factors is called a prime number or a prime. A counting number with more

### Section 19 Integral domains

Section 19 Integral domains Instructor: Yifan Yang Spring 2007 Observation and motivation There are rings in which ab = 0 implies a = 0 or b = 0 For examples, Z, Q, R, C, and Z[x] are all such rings There

### The Fundamental Theorem of Arithmetic

Chapter 1 The Fundamental Theorem of Arithmetic 1.1 Primes Definition 1.1. We say that p N is prime if it has just two factors in N, 1 and p itself. Number theory might be described as the study of the

### [Part 2] Asymmetric-Key Encipherment. Chapter 9. Mathematics of Cryptography. Objectives. Contents. Objectives

[Part 2] Asymmetric-Key Encipherment Mathematics of Cryptography Forouzan, B.A. Cryptography and Network Security (International Edition). United States: McGraw Hill, 2008. Objectives To introduce prime

### Modular Arithmetic. Examples: 17 mod 5 = 2. 5 mod 17 = 5. 8 mod 3 = 1. Some interesting properties of modular arithmetic:

Modular Arithmetic If a mod n = b, then a = c n + b. When you reduce a number a modulo n you usually want 0 b < n. Division Principle [Bar02, pg. 61]: Let n be a positive integer and let a be any integer.

### NONLINEAR CONGRUENCES IN THE THEORY OF NUMBERS

Bulletin of the Marathwada Mathematical Society Vol. 12, No. 2, December 2011, Pages 24-31 NONLINEAR CONGRUENCES IN THE THEORY OF NUMBERS S.R.Joshi 8, Karmayog, Tarak Housing Society, Opp to Ramkrishna

### CHAPTER 1 REAL NUMBERS KEY POINTS

CHAPTER 1 REAL NUMBERS 1. Euclid s division lemma : KEY POINTS For given positive integers a and b there exist unique whole numbers q and r satisfying the relation a = bq + r, 0 r < b. 2. Euclid s division

### ABSTRACT ALGEBRA: A STUDY GUIDE FOR BEGINNERS

ABSTRACT ALGEBRA: A STUDY GUIDE FOR BEGINNERS John A. Beachy Northern Illinois University 2006 2 This is a supplement to Abstract Algebra, Third Edition by John A. Beachy and William D. Blair ISBN 1 57766

### Elementary Number Theory II

Elementary Number Theory II CIS002-2 Computational Alegrba and Number Theory David Goodwin david.goodwin@perisic.com 09:00, Tuesday 1 st November 2011 Contents 1 Divisibility Euclid s Algorithm & Bezout

### In Z: x + 3 = 2 3x = 2 x = 1 No solution In Q: 3x = 2 x 2 = 2. x = 2 No solution. In R: x 2 = 2 x = 0 x = ± 2 No solution Z Q.

THE UNIVERSITY OF NEW SOUTH WALES SCHOOL OF MATHEMATICS AND STATISTICS MATH 1141 HIGHER MATHEMATICS 1A ALGEBRA. Section 1: - Complex Numbers. 1. The Number Systems. Let us begin by trying to solve various

### ACCUPLACER MATH 0310

The University of Teas at El Paso Tutoring and Learning Center ACCUPLACER MATH 00 http://www.academics.utep.edu/tlc MATH 00 Page Linear Equations Linear Equations Eercises 5 Linear Equations Answer to

### MATH 3240Q Introduction to Number Theory Homework 5

The good Christian should beware of mathematicians, and all those who make empty prophecies. The danger already exists that the mathematicians have made a covenant with the devil to darken the spirit and

### An Algorithm for Prime Factorization

An Algorithm for Prime Factorization Fact: If a is the smallest number > 1 that divides n, then a is prime. Proof: By contradiction. (Left to the reader.) A multiset is like a set, except repetitions are

### 3 - Induction and Recursion

November 14, 2017 3 - Induction and Recursion William T. Trotter trotter@math.gatech.edu Using Recurrence Equations (1) Basic Problem How many regions are determined by n lines that intersect in general

### Outline. Some Review: Divisors. Common Divisors. Primes and Factors. b divides a (or b is a divisor of a) if a = mb for some m

Outline GCD and Euclid s Algorithm AIT 682: Network and Systems Security Topic 5.1 Basic Number Theory -- Foundation of Public Key Cryptography Modulo Arithmetic Modular Exponentiation Discrete Logarithms

### This theorem allows us to describe all integer solutions as follows:

On the Diophantine equation a 3 + b 3 + c 3 + d 3 = 0 by RACHEL GAR-EL and LEONID VASERSTEIN (University Park, PA) Introduction. The equation a 3 + b 3 + c 3 + d 3 = 0 (1) has been studied by many mathematicians

### Kevin James. MTHSC 412 Section 3.4 Cyclic Groups

MTHSC 412 Section 3.4 Cyclic Groups Definition If G is a cyclic group and G =< a > then a is a generator of G. Definition If G is a cyclic group and G =< a > then a is a generator of G. Example 1 Z is

### = 5 2 and = 13 2 and = (1) = 10 2 and = 15 2 and = 25 2

BEGINNING ALGEBRAIC NUMBER THEORY Fermat s Last Theorem is one of the most famous problems in mathematics. Its origin can be traced back to the work of the Greek mathematician Diophantus (third century

PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient

University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln MAT Exam Expository Papers Math in the Middle Institute Partnership 7-007 Pythagorean Triples Diane Swartzlander University

### Extending The Natural Numbers. Whole Numbers. Integer Number Set. History of Zero

Whole Numbers Are the whole numbers with the property of addition a group? Extending The Natural Numbers Natural or Counting Numbers {1,2,3 } Extend to Whole Numbers { 0,1,2,3 } to get an additive identity.