# Problem Set 5. AABB = 11k = (10 + 1)k = (10 + 1)XY Z = XY Z0 + XY Z XYZ0 + XYZ AABB

Save this PDF as:

Size: px
Start display at page:

Download "Problem Set 5. AABB = 11k = (10 + 1)k = (10 + 1)XY Z = XY Z0 + XY Z XYZ0 + XYZ AABB"

## Transcription

1 Problem Set 5 1. (a) Four-digit number S = aabb is a square. Find it; (hint: 11 is a factor of S) (b) If n is a sum of two square, so is 2n. (Frank) Solution: (a) Since (A + B) (A + B) = 0, and 11 0, 11 AABB. Thus 11k = AABB for some k Z. Since AABB is a four digit number and 11 is a two digit number, k must be a three digit number. Let k = XY Z, where X, Y and Z are digits. Then: So we have: AABB = 11k = (10 + 1)k = (10 + 1)XY Z = XY Z0 + XY Z XYZ0 + XYZ AABB Hence, Z = B, Y = 0, and X = A. Thus k = XY Z = A0B. Since AABB is a square, we can write AABB = m 2, for some m Z +. Since AABB = 11(A0B), we have 11(A0B) = m 2, so 11 A0B. In order for this to be true, we must have 11 A + B. Since A and B are single digits, their sum must be one of the following: 1, 2, 3,..., 11, 12, 13, 14, 15, 16, 17, 18 Since only one of these is divisible by 11, A + B = 11. Thus A0B must be one of the following: Dividing each of these by 11, we have: 209, 308, 407, 506, 605, 704, 803, , 28, 37, 46, 55, 64, 73, 82 Since AABB = 11(A0B) is a perfect square, A0B/11 must be a perfect square. Only one of the above is a perfect square: 64. Hence, A0B=704. Therefore AABB=7744. (b) If n is the sum of two squares, so is 2n Proof: Consider the following lemma. The set S 2, consisting of the sums of two squares, is closed under multi- Lemma: plication. 1

2 Proof: Let s = a b2 1 and t = a2 2 + b2 2 be elements of S 2, where a 1, b 1, a 2, b 2 Z. Then: st = (a b 2 1)(a b 2 2) = a 2 1a b 2 1a a 2 1b b 2 1b 2 2 = (a 2 1a 2 2 2a 1 a 2 b 1 b 2 + b 2 1b 2 2) + (a 2 1b a 1 a 2 b 1 b 2 + b 2 1a 2 2) = (a 1 a 2 b 1 b 2 ) 2 + (a 1 b 2 + b 1 a 2 ) 2 Since n is a sum of squares, let n = k 2 + m 2. Since 2 = , we have 2n = ( )(k 2 + m 2 ). Since 2n can be expressed as the product of two sums of squares, by the above lemma, 2n is a sum of squares. 2. (a) If n is an even number, then n + 16 n 3 n 1; (hint: factorize 323) (b) If n is an integer, then 9 4 n + 15n 1. (hint: consider cases when n modulus 3) (Ben) Solution. (a) The first thing to note on this problem is that 323 can be factored, that is 323 = Now from number theory we know that if two numbers are relatively prime, and they each individually divide a sum or difference, then their product divides the sum or difference. And gcd(17, 19) = 1, that is, they are relatively prime. Thus it is enough to show that 17 and 19 individually divide (1) 20 n + 16 n 3 n 1. First we will show that 17 divides (1). Now 20 n 3 n (mod17) and 16 n ( 1) n (mod17) And so 20 n + 16 n 3 n 1 3 n + ( 1) n 3 n 1 = 0 (mod17), since n is an even number. That is, n + 16 n 3 n 1. Now, to show 19 divides (1), observe that 20 n 1 n (mod19) and 16 n ( 3) n (mod19) And so 20 n + 16 n 3 n 1 1 n + ( 3) n 3 n 1 = 0 (mod19), since n is an even number. That is, n + 16 n 3 n 1. (b) For this problem, it is easiest to proceed by induction. First, observe that 9 4 n + 15n 1 holds for the case n = 1, because

3 A fact that will be used later (which comes from Number Theory) is the following: If a b and a c, then we can conclude that a b ± c. Also, knowing any two of the statements to be true is enough to conclude the third. Now assume that 9 4 n + 15n 1. We will show that 9 4 n (n + 1) 1. Using the fact from above, it is equivalent to show that 9 [4 n (n + 1) 1] [4 n + 15n 1]. Now, expanding, grouping and simplifying terms, we find [4 n (n + 1) 1] [4 n + 15n 1] = 4 n n n 15n + 1 = [4 n n ] + [15n 15n] + [1 1] + 15 = 4 n (4 1) + 15 = 3 4 n = 3(4 n + 5) Noting that 4 n 1 n = 1 (mod3) and that 5 2 (mod3) yields 3(4 n + 5) 3(1 + 2) (mod3) = 3(3) = 9 Thus 9 [4 n (n + 1) 1] [4 n + 15n 1] as wanted, and so we can conclude that 9 4 n (n + 1) 1, and finally that 9 4 n + 15n (a) If 2n + 1 and 3n + 1 are squares, then 5n + 3 is not a prime; (hint: express 5n + 3 by 2n + 1 and 3n + 1) (b) If 3n + 1 and 4n + 1 are squares, then 56 n. (hint: follow the idea in presentation problem) (Beth) Since both 2n + 1 and 3n + 1 are squares, let 2n+1 = a 2, and let 3n+1 = b 2. 5n + 3 = 4(2n + 1) (3n + 1) so 5n+3 = 4(a 2 ) - (b 2 ) 5n + 1 = (2a) 2 (b 2 ) notice this is a difference of squares, so 5n + 3 = (2a b)(2a + b). Thus 5n + 3 is NOT prime. B. Prove: If 3n+1 and 4n+1 are squares, then 56 n. If we can prove that 8 divides n, and 7 divides n, then we know that 56 divides n. Let 4n+1 = b 2, and let 3n+1 = a 2. Then n = b 2 a 2. We know that 4n+1 is odd so b 2 is odd. 3

4 Let s assume that n is odd. Therefore, n=2m+1. Then we have that b 2 = 8m + 5. If we make a table of b, and b 2 mod 8, then we get: b b and we don t ever get 5 mod 8. Therefore n is even. Thus we have that a 2 is odd, and a is odd. From there we get that (b-a)(b+a) are both even, and then n is divisible by 4. So b and a are either equivalent to 1 or 3 mod 4. There are four cases: Thus 8 n. The case for 7 is much more difficult. solution. a b (b a)(b + a) It involves using Pell s equation to find a Theorem: (x 1, y 1 ) is the smallest solution for x 2 Dy 2 = 1, then x k + Dy k = (x 1 + Dy 1 ) k = (x Dy) k (x + Dy) k is also a solution. So, in our problem, if you set 4n + 1 = b 2, and 3n + 1 = a 2 and multiply the first equation by 4, and the second by 3, you get: 4a 2-3b 2 = 1 x=2a, y=b so x 2-3y 2 = 1. Solving this equation with Pell s equation, then the first two answers you get are (2,1) and (7,4) and this second one shows that n is divisible by (a) If p is a prime, then p 2 1(mod24); (hint: prove 24 p 2 1) (b) Show that if n divides a single Fibonacci number, then it will divide infinitely many Fibonacci numbers. (hint: think Problem Set 2 number 10.) (Tina) Solution: This is only true for primes 5, and from this point all primes are odd. p 2 1(mod 24) = 24 (p 2 1) 24 (p 1)(p + 1) Since p is odd, both (p 1) and (p + 1) are even and therefore divisible by two. Furthermore, since p ±1 (mod 4), either (p 1) or (p + 1) is divisible by 4. Finally, as is true for all prime numbers, p ±1 (mod 3), and therefore, either (p 1) or (p + 1) is divisible by 3. So combining all of these facts, (p 1) or (p + 1) is divisible by 4, the other is divisibly by 2, and one is divisible by 3. (4)(2)(3) = 24 and therefore p 2 1 (mod 24). 4

5 4b) Show that if n divides a single Fibonacci number, then it will divide infinitely many Fibonacci numbers. Solution: Write the Fibonacci numbers F 1, F 2, F 3,... in the following form: (a 1, a 2 ), (a 3, a 4 ), (a 5, a 6 )... where a k = F k (mod n). Obviously, the series will begin (1,1). At some point in this series, (a i, a i+1 ) will equal (a j, a j+1 ). Since the Fibonacci numbers are cyclical, every a i is completely determined by a i 2 and a i1. Therefore, the only way to get (1,1) is if the second number in the pair immediately before it is zero (in other words, divisible by n). So once we find a pair that repeats, it will repeat an infinite number of times, and during each cycle there will be at least one number that is divisible by n. 5. (a) (VT 1979) Show that for all positive integers n, that 14 divides 3 4n n+1 ; (b) (VT 1981) is exactly divisible by what two numbers between 60 and 70? (hint: (a) 14 = 2 7, (b) factorizing) (Lei) Solution. (a) We can rewrite 3 4n n n n 9 81 n n 9 ( 3) n + 5 ( 3) n Which is obviously divisble by 14. (b) Since if k is even. We can start by rewriting mod(14) (±1) k = Since 2 6 = 64 and 64 ±1 mod 65 and 63. So is divisible by 65 and (a) (VT 1982) What is the remainder when X is divided by X 1? Verify your answer (hint: too simple); (b) (MIT training 2 star) Let n be an integer greater than one. Show that n n is not prime. (hint: there is a magic identity due to Sophie Germain: a 4 + 4b 4 = (a 2 + 2b 2 + 2ab)(a 2 + 2b 2 2ab)) (Erin) a) It is pretty easy to see that the answer is 2. Simply do direct long division to obtain the answer. b) We are trying to show that n n is not prime. The first thing to do is note that if n is even we are done. Secondly, since we have the magic identity which says that a 4 + 4b n = (a 2 + 2b 2 + 2ab)(a 2 + 2b 2 2ab) we simply need to match up the two equations. Let a = n, this takes care of the first term. Let 4 n = 4b 4 = 4 n 1 = b 4 = 2 n 1 = b 2 = 2 2m = b 2 (since n is odd) = 2 m = b and we are done. 5

6 7. (VT 1988) Let a be a positive integer. Find all positive integers n such that b = a n satisfying the condition that a 2 +b 2 is divisible by ab+1. (hint: prove that a m +1 a n +1, then m n.) (Brett) Solution. We need to find all positive integers n such that (a 2 +a 2n )/(a n+1 +1) N. Rewrite this to get, a 2 (1 + a 2n 2 ) a n N (a n+1 + 1) (a 2n 2 + 1) (n + 1) (2n 2) n = (m + 2)/(2 m) for some m Z This gives possible n values of 0, 1, or 3. 0 and 1 both fail for a = 2. However, n = 3 gives a result of a 2 N for all values of a. So n must be equal to (Putnam 1972-A5) Show that if n is an integer greater than 1, then n does not divide 2 n 1. (Shelley) Solution: Assume the opposite: 2 n 1.1 (mod p), n = odd because 2 n is even. Take p such that it is the smallest prime dividing n, and m is the smallest divisor of p-1. Then 2 n 1.1 (mod p) is equivalent to 2 m 1 (modp), for m < p. m must be co-prime to n, because p is the smallest prime divisor. So, n = xm + r, 0 < r < m. It follows that 2 r 1(modp). However, m is the smallest divisor of p 1. Contradiction. [ ] (Putnam 1986-A2) What is the units (i.e., rightmost) digit of ? Here [x] + 3 is the greatest integer x. (Richard) Solution. First note 20000/100 = 200, which is the ratio of the numerator to the dominator. Now consider the factorization x 200 y 200 = (x y) (x x 198 y y 199 ) Taking x = and y = 3 in the above factorization shows that the number A = ( )/( ) is an integer. Moreover, ( )/( ) = [ /( )], since /( ) = /( ) < 1 So A is congruent to which is congruent to 3 mod 10. Hence, the units digit is (Putnam 1998-A4) Let A 1 = 0 and A 2 = 1. For n > 2, the number A n is defined by concatenating the decimal expansions of A n 1 and A n 2 from left to right. For example A 3 = A 2 A 1 = 10, A 4 = A 3 A 2 = 101, A 5 = A 4 A 3 = 10110, and so forth. Determine all n such that 11 divides A n. (David Rose) Solution We first define φ(n) as the number of digits in A n. Simple induction shows that φ(n) is odd if n 1, 2(mod3) and even if n 0(mod3). Now note that the definition of A n is equivalent to A n = 10 φ(n 1) A n 1 + A n 2. Now, noting that 10 1(mod11) we see that A n ( 1) φ(n 1) A n 1 + A n 2 (mod11). Now, we claim that 6

7 n 1(mod6) A n 0(mod11) n 2(mod6) A n 1(mod11) n 3(mod6) A n 1(mod11) n 4(mod6) A n 2(mod11) n 5(mod6) A n 1(mod11) n 0(mod6) A n 1(mod11) We will verify using induction. We can check that these hold for n = 1, 2, 3, 4, 5, 6 using our formula above. With this base, we can use circular induction to show that these statements hold for all n. For example, assume the above holds for n 1(mod6) and n + 1 2(mod6). Then A n+2 ( 1) odd (1) + 0 = 1. The rest of the induction follows similarly. 7

### Practice Problems for First Test

Mathematicians have tried in vain to this day to discover some order in the sequence of prime numbers, and we have reason to believe that it is a mystery into which the human mind will never penetrate.-

### Homework until Test #2

MATH31: Number Theory Homework until Test # Philipp BRAUN Section 3.1 page 43, 1. It has been conjectured that there are infinitely many primes of the form n. Exhibit five such primes. Solution. Five such

### Further linear algebra. Chapter I. Integers.

Further linear algebra. Chapter I. Integers. Andrei Yafaev Number theory is the theory of Z = {0, ±1, ±2,...}. 1 Euclid s algorithm, Bézout s identity and the greatest common divisor. We say that a Z divides

### Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

### p 2 1 (mod 6) Adding 2 to both sides gives p (mod 6)

.9. Problems P10 Try small prime numbers first. p p + 6 3 11 5 7 7 51 11 13 Among the primes in this table, only the prime 3 has the property that (p + ) is also a prime. We try to prove that no other

### 10 k + pm pm. 10 n p q = 2n 5 n p 2 a 5 b q = p

Week 7 Summary Lecture 13 Suppose that p and q are integers with gcd(p, q) = 1 (so that the fraction p/q is in its lowest terms) and 0 < p < q (so that 0 < p/q < 1), and suppose that q is not divisible

### Math 319 Problem Set #3 Solution 21 February 2002

Math 319 Problem Set #3 Solution 21 February 2002 1. ( 2.1, problem 15) Find integers a 1, a 2, a 3, a 4, a 5 such that every integer x satisfies at least one of the congruences x a 1 (mod 2), x a 2 (mod

### SYSTEMS OF PYTHAGOREAN TRIPLES. Acknowledgements. I would like to thank Professor Laura Schueller for advising and guiding me

SYSTEMS OF PYTHAGOREAN TRIPLES CHRISTOPHER TOBIN-CAMPBELL Abstract. This paper explores systems of Pythagorean triples. It describes the generating formulas for primitive Pythagorean triples, determines

### Math 139 Problem Set #6 Solution 15 April 2003

Math 139 Problem Set #6 Solution 15 April 2003 A1 (Based on Barr, 41, exercise 3) (a) Let n be an integer greater than 2 Consider the list of numbers n! + 2, n! + 3, n! + 4,, n! + n Explain why all the

### 3 Structure of Modular Multiplication

3 Structure of Modular Multiplication Unlike addition, multiplication in Z M is fairly strange. There may not always be an inverse, you can multiply two non-zero numbers together and get zero, and in general

### Applications of Fermat s Little Theorem and Congruences

Applications of Fermat s Little Theorem and Congruences Definition: Let m be a positive integer. Then integers a and b are congruent modulo m, denoted by a b mod m, if m (a b). Example: 3 1 mod 2, 6 4

### Algebra for Digital Communication

EPFL - Section de Mathématiques Algebra for Digital Communication Fall semester 2008 Solutions for exercise sheet 1 Exercise 1. i) We will do a proof by contradiction. Suppose 2 a 2 but 2 a. We will obtain

### Continued fractions and good approximations.

Continued fractions and good approximations We will study how to find good approximations for important real life constants A good approximation must be both accurate and easy to use For instance, our

### APPLICATIONS OF THE ORDER FUNCTION

APPLICATIONS OF THE ORDER FUNCTION LECTURE NOTES: MATH 432, CSUSM, SPRING 2009. PROF. WAYNE AITKEN In this lecture we will explore several applications of order functions including formulas for GCDs and

### When is a number Fibonacci?

When is a number Fibonacci? Phillip James Department of Computer Science, Swansea University January 5, 009 Abstract This article looks into the importance of the Fibonacci numbers within Computer Science,

### = 2 + 1 2 2 = 3 4, Now assume that P (k) is true for some fixed k 2. This means that

Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without

### Settling a Question about Pythagorean Triples

Settling a Question about Pythagorean Triples TOM VERHOEFF Department of Mathematics and Computing Science Eindhoven University of Technology P.O. Box 513, 5600 MB Eindhoven, The Netherlands E-Mail address:

### 8 Divisibility and prime numbers

8 Divisibility and prime numbers 8.1 Divisibility In this short section we extend the concept of a multiple from the natural numbers to the integers. We also summarize several other terms that express

### 2. Integers and Algorithms Euclidean Algorithm. Euclidean Algorithm. Suppose a and b are integers

2. INTEGERS AND ALGORITHMS 155 2. Integers and Algorithms 2.1. Euclidean Algorithm. Euclidean Algorithm. Suppose a and b are integers with a b > 0. (1) Apply the division algorithm: a = bq + r, 0 r < b.

### Section 5.1 Number Theory: Prime & Composite Numbers

Section 5.1 Number Theory: Prime & Composite Numbers Objectives 1. Determine divisibility. 2. Write the prime factorization of a composite number. 3. Find the greatest common divisor of two numbers. 4.

### Congruences. Robert Friedman

Congruences Robert Friedman Definition of congruence mod n Congruences are a very handy way to work with the information of divisibility and remainders, and their use permeates number theory. Definition

### Exponents and Primes

Exponents and Primes Alexander Remorov alexanderrem@gmail.com 1 Some Fundamentals Fundamental Theorem of Arithmetic: If n is a positive integer, it can be uniquely written as a product of primes: n = p

### INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS

INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS STEVEN HEILMAN Contents 1. Homework 1 1 2. Homework 2 6 3. Homework 3 10 4. Homework 4 16 5. Homework 5 19 6. Homework 6 21 7. Homework 7 25 8. Homework 8 28

### Oh Yeah? Well, Prove It.

Oh Yeah? Well, Prove It. MT 43A - Abstract Algebra Fall 009 A large part of mathematics consists of building up a theoretical framework that allows us to solve problems. This theoretical framework is built

### 9 abcd = dcba. 9 ( b + 10c) = c + 10b b + 90c = c + 10b b = 10c.

In this session, we ll learn how to solve problems related to place value. This is one of the fundamental concepts in arithmetic, something every elementary and middle school mathematics teacher should

### Total Gadha s Complete Book of NUMBER SYSTEM

Total Gadha s Complete Book of NUMBER SYSTEM TYPES OF NUMBERS Natural Numbers The group of numbers starting from 1 and including 1, 2, 3, 4, 5, and so on, are known as natural numbers. Zero, negative numbers,

### Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00

18.781 Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00 Throughout this assignment, f(x) always denotes a polynomial with integer coefficients. 1. (a) Show that e 32 (3) = 8, and write down a list

### Math Workshop October 2010 Fractions and Repeating Decimals

Math Workshop October 2010 Fractions and Repeating Decimals This evening we will investigate the patterns that arise when converting fractions to decimals. As an example of what we will be looking at,

### CONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12

CONTINUED FRACTIONS AND PELL S EQUATION SEUNG HYUN YANG Abstract. In this REU paper, I will use some important characteristics of continued fractions to give the complete set of solutions to Pell s equation.

### MATH 289 PROBLEM SET 1: INDUCTION. 1. The induction Principle The following property of the natural numbers is intuitively clear:

MATH 89 PROBLEM SET : INDUCTION The induction Principle The following property of the natural numbers is intuitively clear: Axiom Every nonempty subset of the set of nonnegative integers Z 0 = {0,,, 3,

### The Fibonacci Numbers

The Fibonacci Numbers The numbers are: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, Each Fibonacci number is the sum of the previous two Fibonacci numbers! Let n any positive integer If F n is

### Homework 5 Solutions

Homework 5 Solutions 4.2: 2: a. 321 = 256 + 64 + 1 = (01000001) 2 b. 1023 = 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = (1111111111) 2. Note that this is 1 less than the next power of 2, 1024, which

### Fractions and Decimals

Fractions and Decimals Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles December 1, 2005 1 Introduction If you divide 1 by 81, you will find that 1/81 =.012345679012345679... The first

### MATH 289 PROBLEM SET 4: NUMBER THEORY

MATH 289 PROBLEM SET 4: NUMBER THEORY 1. The greatest common divisor If d and n are integers, then we say that d divides n if and only if there exists an integer q such that n = qd. Notice that if d divides

### k, then n = p2α 1 1 pα k

Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square

### SOLUTIONS FOR PROBLEM SET 2

SOLUTIONS FOR PROBLEM SET 2 A: There exist primes p such that p+6k is also prime for k = 1,2 and 3. One such prime is p = 11. Another such prime is p = 41. Prove that there exists exactly one prime p such

### Topics in Number Theory

Chapter 8 Topics in Number Theory 8.1 The Greatest Common Divisor Preview Activity 1 (The Greatest Common Divisor) 1. Explain what it means to say that a nonzero integer m divides an integer n. Recall

### Reading 7 : Program Correctness

CS/Math 240: Introduction to Discrete Mathematics Fall 2015 Instructors: Beck Hasti, Gautam Prakriya Reading 7 : Program Correctness 7.1 Program Correctness Showing that a program is correct means that

### Lecture 13 - Basic Number Theory.

Lecture 13 - Basic Number Theory. Boaz Barak March 22, 2010 Divisibility and primes Unless mentioned otherwise throughout this lecture all numbers are non-negative integers. We say that A divides B, denoted

### CHAPTER 5. Number Theory. 1. Integers and Division. Discussion

CHAPTER 5 Number Theory 1. Integers and Division 1.1. Divisibility. Definition 1.1.1. Given two integers a and b we say a divides b if there is an integer c such that b = ac. If a divides b, we write a

### CONTINUED FRACTIONS, PELL S EQUATION, AND TRANSCENDENTAL NUMBERS

CONTINUED FRACTIONS, PELL S EQUATION, AND TRANSCENDENTAL NUMBERS JEREMY BOOHER Continued fractions usually get short-changed at PROMYS, but they are interesting in their own right and useful in other areas

### Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

### Discrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 5

CS 70 Discrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 5 Modular Arithmetic One way to think of modular arithmetic is that it limits numbers to a predefined range {0,1,...,N

### b) Find smallest a > 0 such that 2 a 1 (mod 341). Solution: a) Use succesive squarings. We have 85 =

Problem 1. Prove that a b (mod c) if and only if a and b give the same remainders upon division by c. Solution: Let r a, r b be the remainders of a, b upon division by c respectively. Thus a r a (mod c)

### CS 103X: Discrete Structures Homework Assignment 3 Solutions

CS 103X: Discrete Structures Homework Assignment 3 s Exercise 1 (20 points). On well-ordering and induction: (a) Prove the induction principle from the well-ordering principle. (b) Prove the well-ordering

### 4. Number Theory (Part 2)

4. Number Theory (Part 2) Terence Sim Mathematics is the queen of the sciences and number theory is the queen of mathematics. Reading Sections 4.8, 5.2 5.4 of Epp. Carl Friedrich Gauss, 1777 1855 4.3.

### CS103X: Discrete Structures Homework Assignment 2: Solutions

CS103X: Discrete Structures Homework Assignment 2: Solutions Due February 1, 2008 Exercise 1 (10 Points). Prove or give a counterexample for the following: Use the Fundamental Theorem of Arithmetic to

### 8 Primes and Modular Arithmetic

8 Primes and Modular Arithmetic 8.1 Primes and Factors Over two millennia ago already, people all over the world were considering the properties of numbers. One of the simplest concepts is prime numbers.

### 12 Greatest Common Divisors. The Euclidean Algorithm

Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan 12 Greatest Common Divisors. The Euclidean Algorithm As mentioned at the end of the previous section, we would like to

### There are 8000 registered voters in Brownsville, and 3 8. of these voters live in

Politics and the political process affect everyone in some way. In local, state or national elections, registered voters make decisions about who will represent them and make choices about various ballot

### Theorem (The division theorem) Suppose that a and b are integers with b > 0. There exist unique integers q and r so that. a = bq + r and 0 r < b.

Theorem (The division theorem) Suppose that a and b are integers with b > 0. There exist unique integers q and r so that a = bq + r and 0 r < b. We re dividing a by b: q is the quotient and r is the remainder,

### Elementary Number Theory and Methods of Proof. CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook.

Elementary Number Theory and Methods of Proof CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook.edu/~cse215 1 Number theory Properties: 2 Properties of integers (whole

3. QUADRATIC CONGRUENCES 3.1. Quadratics Over a Finite Field We re all familiar with the quadratic equation in the context of real or complex numbers. The formula for the solutions to ax + bx + c = 0 (where

### LESSON 1 PRIME NUMBERS AND FACTORISATION

LESSON 1 PRIME NUMBERS AND FACTORISATION 1.1 FACTORS: The natural numbers are the numbers 1,, 3, 4,. The integers are the naturals numbers together with 0 and the negative integers. That is the integers

### Prime Numbers. Chapter Primes and Composites

Chapter 2 Prime Numbers The term factoring or factorization refers to the process of expressing an integer as the product of two or more integers in a nontrivial way, e.g., 42 = 6 7. Prime numbers are

### MTH 382 Number Theory Spring 2003 Practice Problems for the Final

MTH 382 Number Theory Spring 2003 Practice Problems for the Final (1) Find the quotient and remainder in the Division Algorithm, (a) with divisor 16 and dividend 95, (b) with divisor 16 and dividend -95,

### Today s Topics. Primes & Greatest Common Divisors

Today s Topics Primes & Greatest Common Divisors Prime representations Important theorems about primality Greatest Common Divisors Least Common Multiples Euclid s algorithm Once and for all, what are prime

### Grade 7/8 Math Circles Fall 2012 Factors and Primes

1 University of Waterloo Faculty of Mathematics Centre for Education in Mathematics and Computing Grade 7/8 Math Circles Fall 2012 Factors and Primes Factors Definition: A factor of a number is a whole

### CHAPTER 5: MODULAR ARITHMETIC

CHAPTER 5: MODULAR ARITHMETIC LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN 1. Introduction In this chapter we will consider congruence modulo m, and explore the associated arithmetic called

### Solutions to Practice Problems

Solutions to Practice Problems March 205. Given n = pq and φ(n = (p (q, we find p and q as the roots of the quadratic equation (x p(x q = x 2 (n φ(n + x + n = 0. The roots are p, q = 2[ n φ(n+ ± (n φ(n+2

### I. GROUPS: BASIC DEFINITIONS AND EXAMPLES

I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called

### Kevin James. MTHSC 412 Section 2.4 Prime Factors and Greatest Comm

MTHSC 412 Section 2.4 Prime Factors and Greatest Common Divisor Greatest Common Divisor Definition Suppose that a, b Z. Then we say that d Z is a greatest common divisor (gcd) of a and b if the following

### It is time to prove some theorems. There are various strategies for doing

CHAPTER 4 Direct Proof It is time to prove some theorems. There are various strategies for doing this; we now examine the most straightforward approach, a technique called direct proof. As we begin, it

### MODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.

MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on

### Discrete Mathematics Lecture 3 Elementary Number Theory and Methods of Proof. Harper Langston New York University

Discrete Mathematics Lecture 3 Elementary Number Theory and Methods of Proof Harper Langston New York University Proof and Counterexample Discovery and proof Even and odd numbers number n from Z is called

### Squares and Square Roots

SQUARES AND SQUARE ROOTS 89 Squares and Square Roots CHAPTER 6 6.1 Introduction You know that the area of a square = side side (where side means the length of a side ). Study the following table. Side

### Number Theory. Proof. Suppose otherwise. Then there would be a finite number n of primes, which we may

Number Theory Divisibility and Primes Definition. If a and b are integers and there is some integer c such that a = b c, then we say that b divides a or is a factor or divisor of a and write b a. Definition

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

### Definition 1 Let a and b be positive integers. A linear combination of a and b is any number n = ax + by, (1) where x and y are whole numbers.

Greatest Common Divisors and Linear Combinations Let a and b be positive integers The greatest common divisor of a and b ( gcd(a, b) ) has a close and very useful connection to things called linear combinations

### Pythagorean Triples Pythagorean triple similar primitive

Pythagorean Triples One of the most far-reaching problems to appear in Diophantus Arithmetica was his Problem II-8: To divide a given square into two squares. Namely, find integers x, y, z, so that x 2

### TRIANGLES ON THE LATTICE OF INTEGERS. Department of Mathematics Rowan University Glassboro, NJ Andrew Roibal and Abdulkadir Hassen

TRIANGLES ON THE LATTICE OF INTEGERS Andrew Roibal and Abdulkadir Hassen Department of Mathematics Rowan University Glassboro, NJ 08028 I. Introduction In this article we will be studying triangles whose

### We can express this in decimal notation (in contrast to the underline notation we have been using) as follows: 9081 + 900b + 90c = 9001 + 100c + 10b

In this session, we ll learn how to solve problems related to place value. This is one of the fundamental concepts in arithmetic, something every elementary and middle school mathematics teacher should

### On the generation of elliptic curves with 16 rational torsion points by Pythagorean triples

On the generation of elliptic curves with 16 rational torsion points by Pythagorean triples Brian Hilley Boston College MT695 Honors Seminar March 3, 2006 1 Introduction 1.1 Mazur s Theorem Let C be a

### A farmer raises rabbits. Each rabbit gives birth when 2 months old and then each month there after.

Fibonacci Problem: A farmer raises rabbits. Each rabbit gives birth when 2 months old and then each month there after. Question: How many rabbits will the farmer have after n months? Try it on small numbers.

### Integer roots of quadratic and cubic polynomials with integer coefficients

Integer roots of quadratic and cubic polynomials with integer coefficients Konstantine Zelator Mathematics, Computer Science and Statistics 212 Ben Franklin Hall Bloomsburg University 400 East Second Street

### NUMBER THEORY. Theorem 1. A number is a square if and only if it has an odd number of factors.

NUMBER THEORY BY THE SPMPS 2013 NUMBER THEORY CLASS Abstract. This paper presents theorems proven by the Number Theory class of the 2013 Summer Program in Mathematical Problem Solving. An appendix is included

### Chapter 6. Number Theory. 6.1 The Division Algorithm

Chapter 6 Number Theory The material in this chapter offers a small glimpse of why a lot of facts that you ve probably nown and used for a long time are true. It also offers some exposure to generalization,

### Chapter 1 Basic Number Concepts

Draft of September 2014 Chapter 1 Basic Number Concepts 1.1. Introduction No problems in this section. 1.2. Factors and Multiples 1. Determine whether the following numbers are divisible by 3, 9, and 11:

### Orders Modulo A Prime

Orders Modulo A Prime Evan Chen March 6, 2015 In this article I develop the notion of the order of an element modulo n, and use it to prove the famous n 2 + 1 lemma as well as a generalization to arbitrary

### Class VI Chapter 3 Playing with Numbers Maths

Exercise 3. Question : Write all the factors of the following numbers: (a) 24 (b) 5 (c) 2 (d) 27 (e) 2 (f) 20 (g) 8 (h) 23 (i) 36 (a) 24 24 = 24 24 = 2 2 24 = 3 8 24 = 4 6 24 = 6 4 Factors of 24 are, 2,

### MATHEMATICAL INDUCTION. Mathematical Induction. This is a powerful method to prove properties of positive integers.

MATHEMATICAL INDUCTION MIGUEL A LERMA (Last updated: February 8, 003) Mathematical Induction This is a powerful method to prove properties of positive integers Principle of Mathematical Induction Let P

### Factoring Algorithms

Factoring Algorithms The p 1 Method and Quadratic Sieve November 17, 2008 () Factoring Algorithms November 17, 2008 1 / 12 Fermat s factoring method Fermat made the observation that if n has two factors

### Lecture 1: Elementary Number Theory

Lecture 1: Elementary Number Theory The integers are the simplest and most fundamental objects in discrete mathematics. All calculations by computers are based on the arithmetical operations with integers

### In a triangle with a right angle, there are 2 legs and the hypotenuse of a triangle.

PROBLEM STATEMENT In a triangle with a right angle, there are legs and the hypotenuse of a triangle. The hypotenuse of a triangle is the side of a right triangle that is opposite the 90 angle. The legs

### The last three chapters introduced three major proof techniques: direct,

CHAPTER 7 Proving Non-Conditional Statements The last three chapters introduced three major proof techniques: direct, contrapositive and contradiction. These three techniques are used to prove statements

### 1 Number System. Introduction. Face Value and Place Value of a Digit. Categories of Numbers

RMAT Success Master 3 1 Number System Introduction In the decimal number system, numbers are expressed by means of symbols 0, 1,, 3, 4, 5, 6, 7, 8, 9, called digits. Here, 0 is called an insignificant

### Tips, tricks and formulae on H.C.F and L.C.M. Follow the steps below to find H.C.F of given numbers by prime factorization method.

Highest Common Factor (H.C.F) Tips, tricks and formulae on H.C.F and L.C.M H.C.F is the highest common factor or also known as greatest common divisor, the greatest number which exactly divides all the

### PROBLEM SET # 2 SOLUTIONS

PROBLEM SET # 2 SOLUTIONS CHAPTER 2: GROUPS AND ARITHMETIC 2. Groups.. Let G be a group and e and e two identity elements. Show that e = e. (Hint: Consider e e and calculate it two ways.) Solution. Since

### Solutions to Homework Set 3 (Solutions to Homework Problems from Chapter 2)

Solutions to Homework Set 3 (Solutions to Homework Problems from Chapter 2) Problems from 21 211 Prove that a b (mod n) if and only if a and b leave the same remainder when divided by n Proof Suppose a

### Mathematics of Cryptography Modular Arithmetic, Congruence, and Matrices. A Biswas, IT, BESU SHIBPUR

Mathematics of Cryptography Modular Arithmetic, Congruence, and Matrices A Biswas, IT, BESU SHIBPUR McGraw-Hill The McGraw-Hill Companies, Inc., 2000 Set of Integers The set of integers, denoted by Z,

### GREATEST COMMON DIVISOR

DEFINITION: GREATEST COMMON DIVISOR The greatest common divisor (gcd) of a and b, denoted by (a, b), is the largest common divisor of integers a and b. THEOREM: If a and b are nonzero integers, then their

### 5-4 Prime and Composite Numbers

5-4 Prime and Composite Numbers Prime and Composite Numbers Prime Factorization Number of Divisorss Determining if a Number is Prime More About Primes Prime and Composite Numbers Students should recognizee

### Playing with Numbers

Playing with Numbers Chapter 3 3.1 Introduction Ramesh has 6 marbles with him. He wants to arrange them in rows in such a way that each row has the same number of marbles. He arranges them in the following

### Algebra. Sample Solutions for Test 1

EPFL - Section de Mathématiques Algebra Fall semester 2008-2009 Sample Solutions for Test 1 Question 1 (english, 30 points) 1) Let n 11 13 17. Find the number of units of the ring Z/nZ. 2) Consider the

### 3 Congruence arithmetic

3 Congruence arithmetic 3.1 Congruence mod n As we said before, one of the most basic tasks in number theory is to factor a number a. How do we do this? We start with smaller numbers and see if they divide

### Playing with Numbers. factors and multiples

3 Playing with Numbers IN T H I S CH A P T E R, Y O U W I L L LEARN: Factors and multiples of a natural number Common factors and common multiples Types of natural numbers even and odd numbers, prime and

### Stanford University Educational Program for Gifted Youth (EPGY) Number Theory. Dana Paquin, Ph.D.

Stanford University Educational Program for Gifted Youth (EPGY) Dana Paquin, Ph.D. paquin@math.stanford.edu Summer 2010 Note: These lecture notes are adapted from the following sources: 1. Ivan Niven,

### Characterizing the Sum of Two Cubes

1 3 47 6 3 11 Journal of Integer Sequences, Vol. 6 (003), Article 03.4.6 Characterizing the Sum of Two Cubes Kevin A. Broughan University of Waikato Hamilton 001 New Zealand kab@waikato.ac.nz Abstract