Practice Midterm 2 Solutions

Size: px
Start display at page:

Download "Practice Midterm 2 Solutions"

Transcription

1 Practice Midterm 2 Solutions May 30, 2013 (1) We want to show that for any odd integer a coprime to 7, a 3 is congruent to 1 or 1 mod 7. In fact, we don t need the assumption that a is odd. By Fermat s theorem, we know a 6 1 (mod 7) since 7 is prime. But this means that (a 3 ) 2 1 (mod 7). But since 7 is prime, this implies a 3 ±1 (mod 7). [Note: it s crucial that 7 is prime. Otherwise, a number other than ±1 may square to 1]. Note: even if you did not see this trick, you could solve this by proving all the cases mod 7. That is: we cube each of the numbers between 1 and 6 and check that the cube is ±1. Note that by symmetry (e.g. 4 3 (mod 7)), we only really need to carry out computations for three numbers and the rest are determined: (mod 7) (mod 7) (mod 7) 4 3 ( 3) 3 (3 3 ) ( 1) 1 (mod 7) 5 3 ( 2) 3 (2 3 ) 1 (mod 7) 6 3 ( 1) 3 1 (mod 7) For the next part: suppose a 3 + b 3 = c 3. As it stands, this is not an equation mod 7, but in particular it follows that a 3 + b 3 c 3 (mod 7). Suppose for contradiction that a, b, c are all coprime to 7. Then from what we proved above, a 3, b 3, c 3 are all ±1. But there is no way of adding two ±1 s to get ±1, a contradiction. Therefore, one of a, b, c is divisible by 7, so abc is divisible by 7. (2) Fermat s Little Theorem: if a is not divisible by a prime p, then a p 1 1 (mod p). Note: 163 is prime. Let b Z 163. Case 1: b = 0. Then b = 0 7. Moreover, this is unique: if 0 = x 7, it follows that 1

2 x = 0. Case 2: b 0. Then b has an inverse, and we may apply Fermat s Little Theorem to b 1 : (b 1 ) 162 = 1 (in Z 163 ). As a shorthand, rewrite this: b 162 = 1. Multiply both sides by b: b 161 = b. But now, note that 7 divides 161, with 161 = 7 23, so we may rewrite the left hand side: (b 23 ) 7 = b. Thus, we have expressed b as the seventh power of b 23. To show uniqueness, suppose b = x 7. The idea is, apply all the reverse operations to conclude x must be b 23. Indeed, we are assuming b is a unit, so if b = x 7, x must be a unit, too. So we may raise both sides to the 23rd power: b 23 = (x 7 ) 23. The right hand side simplifies: b 23 = x 161. But now, we know that, since x is a unit, Fermat s Little Theorem applies and x 162 = 1. So we may multiply the right hand side by x 162 without changing the equation: b 23 = x 161 x 162 = x, as desired. Note: I used equal signs instead of congruence signs because we are working in Z 163 rather than Z, so congruent numbers are actually considered equal. Another note: you probably know this, but Z 163 is the same as the book s Z/163Z. (3) First we must find the prime factorization of 341 in order to apply Euler s Theorem. It is: 341 = So ϕ(341) = = 300. So (mod 341), so (mod 341). So now we must calculate 5 40 mod 341. This is made easier using the binary trick (but is still pretty tedious). Note: 5 40 = We have: (mod 341) (20 + 5) (mod 341). 5 8 (284) 2 ( 57) (50 + 7) ( 21) (mod 341) 5 16 (180) ( 17 20) 5 ( 340) (mod 341) (mod 341). So: 5 40 = (180 5) ( 123) 5 ( 615) 67 (mod 341). I tried to use tricks to make the numbers easy to work with by replacing them with easier congruent numbers. Of course, you don t have to use these tricks. Here is another, probably faster approach: instead of computing the result mod 341 = 11 31, compute it separately for 11 and 31, and then piece it back to- 2

3 gether using the Chinese remainder theorem. Calculations mod 11 and 31 are much quicker than mod 341. I ll let you carry them out yourselves (use Fermat s Little Theorem). You end up getting (mod 11), (mod 31). Using the Chinese remainder theorem, we can conclude what has to be mod 341. In fact, we can just start with a guess of 5 and keep adding multiples of 31 until our number has remainder 1 mod 11. Indeed, = 67 is the answer. (4) Recall a ring R is an integral domain if it has no zero divisors. The ring Z is an example: whenever a b = 0, for a, b Z, we must have a = 0 or b = 0. Thus, Z is an integral domain. However, Z is not a field, since, for example, 2 has no inverse. (5) Note 10 1 (mod 11). So if n = r i=0 a i10 i, then n r i=0 a i( 1) i (mod 11) simply by replacing 10 with the congruent number 1. Since a i are the digits of n, this sum is the alternating sum of the digits of n. If n is divisible by 11, then n 0 (mod 11), so r i=0 a i( 1) i n 0 (mod 11), so the alternating sum of the digits is also divisible by 11. Conversely, if the alternating sum of the digits is divisible by 11, then r i=0 a i( 1) i 0 (mod 11), so n r i=0 a i( 1) i 0 (mod 11), so n is divisible by 11. (6) No. f is multiplicative but not additive. For example, f(1 + 2) = 9 5 = f(1) + f(2). (7) Recall that the kernel of a homomorphism is the set of elements that go to 0. Suppose d m and consider the homomorphism f : Z m Z d defined by f([a] m ) = [a] d. Suppose a is in the kernel of f. Then [a] d = f([a] m ) = 0. By definition, [a] d = 0 means a is divisible by d. Thus the kernel of f consists of all numbers (between 0 and m 1) divisible by d. There are m/d such numbers (including 0). (8)Note: a ring R could technically have only one element, call it, where is both the additive identity 0 and the multiplicative identity 1. We call this ring the trivial ring. It s not very interesting. Also note: it s part of the definition of a field that 0 1, so there is no trivial field. Now, suppose F is a field and R is a ring and f : F R is a homomorphism. Assume R is not the trivial ring. Then I claim f is injective. Recall, to check injectivity of homomorphisms, it suffices to check that if f(x) = 0, then x = 0. Put another way, if x 0, then f(x) 0. Indeed, let x F and suppose x 0. Since F is a field, it follows by definition that x is a unit. So fix y F such that y x = 1. 3

4 But then f(y) f(x) = f(y x) [since f is a homomorphism] = f(1) = 1 [again since f is a homomorphism]. Thus f(y) f(x) = 1. But this implies f(x) 0. Indeed, if f(x) were equal to 0, then 1 = f(y) 0 = 0, which can be true only in the trivial ring, which we assumed we re not in. (9) You could technically just try all the possibilities (you would only have 10 things to check, by symmetry, as discussed in problem 1). Another thing you could try is factorize: x 3 = 1 means x 3 1 = 0, which can be factored as (x 1)(x 2 + x + 1) = 0. But this is messy. The approach I prefer is, as in 3, to reduce this to mod 3 and mod 7 instead of mod 21. Indeed, by the Chinese remainder theorem, x 3 is congruent to 1 mod 21 if and only if it is congruent to 1 mod 3 and it is congruent to 1 mod 7. So we solve these two congruences separately. From problem 1, we know the solutions to x 3 1 (mod 7) are x 1, 2, 4 (mod 7). And for x 3 1 (mod 3), the only solution is x 1 (mod 3). Since a pair of congruences mod 3 and mod 7 gives rise to a unique congruence mod 21 by the Chinese remainder theorem, there are three solutions to x 3 = 1 in Z 21 (we didn t calculate what the solutions are, but we weren t asked to). (10) Suppose m, n are coprime. By Bezout s identity, we can write 1 = rm + sn. But then sn = 1 rm 1 (mod m). Also sn 0 (mod n), so sn is the desired solution. Conversely, I want to show that if we can find a common solution x 1 (mod m), x 0 (mod n), then m and n must be coprime. Indeed, since x 1 (mod m), we can write x = 1 + km for some k Z. Since x 0 (mod n), we can also write x = ln for some l Z. Setting these two expressions equal, we get 1 + km = ln, so 1 = ln km But we know that this implies n and m are coprime. (11) Recall the order of a is the smallest nonzero exponent d such that a d = 1. I will assume we re allowed to use Fermat s Little Theorem. Then we know a p 1 = 1, for a Z p. But we re not yet done. This just tells us d p 1. Now, let r be the remainder you get when you divide p 1 by d. So p 1 = qd + r, 0 r < d. So 1 = a p 1 = a qd+r = (a d ) q a r = 1 q a r [using the fact that d is the order of a] = a r. We just showed a r = 1. But r < d so if r 0, this contradicts the definition of d as the smallest number such that a d = 1. Thus, we must have r = 0. Since r is the remainder when you divide p 1 by d, d must in fact divide p 1, as desired. To find the order of 3 in Z 101, first note 101 is prime. Then by the above, we know the order of 3 must divide = 100. The candidates are : 1, 2, 4, 5, 10, 20, 25 or 100. We can easily see the order is not 1, 2, 4. We compute: 4

5 3 5 = (mod 101) = (3 5 ) 2 (41) ( 1) (mod 101) Since 3 10 ±1, then 3 20 = (3 10 ) = 3 5 (3 10 ) 2 = 41 (65) 2 = = 41 ( ) 41 (42 ( 1) + 25) (41) ( 1) Since 3 25 ±1, then We can stop here and conclude that the only option left is 100, so 3 has order 100. (12) Suppose a Z 163 has order 9. So a 9 = 1 and 9 is the first nonzero number with this property. So, we might guess that a 3 has order 3. Indeed, (a 3 ) 3 = a 9 = 1 (mod 163). We just want to show that 3 is the smallest nonzero number with this property. Indeed, suppose 0 < k < 3 and (a 3 ) k = 1. But this would imply a 3k = 1. However, since 0 < k < 3, it follows that 0 < 3k < 9, which contradicts the assumption that 9 is the order of a. (13) (i). We want to solve the system of equations x 1 (mod 3), x 3 (mod 5), x 0 (mod 12). Notice that the moduli 3, 5, 12 are not relatively prime. Indeed, 3 and 12 have a common divisor: 3. So the first order of business is to check whether the equations we re trying to solve are even consistent. In fact, they re not. If x 0 (mod 12), then it would follow that x 0 (mod 3), which contradicts the requirement x 1 (mod 3). So there are no solutions to this system of equations. (ii) We want to solve the system of equations x 1 (mod 3), x 3 (mod 5), x 4 (mod 12). Now the equations are consistent: indeed, if x 4 (mod 12), then x 1 (mod 3), which is consistent. Note: the equation x 4 (mod 12) already implies x 1 (mod 3), so we don t need the latter. We may throw it away, and reduce our system to two equations: x 3 (mod 5) x 4 (mod 12) Once you got here, you could just solve the equation by trial and error. We start out with a guess which works mod 12: x = 4. This does not work mod 5 (as 4 3 (mod 5). So, we simply keep adding 12 (which preserves the values mod 12) 5

6 until it also works mod = 16 doesn t work, but = 28 does, since 28 3 (mod 5). So 28 is a solution. By the Chinese remainder theorem, which applies since 5 and 12 are relatively prime, 28 must be the unique solution mod 5 12 = 60. On the other hand, adding a multiple of 60 leaves the remainders mod 5 and mod 12 unchanged, so yields another solution. Thus, the general solution is k, k Z. But if you want to solve it in a more methodical way, not involving trial and error, here is how. First you want to find values r and s for which the equation 1 = r 5 + s 12 holds. Now, we know (see the proof of problem 10) that s 12 1 (mod 5) and s 12 0 (mod 12). Hence, s 12 is a number we can add without changing the remainder mod 12, and increasing the remainder mod 5 by 1. However, since we re starting at 4 and want to end up at 3 mod 5, we need to subtract by 1 instead of adding 1. So the answer is: x 4 s 12 (mod 60). Now it remains to find r and s so we can plug them in. We find that r = 5 and s = 2 works. So x 4 ( 2) 12 = 28 (mod 60), as above. 6

An interesting class of problems of a computational nature ask for the standard residue of a power of a number, e.g.,

An interesting class of problems of a computational nature ask for the standard residue of a power of a number, e.g., Binary exponentiation An interesting class of problems of a computational nature ask for the standard residue of a power of a number, e.g., What are the last two digits of the number 2 284? In the absence

More information

Math 127: Equivalence Relations

Math 127: Equivalence Relations Math 127: Equivalence Relations Mary Radcliffe 1 Equivalence Relations Relations can take many forms in mathematics. In these notes, we focus especially on equivalence relations, but there are many other

More information

Calculators will not be permitted on the exam. The numbers on the exam will be suitable for calculating by hand.

Calculators will not be permitted on the exam. The numbers on the exam will be suitable for calculating by hand. Midterm #2: practice MATH 311 Intro to Number Theory midterm: Thursday, Oct 20 Please print your name: Calculators will not be permitted on the exam. The numbers on the exam will be suitable for calculating

More information

The Chinese Remainder Theorem

The Chinese Remainder Theorem The Chinese Remainder Theorem 8-3-2014 The Chinese Remainder Theorem gives solutions to systems of congruences with relatively prime moduli The solution to a system of congruences with relatively prime

More information

SOLUTIONS TO PROBLEM SET 5. Section 9.1

SOLUTIONS TO PROBLEM SET 5. Section 9.1 SOLUTIONS TO PROBLEM SET 5 Section 9.1 Exercise 2. Recall that for (a, m) = 1 we have ord m a divides φ(m). a) We have φ(11) = 10 thus ord 11 3 {1, 2, 5, 10}. We check 3 1 3 (mod 11), 3 2 9 (mod 11), 3

More information

University of British Columbia. Math 312, Midterm, 6th of June 2017

University of British Columbia. Math 312, Midterm, 6th of June 2017 University of British Columbia Math 312, Midterm, 6th of June 2017 Name (please be legible) Signature Student number Duration: 90 minutes INSTRUCTIONS This test has 7 problems for a total of 100 points.

More information

Carmen s Core Concepts (Math 135)

Carmen s Core Concepts (Math 135) Carmen s Core Concepts (Math 135) Carmen Bruni University of Waterloo Week 7 1 Congruence Definition 2 Congruence is an Equivalence Relation (CER) 3 Properties of Congruence (PC) 4 Example 5 Congruences

More information

Calculators will not be permitted on the exam. The numbers on the exam will be suitable for calculating by hand.

Calculators will not be permitted on the exam. The numbers on the exam will be suitable for calculating by hand. Midterm #: practice MATH Intro to Number Theory midterm: Thursday, Nov 7 Please print your name: Calculators will not be permitted on the exam. The numbers on the exam will be suitable for calculating

More information

Applications of Fermat s Little Theorem and Congruences

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

More information

Exam 1 7 = = 49 2 ( ) = = 7 ( ) =

Exam 1 7 = = 49 2 ( ) = = 7 ( ) = Exam 1 Problem 1. a) Define gcd(a, b). Using Euclid s algorithm comute gcd(889, 168). Then find x, y Z such that gcd(889, 168) = x 889 + y 168 (check your answer!). b) Let a be an integer. Prove that gcd(3a

More information

Solutions to Problem Set 6 - Fall 2008 Due Tuesday, Oct. 21 at 1:00

Solutions to Problem Set 6 - Fall 2008 Due Tuesday, Oct. 21 at 1:00 18.781 Solutions to Problem Set 6 - Fall 008 Due Tuesday, Oct. 1 at 1:00 1. (Niven.8.7) If p 3 is prime, how many solutions are there to x p 1 1 (mod p)? How many solutions are there to x p 1 (mod p)?

More information

Solutions for the Practice Questions

Solutions for the Practice Questions Solutions for the Practice Questions Question 1. Find all solutions to the congruence 13x 12 (mod 35). Also, answer the following questions about the solutions to the above congruence. Are there solutions

More information

SOLUTIONS FOR PROBLEM SET 4

SOLUTIONS FOR PROBLEM SET 4 SOLUTIONS FOR PROBLEM SET 4 A. A certain integer a gives a remainder of 1 when divided by 2. What can you say about the remainder that a gives when divided by 8? SOLUTION. Let r be the remainder that a

More information

Math 255 Spring 2017 Solving x 2 a (mod n)

Math 255 Spring 2017 Solving x 2 a (mod n) Math 255 Spring 2017 Solving x 2 a (mod n) Contents 1 Lifting 1 2 Solving x 2 a (mod p k ) for p odd 3 3 Solving x 2 a (mod 2 k ) 5 4 Solving x 2 a (mod n) for general n 9 1 Lifting Definition 1.1. Let

More information

Solutions for the Practice Final

Solutions for the Practice Final Solutions for the Practice Final 1. Ian and Nai play the game of todo, where at each stage one of them flips a coin and then rolls a die. The person who played gets as many points as the number rolled

More information

Wilson s Theorem and Fermat s Theorem

Wilson s Theorem and Fermat s Theorem Wilson s Theorem and Fermat s Theorem 7-27-2006 Wilson s theorem says that p is prime if and only if (p 1)! = 1 (mod p). Fermat s theorem says that if p is prime and p a, then a p 1 = 1 (mod p). Wilson

More information

b) Find all positive integers smaller than 200 which leave remainder 1, 3, 4 upon division by 3, 5, 7 respectively.

b) Find all positive integers smaller than 200 which leave remainder 1, 3, 4 upon division by 3, 5, 7 respectively. Solutions to Exam 1 Problem 1. a) State Fermat s Little Theorem and Euler s Theorem. b) Let m, n be relatively prime positive integers. Prove that m φ(n) + n φ(m) 1 (mod mn). Solution: a) Fermat s Little

More information

1.6 Congruence Modulo m

1.6 Congruence Modulo m 1.6 Congruence Modulo m 47 5. Let a, b 2 N and p be a prime. Prove for all natural numbers n 1, if p n (ab) and p - a, then p n b. 6. In the proof of Theorem 1.5.6 it was stated that if n is a prime number

More information

LECTURE 3: CONGRUENCES. 1. Basic properties of congruences We begin by introducing some definitions and elementary properties.

LECTURE 3: CONGRUENCES. 1. Basic properties of congruences We begin by introducing some definitions and elementary properties. LECTURE 3: CONGRUENCES 1. Basic properties of congruences We begin by introducing some definitions and elementary properties. Definition 1.1. Suppose that a, b Z and m N. We say that a is congruent to

More information

An elementary study of Goldbach Conjecture

An elementary study of Goldbach Conjecture An elementary study of Goldbach Conjecture Denise Chemla 26/5/2012 Goldbach Conjecture (7 th, june 1742) states that every even natural integer greater than 4 is the sum of two odd prime numbers. If we

More information

Constructions of Coverings of the Integers: Exploring an Erdős Problem

Constructions of Coverings of the Integers: Exploring an Erdős Problem Constructions of Coverings of the Integers: Exploring an Erdős Problem Kelly Bickel, Michael Firrisa, Juan Ortiz, and Kristen Pueschel August 20, 2008 Abstract In this paper, we study necessary conditions

More information

MATH 324 Elementary Number Theory Solutions to Practice Problems for Final Examination Monday August 8, 2005

MATH 324 Elementary Number Theory Solutions to Practice Problems for Final Examination Monday August 8, 2005 MATH 324 Elementary Number Theory Solutions to Practice Problems for Final Examination Monday August 8, 2005 Deartment of Mathematical and Statistical Sciences University of Alberta Question 1. Find integers

More information

Permutation Groups. Every permutation can be written as a product of disjoint cycles. This factorization is unique up to the order of the factors.

Permutation Groups. Every permutation can be written as a product of disjoint cycles. This factorization is unique up to the order of the factors. Permutation Groups 5-9-2013 A permutation of a set X is a bijective function σ : X X The set of permutations S X of a set X forms a group under function composition The group of permutations of {1,2,,n}

More information

6. Find an inverse of a modulo m for each of these pairs of relatively prime integers using the method

6. Find an inverse of a modulo m for each of these pairs of relatively prime integers using the method Exercises Exercises 1. Show that 15 is an inverse of 7 modulo 26. 2. Show that 937 is an inverse of 13 modulo 2436. 3. By inspection (as discussed prior to Example 1), find an inverse of 4 modulo 9. 4.

More information

Goldbach Conjecture (7 th june 1742)

Goldbach Conjecture (7 th june 1742) Goldbach Conjecture (7 th june 1742) We note P the odd prime numbers set. P = {p 1 = 3, p 2 = 5, p 3 = 7, p 4 = 11,...} n 2N\{0, 2, 4}, p P, p n/2, q P, q n/2, n = p + q We call n s Goldbach decomposition

More information

Introduction to Modular Arithmetic

Introduction to Modular Arithmetic 1 Integers modulo n 1.1 Preliminaries Introduction to Modular Arithmetic Definition 1.1.1 (Equivalence relation). Let R be a relation on the set A. Recall that a relation R is a subset of the cartesian

More information

Modular Arithmetic. Kieran Cooney - February 18, 2016

Modular Arithmetic. Kieran Cooney - February 18, 2016 Modular Arithmetic Kieran Cooney - kieran.cooney@hotmail.com February 18, 2016 Sums and products in modular arithmetic Almost all of elementary number theory follows from one very basic theorem: Theorem.

More information

Solutions for the 2nd Practice Midterm

Solutions for the 2nd Practice Midterm Solutions for the 2nd Practice Midterm 1. (a) Use the Euclidean Algorithm to find the greatest common divisor of 44 and 17. The Euclidean Algorithm yields: 44 = 2 17 + 10 17 = 1 10 + 7 10 = 1 7 + 3 7 =

More information

6.2 Modular Arithmetic

6.2 Modular Arithmetic 6.2 Modular Arithmetic Every reader is familiar with arithmetic from the time they are three or four years old. It is the study of numbers and various ways in which we can combine them, such as through

More information

Launchpad Maths. Arithmetic II

Launchpad Maths. Arithmetic II Launchpad Maths. Arithmetic II LAW OF DISTRIBUTION The Law of Distribution exploits the symmetries 1 of addition and multiplication to tell of how those operations behave when working together. Consider

More information

Introduction. and Z r1 Z rn. This lecture aims to provide techniques. CRT during the decription process in RSA is explained.

Introduction. and Z r1 Z rn. This lecture aims to provide techniques. CRT during the decription process in RSA is explained. THE CHINESE REMAINDER THEOREM INTRODUCED IN A GENERAL KONTEXT Introduction The rst Chinese problem in indeterminate analysis is encountered in a book written by the Chinese mathematician Sun Tzi. The problem

More information

Quadratic Residues. Legendre symbols provide a computational tool for determining whether a quadratic congruence has a solution. = a (p 1)/2 (mod p).

Quadratic Residues. Legendre symbols provide a computational tool for determining whether a quadratic congruence has a solution. = a (p 1)/2 (mod p). Quadratic Residues 4--015 a is a quadratic residue mod m if x = a (mod m). Otherwise, a is a quadratic nonresidue. Quadratic Recirocity relates the solvability of the congruence x = (mod q) to the solvability

More information

LECTURE 7: POLYNOMIAL CONGRUENCES TO PRIME POWER MODULI

LECTURE 7: POLYNOMIAL CONGRUENCES TO PRIME POWER MODULI LECTURE 7: POLYNOMIAL CONGRUENCES TO PRIME POWER MODULI 1. Hensel Lemma for nonsingular solutions Although there is no analogue of Lagrange s Theorem for prime power moduli, there is an algorithm for determining

More information

CHAPTER 2. Modular Arithmetic

CHAPTER 2. Modular Arithmetic CHAPTER 2 Modular Arithmetic In studying the integers we have seen that is useful to write a = qb + r. Often we can solve problems by considering only the remainder, r. This throws away some of the information,

More information

Fermat s little theorem. RSA.

Fermat s little theorem. RSA. .. Computing large numbers modulo n (a) In modulo arithmetic, you can always reduce a large number to its remainder a a rem n (mod n). (b) Addition, subtraction, and multiplication preserve congruence:

More information

To be able to determine the quadratic character of an arbitrary number mod p (p an odd prime), we. The first (and most delicate) case concerns 2

To be able to determine the quadratic character of an arbitrary number mod p (p an odd prime), we. The first (and most delicate) case concerns 2 Quadratic Reciprocity To be able to determine the quadratic character of an arbitrary number mod p (p an odd prime), we need to be able to evaluate q for any prime q. The first (and most delicate) case

More information

and problem sheet 7

and problem sheet 7 1-18 and 15-151 problem sheet 7 Solutions to the following five exercises and optional bonus problem are to be submitted through gradescope by 11:30PM on Friday nd November 018. Problem 1 Let A N + and

More information

MODULAR ARITHMETIC II: CONGRUENCES AND DIVISION

MODULAR ARITHMETIC II: CONGRUENCES AND DIVISION MODULAR ARITHMETIC II: CONGRUENCES AND DIVISION MATH CIRCLE (BEGINNERS) 02/05/2012 Modular arithmetic. Two whole numbers a and b are said to be congruent modulo n, often written a b (mod n), if they give

More information

Collection of rules, techniques and theorems for solving polynomial congruences 11 April 2012 at 22:02

Collection of rules, techniques and theorems for solving polynomial congruences 11 April 2012 at 22:02 Collection of rules, techniques and theorems for solving polynomial congruences 11 April 2012 at 22:02 Public Polynomial congruences come up constantly, even when one is dealing with much deeper problems

More information

Rational Points On Elliptic Curves - Solutions. (i) Throughout, we ve been looking at elliptic curves in the general form. y 2 = x 3 + Ax + B

Rational Points On Elliptic Curves - Solutions. (i) Throughout, we ve been looking at elliptic curves in the general form. y 2 = x 3 + Ax + B Rational Points On Elliptic Curves - Solutions (Send corrections to cbruni@uwaterloo.ca) (i) Throughout, we ve been looking at elliptic curves in the general form y 2 = x 3 + Ax + B However we did claim

More information

ALGEBRA: Chapter I: QUESTION BANK

ALGEBRA: Chapter I: QUESTION BANK 1 ALGEBRA: Chapter I: QUESTION BANK Elements of Number Theory Congruence One mark questions: 1 Define divisibility 2 If a b then prove that a kb k Z 3 If a b b c then PT a/c 4 If a b are two non zero integers

More information

Public Key Encryption

Public Key Encryption Math 210 Jerry L. Kazdan Public Key Encryption The essence of this procedure is that as far as we currently know, it is difficult to factor a number that is the product of two primes each having many,

More information

MAT Modular arithmetic and number theory. Modular arithmetic

MAT Modular arithmetic and number theory. Modular arithmetic Modular arithmetic 1 Modular arithmetic may seem like a new and strange concept at first The aim of these notes is to describe it in several different ways, in the hope that you will find at least one

More information

Primitive Roots. Chapter Orders and Primitive Roots

Primitive Roots. Chapter Orders and Primitive Roots Chapter 5 Primitive Roots The name primitive root applies to a number a whose powers can be used to represent a reduced residue system modulo n. Primitive roots are therefore generators in that sense,

More information

Solutions to Exam 1. Problem 1. a) State Fermat s Little Theorem and Euler s Theorem. b) Let m, n be relatively prime positive integers.

Solutions to Exam 1. Problem 1. a) State Fermat s Little Theorem and Euler s Theorem. b) Let m, n be relatively prime positive integers. Solutions to Exam 1 Problem 1. a) State Fermat s Little Theorem and Euler s Theorem. b) Let m, n be relatively rime ositive integers. Prove that m φ(n) + n φ(m) 1 (mod mn). c) Find the remainder of 1 008

More information

The congruence relation has many similarities to equality. The following theorem says that congruence, like equality, is an equivalence relation.

The congruence relation has many similarities to equality. The following theorem says that congruence, like equality, is an equivalence relation. Congruences A congruence is a statement about divisibility. It is a notation that simplifies reasoning about divisibility. It suggests proofs by its analogy to equations. Congruences are familiar to us

More information

Two congruences involving 4-cores

Two congruences involving 4-cores Two congruences involving 4-cores ABSTRACT. The goal of this paper is to prove two new congruences involving 4- cores using elementary techniques; namely, if a 4 (n) denotes the number of 4-cores of n,

More information

Sheet 1: Introduction to prime numbers.

Sheet 1: Introduction to prime numbers. Option A Hand in at least one question from at least three sheets Sheet 1: Introduction to prime numbers. [provisional date for handing in: class 2.] 1. Use Sieve of Eratosthenes to find all prime numbers

More information

Algorithmic Number Theory and Cryptography (CS 303)

Algorithmic Number Theory and Cryptography (CS 303) Algorithmic Number Theory and Cryptography (CS 303) Modular Arithmetic Jeremy R. Johnson 1 Introduction Objective: To become familiar with modular arithmetic and some key algorithmic constructions that

More information

Distribution of Primes

Distribution of Primes Distribution of Primes Definition. For positive real numbers x, let π(x) be the number of prime numbers less than or equal to x. For example, π(1) = 0, π(10) = 4 and π(100) = 25. To use some ciphers, we

More information

Number Theory - Divisibility Number Theory - Congruences. Number Theory. June 23, Number Theory

Number Theory - Divisibility Number Theory - Congruences. Number Theory. June 23, Number Theory - Divisibility - Congruences June 23, 2014 Primes - Divisibility - Congruences Definition A positive integer p is prime if p 2 and its only positive factors are itself and 1. Otherwise, if p 2, then p

More information

p 1 MAX(a,b) + MIN(a,b) = a+b n m means that m is a an integer multiple of n. Greatest Common Divisor: We say that n divides m.

p 1 MAX(a,b) + MIN(a,b) = a+b n m means that m is a an integer multiple of n. Greatest Common Divisor: We say that n divides m. Great Theoretical Ideas In Computer Science Steven Rudich CS - Spring Lecture Feb, Carnegie Mellon University Modular Arithmetic and the RSA Cryptosystem p- p MAX(a,b) + MIN(a,b) = a+b n m means that m

More information

Modular Arithmetic. claserken. July 2016

Modular Arithmetic. claserken. July 2016 Modular Arithmetic claserken July 2016 Contents 1 Introduction 2 2 Modular Arithmetic 2 2.1 Modular Arithmetic Terminology.................. 2 2.2 Properties of Modular Arithmetic.................. 2 2.3

More information

Discrete Math Class 4 ( )

Discrete Math Class 4 ( ) Discrete Math 37110 - Class 4 (2016-10-06) 41 Division vs congruences Instructor: László Babai Notes taken by Jacob Burroughs Revised by instructor DO 41 If m ab and gcd(a, m) = 1, then m b DO 42 If gcd(a,

More information

Number Theory. Konkreetne Matemaatika

Number Theory. Konkreetne Matemaatika ITT9131 Number Theory Konkreetne Matemaatika Chapter Four Divisibility Primes Prime examples Factorial Factors Relative primality `MOD': the Congruence Relation Independent Residues Additional Applications

More information

Number Theory/Cryptography (part 1 of CSC 282)

Number Theory/Cryptography (part 1 of CSC 282) Number Theory/Cryptography (part 1 of CSC 282) http://www.cs.rochester.edu/~stefanko/teaching/11cs282 1 Schedule The homework is due Sep 8 Graded homework will be available at noon Sep 9, noon. EXAM #1

More information

Assignment 2. Due: Monday Oct. 15, :59pm

Assignment 2. Due: Monday Oct. 15, :59pm Introduction To Discrete Math Due: Monday Oct. 15, 2012. 11:59pm Assignment 2 Instructor: Mohamed Omar Math 6a For all problems on assignments, you are allowed to use the textbook, class notes, and other

More information

Multiples and Divisibility

Multiples and Divisibility Multiples and Divisibility A multiple of a number is a product of that number and an integer. Divisibility: A number b is said to be divisible by another number a if b is a multiple of a. 45 is divisible

More information

Math 412: Number Theory Lecture 6: congruence system and

Math 412: Number Theory Lecture 6: congruence system and Math 412: Number Theory Lecture 6: congruence system and classes Gexin Yu gyu@wm.edu College of William and Mary Chinese Remainder Theorem Chinese Remainder Theorem: let m 1, m 2,..., m k be pairwise coprimes.

More information

Mathematics Explorers Club Fall 2012 Number Theory and Cryptography

Mathematics Explorers Club Fall 2012 Number Theory and Cryptography Mathematics Explorers Club Fall 2012 Number Theory and Cryptography Chapter 0: Introduction Number Theory enjoys a very long history in short, number theory is a study of integers. Mathematicians over

More information

Modular Arithmetic and Doomsday

Modular Arithmetic and Doomsday Modular Arithmetic and Doomsday Blake Thornton Much of this is due directly to Joshua Zucker and Paul Zeitz. 1. Subtraction Magic Trick. While blindfolded, a magician asks a member from the audience to

More information

Lecture 32. Handout or Document Camera or Class Exercise. Which of the following is equal to [53] [5] 1 in Z 7? (Do not use a calculator.

Lecture 32. Handout or Document Camera or Class Exercise. Which of the following is equal to [53] [5] 1 in Z 7? (Do not use a calculator. Lecture 32 Instructor s Comments: This is a make up lecture. You can choose to cover many extra problems if you wish or head towards cryptography. I will probably include the square and multiply algorithm

More information

Linear Congruences. The solutions to a linear congruence ax b (mod m) are all integers x that satisfy the congruence.

Linear Congruences. The solutions to a linear congruence ax b (mod m) are all integers x that satisfy the congruence. Section 4.4 Linear Congruences Definition: A congruence of the form ax b (mod m), where m is a positive integer, a and b are integers, and x is a variable, is called a linear congruence. The solutions

More information

LUCAS-SIERPIŃSKI AND LUCAS-RIESEL NUMBERS

LUCAS-SIERPIŃSKI AND LUCAS-RIESEL NUMBERS LUCAS-SIERPIŃSKI AND LUCAS-RIESEL NUMBERS DANIEL BACZKOWSKI, OLAOLU FASORANTI, AND CARRIE E. FINCH Abstract. In this paper, we show that there are infinitely many Sierpiński numbers in the sequence of

More information

Public Key Cryptography Great Ideas in Theoretical Computer Science Saarland University, Summer 2014

Public Key Cryptography Great Ideas in Theoretical Computer Science Saarland University, Summer 2014 7 Public Key Cryptography Great Ideas in Theoretical Computer Science Saarland University, Summer 2014 Cryptography studies techniques for secure communication in the presence of third parties. A typical

More information

16.1 Introduction Numbers in General Form

16.1 Introduction Numbers in General Form 16.1 Introduction You have studied various types of numbers such as natural numbers, whole numbers, integers and rational numbers. You have also studied a number of interesting properties about them. In

More information

Lecture 8. Outline. 1. Modular Arithmetic. Clock Math!!! 2. Inverses for Modular Arithmetic: Greatest Common Divisor. 3. Euclid s GCD Algorithm

Lecture 8. Outline. 1. Modular Arithmetic. Clock Math!!! 2. Inverses for Modular Arithmetic: Greatest Common Divisor. 3. Euclid s GCD Algorithm Lecture 8. Outline. 1. Modular Arithmetic. Clock Math!!! 2. Inverses for Modular Arithmetic: Greatest Common Divisor. 3. Euclid s GCD Algorithm Clock Math If it is 1:00 now. What time is it in 5 hours?

More information

Study Material. For. Shortcut Maths

Study Material. For. Shortcut Maths N ew Shortcut Maths Edition 2015 Study Material For Shortcut Maths Regd. Office :- A-202, Shanti Enclave, Opp.Railway Station, Mira Road(E), Mumbai. bankpo@laqshya.in (Not For Sale) (For Private Circulation

More information

CMath 55 PROFESSOR KENNETH A. RIBET. Final Examination May 11, :30AM 2:30PM, 100 Lewis Hall

CMath 55 PROFESSOR KENNETH A. RIBET. Final Examination May 11, :30AM 2:30PM, 100 Lewis Hall CMath 55 PROFESSOR KENNETH A. RIBET Final Examination May 11, 015 11:30AM :30PM, 100 Lewis Hall Please put away all books, calculators, cell phones and other devices. You may consult a single two-sided

More information

SYMMETRIES OF FIBONACCI POINTS, MOD m

SYMMETRIES OF FIBONACCI POINTS, MOD m PATRICK FLANAGAN, MARC S. RENAULT, AND JOSH UPDIKE Abstract. Given a modulus m, we examine the set of all points (F i,f i+) Z m where F is the usual Fibonacci sequence. We graph the set in the fundamental

More information

MATH 13150: Freshman Seminar Unit 15

MATH 13150: Freshman Seminar Unit 15 MATH 1310: Freshman Seminar Unit 1 1. Powers in mod m arithmetic In this chapter, we ll learn an analogous result to Fermat s theorem. Fermat s theorem told us that if p is prime and p does not divide

More information

Is 1 a Square Modulo p? Is 2?

Is 1 a Square Modulo p? Is 2? Chater 21 Is 1 a Square Modulo? Is 2? In the revious chater we took various rimes and looked at the a s that were quadratic residues and the a s that were nonresidues. For examle, we made a table of squares

More information

MA/CSSE 473 Day 9. The algorithm (modified) N 1

MA/CSSE 473 Day 9. The algorithm (modified) N 1 MA/CSSE 473 Day 9 Primality Testing Encryption Intro The algorithm (modified) To test N for primality Pick positive integers a 1, a 2,, a k < N at random For each a i, check for a N 1 i 1 (mod N) Use the

More information

NUMBER THEORY AMIN WITNO

NUMBER THEORY AMIN WITNO NUMBER THEORY AMIN WITNO.. w w w. w i t n o. c o m Number Theory Outlines and Problem Sets Amin Witno Preface These notes are mere outlines for the course Math 313 given at Philadelphia

More information

Numbers (8A) Young Won Lim 5/24/17

Numbers (8A) Young Won Lim 5/24/17 Numbers (8A Copyright (c 2017 Young W. Lim. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version

More information

12. Let Rm = {0,1,2,..., m 1} be a complete residue system modulo ra. Let a be an integer. When is a Rm = {0,1 a, 2 a,...

12. Let Rm = {0,1,2,..., m 1} be a complete residue system modulo ra. Let a be an integer. When is a Rm = {0,1 a, 2 a,... 12. Let Rm = {0,1,2,..., m 1} be a complete residue system modulo ra. Let a be an integer. When is a Rm = {0,1 a, 2 a,..., a (ra - 1)} a complete residue system modulo m? Prove your conjecture. (Try m

More information

Numbers (8A) Young Won Lim 6/21/17

Numbers (8A) Young Won Lim 6/21/17 Numbers (8A Copyright (c 2017 Young W. Lim. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version

More information

2. Nine points are distributed around a circle in such a way that when all ( )

2. Nine points are distributed around a circle in such a way that when all ( ) 1. How many circles in the plane contain at least three of the points (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)? Solution: There are ( ) 9 3 = 8 three element subsets, all

More information

CMPSCI 250: Introduction to Computation. Lecture #14: The Chinese Remainder Theorem David Mix Barrington 4 October 2013

CMPSCI 250: Introduction to Computation. Lecture #14: The Chinese Remainder Theorem David Mix Barrington 4 October 2013 CMPSCI 250: Introduction to Computation Lecture #14: The Chinese Remainder Theorem David Mix Barrington 4 October 2013 The Chinese Remainder Theorem Infinitely Many Primes Reviewing Inverses and the Inverse

More information

Final exam. Question Points Score. Total: 150

Final exam. Question Points Score. Total: 150 MATH 11200/20 Final exam DECEMBER 9, 2016 ALAN CHANG Please present your solutions clearly and in an organized way Answer the questions in the space provided on the question sheets If you run out of room

More information

Congruence. Solving linear congruences. A linear congruence is an expression in the form. ax b (modm)

Congruence. Solving linear congruences. A linear congruence is an expression in the form. ax b (modm) Congruence Solving linear congruences A linear congruence is an expression in the form ax b (modm) a, b integers, m a positive integer, x an integer variable. x is a solution if it makes the congruence

More information

arxiv: v3 [math.co] 4 Dec 2018 MICHAEL CORY

arxiv: v3 [math.co] 4 Dec 2018 MICHAEL CORY CYCLIC PERMUTATIONS AVOIDING PAIRS OF PATTERNS OF LENGTH THREE arxiv:1805.05196v3 [math.co] 4 Dec 2018 MIKLÓS BÓNA MICHAEL CORY Abstract. We enumerate cyclic permutations avoiding two patterns of length

More information

Mark Kozek. December 7, 2010

Mark Kozek. December 7, 2010 : in : Whittier College December 7, 2010 About. : in Hungarian mathematician, 1913-1996. Interested in combinatorics, graph theory, number theory, classical analysis, approximation theory, set theory,

More information

Math 319 Problem Set #7 Solution 18 April 2002

Math 319 Problem Set #7 Solution 18 April 2002 Math 319 Problem Set #7 Solution 18 April 2002 1. ( 2.4, problem 9) Show that if x 2 1 (mod m) and x / ±1 (mod m) then 1 < (x 1, m) < m and 1 < (x + 1, m) < m. Proof: From x 2 1 (mod m) we get m (x 2 1).

More information

Class 8 Cubes and Cube Root

Class 8 Cubes and Cube Root ID : in-8-cubes-and-cube-root [1] Class 8 Cubes and Cube Root For more such worksheets visit www.edugain.com Answer the questions (1) Find the value of A if (2) If you subtract a number x from 15 times

More information

ON THE EQUATION a x x (mod b) Jam Germain

ON THE EQUATION a x x (mod b) Jam Germain ON THE EQUATION a (mod b) Jam Germain Abstract. Recently Jimenez and Yebra [3] constructed, for any given a and b, solutions to the title equation. Moreover they showed how these can be lifted to higher

More information

Algorithmic Number Theory and Cryptography (CS 303)

Algorithmic Number Theory and Cryptography (CS 303) Algorithmic Number Theory and Cryptography (CS 303) Modular Arithmetic and the RSA Public Key Cryptosystem Jeremy R. Johnson 1 Introduction Objective: To understand what a public key cryptosystem is and

More information

Variations on a Theme of Sierpiński

Variations on a Theme of Sierpiński 1 2 3 47 6 23 11 Journal of Integer Sequences, Vol. 10 (2007), Article 07.4.4 Variations on a Theme of Sierpiński Lenny Jones Department of Mathematics Shippensburg University Shippensburg, Pennsylvania

More information

SESAME Modular Arithmetic. MurphyKate Montee. March 2018 IN,Z, We think numbers should satisfy certain rules, which we call axioms:

SESAME Modular Arithmetic. MurphyKate Montee. March 2018 IN,Z, We think numbers should satisfy certain rules, which we call axioms: SESAME Modular Arithmetic MurphyKate Montee March 08 What is a Number? Examples of Number Systems: We think numbers should satisfy certain rules which we call axioms: Commutivity Associativity 3 Existence

More information

Goldbach conjecture (1742, june, the 7 th )

Goldbach conjecture (1742, june, the 7 th ) Goldbach conjecture (1742, june, the 7 th ) We note P the prime numbers set. P = {p 1 = 2, p 2 = 3, p 3 = 5, p 4 = 7, p 5 = 11,...} remark : 1 P Statement : Each even number greater than 2 is the sum of

More information

Discrete Square Root. Çetin Kaya Koç Winter / 11

Discrete Square Root. Çetin Kaya Koç  Winter / 11 Discrete Square Root Çetin Kaya Koç koc@cs.ucsb.edu Çetin Kaya Koç http://koclab.cs.ucsb.edu Winter 2017 1 / 11 Discrete Square Root Problem The discrete square root problem is defined as the computation

More information

Asymptotic Results for the Queen Packing Problem

Asymptotic Results for the Queen Packing Problem Asymptotic Results for the Queen Packing Problem Daniel M. Kane March 13, 2017 1 Introduction A classic chess problem is that of placing 8 queens on a standard board so that no two attack each other. This

More information

AwesomeMath Admission Test A

AwesomeMath Admission Test A 1 (Before beginning, I d like to thank USAMTS for the template, which I modified to get this template) It would be beneficial to assign each square a value, and then make a few equalities. a b 3 c d e

More information

Example Enemy agents are trying to invent a new type of cipher. They decide on the following encryption scheme: Plaintext converts to Ciphertext

Example Enemy agents are trying to invent a new type of cipher. They decide on the following encryption scheme: Plaintext converts to Ciphertext Cryptography Codes Lecture 3: The Times Cipher, Factors, Zero Divisors, and Multiplicative Inverses Spring 2015 Morgan Schreffler Office: POT 902 http://www.ms.uky.edu/~mschreffler New Cipher Times Enemy

More information

MAT 302: ALGEBRAIC CRYPTOGRAPHY. Department of Mathematical and Computational Sciences University of Toronto, Mississauga.

MAT 302: ALGEBRAIC CRYPTOGRAPHY. Department of Mathematical and Computational Sciences University of Toronto, Mississauga. MAT 302: ALGEBRAIC CRYPTOGRAPHY Department of Mathematical and Computational Sciences University of Toronto, Mississauga February 27, 2013 Mid-term Exam INSTRUCTIONS: The duration of the exam is 100 minutes.

More information

Proof that Mersenne Prime Numbers are Infinite and that Even Perfect Numbers are Infinite

Proof that Mersenne Prime Numbers are Infinite and that Even Perfect Numbers are Infinite Proof that Mersenne Prime Numbers are Infinite and that Even Perfect Numbers are Infinite Stephen Marshall 7 November 208 Abstract Mersenne prime is a prime number that is one less than a power of two.

More information

Logarithms ID1050 Quantitative & Qualitative Reasoning

Logarithms ID1050 Quantitative & Qualitative Reasoning Logarithms ID1050 Quantitative & Qualitative Reasoning History and Uses We noticed that when we multiply two numbers that are the same base raised to different exponents, that the result is the base raised

More information

MST125. Essential mathematics 2. Number theory

MST125. Essential mathematics 2. Number theory MST125 Essential mathematics 2 Number theory This publication forms part of the Open University module MST125 Essential mathematics 2. Details of this and other Open University modules can be obtained

More information

MATH 135 Algebra, Solutions to Assignment 7

MATH 135 Algebra, Solutions to Assignment 7 MATH 135 Algebra, Solutions to Assignment 7 1: (a Find the smallest non-negative integer x such that x 41 (mod 9. Solution: The smallest such x is the remainder when 41 is divided by 9. We have 41 = 9

More information

Numbers (8A) Young Won Lim 5/22/17

Numbers (8A) Young Won Lim 5/22/17 Numbers (8A Copyright (c 2017 Young W. Lim. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version

More information