MTH 3527 Number Theory Quiz 10 (Some problems that might be on the quiz and some solutions.) 1. Euler φ-function. Desribe all integers n such that:
|
|
- Margery McDonald
- 6 years ago
- Views:
Transcription
1 MTH 7 Number Theory Quiz 10 (Some roblems that might be on the quiz and some solutions.) 1. Euler φ-function. Desribe all integers n such that: (a) φ(n) = Solution: n = 4,, 6 since φ( ) = ( 1) =, φ() = 1 =, φ(6) = φ()φ() = 1 = (b) φ(n) = 4 Solution: n = 8, 1,, 10 since φ( ) = () = 4, φ( ) = = 4, φ() = = 4, φ( ) = 1 ( 1) = 4. (c) φ(n) = 8 Solution: n = 16, 4, 0, 1, 0 (d) φ(n) = 16 Solution: n = 17, 4, 60, 40, 48,, which is the same as: n = 17, 17,,, 4,, (e) φ(n) = (f) φ(n) =. Solution: No such n exists by (m,n,o) arts of this roblem. (g) φ(n) = 9. Solution: No such n exists by (m,n,o) arts of this roblem. (h) φ(n) = 6 Solution: n = 7, 7, 9, 9 (i) φ(n) =. Solution: No such n exists by (m,n,o) arts of this roblem. (j) φ(n) = 10 Solution: n =, (k) φ(n) = 0 Solution: n =,,,, (l) φ(n) = 40 Solution: n = 41, 8,, 0, 1, 88, 7, 10, 100 n = 41, 41,,,,,,, (m) Prove that φ() is even integer for all odd rimes. Solution: If is rime, then φ() = 1. If is odd, then 1 is even. Hence φ() is even. (n) Prove that φ( k ) is even integer for all odd rimes. Solution: If is rime, then φ( k ) = k ( 1). So, if is odd, then 1 is even, hence φ( k ) is even. 1
2 (o) Prove that φ(n) is even integer for all n >. Solution: If a, b are relatively rime integers we have φ(ab) = φ(a) φ(b). Case 1. There is an odd rime q, such that q k aears in the rime factorization of n. Then (q 1) φ(q k ) φ(n). Hence φ(n) is even. Case. n = k. From the assumtion n >, we have that k. So k = φ( k ) () Prove that the only integers n, such that φ(n) = 0 are n = 1, 6. (q) Prove that there are no integers n such that φ(n) = 0.. Roots of olynomials (mod ): Theorem: Let be a rime. If n is a ositive integer, such that n (), then x n 0 (mod) has exactly n solutions with 0 x <. (a) Show that x 1 0(mod 1) has non-congruent roots. Solution: Since 0, we may aly the above theorem. So there are solutions with 0 x < 0. Since all integers {0, 1,..., 0} are non-congruent modulo 1, these solutions are non-congruent roots modulo 1. (b) Show that x 1 0(mod ) does not have non-congruent roots. It has only one root. Solution: 0 0(mod ), 1 1(mod ), (mod ), (mod ), 4 4(mod ). Therefore x = 1 is the only root for x 1 0(mod ) with 0 x < 4. (c) Show that x + x + 1 has roots (mod 1). Solution: x 1 0(mod 1) has non-congruent roots, since 0. x 1 0(mod 1) has 1 (non-congruent) root since 1 0. x 1 = (x 1)(x + x + 1) 0(mod 1) has non-congruent roots, x 1 has one root, imlies that (x + x + 1) 0(mod 1) has roots. (d) Show that x + x + 1 0(mod ) does not have any roots. Solution: (mod ), (mod ), (mod ), (mod ), (mod ). Therefore no x is a root for x + x + 1 0(mod ). (e) Show that x 1 has non-congruent roots (mod 1). (f) Show that x 1 has non-congruent roots (mod ). (g) Show that x 1 has non-congruent roots (mod 101). (h) Show that x 0 1 has 0 non-congruent roots (mod 101). Solution: Since 0 (101 1), we may aly the above theorem and get 0 non-congruent solutions. (i) Show that x 4 + x + x + x + 1 has 4 roots (mod 1). Solution: x 1 0(mod 1) has non-congruent roots, since 0. x 1 0(mod 1) has 1 (non-congruent) root since 1 0. x 1 = (x 1)(x 4 + x + x + x + 1) 0(mod 1) has non-congruent roots, x 1 has one root, imlies that (x 4 + x + x + x + 1) 0(mod 1) has 4 roots.
3 (j) Show that x 4 + x + x + x + 1 has 4 roots (mod ). (k) Show that x 4 + x + x + x + 1 has 4 roots (mod 101). (l) Show that x 10 1 has 10 roots (mod ). (m) Show that x + 1 has roots (mod ). Solution: x (mod ) has 10 non-congruent roots, since 10 ( 1). x 1 0(mod ) has (non-congruent) roots since 10. x 10 1 = (x 1)(x + 1) 0(mod ) has 10 non-congruent roots, x 1 has roots, imlies that (x + 1) 0(mod ) has roots. (n) Show that x 8 + x 6 + x 4 + x + 1 has 8 roots (mod ). Solution: Use the following factorization, and arguments similar to the revious roblems. x 10 1 = (x ) 1 = (x 1)((x ) 4 + (x ) + (x ) + x + 1) 0(mod ). Most of you did not state the Rabin-Miller test correctly. So here are some roblems to clarify how the test works (or doesn t). (a) State Rabin-Miller test. (b) Use a = as a Rabin-Miller witness that 1 is comosite. (c) Use a = as a Rabin-Miller witness that 1 is comosite. (d) Use a = 4 as a Rabin-Miller witness that 1 is comosite. (e) Use a = as a Rabin-Miller witness that 7 is comosite. (f) Use a = as a Rabin-Miller witness that 7 is comosite. (g) Use a = as a Rabin-Miller witness that 81 is comosite. (h) Can you use a = as a Rabin-Miller witness that 0 is comosite? Exlain. (i) Can you use a = as a Rabin-Miller witness that 0 is comosite? Exlain. (j) Use a = in Rabin-Miller test to check if 7 is comosite. (k) Use a = in Rabin-Miller test to check if 7 is comosite. (l) What haens if you use a = in Rabin-Miller test to check if 101 is comosite? (m) What haens if you use a = in Rabin-Miller test to check if 101 is comosite? a 4. Legendre symbol:. Find the following: (a) Solution: = since = when the rime (mod 4).
4 (b) = Solution: = 1 since = 1 when the rime 1(mod 4). 1 1 (c) = Solution: = 1 since = 1 when the rime 1, 7(mod 8) (d) = Solution: = since = when the rime, (mod 8) 1 1 (e) 101 (f) = Solution: = = ()() = 1, since the rime (mod 4) and (mod 8) (g) (h) (i) (j) (k) (l) = Solution: = 1 9 = 10 = = 10 1 = since (mod 4) and 1 (mod 4) = since 1 1(mod ). 1 = 1 since 1 is quadratic residue (mod), which is true since 1 1 (mod). 4
5 . For each of the following congruences determine if it has a solution: (a) x (mod 987). (987 is rime.) Solution: The congruence x (mod 987) has a solution iff is a square modulo 987 (which is rime) iff is quadratic residue modulo 987 iff = 1 iff (mod4). Since 987 (mod4) it follows that is not a square modulo rime 987 and therefore there is no solution to x (mod 987). (b) x (mod 987). (987 is rime.) (c) x (mod 987). (987 is rime.) (d) x (mod 987). (987 is rime.) (e) x 10(mod 987). (987 is rime.) 6. For each of the following integers determine if it is a quadratic residue modulo, i.e. QR : (a) (mod 887). (887 is rime.) (b) (mod 887). (887 is rime.) (c) (mod 16). (16 is rime.) (d) (mod 167). (167 is rime.) (e) 10(mod 8). (8 is rime.) Solution: 10 is QR 8 iff the Legendre symbol is = Comute = = () = () = () = () = ()() = 1. Therefore 10 is QR 8.
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 informationExam 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 informationMATH 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 informationIs 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 informationQuadratic 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 informationFoundations of Cryptography
Foundations of Cryptography Ville Junnila viljun@utu.fi Department of Mathematics and Statistics University of Turku 2015 Ville Junnila viljun@utu.fi Lecture 10 1 of 17 The order of a number (mod n) Definition
More informationMath 124 Homework 5 Solutions
Math 12 Homework 5 Solutions by Luke Gustafson Fall 2003 1. 163 1 2 (mod 2 gives = 2 the smallest rime. 2a. First, consider = 2. We know 2 is not a uadratic residue if and only if 3, 5 (mod 8. By Dirichlet
More informationb) 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 informationTo 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 informationIntroduction to Number Theory 2. c Eli Biham - November 5, Introduction to Number Theory 2 (12)
Introduction to Number Theory c Eli Biham - November 5, 006 345 Introduction to Number Theory (1) Quadratic Residues Definition: The numbers 0, 1,,...,(n 1) mod n, are called uadratic residues modulo n.
More informationUNIVERSITY OF MANITOBA DATE: December 7, FINAL EXAMINATION TITLE PAGE TIME: 3 hours EXAMINER: M. Davidson
TITLE PAGE FAMILY NAME: (Print in ink) GIVEN NAME(S): (Print in ink) STUDENT NUMBER: SEAT NUMBER: SIGNATURE: (in ink) (I understand that cheating is a serious offense) INSTRUCTIONS TO STUDENTS: This is
More informationSOLUTIONS 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 informationMT 430 Intro to Number Theory MIDTERM 2 PRACTICE
MT 40 Intro to Number Theory MIDTERM 2 PRACTICE Material covered Midterm 2 is comrehensive but will focus on the material of all the lectures from February 9 u to Aril 4 Please review the following toics
More information6. 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 informationCollection 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 informationSolutions 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 informationDiscrete 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 informationApplications 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 informationSolutions 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 informationx 8 (mod 15) x 8 3 (mod 5) eli 2 2y 6 (mod 10) y 3 (mod 5) 6x 9 (mod 11) y 3 (mod 11) So y = 3z + 3u + 3w (mod 990) z = (990/9) (990/9) 1
Exercise help set 6/2011 Number Theory 1. x 2 0 (mod 2) x 2 (mod 6) x 2 (mod 3) a) x 5 (mod 7) x 5 (mod 7) x 8 (mod 15) x 8 3 (mod 5) (x 8 2 (mod 3)) So x 0y + 2z + 5w + 8u (mod 210). y is not needed.
More informationThe Chinese Remainder Theorem
The Chinese Remainder Theorem Theorem. Let m and n be two relatively prime positive integers. Let a and b be any two integers. Then the two congruences x a (mod m) x b (mod n) have common solutions. Any
More informationThe Chinese Remainder Theorem
The Chinese Remainder Theorem Theorem. Let n 1,..., n r be r positive integers relatively prime in pairs. (That is, gcd(n i, n j ) = 1 whenever 1 i < j r.) Let a 1,..., a r be any r integers. Then the
More informationLECTURE 9: QUADRATIC RESIDUES AND THE LAW OF QUADRATIC RECIPROCITY
LECTURE 9: QUADRATIC RESIDUES AND THE LAW OF QUADRATIC RECIPROCITY 1. Bsic roerties of qudrtic residues We now investigte residues with secil roerties of lgebric tye. Definition 1.1. (i) When (, m) 1 nd
More informationOn the Fibonacci Sequence. By: Syrous Marivani LSUA. Mathematics Department. Alexandria, LA 71302
On the Fibonacci Sequence By: Syrous Marivani LSUA Mathematics Deartment Alexandria, LA 70 The so-called Fibonacci sequence {(n)} n 0 given by: (n) = (n ) + (n ), () where (0) = 0, and () =. The ollowing
More informationMATH 118 PROBLEM SET 6
MATH 118 PROBLEM SET 6 WASEEM LUTFI, GABRIEL MATSON, AND AMY PIRCHER Section 1 #16: Show tht if is qudrtic residue modulo m, nd b 1 (mod m, then b is lso qudrtic residue Then rove tht the roduct of the
More informationAssignment 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 informationMath 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 informationNUMBER 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 informationDegree project NUMBER OF PERIODIC POINTS OF CONGRUENTIAL MONOMIAL DYNAMICAL SYSTEMS
Degree project NUMBER OF PERIODIC POINTS OF CONGRUENTIAL MONOMIAL DYNAMICAL SYSTEMS Author: MD.HASIRUL ISLAM NAZIR BASHIR Supervisor: MARCUS NILSSON Date: 2012-06-15 Subject: Mathematics and Modeling Level:
More informationUniversity 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 informationPrimitive 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 informationCarmen 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 informationMA/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 informationON 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 informationNumber 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 informationMAT 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 informationCalculators 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 informationNumber 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 informationSOLUTIONS 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 informationp 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 informationMath 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 informationSIZE OF THE SET OF RESIDUES OF INTEGER POWERS OF FIXED EXPONENT
SIZE OF THE SET OF RESIDUES OF INTEGER POWERS OF FIXED EXPONENT RICHARD J. MATHAR Abstract. The ositive integers corime to some integer m generate the abelian grou (Z/nZ) of multilication modulo m. Admitting
More informationCryptography CS 555. Topic 20: Other Public Key Encryption Schemes. CS555 Topic 20 1
Cryptography CS 555 Topic 20: Other Public Key Encryption Schemes Topic 20 1 Outline and Readings Outline Quadratic Residue Rabin encryption Goldwasser-Micali Commutative encryption Homomorphic encryption
More informationTHE NUMBER OF PERMUTATIONS WHICH FORM ARITHMETIC PROGRESSIONS MODULO m
ANALELE ŞTIINŢIFICE ALE UNIVERSITĂŢII AL.I. CUZA DIN IAŞI (S.N.) MATEMATICĂ, Tomul LXI, 2015, f.2 THE NUMBER OF PERMUTATIONS WHICH FORM ARITHMETIC PROGRESSIONS MODULO m BY FLORIAN LUCA and AUGUSTINE O.
More informationLECTURE 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 informationData security (Cryptography) exercise book
University of Debrecen Faculty of Informatics Data security (Cryptography) exercise book 1 Contents 1 RSA 4 1.1 RSA in general.................................. 4 1.2 RSA background.................................
More informationDUBLIN CITY UNIVERSITY
DUBLIN CITY UNIVERSITY SEMESTER ONE EXAMINATIONS 2013 MODULE: (Title & Code) CA642 Cryptography and Number Theory COURSE: M.Sc. in Security and Forensic Computing YEAR: 1 EXAMINERS: (Including Telephone
More information1.6 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 informationCHAPTER 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 informationConjectures and Results on Super Congruences
Conjectures and Results on Suer Congruences Zhi-Wei Sun Nanjing University Nanjing 210093, P. R. China zwsun@nju.edu.cn htt://math.nju.edu.cn/ zwsun Feb. 8, 2010 Part A. Previous Wor by Others What are
More informationA REMARK ON A PAPER OF LUCA AND WALSH 1. Zhao-Jun Li Department of Mathematics, Anhui Normal University, Wuhu, China. Min Tang 2.
#A40 INTEGERS 11 (2011) A REMARK ON A PAPER OF LUCA AND WALSH 1 Zhao-Jun Li Department of Mathematics, Anhui Normal University, Wuhu, China Min Tang 2 Department of Mathematics, Anhui Normal University,
More informationPT. Primarity Tests Given an natural number n, we want to determine if n is a prime number.
PT. Primarity Tests Given an natural number n, we want to determine if n is a prime number. (PT.1) If a number m of the form m = 2 n 1, where n N, is a Mersenne number. If a Mersenne number m is also a
More informationPractice Midterm 2 Solutions
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
More informationL29&30 - RSA Cryptography
L29&30 - RSA Cryptography CSci/Math 2112 20&22 July 2015 1 / 13 Notation We write a mod n for the integer b such that 0 b < n and a b (mod n). 2 / 13 Calculating Large Powers Modulo n Example 1 What is
More informationModular 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 informationTwo 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 informationWilson 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 informationLECTURE 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 informationMath 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 informationDiscrete 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 informationMathematics 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 informationOutline Introduction Big Problems that Brun s Sieve Attacks Conclusions. Brun s Sieve. Joe Fields. November 8, 2007
Big Problems that Attacks November 8, 2007 Big Problems that Attacks The Sieve of Eratosthenes The Chinese Remainder Theorem picture Big Problems that Attacks Big Problems that Attacks Eratosthene s Sieve
More informationDiffie-Hellman key-exchange protocol
Diffie-Hellman key-exchange protocol This protocol allows two users to choose a common secret key, for DES or AES, say, while communicating over an insecure channel (with eavesdroppers). The two users
More informationProblem Set 6 Solutions Math 158, Fall 2016
All exercise numbers from the textbook refer to the second edition. 1. (a) Textbook exercise 3.3 (this shows, as we mentioned in class, that RSA decryption always works when the modulus is a product of
More information6.2 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 informationSolutions 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#A3 INTEGERS 17 (2017) A NEW CONSTRAINT ON PERFECT CUBOIDS. Thomas A. Plick
#A3 INTEGERS 17 (2017) A NEW CONSTRAINT ON PERFECT CUBOIDS Thomas A. Plick tomplick@gmail.com Received: 10/5/14, Revised: 9/17/16, Accepted: 1/23/17, Published: 2/13/17 Abstract We show that out of the
More informationThe 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 information30 HWASIN PARK, JOONGSOO PARK AND DAEYEOUL KIM Lemma 1.1. Let =2 k q +1, k 2 Z +. Then the set of rimitive roots modulo is the set of quadratic non-re
J. KSIAM Vol.4, No.1, 29-38, 2000 A CRITERION ON PRIMITIVE ROOTS MODULO Hwasin Park, Joongsoo Park and Daeyeoul Kim Abstract. In this aer, we consider a criterion on rimitive roots modulo where is the
More informationImplementation / Programming: Random Number Generation
Introduction to Modeling and Simulation Implementation / Programming: Random Number Generation OSMAN BALCI Professor Department of Computer Science Virginia Polytechnic Institute and State University (Virginia
More informationIntroduction 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 informationQ(173)Q(177)Q(188)Q(193)Q(203)
MATH 313: SOLUTIONS HW3 Problem 1 (a) 30941 We use the Miller-Rabin test to check if it prime. We know that the smallest number which is a strong pseudoprime both base 2 and base 3 is 1373653; hence, if
More informationNumber 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 informationEE 418: Network Security and Cryptography
EE 418: Network Security and Cryptography Homework 3 Solutions Assigned: Wednesday, November 2, 2016, Due: Thursday, November 10, 2016 Instructor: Tamara Bonaci Department of Electrical Engineering University
More informationIntroduction. 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 informationFermat 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 informationCMPSCI 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 informationPRIMES IN SHIFTED SUMS OF LUCAS SEQUENCES. Lenny Jones Department of Mathematics, Shippensburg University, Shippensburg, Pennsylvania
#A52 INTEGERS 17 (2017) PRIMES IN SHIFTED SUMS OF LUCAS SEQUENCES Lenny Jones Department of Mathematics, Shippensburg University, Shippensburg, Pennsylvania lkjone@ship.edu Lawrence Somer Department of
More informationMODULAR 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 informationCongruence properties of the binary partition function
Congruence properties of the binary partition function 1. Introduction. We denote by b(n) the number of binary partitions of n, that is the number of partitions of n as the sum of powers of 2. As usual,
More informationPublic-Key Cryptosystem Based on Composite Degree Residuosity Classes. Paillier Cryptosystem. Harmeet Singh
Public-Key Cryptosystem Based on Composite Degree Residuosity Classes aka Paillier Cryptosystem Harmeet Singh Harmeet Singh Winter 2018 1 / 26 Background s Background Foundation of public-key encryption
More informationZhanjiang , People s Republic of China
Math. Comp. 78(2009), no. 267, 1853 1866. COVERS OF THE INTEGERS WITH ODD MODULI AND THEIR APPLICATIONS TO THE FORMS x m 2 n AND x 2 F 3n /2 Ke-Jian Wu 1 and Zhi-Wei Sun 2, 1 Department of Mathematics,
More informationRational 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 informationExample: Modulo 11: Since Z p is cyclic, there is a generator. Let g be a generator of Z p.
Qudrtic Residues Defiitio: The umbers 0, 1,,, ( mod, re clled udrtic residues modulo Numbers which re ot udrtic residues modulo re clled udrtic o-residues modulo Exmle: Modulo 11: Itroductio to Number
More informationGoldbach 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 informationPublic 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 informationON SPLITTING UP PILES OF STONES
ON SPLITTING UP PILES OF STONES GREGORY IGUSA Abstract. In this paper, I describe the rules of a game, and give a complete description of when the game can be won, and when it cannot be won. The first
More informationComputational Complexity of Generalized Push Fight
Comutational Comlexity of Generalized Push Fight Jeffrey Bosboom Erik D. Demaine Mikhail Rudoy Abstract We analyze the comutational comlexity of otimally laying the two-layer board game Push Fight, generalized
More informationAlgorithmic 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 informationELEMENTS OF NUMBER THEORY & CONGRUENCES. Lagrange, Legendre and Gauss. Mth Mathematicst
ELEMENTS OF NUMBER THEORY & CONGRUENCES Lagrange, Legendre and Gauss ELEMENTS OF NUMBER THEORY & CONGRUENCES 1) If a 0, b 0 Z and a/b, b/a then 1) a=b 2) a=1 3) b=1 4) a=±b Ans : is 4 known result. If
More informationMassachusetts Institute of Technology 6.042J/18.062J, Spring 04: Mathematics for Computer Science April 16 Prof. Albert R. Meyer and Dr.
Massachusetts Institute of Technology 6.042J/18.062J, Spring 04: Mathematics for Computer Science April 16 Prof. Albert R. Meyer and Dr. Eric Lehman revised April 16, 2004, 202 minutes Solutions to Quiz
More informationCaltech Harvey Mudd Mathematics Competition February 20, 2010
Mixer Round Solutions Caltech Harvey Mudd Mathematics Competition February 0, 00. (Ying-Ying Tran) Compute x such that 009 00 x (mod 0) and 0 x < 0. Solution: We can chec that 0 is prime. By Fermat s Little
More informationAn 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 informationA4M33PAL, ZS , FEL ČVUT
Pseudorandom numbers John von Neumann: Any one who considers arithmetical methods of producing random digits is, of course, in a state of sin. For, as has been pointed out several times, there is no such
More informationPublic-key Cryptography: Theory and Practice
Public-key Cryptography Theory and Practice Department of Computer Science and Engineering Indian Institute of Technology Kharagpur Chapter 5: Cryptographic Algorithms Common Encryption Algorithms RSA
More informationAn 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 informationRamanujan-type Congruences for Overpartitions Modulo 5. Nankai University, Tianjin , P. R. China
Ramanujan-type Congruences for Overpartitions Modulo 5 William Y.C. Chen a,b, Lisa H. Sun a,, Rong-Hua Wang a and Li Zhang a a Center for Combinatorics, LPMC-TJKLC Nankai University, Tianjin 300071, P.
More informationMath 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 informationand 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 informationGoldbach 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