Math 124 Homework 5 Solutions
|
|
- Martin Jordan
- 5 years ago
- Views:
Transcription
1 Math 12 Homework 5 Solutions by Luke Gustafson Fall (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 s theorem, there are infinitely many rimes of the form 3 + 8n, so there are infinitely many rimes with = 2 not a uadratic residue. Now consider an odd rime. Let a be a uadratic non-residue modulo. By Dirichlet s theorem, there exists an odd rime r = a + n for some n N. Thus r a (mod, so r is a uadratic non-residue modulo. If r 1 (mod, consider rimes of the form = r + n. If r 3 (mod, then consider = r n (notice that r + 2 and must be relatively rime. In each case, it is easy to check 1 (mod. By Dirichlet s theorem, there are infinitely many rimes of this form. Then, use uadratic recirocity to get = ( 1 ( 1( 1 Use the fact that 1 (mod and r (mod : r = (1 = 1 That roves there are infinitely many such that is a uadratic non-residue. 2b. From art (a, we know that for odd rimes 1, the condition that α (mod 1, where α = r if r 1 (mod and α = r if r 3 (mod, is sufficient to ensure 1 is a uadratic non-residue modulo. We may slit this congruence into the euivalent system α (mod and α (mod 1. Also shown in art (a is that α 1 (mod, so our system becomes 1 (mod α (mod 1 1
2 In the case that 1 = 2, we will use the condition 5 (mod 8 to guarantee that 2 is a uadratic non-residue modulo. In the case that i = 2 for i > 1, we will use the condition 1 (mod 8 to guarantee that 2 is a uadratic residue modulo. For odd rimes i, consider rimes of the form = 1+ i n, where n N. Then we have i = ( 1 ( 1( i 1 Use the fact that 1 (mod and 1 (mod i : 1 = (1 = 1 Hence, it suffices to imose the restriction 1 (mod i to ensure i is a uadratic residue modulo. Euivalently, 1 (mod and 1 (mod i. Now we combine all these conditions on to get a with the desired roerties. We consider the following system of congruences: (1 1 (mod if i 2 for all i (2 5 (mod 8 if 1 = 2 (3 1 (mod 8 if i = 2, i > 1 ( α (mod 1 (5 1 (mod i for i > 1. Notice that conditions (2 and (3 cannot occur simultaneously since we insist all the i are distinct. We show that all rimes satisfying the above system satisfy the conditions of the roblem. By the receding arguments, if i = 2 for some i, then conditions (2 and (3 guarantee that 2 is aroriately a uadratic residue or non-residue. Observe that conditions (1, (2, and (3 guarantee that 1 (mod. Using this fact lus conditions ( and (5, our receding arguments have shown that 1 is a uadratic non-residue and i is a uadratic residue for i > 2, as desired. Now we show that there are infinitely many rimes satisfying our system of congruences. Since the i are distinct, all the moduli are relatively rime. Thus, we aly the Chinese Remainder Theorem to rewrite the system as i x (mod N where N is the roduct of the moduli and x is the uniue solution. Suose, for sake of contradiction, that x and N are not relatively rime. Then there exists a rime dividing x and N. Since divides N, it is eual to 2 or some i. However, our system of congruences guarantees that x is not divisible by 2 (by conditions (1, (2, and (3 or any of the i (by conditions ( and (5. Therefore, x and N are relatively rime. Using Dirichlet s theorem, i 2
3 there are infinitely many rimes of the form x (mod N, which roves the desired result. 2c. Factor m into rimes a 1 1 a 2 2 ar r. Write a i = 2b i +c i where c i {0, 1}, 1 i r. Then we have m = 2b 1+c 1 1 2b 2+c 2 2 2br+cr r = ( b 1 1 br r 2 c 1 1 cr r Let n = b 1 1 br r. Let 1 2 k = c 1 1 cr r by making the j eual to the i for which c i = 1. The j are distinct because the i are. That gives m in the desired form. 2d. Use art (c to write m in the form n 2 1 k. If m is suare, then it is a uadratic residue for all rimes. If m is not suare, then k 1. By art b, there exists infinitely many ( rimes for which 1 k = 1 2 ( exist infinitely many rimes for which desired. k n 2 1 k = 1. It follows that there = = 1 as n 2 1 k 3a. Let g be a rimitive root modulo. Consider ζ g 1 (mod. Then ζ k g k k (mod, so the order of ζ is the smallest k such that 1 (mod. Letting k < gives an exonent less than 1, which is the smallest ositive exonent for which g is one (since g is a rimitive root. Hence, ζ k 1 (mod when k <. Letting k = gives ζ g 1 1 (mod, so the order of ζ is exactly. g k k 3b. We have ζ 3 1 (mod so (ζ 1(ζ 2 + ζ (mod. Now, ζ 1 (mod because its order is exactly 3. Hence ζ 1 0 (mod, and so ζ 2 + ζ (mod. Therefore, ζ 2 + ζ (mod (2ζ (mod.. First, consider the solutions to x 2 1 (mod a for an odd rime. x is a solution if and only if a x 2 1 a (x 1(x + 1. If x 1, then does not divide x + 1, and vice versa. Hence, either a x 1 or a x + 1. Euivalently, x 1 (mod a or x 1 (mod a. That gives us exactly two solutions to the euation x 2 1 (mod a. Now consider the solutions to x 2 1 (mod 2 a. If a = 1, then x 1 (mod 2 is the only solution. If a = 2, one finds the two solutions x ±1 (mod. Next, consider a 3. x is a solution if and only if 2 a (x 1(x + 1. Hence x 1 and x + 1 are consecutive even numbers. That means one of them is divisible by 2 but not. Then, the other number must be divisible by 2 a 1. So, the only ossible solutions are x 1 0 (mod 2 a 1 and x (mod 2 a 1 ; i.e. x ±1 (mod 2 a 1. That gives us four distinct 3
4 ossibilities modulo 2 a : x ±1 (mod 2 a and x ±1 + 2 a 1 (mod 2 a. Checking, we find that all four of these solutions satisfy x 2 1 (mod 2 a. Hence, there are exactly solutions when a 3. Factor m = a 1 1 a 2 2 a 3 3 ar r. Using the Chinese Remainder Theorem, x 2 1 (mod m if and only if x 2 1 (mod a i i for 1 i r. Moreover, if x is different modulo any i, then x is different modulo m. This establishes an euivalence between the distinct solutions to x 2 1 (mod a i i for each i and the distinct solutions to x 2 1 (mod m. Therefore, the number of solutions to x 2 1 (mod m is r i=1 s i where s i is the number of solutions to x 2 1 (mod a i i. The receding arguments rove that each s i is a ower of two, so it follows that the total number of solutions to x 2 1 (mod m is a ower of two. 5. Define A i to be the ith symmetric sum of the numbers {1, 2,..., 1}; that is, A i = j 1 j 2 j i 1 j 1 <j 2 < <j i 1 The olynomial x 1 1 has solutions 1, 2,..., 1 (mod by Fermat s theorem. Hence, x 1 1 (x 1(x 2 (x +1 (mod. Exanding the right side gives x 1 1 x 1 A 1 x 2 + A 2 x 3 + A 1 (mod (the last term is ositive since 1 is even. Comaring coefficients gives A i 0 (mod for 1 i 2, and A 1 1 (mod. Next, consider the identity (x 1(x 2 (x + 1 = x 1 A 1 x 2 + A 2 x 3 + A 1 Substitute x = to get ( 1! = 1 A A 2 + A 1 Since A 1 = ( 1! it follows that 0 = 1 A A 2 0 = 2 A A 2 We have 2 2 because 5, and i 2 and A i for 1 i 3. Thus, 2 A A 3 is divisible by 2, which imlies 2 A 2. Now substitute x = 2 into the revious identity. We get (2 1 ( + 1 = (2 1 A 1 ( A 2 + A 1 We have 3 (2 1, 2 (2 i 1 and A i for 1 i 3, and 3 2A 2. Hence, 3 divides all terms on the right side excet A 1. Modulo 3, we get (2 1 ( + 1 A 1 (mod 3
5 so We have so finally, ( 2 (2 1 ( + 1 ( 1 1 ( 2 = 1 (mod 3 2(2 1 ( + 1 ( 1 1 (2 1 ( + 1 = 2 ( 1 1 (2 1 ( ( (mod 3 (mod 3 5
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 informationSolutions 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 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 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 informationMTH 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:
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() =
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 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 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 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 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 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 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 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 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 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 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 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 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 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 informationConstructions 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 informationLUCAS-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 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 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 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 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 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 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 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 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 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 informationThe covering congruences of Paul Erdős. Carl Pomerance Dartmouth College
The covering congruences of Paul Erdős Carl Pomerance Dartmouth College Conjecture (Erdős, 1950): For each number B, one can cover Z with finitely many congruences to distinct moduli all > B. Erdős (1995):
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 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 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 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 informationby Michael Filaseta University of South Carolina
by Michael Filaseta University of South Carolina Background: A covering of the integers is a system of congruences x a j (mod m j, j =, 2,..., r, with a j and m j integral and with m j, such that every
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 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 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 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 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 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 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 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 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 informationSQUARING THE MAGIC SQUARES OF ORDER 4
Journal of lgebra Number Theory: dvances and lications Volume 7 Number Pages -6 SQURING THE MGIC SQURES OF ORDER STEFNO BRBERO UMBERTO CERRUTI and NDIR MURRU Deartment of Mathematics University of Turin
More informationMark 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 informationCMPSCI 250: Introduction to Computation. Lecture #14: The Chinese Remainder Theorem David Mix Barrington 24 February 2012
CMPSCI 250: Introduction to Computation Lecture #14: The Chinese Remainder Theorem David Mix Barrington 24 February 2012 The Chinese Remainder Theorem Infinitely Many Primes Reviewing Inverses and the
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 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 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 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 informationThe 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 informationThe Strong Finiteness of Double Mersenne Primes and the Infinity of Root Mersenne Primes and Near-square Primes of Mersenne Primes
The Strong Finiteness of Double Mersenne Primes and the Infinity of Root Mersenne Primes and Near-square Primes of Mersenne Primes Pingyuan Zhou E-mail:zhoupingyuan49@hotmail.com Abstract In this paper
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 information12. 6 jokes are minimal.
Pigeonhole Principle Pigeonhole Principle: When you organize n things into k categories, one of the categories has at least n/k things in it. Proof: If each category had fewer than n/k things in it then
More informationVariations 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 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 informationALGEBRA: 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 informationModular 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 information12. 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 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 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 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 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 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 informationDiscrete Mathematics. Spring 2017
Discrete Mathematics Spring 2017 Previous Lecture Binomial Coefficients Pascal s Triangle The Pigeonhole Principle If a flock of 20 pigeons roosts in a set of 19 pigeonholes, one of the pigeonholes must
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 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 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 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 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 informationSome Fine Combinatorics
Some Fine Combinatorics David P. Little Department of Mathematics Penn State University University Park, PA 16802 Email: dlittle@math.psu.edu August 3, 2009 Dedicated to George Andrews on the occasion
More informationSolutions 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 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 informationNON-OVERLAPPING PERMUTATION PATTERNS. To Doron Zeilberger, for his Sixtieth Birthday
NON-OVERLAPPING PERMUTATION PATTERNS MIKLÓS BÓNA Abstract. We show a way to compute, to a high level of precision, the probability that a randomly selected permutation of length n is nonoverlapping. As
More informationSolutions to Exercises Chapter 6: Latin squares and SDRs
Solutions to Exercises Chapter 6: Latin squares and SDRs 1 Show that the number of n n Latin squares is 1, 2, 12, 576 for n = 1, 2, 3, 4 respectively. (b) Prove that, up to permutations of the rows, columns,
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 informationUplink Scheduling in Wireless Networks with Successive Interference Cancellation
1 Ulink Scheduling in Wireless Networks with Successive Interference Cancellation Majid Ghaderi, Member, IEEE, and Mohsen Mollanoori, Student Member, IEEE, Abstract In this aer, we study the roblem of
More informationcopyright amberpasillas2010 What is Divisibility? Divisibility means that after dividing, there will be No remainder.
What is Divisibility? Divisibility means that after dividing, there will be No remainder. 1 356,821 Can you tell by just looking at this number if it is divisible by 2? by 5? by 10? by 3? by 9? By 6? The
More informationON MODULI FOR WHICH THE FIBONACCI SEQUENCE CONTAINS A COMPLETE SYSTEM OF RESIDUES S. A. BURR Belt Telephone Laboratories, Inc., Whippany, New Jersey
ON MODULI FOR WHICH THE FIBONACCI SEQUENCE CONTAINS A COMPLETE SYSTEM OF RESIDUES S. A. BURR Belt Telephone Laboratories, Inc., Whippany, New Jersey Shah [1] and Bruckner [2] have considered the problem
More informationTilings with T and Skew Tetrominoes
Quercus: Linfield Journal of Undergraduate Research Volume 1 Article 3 10-8-2012 Tilings with T and Skew Tetrominoes Cynthia Lester Linfield College Follow this and additional works at: http://digitalcommons.linfield.edu/quercus
More information(1) 2 x 6. (2) 5 x 8. (3) 9 x 12. (4) 11 x 14. (5) 13 x 18. Soln: Initial quantity of rice is x. After 1st customer, rice available In the Same way
1. A shop stores x kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys
More informationDistribution 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 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 informationCalculators 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 informationPublic 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 informationIn this paper, we discuss strings of 3 s and 7 s, hereby dubbed dreibens. As a first step
Dreibens modulo A New Formula for Primality Testing Arthur Diep-Nguyen In this paper, we discuss strings of s and s, hereby dubbed dreibens. As a first step towards determining whether the set of prime
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 informationRESIDUE NUMBER SYSTEM. (introduction to hardware aspects) Dr. Danila Gorodecky
RESIDUE NUMBER SYSTEM (introduction to hardware asects) Dr. Danila Gorodecky danila.gorodecky@gmail.com Terminology Residue number system (RNS) (refers to Chinese remainder theorem) Residue numeral system
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 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 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 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 informationMultiples 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 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 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 informationPropagation of EM Waves in material media
Propagation of EM Waves in material media S.M.Lea 017 1 Wave propagation As usual, we start with Maxwell s euations with no free charges: =0 =0 = = + If we now assume that each field has the plane wave
More information