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

Size: px
Start display at page:

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

Transcription

1 Quadratic Residues 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 of the congruence x = q (mod ), where and q are distinct odd rimes. If is an odd rime, there are equal numbers of quadratic residues and quadratic nonresidues among {1,,..., 1}. If is an odd rime, a > 0, and (a,) = 1, the Legendre symbol a is defined by a { 1 if a is a quadratic residue mod = 1 if a is a quadratic nonresidue mod. Legendre symbols rovide a comutational tool for determining whether a quadratic congruence has a solution. Euler s theorem says that if is an odd rime, a > 0, and (a,) = 1, then a = a ( 1)/ (mod ). Gauss considered the roofs he gave of quadratic recirocity one of his crowning achievements; in fact, he gave 6 distinct roofs during his lifetime. Recirocity is a dee result: Proofs eluded both Euler and Legendre. The recirocitylaw is simle to state. Forand q odd rimes, it relatessolutions to the two congruences x = (mod q) and x = q (mod ). (Note how and q switch laces: This exlains why it s called a recirocity law.) The law of quadratic recirocity says: The congruences are either both solvable or both unsolvable, unless both rimes are congruent to 3 mod 4. In that case, one is solvable while the other is not. Gauss first gave a roof of this when he was 19! Gauss s masterwork, the Disquisitiones Arithmeticae, was ublished in 1801 when Gauss was 4. It changed the course of number theory, collecting scattered results into a unified theory. Definition. Let (a,m) = 1, m > 0. a is a quadratic residue mod m if the following equation has a solution: x = a (mod m). Otherwise, a is a quadratic nonresidue mod m. Examle. 8 is a quadratic residue mod 17, since 5 = 8 (mod 17). 1

2 However, 8 is a quadratic nonresidue mod 11, because x = 8 (mod 11) has no solutions. n n (mod 11) As the table shows, 1, 3, 4, 5, and 9 are quadratic residues mod 11. (0 is not considered a quadratic residue, since (0,11) = 11 1.) But 8 is a quadratic nonresidue mod 11. Notice the symmetry in the nonzero elements of the table. Do you see why this is haening? Lemma. Let be an odd rime. The congruence has: (a) Only the solution x = 0 if a = 0. (b) Exactly 0 or solutions if a. x = a (mod ) Proof. x = 0 solves x = 0 (mod ). Conversely, if x = 0 (mod ), then x, so x, and hence x = 0 (mod ). Suose a. To show there are 0 or solutions, suose there is at least one solution b. Then b = a (mod ), so ( b) = a (mod ). I claim that b and b are distinct. If not, then b = b (mod ), so b. is an odd rime, so. Therefore, b, b = 0 (mod ), b = 0 (mod ), and finally a = 0 (mod ) contradicting a. Hence, b b (mod ). Now I have two distinct solutions; since a quadratic equation mod has at most two solutions (Prove it!), there are exactly two. Examle. x = 8 (mod 17) has 5 and 1 as solutions, and 5 = 1 (mod 17). But note that the result is false if = : x = 1 (mod ) has exactly one solution (x = 1 (mod )). Corollary. Let be an odd rime. There are 1 in {1,..., 1}. quadratic residues and 1 quadratic nonresidues mod Proof. k and k = k have the same square mod. That is, 1 and 1 have the same square, and have the same square,..., and 1 and 1 +1 have the same square. Thus, the number of different squares is 1 these squares are the quadratic residues, and the other 1 numbers in {1,,..., 1} are quadratic nonresidues. The fact observed in the first sentence of the roof exlains the symmetries in the table of squares mod 11 and mod 7 that I gave above. Definition. Let be an odd rime, and let (a,) = 1. The Legendre symbol a is defined by a { 1 if a is a quadratic residue mod = 1 if a is a quadratic nonresidue mod Note that a = 0 is disallowed (since (0,) = 1) even though x = 0 (mod ) has a solution.

3 Examle. (5,11) = 1. 5 = 1, since 4 = 5 (mod 11). Likewise, 11 = 1, since 6 = 11 (mod 5) Note that 5 is congruent to 1 mod 4; as redicted by recirocity, both of the following the congruences have solutions: x = 5 (mod 11) and x = 11 (mod 5). You might wonder about the case where =, or the case where the modulus is comosite. For =, there are only two quadratic congruences: x = 0 (mod ) and x = 1 (mod ). These have the solutions x = 0 (mod ) and x = 1 (mod ) nothing much is going on. If the modulus has rime factorization n = r1 1 r k k, then relative rimality imlies that it s enough to solve the congruences x = a (mod ri i ) for each i. It turns out that solving such a congruence reduces to determining whether a is a quadratic residue mod i. Therefore, there is little harm in concentrating on the case of a single rime. Examle. Solve the congruence I ll solve the congruences x = 79 (mod 91). x = 79 (mod 7) and x = 79 (mod 13). x = 79 (mod 7) reduces to x = (mod 7). Making a table of squares mod 7, I find that the solutions are x = 3 and x = 4 mod 7. x = 79 (mod 13) reduces to x = 1 (mod 13). The solutions are x = 1 and x = 1 = 1 mod 13. I ll consider the = 4 ossibilities, solving using the Chinese Remainder Theorem. But note that since m = ( m), the solutions will come in airs. So once I find a solution m, I know that m is also a solution. Consider x = 3 (mod 7) x = 1 (mod 13) m s = 1 (mod m) 6 a 3 1 x = = 48 = 66 (mod 91). Then x = 66 = 5 (mod 91) is another solution. Consider x = 3 (mod 7) x = 1 (mod 13) m s = 1 (mod m) 6 a 3 1 3

4 x = = 40 = 38 (mod 91). Then x = 38 = 53 (mod 91) is another solution. It s ossible that the second comutation might have given me 5, the solution I got earlier. In that case, I d have to move on to one of the other two cases. I got lucky and had to only do two cases, instead of three. Here are some tools for comuting Legendre symbols. Theorem. (Euler) Let be an odd rime, a > 0, (a,) = 1. Then a = a ( 1)/ (mod ). Proof. There are two cases. Suose that a = 1. Then there is a number b such that b = a (mod ). So (b ) ( 1)/ = a ( 1)/ (mod ) b 1 = a ( 1)/ (mod ) If b, then b = a, a contradiction. So b, and Fermat s theorem imlies that b 1 = 1 (mod ). So a ( 1)/ = 1 (mod ), and a = a ( 1)/ (mod ). The other ossibility is a = 1. In this case, consider the set {1,,..., 1}. I claim that these integers occur in airs s, t, such that st = a. First, if s {1,,..., 1}, then s is invertible mod. So I can write s(s 1 a) = a, and the air s, s 1 a, multilies to a. Moreover, s and s 1 a are distinct. If not, s = s 1 a, or s = a, which contradicts a = 1. Since the integers {1,,..., 1} divide u into airs, each multilying to a, and since there are 1 airs, I have 1 ( 1) = a ( 1)/ (mod ). By Wilson s theorem, 1 = a ( 1)/ (mod ) a = a ( 1)/ (mod ) Examle. Suose = 13 and a = 10. Then a ( 1)/ = 10 6 = 1 (mod 13). Hence, 10 = 1, and x = 10 (mod 13) should have a solution. Indeed, 13 7 = 49 = 10 (mod 13). 4

5 Lemma. If a = b (mod ), then a = b Proof. If a = b (mod ), then x = a (mod ) if and only if x = b (mod ). Thus, one of these equations is solvable or not solvable if and only if the same is true for the other which means a = b Note that I can use this result to aly Euler s formula to a for a < 0 by simly relacing a with b > 0 such that a = b (mod ). Lemma. Let be an odd rime, a,b > 0, (a,) = (b,) = 1. Then a b = ab Proof. By Euler, a b = a ( 1)/ b ( 1)/ (mod ), and ab = (ab) ( 1)/ (mod ). Therefore, a b = ab (mod ). The two sides of this equation are ±1. Since is an odd rime, the two sides can t differ by. Hence, they must be equal as integers: a b = ab Corollary. Let be an odd rime, a > 0, (a,) = 1. Then a = 1. You can use the results above to comute a for secific values of a and arbitrary. Lemma. { 1 1 if = 4k+1 = 1 if = 4k + 3. Proof. By Euler s formula, 1 = 1 = ( 1) ( 1)/ = ( 1) ( 1)/ = { { ( 1) k if = 4k +1 1 if = 4k +1 ( 1) k+1 if = 4k + 3 = 1 if = 4k + 3. Using Gauss s lemma, which I ll rove shortly, you can also show that = ( 1) ( 1)/8. Note that the exonent on the right is actually an integer: Since = k +1, 1 = 4k(k +1). And 4k(k +1) is divisible by 8, because one of k, k +1, must be even. 5

6 Examle. 1 = 1, because 13 = Thus, x = 1 (mod 13) has solutions. And in fact, 13 5 = 5 = 1 = 1 (mod 13). Likewise, 1 = 1, because 3 = Hence, x = 1 (mod 3) has no solutions. 3 Finally, = ( 1) (7 1)/8 = 1. 7 Therefore, x = (mod 7) has solutions. x = 3 works, for instance. c 015 by Bruce Ikenaga 6

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

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

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

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

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:

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: 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 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

MT 430 Intro to Number Theory MIDTERM 2 PRACTICE

MT 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 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

On the Fibonacci Sequence. By: Syrous Marivani LSUA. Mathematics Department. Alexandria, LA 71302

On 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 information

Introduction to Number Theory 2. c Eli Biham - November 5, Introduction to Number Theory 2 (12)

Introduction 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 information

Math 124 Homework 5 Solutions

Math 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 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

30 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

30 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 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

Practice Midterm 2 Solutions

Practice 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 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

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

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

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

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

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

LECTURE 9: QUADRATIC RESIDUES AND THE LAW OF QUADRATIC RECIPROCITY

LECTURE 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 information

UNIVERSITY OF MANITOBA DATE: December 7, FINAL EXAMINATION TITLE PAGE TIME: 3 hours EXAMINER: M. Davidson

UNIVERSITY 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 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

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

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

The Chinese Remainder Theorem

The 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 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

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

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

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

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

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

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

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

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

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

#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 #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 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

MATH 118 PROBLEM SET 6

MATH 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 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

SIZE OF THE SET OF RESIDUES OF INTEGER POWERS OF FIXED EXPONENT

SIZE 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 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

Computational Complexity of Generalized Push Fight

Computational 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 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

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

ON SPLITTING UP PILES OF STONES

ON 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 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

Conjectures and Results on Super Congruences

Conjectures 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 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

A REMARK ON A PAPER OF LUCA AND WALSH 1. Zhao-Jun Li Department of Mathematics, Anhui Normal University, Wuhu, China. Min Tang 2.

A 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 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

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

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

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

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

The Chinese Remainder Theorem

The 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 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

Degree project NUMBER OF PERIODIC POINTS OF CONGRUENTIAL MONOMIAL DYNAMICAL SYSTEMS

Degree 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 information

The covering congruences of Paul Erdős. Carl Pomerance Dartmouth College

The 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 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

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

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

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

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

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

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

Zhanjiang , People s Republic of China

Zhanjiang , 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 information

SQUARING THE MAGIC SQUARES OF ORDER 4

SQUARING 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 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

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

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

Arithmetic of Remainders (Congruences)

Arithmetic of Remainders (Congruences) Arithmetic of Remainders (Congruences) Donald Rideout, Memorial University of Newfoundland 1 Divisibility is a fundamental concept of number theory and is one of the concepts that sets it apart from other

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

x 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

x 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 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

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

Congruence properties of the binary partition function

Congruence 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 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

PRIMES IN SHIFTED SUMS OF LUCAS SEQUENCES. Lenny Jones Department of Mathematics, Shippensburg University, Shippensburg, Pennsylvania

PRIMES 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 information

Ramanujan-type Congruences for Overpartitions Modulo 5. Nankai University, Tianjin , P. R. China

Ramanujan-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 information

Outline Introduction Big Problems that Brun s Sieve Attacks Conclusions. Brun s Sieve. Joe Fields. November 8, 2007

Outline 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 information

Minimal tilings of a unit square

Minimal tilings of a unit square arxiv:1607.00660v1 [math.mg] 3 Jul 2016 Minimal tilings of a unit square Iwan Praton Franklin & Marshall College Lancaster, PA 17604 Abstract Tile the unit square with n small squares. We determine the

More information

Arithmetic Properties of Combinatorial Quantities

Arithmetic Properties of Combinatorial Quantities A tal given at the National Center for Theoretical Sciences (Hsinchu, Taiwan; August 4, 2010 Arithmetic Properties of Combinatorial Quantities Zhi-Wei Sun Nanjing University Nanjing 210093, P. R. China

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

ON 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 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 information

Foundations of Cryptography

Foundations 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 information

Permutation group and determinants. (Dated: September 19, 2018)

Permutation group and determinants. (Dated: September 19, 2018) Permutation group and determinants (Dated: September 19, 2018) 1 I. SYMMETRIES OF MANY-PARTICLE FUNCTIONS Since electrons are fermions, the electronic wave functions have to be antisymmetric. This chapter

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

Solutions to Exercises Chapter 6: Latin squares and SDRs

Solutions 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 information

by Michael Filaseta University of South Carolina

by 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 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

Computational Complexity of Generalized Push Fight

Computational Complexity of Generalized Push Fight Comutational Comlexity of Generalized Push Fight Jeffrey Bosboom MIT CSAIL, 32 Vassar Street, Cambridge, MA 2139, USA jbosboom@csail.mit.edu Erik D. Demaine MIT CSAIL, 32 Vassar Street, Cambridge, MA 2139,

More information

A Quick Introduction to Modular Arithmetic

A Quick Introduction to Modular Arithmetic A Quick Introduction to Modular Arithmetic Art Duval University of Texas at El Paso November 16, 2004 1 Idea Here are a few quick motivations for modular arithmetic: 1.1 Sorting integers Recall how you

More information

BAND SURGERY ON KNOTS AND LINKS, III

BAND SURGERY ON KNOTS AND LINKS, III BAND SURGERY ON KNOTS AND LINKS, III TAIZO KANENOBU Abstract. We give two criteria of links concerning a band surgery: The first one is a condition on the determinants of links which are related by a band

More information

How to Become a Mathemagician: Mental Calculations and Math Magic

How to Become a Mathemagician: Mental Calculations and Math Magic How to Become a Mathemagician: Mental Calculations and Math Magic Adam Gleitman (amgleit@mit.edu) Splash 2012 A mathematician is a conjurer who gives away his secrets. John H. Conway This document describes

More information

THE NUMBER OF PERMUTATIONS WHICH FORM ARITHMETIC PROGRESSIONS MODULO m

THE 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 information

A Study of Relationship Among Goldbach Conjecture, Twin prime and Fibonacci number

A Study of Relationship Among Goldbach Conjecture, Twin prime and Fibonacci number A Study of Relationship Among Goldbach Conjecture, Twin and Fibonacci number Chenglian Liu Department of Computer Science, Huizhou University, China chenglianliu@gmailcom May 4, 015 Version 48 1 Abstract

More information

RESIDUE NUMBER SYSTEM. (introduction to hardware aspects) Dr. Danila Gorodecky

RESIDUE 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 information