# Elementary factoring algorithms

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 Math 5330 Spring 013 Elementary factoring algorithms The RSA cryptosystem is founded on the idea that, in general, factoring is hard. Where as with Fermat s Little Theorem and some related ideas, one can usually tell very quickly if a composite number is, in fact, composite, actually producing a factorization of a composite number is a very different thing. Currently, the only method at our disposal is trial division. For small numbers, trial division is the method of choice. If you wish to factor a number n ă 10 10, you should probably use trial division. But what if you want to factor a large number? Trial division still has a part to play. If you have a number of size roughly 10 30, then you would need to be very lucky to factor it with trial division. If the number were to be the product of two nearly equal primes (or if the number itself were prime) then you would have to perform trial division up to about to see this. To put this in perspective, there are roughly 9,000,000,000,000 primes up to 10 15, and even if we could perform 10 6 multi precision divisions a second, it would take 9,000,000 seconds to try them all. That is, trial division could take about a year. So what to do with a 30-digit or larger number? First, one usually uses trial division for a while. After all, we know how to factor any even number. At some point, it is useful to now that the number actually is composite, so after some trial division, if m is the current unfactored part, calculate m 1 pmod mq. If it is not 1, then m is composite. Usually one does some more trial division (try, say, all primes p ă 10 6.) But after that, switch to some other factoring method. What other factoring methods are there? Here I will present several other fairly simple factoring methods. The first dates back to Fermat, the rest are less than 50 years old. Fermat s Factoring Method Our first method is based on the idea that if n x y, then n px yqpx ` yq. That is, we will try to represent n as the difference of two squares, and use that representation to factor n. To do this, we start with a number x 0 r? ns, and calculate px 0 ` kq n, for k 0, 1,,..., stopping when a square is returned. There is a trick to speed up the calculations for px 0 ` kq n, and that is that two successive values are related. That is, px 0 ` k ` 1q n rpx 0 ` kq ns ` px 0 ` kq ` 1, so we only have to calculate one square. For example, if n 3977, then x 0 64, and we need to calculate x 0 n To calculate we don t even have to square 65, we just add ˆ 64 ` 1 to 119 to get 48. Moreover, these numbers, px 0 ` kq ` 1 grow by each time, so we don t even need to recalculate them, we just add to the previous value. Here is a table for these calculations.

2 k x 0 ` k px 0 ` kq ` 1 px 0 ` kq n = 8 What this tells us is that which we rearrange as p69 8qp69 ` 8q 41 ˆ 97. Each iteration in the table goes very fast on a computer, the most difficult step of which is to determine if px 0 ` kq n is a square. Fermat s factoring method works reasonably well for small numbers n and for numbers n pq where p and q are nearly equal. An example I ve come across is in trying to factor n 10 ` 1. If you use trial division for a while, you find factors 89 and 101, leaving a 19-digit number, 1,11,470,797,641, 561, 909. If you try Fermat s method on this number, you fairly quickly find 1, 11, 470, 797, 641, 561, ˆ How good is Fermat s method? For small numbers, it is a reasonable thing to try. But in fact, it is worse than trial division in general! The worst case of Fermat s method is where n is prime. In this case, n factors as n 1, so we need x ` y n, x y 1. This means x n ` 1 and y n 1. Now the x here is x 0 `k, where x 0 is roughly? n. That is, we need? n ` 1 n ` k, so k «n ` 1?n steps before concluding that n is prime. To see what this means, suppose we have an n around This is a very small number, as factoring goes. If n is prime, it will take about? n steps or 10 5 steps to show this by trial division. With Fermat s method, it will take steps. Thus, trial divisor takes about 100,000 steps, Fermat s method takes 4,999,900,000 steps. On average, one expects to find a composite number n to have a prime divisor of size n.63, and coprime part of size about n 37. If the coprime part is actually prime, then trial divisor will find the factorization of n in about n.37 steps. Fermat s method will take something like 1 n.63 steps, so again trial division wins. Thus, in general, one should never use Fermat s method to completion. You can try several million steps, maybe, hoping to get lucky, but then switch to something else. Before moving on to the next method, I should mention that many approaches can be improved, or are more advantageous in some situations than in others. We already know, for example, that if n p, then the only possible divisors of n are primes q 1 pmod pq, so we can skip most numbers when using trial division on such numbers. With Fermat s method, there is another way to speed things up. Paradoxically, it is to try to factor a number larger than n rather than factoring n. Pick some appropriate number, m, and try to factor mn Page

3 rather than n. The idea is that mn might factor into two nearly equal parts. Here is a simple example. If we wish to factor 107 with Fermat s method, then x 0 35 and after 10 steps, we get x 0 ` 9 44, with p44 ` 7qp44 7q 71 ˆ 17. If, on the other hand, we first multiply n by 3, and use Fermat s method on 361, then x 0 61 and already we have Here, we have ˆ 51, and looking for the factor divisible by 3, we recover ˆ 17. In general, one multiplies n by some number with lots of factors, like ˆ 5 ˆ 7 on the hopes that some factors multiplying p with others multiplying q producing nearly equal numbers. For example, suppose we wish to use Fermat s method to factor 741. This would require 35 steps with Fermat s method: x 0 87, , ,..., p87 ` 4q If, instead, we multiply n by 315 and try to factor , then four steps are required: x 0 159, , 1530 Ñ 385, 1531 Ñ 6346, 153 Ñ The reason: ˆ 181, and these primes are far apart. However, multiplying by 315 gave the factorization 315ˆ ˆ1435 p9ˆ181qp35ˆ41q. Multiplying by a number m CAN make Fermat s method worse. I believe there is an algorithm for picking a sequence of numbers m to multiply by n. One tries Fermat s method on each mn for some prescribed period of time, and in the end, you can factor n in something under 3? n steps rather than? n steps as required by trial division. I do not know the details. The next two methods were both devised by a mathematician by the name of John Pollard. They are both considerably better than trial division. However, before using them, one should check that n ı pmod nq, so one knows n is composite. Pollard s rho method (1975) This method uses an iterated functions approach. Let fpxq x `1 (lots of other functions could be used instead of this one), and consider the sequence fp1q, fpfp1qq, fpfpfp1qqq,.... pmod pq. This sequence will be eventually periodic. This means that after a while, a periodic pattern will present itself. For example, if p 3, the sequence is 1,, 5, 3, 10, 9, 13, 9, 13, 9,.... We call 1,, 3, 4, 10 the tail of this eventually periodic pattern. If we let f m pxq represent the m-fold composition fpfp fpxq qq, then for any prime p there are integers k m for which f k paq f m p1q pmod pq. This is because there are only p possible remainders when a number is divided by p, but there are infinitely many m. Once we have an m and a k, then f k`1 p1q f m`1 p1q, f k` p1q f m` p1q, and so on. This means that if p is some unknown divisor of n, and if we could find the right m and k, then we might be able to find p because p would be a divisor of gcdpf m p1q f k p1q, nq. How do we find m and k when we don t even know p? We use a method called Floyd s Cylce Finding Algorithm. The algorithm works like this: Suppose we have a sequence a 0, a 1, a,... which is eventually periodic. Then a m a m for some integer m. We can use this to form a factoring algorithm: To factor n, for k 1,, 3,..., calculate gcdpf k p1q f k p1q, nq. In fact, what we do is calculate a sequence f k p1q pmod nq, to keep the numbers from getting Page 3

4 too large, and for even values of k, we calculate gcdpf k p1q f k{ p1q, nq. As an example, let n We have k f k f k{ difference gcd and so, 3 is a divisor of The reason this works should be made clear if we just do things modulo 3: k f k f k{ difference That is, f 1 p1q f 6 is divisible by 3, so it is at the stage k 1 that the prime 3 is discovered by Pollard s rho algorithm. How fast is the rho method? Certainly it has to find a prime p in at most p steps. This does not sound very good: trial division will find p in exactly p steps. However, there is reason to believe the rho method finds p much faster than p steps. Suppose, instead of numbers f m p1q, we just produced random numbers. How long would it take before two of our random numbers agreed modulo p? The is a variation of the birthday problem in probability: If you pick k things (with replacement) from n types of things, what is the probability of getting two of the same thing? The probability that the are all different is ˆ npn 1qpn q pn k ` 1q ˆ 1 ˆ 1 k 1. n k n n n Page 4

5 Let s ask a different question: When is the probability of finding a match 1? To approximate the probability, take the logarithm. We want ln k 1 ÿ j 1 Using the approximation lnp1 xq «x, we want ln «1 n ` n ` ` k 1 n ˆ ln 1 j. n kpk 1q n «k n. This means we want k «a n lnpq «1.177? n. For example, with the birthday problem (how many people do you need in a room to have a chance that two have a birthday in common?), this says you would need about 1.177? 365 «.5 people. What this means for the rho method: If the numbers f m p1q act random enough, then we expect to find a prime p not in p steps, but more like 1.177? p steps. Numerical evidence supports this, so for simplicity, we say the rho method probably finds a factor p in? p ă n 1{4 steps. More is known. If we used a simpler function for fpxq, say fpxq ax ` b, a linear function rather than a quadratic, then the iterates do not seem random enough, and we get something more like p steps again. But using most quadratic or higher degree polynomials, the iterates do appear to act like random numbers. Pollard s p 1 method (1974) Recall Fermat s Little Theorem yet again: For any prime p, and any number a with p ffl a, then a p 1 1 pmod pq. In particular, if p ą, then p 1 1 pmod pq. If m is a multiple of p 1, say m kpp 1q, then m p p 1 q k 1 k 1 pmod pq. This means that p m 1 for any m where pp 1q m. For example, if p 7, then p 1 6 so 7 m 1 for any m divisible by 6. For example, ˆ 585. We can turn this into a factoring algorithm as follows: take a sequence of m s with lots of small factors (we will use the sequence m k k!, but other sequences would work as well.) For each term in the sequence, we calculate gcdpn, m k 1q, and stop when the gcd returns a number larger than 1. This method will find a prime divisor p of n if p 1 m k. This method works very well if p 1 has all small prime divisors. The Maple command ifactor(n, easy) does the following: It uses trial division up to some limit, and then uses some fixed number of iterations of the p 1 method. For example, ifactor( , easy) returns p3q c8 p q. What this means is that it found 9 and 47,69,103 as factors of , leaving a 8-digit number that it knew to be composite (the meaning of the c). The factor was found by the p 1 method. It was successful because p 1 ˆ 3 ˆ 37 ˆ 41 ˆ 61 ˆ 3 Page 5

6 has all small divisors. In particular, it did NOT find the smaller prime divisor q because q 1 ˆ 37 ˆ 7407, and it did not do enough iterations so that 7407 m. As a simple example of the p 1 method, let s factor n As with the rho method, we form a table: k k! pmod 3811q gcdp k! 1, 3811q and ˆ 103. We found 37 after 6 steps because 37-1 = 36, a divisor of 6!. Some notes on this table: We did not calculate k!, but k! pmod nq. Also, one can calculate pk`1q! by using the formula pk`1q! p k! q k`1, using the binary squaring algorithm. That is, once we know 5! 17 pmod 3811q, we calculate 6! pmod 3811q by calculating instead, 17 6 pmod 3811q. In real life, back in the late 70 s, the p 1 method was used to show that is divisible by p In fact, this prime was found fairly quickly because p 1 ˆ 3 ˆ 11 ˆ 53 ˆ 179 ˆ 1553 ˆ 3557 ˆ 8941, which has all of its prime divisors less than 10,000. Page 6

### Factoring Algorithms

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

### FACTORING. n = 2 25 + 1. fall in the arithmetic sequence

FACTORING The claim that factorization is harder than primality testing (or primality certification) is not currently substantiated rigorously. As some sort of backward evidence that factoring is hard,

### Factoring & Primality

Factoring & Primality Lecturer: Dimitris Papadopoulos In this lecture we will discuss the problem of integer factorization and primality testing, two problems that have been the focus of a great amount

### FACTORING LARGE NUMBERS, A GREAT WAY TO SPEND A BIRTHDAY

FACTORING LARGE NUMBERS, A GREAT WAY TO SPEND A BIRTHDAY LINDSEY R. BOSKO I would like to acknowledge the assistance of Dr. Michael Singer. His guidance and feedback were instrumental in completing this

### U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009. Notes on Algebra

U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009 Notes on Algebra These notes contain as little theory as possible, and most results are stated without proof. Any introductory

### An Overview of Integer Factoring Algorithms. The Problem

An Overview of Integer Factoring Algorithms Manindra Agrawal IITK / NUS The Problem Given an integer n, find all its prime divisors as efficiently as possible. 1 A Difficult Problem No efficient algorithm

### Integer Factorization using the Quadratic Sieve

Integer Factorization using the Quadratic Sieve Chad Seibert* Division of Science and Mathematics University of Minnesota, Morris Morris, MN 56567 seib0060@morris.umn.edu March 16, 2011 Abstract We give

### Lecture 13: Factoring Integers

CS 880: Quantum Information Processing 0/4/0 Lecture 3: Factoring Integers Instructor: Dieter van Melkebeek Scribe: Mark Wellons In this lecture, we review order finding and use this to develop a method

### JUST THE MATHS UNIT NUMBER 1.8. ALGEBRA 8 (Polynomials) A.J.Hobson

JUST THE MATHS UNIT NUMBER 1.8 ALGEBRA 8 (Polynomials) by A.J.Hobson 1.8.1 The factor theorem 1.8.2 Application to quadratic and cubic expressions 1.8.3 Cubic equations 1.8.4 Long division of polynomials

### (x + a) n = x n + a Z n [x]. Proof. If n is prime then the map

22. A quick primality test Prime numbers are one of the most basic objects in mathematics and one of the most basic questions is to decide which numbers are prime (a clearly related problem is to find

### Homework 5 Solutions

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

### Determining the Optimal Combination of Trial Division and Fermat s Factorization Method

Determining the Optimal Combination of Trial Division and Fermat s Factorization Method Joseph C. Woodson Home School P. O. Box 55005 Tulsa, OK 74155 Abstract The process of finding the prime factorization

### Factorization Methods: Very Quick Overview

Factorization Methods: Very Quick Overview Yuval Filmus October 17, 2012 1 Introduction In this lecture we introduce modern factorization methods. We will assume several facts from analytic number theory.

### RSA Question 2. Bob thinks that p and q are primes but p isn t. Then, Bob thinks Φ Bob :=(p-1)(q-1) = φ(n). Is this true?

RSA Question 2 Bob thinks that p and q are primes but p isn t. Then, Bob thinks Φ Bob :=(p-1)(q-1) = φ(n). Is this true? Bob chooses a random e (1 < e < Φ Bob ) such that gcd(e,φ Bob )=1. Then, d = e -1

### Breaking The Code. Ryan Lowe. Ryan Lowe is currently a Ball State senior with a double major in Computer Science and Mathematics and

Breaking The Code Ryan Lowe Ryan Lowe is currently a Ball State senior with a double major in Computer Science and Mathematics and a minor in Applied Physics. As a sophomore, he took an independent study

### Computing exponents modulo a number: Repeated squaring

Computing exponents modulo a number: Repeated squaring How do you compute (1415) 13 mod 2537 = 2182 using just a calculator? Or how do you check that 2 340 mod 341 = 1? You can do this using the method

Family Name:... First Name:... Section:... Advanced Cryptography Final Exam July 18 th, 2006 Start at 9:15, End at 12:00 This document consists of 12 pages. Instructions Electronic devices are not allowed.

### Cryptography: RSA and the discrete logarithm problem

Cryptography: and the discrete logarithm problem R. Hayden Advanced Maths Lectures Department of Computing Imperial College London February 2010 Public key cryptography Assymmetric cryptography two keys:

### Primality - Factorization

Primality - Factorization Christophe Ritzenthaler November 9, 2009 1 Prime and factorization Definition 1.1. An integer p > 1 is called a prime number (nombre premier) if it has only 1 and p as divisors.

### Factoring pq 2 with Quadratic Forms: Nice Cryptanalyses

Factoring pq 2 with Quadratic Forms: Nice Cryptanalyses Phong Nguyễn http://www.di.ens.fr/~pnguyen & ASIACRYPT 2009 Joint work with G. Castagnos, A. Joux and F. Laguillaumie Summary Factoring A New Factoring

### MATH 168: FINAL PROJECT Troels Eriksen. 1 Introduction

MATH 168: FINAL PROJECT Troels Eriksen 1 Introduction In the later years cryptosystems using elliptic curves have shown up and are claimed to be just as secure as a system like RSA with much smaller key

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

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

### Primality Testing and Factorization Methods

Primality Testing and Factorization Methods Eli Howey May 27, 2014 Abstract Since the days of Euclid and Eratosthenes, mathematicians have taken a keen interest in finding the nontrivial factors of integers,

### The Quadratic Sieve Factoring Algorithm

The Quadratic Sieve Factoring Algorithm Eric Landquist MATH 488: Cryptographic Algorithms December 14, 2001 1 Introduction Mathematicians have been attempting to find better and faster ways to factor composite

### Cryptography and Network Security Chapter 8

Cryptography and Network Security Chapter 8 Fifth Edition by William Stallings Lecture slides by Lawrie Brown (with edits by RHB) Chapter 8 Introduction to Number Theory The Devil said to Daniel Webster:

### Revised Version of Chapter 23. We learned long ago how to solve linear congruences. ax c (mod m)

Chapter 23 Squares Modulo p Revised Version of Chapter 23 We learned long ago how to solve linear congruences ax c (mod m) (see Chapter 8). It s now time to take the plunge and move on to quadratic equations.

### CLASS 3, GIVEN ON 9/27/2010, FOR MATH 25, FALL 2010

CLASS 3, GIVEN ON 9/27/2010, FOR MATH 25, FALL 2010 1. Greatest common divisor Suppose a, b are two integers. If another integer d satisfies d a, d b, we call d a common divisor of a, b. Notice that as

### Factoring Algorithms

Institutionen för Informationsteknologi Lunds Tekniska Högskola Department of Information Technology Lund University Cryptology - Project 1 Factoring Algorithms The purpose of this project is to understand

### 3. Applications of Number Theory

3. APPLICATIONS OF NUMBER THEORY 163 3. Applications of Number Theory 3.1. Representation of Integers. Theorem 3.1.1. Given an integer b > 1, every positive integer n can be expresses uniquely as n = a

### Notes on Factoring. MA 206 Kurt Bryan

The General Approach Notes on Factoring MA 26 Kurt Bryan Suppose I hand you n, a 2 digit integer and tell you that n is composite, with smallest prime factor around 5 digits. Finding a nontrivial factor

### Some Notes on Taylor Polynomials and Taylor Series

Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited

### MA2C03 Mathematics School of Mathematics, Trinity College Hilary Term 2016 Lecture 59 (April 1, 2016) David R. Wilkins

MA2C03 Mathematics School of Mathematics, Trinity College Hilary Term 2016 Lecture 59 (April 1, 2016) David R. Wilkins The RSA encryption scheme works as follows. In order to establish the necessary public

### Elements of Applied Cryptography Public key encryption

Network Security Elements of Applied Cryptography Public key encryption Public key cryptosystem RSA and the factorization problem RSA in practice Other asymmetric ciphers Asymmetric Encryption Scheme Let

### Intermediate Math Circles March 7, 2012 Linear Diophantine Equations II

Intermediate Math Circles March 7, 2012 Linear Diophantine Equations II Last week: How to find one solution to a linear Diophantine equation This week: How to find all solutions to a linear Diophantine

### CONTINUED FRACTIONS AND FACTORING. Niels Lauritzen

CONTINUED FRACTIONS AND FACTORING Niels Lauritzen ii NIELS LAURITZEN DEPARTMENT OF MATHEMATICAL SCIENCES UNIVERSITY OF AARHUS, DENMARK EMAIL: niels@imf.au.dk URL: http://home.imf.au.dk/niels/ Contents

### SOLVING POLYNOMIAL EQUATIONS

C SOLVING POLYNOMIAL EQUATIONS We will assume in this appendix that you know how to divide polynomials using long division and synthetic division. If you need to review those techniques, refer to an algebra

### 8 Primes and Modular Arithmetic

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

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

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

### 3.6. The factor theorem

3.6. The factor theorem Example 1. At the right we have drawn the graph of the polynomial y = x 4 9x 3 + 8x 36x + 16. Your problem is to write the polynomial in factored form. Does the geometry of the

### I. Introduction. MPRI Cours 2-12-2. Lecture IV: Integer factorization. What is the factorization of a random number? II. Smoothness testing. F.

F. Morain École polytechnique MPRI cours 2-12-2 2013-2014 3/22 F. Morain École polytechnique MPRI cours 2-12-2 2013-2014 4/22 MPRI Cours 2-12-2 I. Introduction Input: an integer N; logox F. Morain logocnrs

### Lecture 13 - Basic Number Theory.

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

### CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY

January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.

### The application of prime numbers to RSA encryption

The application of prime numbers to RSA encryption Prime number definition: Let us begin with the definition of a prime number p The number p, which is a member of the set of natural numbers N, is considered

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

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

### Week 4: Gambler s ruin and bold play

Week 4: Gambler s ruin and bold play Random walk and Gambler s ruin. Imagine a walker moving along a line. At every unit of time, he makes a step left or right of exactly one of unit. So we can think that

### Factoring Polynomials

Factoring Polynomials Hoste, Miller, Murieka September 12, 2011 1 Factoring In the previous section, we discussed how to determine the product of two or more terms. Consider, for instance, the equations

### International Journal of Information Technology, Modeling and Computing (IJITMC) Vol.1, No.3,August 2013

FACTORING CRYPTOSYSTEM MODULI WHEN THE CO-FACTORS DIFFERENCE IS BOUNDED Omar Akchiche 1 and Omar Khadir 2 1,2 Laboratory of Mathematics, Cryptography and Mechanics, Fstm, University of Hassan II Mohammedia-Casablanca,

### Arithmetic algorithms for cryptology 5 October 2015, Paris. Sieves. Razvan Barbulescu CNRS and IMJ-PRG. R. Barbulescu Sieves 0 / 28

Arithmetic algorithms for cryptology 5 October 2015, Paris Sieves Razvan Barbulescu CNRS and IMJ-PRG R. Barbulescu Sieves 0 / 28 Starting point Notations q prime g a generator of (F q ) X a (secret) integer

### The Prime Numbers. Definition. A prime number is a positive integer with exactly two positive divisors.

The Prime Numbers Before starting our study of primes, we record the following important lemma. Recall that integers a, b are said to be relatively prime if gcd(a, b) = 1. Lemma (Euclid s Lemma). If gcd(a,

### 9 Modular Exponentiation and Cryptography

9 Modular Exponentiation and Cryptography 9.1 Modular Exponentiation Modular arithmetic is used in cryptography. In particular, modular exponentiation is the cornerstone of what is called the RSA system.

### MATH 289 PROBLEM SET 4: NUMBER THEORY

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

### 1 Lecture: Integration of rational functions by decomposition

Lecture: Integration of rational functions by decomposition into partial fractions Recognize and integrate basic rational functions, except when the denominator is a power of an irreducible quadratic.

### The Mathematics of the RSA Public-Key Cryptosystem

The Mathematics of the RSA Public-Key Cryptosystem Burt Kaliski RSA Laboratories ABOUT THE AUTHOR: Dr Burt Kaliski is a computer scientist whose involvement with the security industry has been through

### Mathematics of Cryptography Part I

CHAPTER 2 Mathematics of Cryptography Part I (Solution to Odd-Numbered Problems) Review Questions 1. The set of integers is Z. It contains all integral numbers from negative infinity to positive infinity.

### http://wrap.warwick.ac.uk/

Original citation: Hart, William B.. (2012) A one line factoring algorithm. Journal of the Australian Mathematical Society, Volume 92 (Number 1). pp. 61-69. ISSN 1446-7887 Permanent WRAP url: http://wrap.warwick.ac.uk/54707/

### Computer and Network Security

MIT 6.857 Computer and Networ Security Class Notes 1 File: http://theory.lcs.mit.edu/ rivest/notes/notes.pdf Revision: December 2, 2002 Computer and Networ Security MIT 6.857 Class Notes by Ronald L. Rivest

### Library (versus Language) Based Parallelism in Factoring: Experiments in MPI. Dr. Michael Alexander Dr. Sonja Sewera.

Library (versus Language) Based Parallelism in Factoring: Experiments in MPI Dr. Michael Alexander Dr. Sonja Sewera Talk 2007-10-19 Slide 1 of 20 Primes Definitions Prime: A whole number n is a prime number

### 9. POLYNOMIALS. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11.

9. POLYNOMIALS 9.1. Definition of a Polynomial A polynomial is an expression of the form: a(x) = a n x n + a n-1 x n-1 +... + a 1 x + a 0. The symbol x is called an indeterminate and simply plays the role

### Factoring. Factoring 1

Factoring Factoring 1 Factoring Security of RSA algorithm depends on (presumed) difficulty of factoring o Given N = pq, find p or q and RSA is broken o Rabin cipher also based on factoring Factoring like

### Section 6.1 Factoring Expressions

Section 6.1 Factoring Expressions The first method we will discuss, in solving polynomial equations, is the method of FACTORING. Before we jump into this process, you need to have some concept of what

### Theorem3.1.1 Thedivisionalgorithm;theorem2.2.1insection2.2 If m, n Z and n is a positive

Chapter 3 Number Theory 159 3.1 Prime Numbers Prime numbers serve as the basic building blocs in the multiplicative structure of the integers. As you may recall, an integer n greater than one is prime

### The RSA Algorithm. Evgeny Milanov. 3 June 2009

The RSA Algorithm Evgeny Milanov 3 June 2009 In 1978, Ron Rivest, Adi Shamir, and Leonard Adleman introduced a cryptographic algorithm, which was essentially to replace the less secure National Bureau

### Runtime and Implementation of Factoring Algorithms: A Comparison

Runtime and Implementation of Factoring Algorithms: A Comparison Justin Moore CSC290 Cryptology December 20, 2003 Abstract Factoring composite numbers is not an easy task. It is classified as a hard algorithm,

### Announcements. CS243: Discrete Structures. More on Cryptography and Mathematical Induction. Agenda for Today. Cryptography

Announcements CS43: Discrete Structures More on Cryptography and Mathematical Induction Işıl Dillig Class canceled next Thursday I am out of town Homework 4 due Oct instead of next Thursday (Oct 18) Işıl

### ALGEBRAIC APPROACH TO COMPOSITE INTEGER FACTORIZATION

ALGEBRAIC APPROACH TO COMPOSITE INTEGER FACTORIZATION Aldrin W. Wanambisi 1* School of Pure and Applied Science, Mount Kenya University, P.O box 553-50100, Kakamega, Kenya. Shem Aywa 2 Department of Mathematics,

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

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

### Lecture 10: Distinct Degree Factoring

CS681 Computational Number Theory Lecture 10: Distinct Degree Factoring Instructor: Piyush P Kurur Scribe: Ramprasad Saptharishi Overview Last class we left of with a glimpse into distant degree factorization.

### The Factor Theorem and a corollary of the Fundamental Theorem of Algebra

Math 421 Fall 2010 The Factor Theorem and a corollary of the Fundamental Theorem of Algebra 27 August 2010 Copyright 2006 2010 by Murray Eisenberg. All rights reserved. Prerequisites Mathematica Aside

### 6.1 The Greatest Common Factor; Factoring by Grouping

386 CHAPTER 6 Factoring and Applications 6.1 The Greatest Common Factor; Factoring by Grouping OBJECTIVES 1 Find the greatest common factor of a list of terms. 2 Factor out the greatest common factor.

### Faster deterministic integer factorisation

David Harvey (joint work with Edgar Costa, NYU) University of New South Wales 25th October 2011 The obvious mathematical breakthrough would be the development of an easy way to factor large prime numbers

### Grade 6 Math Circles March 10/11, 2015 Prime Time Solutions

Faculty of Mathematics Waterloo, Ontario N2L 3G1 Centre for Education in Mathematics and Computing Lights, Camera, Primes! Grade 6 Math Circles March 10/11, 2015 Prime Time Solutions Today, we re going

### Discrete Mathematics, Chapter 4: Number Theory and Cryptography

Discrete Mathematics, Chapter 4: Number Theory and Cryptography Richard Mayr University of Edinburgh, UK Richard Mayr (University of Edinburgh, UK) Discrete Mathematics. Chapter 4 1 / 35 Outline 1 Divisibility

### Outline. Computer Science 418. Digital Signatures: Observations. Digital Signatures: Definition. Definition 1 (Digital signature) Digital Signatures

Outline Computer Science 418 Digital Signatures Mike Jacobson Department of Computer Science University of Calgary Week 12 1 Digital Signatures 2 Signatures via Public Key Cryptosystems 3 Provable 4 Mike

### 1. LINEAR EQUATIONS. A linear equation in n unknowns x 1, x 2,, x n is an equation of the form

1. LINEAR EQUATIONS A linear equation in n unknowns x 1, x 2,, x n is an equation of the form a 1 x 1 + a 2 x 2 + + a n x n = b, where a 1, a 2,..., a n, b are given real numbers. For example, with x and

### The Method of Partial Fractions Math 121 Calculus II Spring 2015

Rational functions. as The Method of Partial Fractions Math 11 Calculus II Spring 015 Recall that a rational function is a quotient of two polynomials such f(x) g(x) = 3x5 + x 3 + 16x x 60. The method

### RSA Attacks. By Abdulaziz Alrasheed and Fatima

RSA Attacks By Abdulaziz Alrasheed and Fatima 1 Introduction Invented by Ron Rivest, Adi Shamir, and Len Adleman [1], the RSA cryptosystem was first revealed in the August 1977 issue of Scientific American.

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

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

### Chapter 11 Number Theory

Chapter 11 Number Theory Number theory is one of the oldest branches of mathematics. For many years people who studied number theory delighted in its pure nature because there were few practical applications

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

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

### Factoring Polynomials

Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent

### The cyclotomic polynomials

The cyclotomic polynomials Notes by G.J.O. Jameson 1. The definition and general results We use the notation e(t) = e 2πit. Note that e(n) = 1 for integers n, e(s + t) = e(s)e(t) for all s, t. e( 1 ) =

### On Generalized Fermat Numbers 3 2n +1

Applied Mathematics & Information Sciences 4(3) (010), 307 313 An International Journal c 010 Dixie W Publishing Corporation, U. S. A. On Generalized Fermat Numbers 3 n +1 Amin Witno Department of Basic

### Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize

### Modern Factoring Algorithms

Modern Factoring Algorithms Kostas Bimpikis and Ragesh Jaiswal University of California, San Diego... both Gauss and lesser mathematicians may be justified in rejoicing that there is one science [number

### 2 Primality and Compositeness Tests

Int. J. Contemp. Math. Sciences, Vol. 3, 2008, no. 33, 1635-1642 On Factoring R. A. Mollin Department of Mathematics and Statistics University of Calgary, Calgary, Alberta, Canada, T2N 1N4 http://www.math.ucalgary.ca/

### SOLUTIONS FOR PROBLEM SET 2

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

### The Math. P (x) = 5! = 1 2 3 4 5 = 120.

The Math Suppose there are n experiments, and the probability that someone gets the right answer on any given experiment is p. So in the first example above, n = 5 and p = 0.2. Let X be the number of correct

### CIS 5371 Cryptography. 8. Encryption --

CIS 5371 Cryptography p y 8. Encryption -- Asymmetric Techniques Textbook encryption algorithms In this chapter, security (confidentiality) is considered in the following sense: All-or-nothing secrecy.

### ELLIPTIC CURVES AND LENSTRA S FACTORIZATION ALGORITHM

ELLIPTIC CURVES AND LENSTRA S FACTORIZATION ALGORITHM DANIEL PARKER Abstract. This paper provides a foundation for understanding Lenstra s Elliptic Curve Algorithm for factoring large numbers. We give

### RSA Encryption. Grade Levels. Objectives and Topics. Introduction and Outline (54)(2)(2) (27)(2)(2)(2) (9)(3)(2)(2)(2) (3)(3)(3)(2)(2)(2)

RSA Encryption Grade Levels This activity is intended for high schools students, grades 10 12. Objectives and Topics One of the classical examples of applied mathematics is encryption. Through this activity,

### LUC: A New Public Key System

LUC: A New Public Key System Peter J. Smith a and Michael J. J. Lennon b a LUC Partners, Auckland UniServices Ltd, The University of Auckland, Private Bag 92019, Auckland, New Zealand. b Department of

### Factoring a semiprime n by estimating φ(n)

Factoring a semiprime n by estimating φ(n) Kyle Kloster May 7, 2010 Abstract A factoring algorithm, called the Phi-Finder algorithm, is presented that factors a product of two primes, n = pq, by determining

### Public-key cryptography RSA

Public-key cryptography RSA NGUYEN Tuong Lan LIU Yi Master Informatique University Lyon 1 Objective: Our goal in the study is to understand the algorithm RSA, some existence attacks and implement in Java.

### Public Key Cryptography: RSA and Lots of Number Theory

Public Key Cryptography: RSA and Lots of Number Theory Public vs. Private-Key Cryptography We have just discussed traditional symmetric cryptography: Uses a single key shared between sender and receiver

### 5544 = 2 2772 = 2 2 1386 = 2 2 2 693. Now we have to find a divisor of 693. We can try 3, and 693 = 3 231,and we keep dividing by 3 to get: 1

MATH 13150: Freshman Seminar Unit 8 1. Prime numbers 1.1. Primes. A number bigger than 1 is called prime if its only divisors are 1 and itself. For example, 3 is prime because the only numbers dividing

### Cryptography and Network Security Number Theory

Cryptography and Network Security Number Theory Xiang-Yang Li Introduction to Number Theory Divisors b a if a=mb for an integer m b a and c b then c a b g and b h then b (mg+nh) for any int. m,n Prime

### A Comparison Of Integer Factoring Algorithms. Keyur Anilkumar Kanabar

A Comparison Of Integer Factoring Algorithms Keyur Anilkumar Kanabar Batchelor of Science in Computer Science with Honours The University of Bath May 2007 This dissertation may be made available for consultation

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

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