# Winter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov

Size: px
Start display at page:

Download "Winter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov alexanderrem@gmail.com"

Transcription

1 Polynomials Alexander Remorov Warm-up Problem 1: Let f(x) be a quadratic polynomial. Prove that there exist quadratic polynomials g(x) and h(x) such that f(x)f(x + 1) = g(h(x)). (University of Toronto Math Competition 2010) Solution: The standard approach would be to write f(x) = ax 2 + bx + c and play around with the coefficients of f(x)f(x + 1). It is doable, but quite messy. Let us look at the roots. Let f(x) = a(x r)(x s), then: f(x)f(x + 1) = a 2 (x r)(x s + 1) (x s)(x r + 1) = = a 2 ([x 2 (r + s 1)x + rs] r)([x 2 (r + s 1)x + rs] s) and we are done by setting g(x) = a 2 (x r)(x s), h(x) = x 2 (r + s 1)x + rs. Warm-up Problem 2: The polynomial f(x) = x n + a 1 x n 1 + a 2 x n a n 1 x + a n = 0 with integer non-zero coefficients has n distinct integer roots. Prove that if the roots are pairwise coprime, then a n 1 and a n are coprime. (Russian Math Olympiad 2004) Solution: Assume gcd(a n 1, a n ) 1, then both a n 1 and a n are divisible by some prime p. Let the roots of the polynomial be r 1, r 2,, r n. Then r 1 r 2 r n = ( 1) n a n. This is divisible p, so at least one of the roots, wolog r 1, is divisible by p. We also have: r 1 r 2 r n 1 + r 1 r 3 r 4 r n r 2 r 3 r n = ( 1) n 1 a n 1 0 mod p All terms containing r 1 are divisible by p, hence r 2 r 3 r n is divisible by p. Hence gcd(r 1, r 2 r 3 r n ) is divisible by p contradicting the fact that the roots are pairwise coprime. The result follows. 1 Algebra Fundamental Theorem of Algebra: A polynomial P (x) of degree n with complex coefficients has n complex roots. It can be uniquely factored as: P (x) = a(x r 1 )(x r 2 ) (x r n ) Vieta s Formulas: Let P (x) = a n x n + a n 1 x n a 1 x + a 0 with complex coefficients have roots r 1, r 2,, r n. Then: n i=1 r i = ( 1) 1 a n 1 a n ; i<j r i r j = ( 1) 2 a n 2 a n ; r 1 r 2 r n = ( 1) n a 0 a n 1

2 Bezout s Theorem: A polynomial P (x) is divisible by (x a) iff P (a) = 0. Lagrange Interpolation: Given n points (x 1, y 1 ),, (x n, y n ), there is a unique polynomial P (x) satisfying P (x i ) = y i. Its explicit formula is: P (x) = n i=1 y i 1 j n,j i x x j x i x j A few general tricks related to polynomials: Look at the roots. If you want to show a polynomial is identically 0, it is sometimes useful to look at an arbitrary root r of this polynomial, and then show the polynomial must have another root, e.g. r + 1, thus producing a sequence of infinitely many roots. Look at the coefficients. This is particularly useful when the coefficients are integers. It is often a good idea to look at the leading coefficient and the constant term. Consider the degrees of polynomials. If P (x) is divisible by Q(x) where P, Q are polynomials, then deg(q) deg(p ). A straight-forward fact, yet a useful one. Perform clever algebraic manipulations, such as factoring, expanding, introducing new polynomials, substituting other values for x, e.g. x + 1, 1 x, etc. 1.1 Warm-up 1. Let P (x) and Q(x) be polynomials with real coefficients such that P (x) = Q(x) for all real values of x. Prove that P (x) = Q(x) for all complex values of x. 2. (a) Determine all polynomials P (x) with real coefficients such that P (x 2 ) = P 2 (x). (b) Determine all polynomials P (x) with real coefficients such that P (x 2 ) = P (x)p (x + 1). (c) Suppose P (x) is a polynomial such that P (x 1) + P (x + 1) = 2P (x) for all real x. Prove that P (x) has degree at most (USAMO 1975) A polynomial P (x) of degree n satisfies P (k) = k k + 1 Find P (n + 1). for k = 0, 1, 2,..., n. 1.2 Problems 1. (Brazil 2007) Let P (x) = x x+1. Prove that for every positive integer n, the equation P (P (... (P (x))...)) = 0 has at least one real solution, where the composition is performed n times. 2. (Russia 2002) Among the polynomials P (x), Q(x), R(x) with real coefficients at least one has degree two and one has degree three. If P 2 (x) + Q 2 (x) = R 2 (x) prove that one of the polynomials of degree three has three real roots. 3. Let P (x) = a n x n + a n 1 x n a 1 x + a 0 be a polynomial with integer coefficients such that a 0 is prime and a 0 > a 1 + a a n. Prove that P (x) is irreducible (that is, cannot be factored into two polynomials with integer coefficients of degree at least 1). 2

3 4. (Russia 2003) The side lengths of a triangle are the roots of a cubic equation with rational coefficients. Prove that the altitudes are the roots of a degree six equation with rational coefficients. 5. (Russia 1997) Does there exist a set S of non-zero real numbers such that for any positive integer n there exists a polynomial P (x) with degree at least n, all the roots and all the coefficients of which are from S? 6. (Putnam 2010) Find all polynomials P (x), Q(x) with real coefficients such that P (x)q(x + 1) P (x + 1)Q(x) = (IMO SL 2005) Let a, b, c, d, e, f be positive integers. Suppose that S = a + b + c + d + e + f divides both abc + def and ab + bc + ca de ef fd. Prove that S is composite. 8. (USAMO 2002) Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree n with real coefficients can be written as the average of two monic polynomials of degree n with n real roots. 9. (Iran TST 2010) Find all two-variable polynomials P (x, y) such that for any real numbers a, b, c: P (ab, c 2 + 1) + P (bc, a 2 + 1) + P (ca, b 2 + 1) = (China TST 2007) Prove that for any positive integer n, there exists exactly one polynomial P (x) of degree n with real coefficients, such that P (0) = 1 and (x + 1)(P (x)) 2 1 is an odd function. (A function f(x) is odd if f(x) = f( x) for all x). 2 Number Theory By Z[x] we denote all the polynomials of one variable with integer coefficients. Arguably the most useful property when it comes to polynomials and integers is: If P (x) Z[x], and a, b are integers, then (a b) (P (a) P (b)) Recall that polynomial in Z[x] is irreducible over the integers if it cannot be factored into two polynomials with integer coefficients. Eisenstein s Criterion: Let P (x) = a n x n + a n 1 x n a 1 x + a 0 Z[x] be a polynomial and p be a prime dividing a 0, a 1,..., a n 1, such that p a n and p 2 a 0. Then P (x) is irreducible. Proof: Assume P (x) = Q(x)R(x), where Q(x) = b k x k + b k 1 x k b 1 x + b 0, R(x) = c l x l + c l 1 x l c 1 x + c 0. Then b 0 c 0 is divisible by p but not p 2. Wolog p b 0, p c 0. Since p a 1 = b 1 c 0 + b 0 c 1 it follows that p b 1. Since p a 2 = b 2 c 0 + b 1 c 1 + b 0 c 2 it follows that p b 2. By induction it follows that p b k which implies that p a n, a contradiction. Lemma [Schur] Let P (x) Z[x] be a non-constant polynomial. Then there are infinitely many primes dividing at least one of the non-zero terms in the sequence P (1), P (2), P (3),... Proof: Assume first that P (0) = 1. There exists an integer M such that P (n) 1 for all n > M (or else P (x) 1 has infinitely many roots and therefore is constant). We also have P (n!) 1( mod n!), and by taking arbitrarily large integers n we can generate arbitrarily large primes dividing P (n!). 3

4 If P (0) = 0, the result is obvious. Otherwise consider Q(x) = of reasoning to Q(x); the result follows. P (xp (0)) P (0) and apply the same line For polynomials in Z[x] it is often useful to work modulo a positive integer k. If P (x) = n i=0 a ix i Z[x] and k is a positive integer we call P (x) = n i=0 a ix i the reduction of P (x) (mod k), where a i = a i (mod k). Some useful facts about reduced polynomials: 1. Let P (x), Q(x), R(x), S(x) Z[x], such that P (x) = (Q(x) + R(x))S(x). Then P (x) = (Q(x) + R(x))(S(x)). 2. Let p be a prime and P (x) Z[x]. Then the factorization of P (x) is unique modulo p (more formally, in F p [x] up to permutation.) Note that this result does not hold when p is not a prime. For example, x = (2x + 3)(3x + 2) mod 6 if x, 2x + 3, 3x + 2 are prime. Also remember that all roots of P (x) modulo p are in the set {0, 1,..., p 1}. 2.1 Warm-Up 1. (a) Let p be a prime number. Prove that P (x) = x p 1 + x p x + 1 is irreducible. (b) Prove Eisenstein s Criterion by considering a reduction modulo p. 2. (Iran 2007) Does there exist a sequence of integers a 0, a 1, a 2,... such that gcd(a i, a j ) = 1 for n i j, and for every positive integer n, the polynomial a i x i is irreducible? 3. (a) (Bezout) Let P (x), Q(x) be polynomials with integer coefficients such that P (x), Q(x) do not have any roots in common. Prove that there exist polynomials A(x), B(x) and an integer N such that A(x)P (x) + B(x)Q(x) = N. (b) Let P (x), Q(x) be monic non-constant irreducible polynomials with integer coefficients. For all sufficiently large n, P (n) and Q(n) have the same prime divisors. Prove that P (x) Q(x). 2.2 Problems 1. (a) (USAMO 1974)Let a, b, c be three distinct integers. Prove that there does not exist a polynomial P (x) with integer coefficients such that P (a) = b, P (b) = c, P (c) = a. (b) (IMO 2006) Let P (x) be a polynomial of degree n > 1 with integer coefficients and let k be a positive integer. Let Q(x) = P (P (... P (P (x))...)), where the polynomial P is composed k times. Prove that there are at most n integers t such that Q(t) = t. 2. (Romania TST 2007) Let P (x) = x n + a n 1 x n a 1 x + a 0 be a polynomial of degree n 3 with integer coefficients such that P (m) is even for all even integers m. Furthermore, a 0 is even, and a k + a n k is even for k = 1, 2,..., n 1. Suppose P (x) = Q(x)R(x) where Q(x), R(x) are polynomials with integer coefficients, deg Q deg R, and all coefficients of R(x) are odd. Prove that P (x) has an integer root. 3. (USA TST 2010) Let P (x) be a polynomial with integer coefficients such that P (0) = 0 and gcd(p (0), P (1), P (2),...) = 1. Prove that there are infinitely many positive integers n such that gcd(p (n) P (0), P (n + 1) P (1), P (n + 2) P (2),...) = n. i=0 4

5 4. (Iran TST 2004) Let P (x) be a polynomial with integer coefficients such that P (n) > n for every positive integer n. Define the sequence x k by x 1 = 1, x i+1 = P (x i ) for i 1. For every positive integer m, there exists a term in this sequence divisible by m. Prove that P (x) = x (China TST 2006) Prove that for any n 2, there exists a polynomial P (x) = x n +a n 1 x n a 1 x + a 0 such that: (a) a 0, a 1,..., a n 1 all are non-zero. (b) P (x) is irreducible. (c) For any integer x, P (x) is not prime. 6. (Russia 2006) A polynomial (x + 1) n 1 is divisible by a polynomial P (x) = x k + a k 1 x k a 1 x + a 0 of even degree k, such that all of its coefficients are odd integers. Prove that n is divisible by k (USAMO 2006) For an integer m, let p(m) be the greatest prime divisor of m. By convention, we set p(±1) = 1 and p(0) =. Find all polynomials f with integer coefficients such that the sequence {p ( f ( n 2)) 2n} n 0 is bounded above. (In particular, f ( n 2) 0 for n 0.) 8. Find all non-constant polynomials P (x) with integer coefficients, such that for any relatively prime integers a, b, the sequence {f(an + b)} n 1 contains an infinite number of terms and any two of which are relatively prime. 9. (IMO SL 2009) Let P (x) be a non-constant polynomial with integer coefficients. Prove that there is no function T from the set of integers into the set of integers such that the number of integers x with T n (x) = x is equal to P (n) for every positive integer n, where T n denotes the n-fold application of T. 10. (USA TST 2008) Let n be a positive integer. Given polynomial P (x) with integer coefficients, define its signature modulo n to be the (ordered) sequence P (1),..., P (n) modulo n. Of the n n such n-term sequences of integers modulo n, how many are the signature of some polynomial P (x) if: (a) n is a positive integer not divisible by the square of a prime. (b) n is a positive integer not divisible by the cube of a prime. 5

6 3 Hints to Selected Problems 3.1 Algebra 2. Difference of squares. 3. Prove that for any complex root r of P (x), we have r > Use Heron s formula to prove the square of the area is a rational number. 5. Look at the smallest and the largest numbers in S by absolute value. Use Vieta s thoerem. 7. The solution involves polynomials. 8. A polynomial has n real roots iff it changes sign n + 1 times. Define one of the polynomials as kq(x) where Q(x) has n roots and k is a constant. 9. Prove that P (x, y) is divisible by x 2 (y 1). 10. Let P (x) = Q(x) + R(x) where Q is an even function and R is an odd function. 3.2 Number Theory 2. Reduce modulo 2. Prove that deg R(x) = Let P (x) = x k Q(x) with Q(x) 0. Consider prime n = p k where p is prime. 4. Prove that x k+1 x k x k+2 x k Reduce modulo Use Eisenstein s Criterion. 7. Look at the irreducibe factors of f. Prove they are of form 4x k What can you say about gcd(n, f(n))? 9. For k N, look at a k, the number of integers x, such that k is the smallest integer for which T k (x) = x. 10. First solve the problem if n is a prime. 6

### MOP 2007 Black Group Integer Polynomials Yufei Zhao. Integer Polynomials. June 29, 2007 Yufei Zhao yufeiz@mit.edu

Integer Polynomials June 9, 007 Yufei Zhao yufeiz@mit.edu We will use Z[x] to denote the ring of polynomials with integer coefficients. We begin by summarizing some of the common approaches used in dealing

### The Division Algorithm for Polynomials Handout Monday March 5, 2012

The Division Algorithm for Polynomials Handout Monday March 5, 0 Let F be a field (such as R, Q, C, or F p for some prime p. This will allow us to divide by any nonzero scalar. (For some of the following,

### 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

### 3 1. Note that all cubes solve it; therefore, there are no more

Math 13 Problem set 5 Artin 11.4.7 Factor the following polynomials into irreducible factors in Q[x]: (a) x 3 3x (b) x 3 3x + (c) x 9 6x 6 + 9x 3 3 Solution: The first two polynomials are cubics, so if

### 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

### PROBLEM SET 6: POLYNOMIALS

PROBLEM SET 6: POLYNOMIALS 1. introduction In this problem set we will consider polynomials with coefficients in K, where K is the real numbers R, the complex numbers C, the rational numbers Q or any other

### 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.

### calculating the result modulo 3, as follows: p(0) = 0 3 + 0 + 1 = 1 0,

Homework #02, due 1/27/10 = 9.4.1, 9.4.2, 9.4.5, 9.4.6, 9.4.7. Additional problems recommended for study: (9.4.3), 9.4.4, 9.4.9, 9.4.11, 9.4.13, (9.4.14), 9.4.17 9.4.1 Determine whether the following polynomials

### PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.

PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include

### Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2)

Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2) Kevin Broughan University of Waikato, Hamilton, New Zealand May 13, 2010 Remainder and Factor Theorem 15 Definition of factor If f (x)

### minimal polyonomial Example

Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We

### it is easy to see that α = a

21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UF. Therefore

### 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

### Quotient Rings and Field Extensions

Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.

### 7. Some irreducible polynomials

7. Some irreducible polynomials 7.1 Irreducibles over a finite field 7.2 Worked examples Linear factors x α of a polynomial P (x) with coefficients in a field k correspond precisely to roots α k [1] of

### (a) Write each of p and q as a polynomial in x with coefficients in Z[y, z]. deg(p) = 7 deg(q) = 9

Homework #01, due 1/20/10 = 9.1.2, 9.1.4, 9.1.6, 9.1.8, 9.2.3 Additional problems for study: 9.1.1, 9.1.3, 9.1.5, 9.1.13, 9.2.1, 9.2.2, 9.2.4, 9.2.5, 9.2.6, 9.3.2, 9.3.3 9.1.1 (This problem was not assigned

### The Chinese Remainder Theorem

The Chinese Remainder Theorem Evan Chen evanchen@mit.edu February 3, 2015 The Chinese Remainder Theorem is a theorem only in that it is useful and requires proof. When you ask a capable 15-year-old why

### 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

### 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

### 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,

### POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS

POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS RUSS WOODROOFE 1. Unique Factorization Domains Throughout the following, we think of R as sitting inside R[x] as the constant polynomials (of degree 0).

### PYTHAGOREAN TRIPLES KEITH CONRAD

PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient

### Math 345-60 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field

Math 345-60 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field 1. Throughout this section, F is a field and F [x] is the ring of polynomials with coefficients in F. We will

### I. GROUPS: BASIC DEFINITIONS AND EXAMPLES

I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called

### FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z

FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization

### Introduction to Finite Fields (cont.)

Chapter 6 Introduction to Finite Fields (cont.) 6.1 Recall Theorem. Z m is a field m is a prime number. Theorem (Subfield Isomorphic to Z p ). Every finite field has the order of a power of a prime number

### Unique Factorization

Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon

### a 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4)

ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x

### ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS

ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS DANIEL RABAYEV AND JACK SONN Abstract. Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q p for

### Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013

Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents A Cyclotomic Polynomials 79 A.1 Minimum Polynomials of Roots of

### On the generation of elliptic curves with 16 rational torsion points by Pythagorean triples

On the generation of elliptic curves with 16 rational torsion points by Pythagorean triples Brian Hilley Boston College MT695 Honors Seminar March 3, 2006 1 Introduction 1.1 Mazur s Theorem Let C be a

### 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

### H/wk 13, Solutions to selected problems

H/wk 13, Solutions to selected problems Ch. 4.1, Problem 5 (a) Find the number of roots of x x in Z 4, Z Z, any integral domain, Z 6. (b) Find a commutative ring in which x x has infinitely many roots.

### 3 Factorisation into irreducibles

3 Factorisation into irreducibles Consider the factorisation of a non-zero, non-invertible integer n as a product of primes: n = p 1 p t. If you insist that primes should be positive then, since n could

### k, then n = p2α 1 1 pα k

Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square

### THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS

THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS KEITH CONRAD 1. Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear

### 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

### Factorization Algorithms for Polynomials over Finite Fields

Degree Project Factorization Algorithms for Polynomials over Finite Fields Sajid Hanif, Muhammad Imran 2011-05-03 Subject: Mathematics Level: Master Course code: 4MA11E Abstract Integer factorization is

### 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 ) =

### Die ganzen zahlen hat Gott gemacht

Die ganzen zahlen hat Gott gemacht Polynomials with integer values B.Sury A quote attributed to the famous mathematician L.Kronecker is Die Ganzen Zahlen hat Gott gemacht, alles andere ist Menschenwerk.

### 1 = (a 0 + b 0 α) 2 + + (a m 1 + b m 1 α) 2. for certain elements a 0,..., a m 1, b 0,..., b m 1 of F. Multiplying out, we obtain

Notes on real-closed fields These notes develop the algebraic background needed to understand the model theory of real-closed fields. To understand these notes, a standard graduate course in algebra is

### 8 Divisibility and prime numbers

8 Divisibility and prime numbers 8.1 Divisibility In this short section we extend the concept of a multiple from the natural numbers to the integers. We also summarize several other terms that express

### 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.

### Today s Topics. Primes & Greatest Common Divisors

Today s Topics Primes & Greatest Common Divisors Prime representations Important theorems about primality Greatest Common Divisors Least Common Multiples Euclid s algorithm Once and for all, what are prime

### Homework until Test #2

MATH31: Number Theory Homework until Test # Philipp BRAUN Section 3.1 page 43, 1. It has been conjectured that there are infinitely many primes of the form n. Exhibit five such primes. Solution. Five such

### Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

### MATH10040 Chapter 2: Prime and relatively prime numbers

MATH10040 Chapter 2: Prime and relatively prime numbers Recall the basic definition: 1. Prime numbers Definition 1.1. Recall that a positive integer is said to be prime if it has precisely two positive

### 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.

### 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

### 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

### Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any.

Algebra 2 - Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers - {1,2,3,4,...}

### 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

### Real Roots of Univariate Polynomials with Real Coefficients

Real Roots of Univariate Polynomials with Real Coefficients mostly written by Christina Hewitt March 22, 2012 1 Introduction Polynomial equations are used throughout mathematics. When solving polynomials

### RESULTANT AND DISCRIMINANT OF POLYNOMIALS

RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results

### 15. Symmetric polynomials

15. Symmetric polynomials 15.1 The theorem 15.2 First examples 15.3 A variant: discriminants 1. The theorem Let S n be the group of permutations of {1,, n}, also called the symmetric group on n things.

### 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

### Irreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients

DOI: 10.2478/auom-2014-0007 An. Şt. Univ. Ovidius Constanţa Vol. 221),2014, 73 84 Irreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients Anca

### Putnam Notes Polynomials and palindromes

Putnam Notes Polynomials and palindromes Polynomials show up one way or another in just about every area of math. You will hardly ever see any math competition without at least one problem explicitly concerning

### 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

### MATH 537 (Number Theory) FALL 2016 TENTATIVE SYLLABUS

MATH 537 (Number Theory) FALL 2016 TENTATIVE SYLLABUS Class Meetings: MW 2:00-3:15 pm in Physics 144, September 7 to December 14 [Thanksgiving break November 23 27; final exam December 21] Instructor:

### A number field is a field of finite degree over Q. By the Primitive Element Theorem, any number

Number Fields Introduction A number field is a field of finite degree over Q. By the Primitive Element Theorem, any number field K = Q(α) for some α K. The minimal polynomial Let K be a number field and

### SUM OF TWO SQUARES JAHNAVI BHASKAR

SUM OF TWO SQUARES JAHNAVI BHASKAR Abstract. I will investigate which numbers can be written as the sum of two squares and in how many ways, providing enough basic number theory so even the unacquainted

### Elementary Number Theory and Methods of Proof. CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook.

Elementary Number Theory and Methods of Proof CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook.edu/~cse215 1 Number theory Properties: 2 Properties of integers (whole

### Section 4.2: The Division Algorithm and Greatest Common Divisors

Section 4.2: The Division Algorithm and Greatest Common Divisors The Division Algorithm The Division Algorithm is merely long division restated as an equation. For example, the division 29 r. 20 32 948

### Integer roots of quadratic and cubic polynomials with integer coefficients

Integer roots of quadratic and cubic polynomials with integer coefficients Konstantine Zelator Mathematics, Computer Science and Statistics 212 Ben Franklin Hall Bloomsburg University 400 East Second Street

### The last three chapters introduced three major proof techniques: direct,

CHAPTER 7 Proving Non-Conditional Statements The last three chapters introduced three major proof techniques: direct, contrapositive and contradiction. These three techniques are used to prove statements

### 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

### 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

### 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

### BX in ( u, v) basis in two ways. On the one hand, AN = u+

1. Let f(x) = 1 x +1. Find f (6) () (the value of the sixth derivative of the function f(x) at zero). Answer: 7. We expand the given function into a Taylor series at the point x = : f(x) = 1 x + x 4 x

### Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given.

Polynomials (Ch.1) Study Guide by BS, JL, AZ, CC, SH, HL Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given. Sasha s method

### Factoring Polynomials

Factoring Polynomials Factoring Factoring is the process of writing a polynomial as the product of two or more polynomials. The factors of 6x 2 x 2 are 2x + 1 and 3x 2. In this section, we will be factoring

### GREATEST COMMON DIVISOR

DEFINITION: GREATEST COMMON DIVISOR The greatest common divisor (gcd) of a and b, denoted by (a, b), is the largest common divisor of integers a and b. THEOREM: If a and b are nonzero integers, then their

### PROOFS BY DESCENT KEITH CONRAD

PROOFS BY DESCENT KEITH CONRAD As ordinary methods, such as are found in the books, are inadequate to proving such difficult propositions, I discovered at last a most singular method... that I called the

### = 2 + 1 2 2 = 3 4, Now assume that P (k) is true for some fixed k 2. This means that

Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without

### Mathematics Review for MS Finance Students

Mathematics Review for MS Finance Students Anthony M. Marino Department of Finance and Business Economics Marshall School of Business Lecture 1: Introductory Material Sets The Real Number System Functions,

### Math Review. for the Quantitative Reasoning Measure of the GRE revised General Test

Math Review for the Quantitative Reasoning Measure of the GRE revised General Test www.ets.org Overview This Math Review will familiarize you with the mathematical skills and concepts that are important

### r + s = i + j (q + t)n; 2 rs = ij (qj + ti)n + qtn.

Chapter 7 Introduction to finite fields This chapter provides an introduction to several kinds of abstract algebraic structures, particularly groups, fields, and polynomials. Our primary interest is in

### SOLVING QUADRATIC EQUATIONS OVER POLYNOMIAL RINGS OF CHARACTERISTIC TWO

Publicacions Matemàtiques, Vol 42 (1998), 131 142. SOLVING QUADRATIC EQUATIONS OVER POLYNOMIAL RINGS OF CHARACTERISTIC TWO Jørgen Cherly, Luis Gallardo, Leonid Vaserstein and Ethel Wheland Abstract We

### 3.6 The Real Zeros of a Polynomial Function

SECTION 3.6 The Real Zeros of a Polynomial Function 219 3.6 The Real Zeros of a Polynomial Function PREPARING FOR THIS SECTION Before getting started, review the following: Classification of Numbers (Appendix,

### Projective Geometry - Part 2

Projective Geometry - Part 2 Alexander Remorov alexanderrem@gmail.com Review Four collinear points A, B, C, D form a harmonic bundle (A, C; B, D) when CA : DA CB DB = 1. A pencil P (A, B, C, D) is the

### Tim Kerins. Leaving Certificate Honours Maths - Algebra. Tim Kerins. the date

Leaving Certificate Honours Maths - Algebra the date Chapter 1 Algebra This is an important portion of the course. As well as generally accounting for 2 3 questions in examination it is the basis for many

### 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.

### Zeros of a Polynomial Function

Zeros of a Polynomial Function An important consequence of the Factor Theorem is that finding the zeros of a polynomial is really the same thing as factoring it into linear factors. In this section we

### Congruence properties of binary partition functions

Congruence properties of binary partition functions Katherine Anders, Melissa Dennison, Jennifer Weber Lansing and Bruce Reznick Abstract. Let A be a finite subset of N containing 0, and let f(n) denote

### The van Hoeij Algorithm for Factoring Polynomials

The van Hoeij Algorithm for Factoring Polynomials Jürgen Klüners Abstract In this survey we report about a new algorithm for factoring polynomials due to Mark van Hoeij. The main idea is that the combinatorial

### SMT 2014 Algebra Test Solutions February 15, 2014

1. Alice and Bob are painting a house. If Alice and Bob do not take any breaks, they will finish painting the house in 20 hours. If, however, Bob stops painting once the house is half-finished, then the

### 5. Factoring by the QF method

5. Factoring by the QF method 5.0 Preliminaries 5.1 The QF view of factorability 5.2 Illustration of the QF view of factorability 5.3 The QF approach to factorization 5.4 Alternative factorization by the

### p e i 1 [p e i i ) = i=1

Homework 1 Solutions - Sri Raga Velagapudi Algebra Section 1. Show that if n Z then for every integer a with gcd(a, n) = 1, there exists a unique x mod n such that ax = 1 mod n. By the definition of gcd,

### Prime Numbers and Irreducible Polynomials

Prime Numbers and Irreducible Polynomials M. Ram Murty The similarity between prime numbers and irreducible polynomials has been a dominant theme in the development of number theory and algebraic geometry.

### FACTORING SPARSE POLYNOMIALS

FACTORING SPARSE POLYNOMIALS Theorem 1 (Schinzel): Let r be a positive integer, and fix non-zero integers a 0,..., a r. Let F (x 1,..., x r ) = a r x r + + a 1 x 1 + a 0. Then there exist finite sets S

### Factoring Polynomials

UNIT 11 Factoring Polynomials You can use polynomials to describe framing for art. 396 Unit 11 factoring polynomials A polynomial is an expression that has variables that represent numbers. A number can

### FACTORING IN QUADRATIC FIELDS. 1. Introduction. This is called a quadratic field and it has degree 2 over Q. Similarly, set

FACTORING IN QUADRATIC FIELDS KEITH CONRAD For a squarefree integer d other than 1, let 1. Introduction K = Q[ d] = {x + y d : x, y Q}. This is called a quadratic field and it has degree 2 over Q. Similarly,

### Galois Theory. Richard Koch

Galois Theory Richard Koch April 2, 2015 Contents 1 Preliminaries 4 1.1 The Extension Problem; Simple Groups.................... 4 1.2 An Isomorphism Lemma............................. 5 1.3 Jordan Holder...................................

### ALGEBRAIC NUMBER THEORY AND QUADRATIC RECIPROCITY

ALGEBRAIC NUMBER THEORY AND QUADRATIC RECIPROCITY HENRY COHN, JOSHUA GREENE, JONATHAN HANKE 1. Introduction These notes are from a series of lectures given by Henry Cohn during MIT s Independent Activities

### Handout NUMBER THEORY

Handout of NUMBER THEORY by Kus Prihantoso Krisnawan MATHEMATICS DEPARTMENT FACULTY OF MATHEMATICS AND NATURAL SCIENCES YOGYAKARTA STATE UNIVERSITY 2012 Contents Contents i 1 Some Preliminary Considerations

### Factoring of Prime Ideals in Extensions

Chapter 4 Factoring of Prime Ideals in Extensions 4. Lifting of Prime Ideals Recall the basic AKLB setup: A is a Dedekind domain with fraction field K, L is a finite, separable extension of K of degree

### Solutions to TOPICS IN ALGEBRA I.N. HERSTEIN. Part II: Group Theory

Solutions to TOPICS IN ALGEBRA I.N. HERSTEIN Part II: Group Theory No rights reserved. Any part of this work can be reproduced or transmitted in any form or by any means. Version: 1.1 Release: Jan 2013