Mathematical Olympiad Training Polynomials


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1 Mathematical Olympiad Training Polynomials Definition A polynomial over a ring R(Z, Q, R, C) in x is an expression of the form p(x) = a n x n + a n 1 x n a 1 x + a 0, a i R, for 0 i n. If a n 0, then n = deg p(x) is called the degree of p(x). A nonzero element r R is a polynomial of degree 0. The zero 0 R is a polynomial and its degree is negative infinity or undefined. The set of all polynomials over R in x is denoted by R[x]. For any f(x), g(x) R[x]. 1. deg(f(x) + g(x)) max{deg f(x), deg g(x)}. deg f(x)g(x) = deg f(x) + deg g(x) Many properties of integers have analogues for polynomials. Properties 1. Sum, difference and product of polynomials are polynomials.. Let f(x), g(x) R[x], we say that f(x) divides g(x) if there exists nonzero q(x) R[x] such that g(x) = f(x)q(x). If f(x) divids g(x), we say that f(x) is a divisor of g(x) and write f(x) g(x). 3. Let f(x), g(x) R[x], then there exists q(x), r(x) R[x] such that f(x) = q(x)g(x) + r(x) and deg r(x) < deg g(x). (We say that R[x] is a Euclidean domain.) 1
2 4. A polynomial p(x) is said to be irreducible if it cannot be factorized into product of polynomials of positive degree. It is said to be reducible if it is not irreducible.. A polynomial p(x) is called a prime polynomial if p(x) f(x)g(x) implies p(x) f(x) or p(x) g(x). It is easy to see that a prime polynomial is irreducible and the converse is true, but less obvious, only when R is a UFD. 6. Fix f(x), g(x) R[x], the following statements for d(x) R[x] are equivalent. (a) For any nonzero p(x) R[x], p(x) f(x) and p(x) g(x) imply p(x) d(x). (b) d(x) is a polynomial of maximal degree satisfying the properties that d(x) f(x) and d(x) g(x). (c) d(x) is a nonzero polynomial of minimal degree such that there exists a(x), b(x) R[x] with d(x) = p(x)f(x) + q(x)g(x). If d(x) satisfies one, hence all, of the above properties, we say that d(x) is a greatest common divisor (GCD) of f(x) and g(x) and write d(x) = (f(x), g(x)). GCD always exists and is unique up to a unit (an invertible element in R) for every nonzero polynomials f(x) and g(x). 7. Any nonzero p(x) R[x] can be factorized uniquely (up to unit and permutation) into product of irreducible polynomials. (We say that R[x] is a unique factorization domain (UFD). To be more precise, R[x] is a UFD if R is a UFD and it is well known that Z, Q, R, C are all UFD.)
3 Theorem (Remainder Theorem) When p(x) R[x] is divided by x a, the remainder is p(a). In particular x a divides p(x) if and only if p(a) = 0. The notion of reducibility of polynomial depends on the ring of coefficients R. For example, x is irreducible over Z but is reducible over R and x + 1 is irreducible over R but is reducible over C. Proposition 1. (Gauss Lemma) If a polynomial f(x) Z[x] is reducible over Q, i.e. there exists p(x), q(x) Q[x] of positive degree such that f(x) = p(x)q(x), then f(x) is reducible over Z.. (Eisenstein Criterion) Let f(x) = a n x n + a n 1 x n a 1 x + a 0 Z and p be a prime number. Suppose (a) p a n (b) p a k for 0 k n 1 (c) p a 0 Then f(x) is irreducible over Q. The most important theorem about polynomials is the following. Theorem (Fundamental Theorem of Algebra) A polynomial of degree n over C has n zeros on C counting multiplicity. 3
4 Another way of stating Fundamental Theorem of Algebra is every complex polynomial p(x) C[x] of degree n can be factorized into product of linear polynomials, i.e. there exists a, α 1, α,, α n C such that p(x) = a(x α 1 )(x α ) (x α n ). Corollary 1. If a polynomial of degree not greater than n has n + 1 distinct zeros, then it is the zero polynomial.. Two polynomials of degree not greater than n are equal if they have the same value at n + 1 distinct numbers. For polynomials with real coefficients p(x) R[x], we have Propositions 1. Let p(x) R[x] and α C. If p(α) = 0, then p(ᾱ) = 0, where ᾱ denotes the complex conjugate of α.. Any p(x) R[x] can be factorized into product of quadratic and linear polynomials over R. A polynomial p(x) R[x] can also be considered as a function p : R R. One can always find a unique polynomial of degree n with n + 1 prescribed values. Lagrange Interpolation Formula Given any distinct x 0, x 1,, x n R and any y 0, y 1,, y n R. There exists 4
5 unique polynomial p(x) R[x] of degree n such that p(x i ) = y i for all i = 0, 1,, n. In fact we have p(x) = where the notation n i=0 (x x 0 )(x x 1 ) (x xi ) (x x n )y i, (x i x 0 )(x i x 1 ) (xi x i ) (x i x n ) (x x i ) means that (x x i ) is absent. Another way of writing the interpolation formula is y x n x x 1 n y 0 x 0 x 0 x 0 1 n y 1 x 1 x 1 x 1 1 = n y n x n x n x n 1 Proposition A rational polynomial p(x) Q[x] takes integer values on all integers if and only if ) ) ) ) p(x) = a 0 ( x 0 + a 1 ( x 1 + a ( x + + a n ( x n for some a 0, a 1, a,, a n Z, where ( ) x k = x(x 1)(x ) (x k+1), k 0 and ( x k! 0) = 1. Symmetric Polynomials 1. If a polynomial p(x 1, x,, x n ) in n variables satisfies p(x 1,, x i,, x j,, x n ) = p(x 1,, x j,, x i,, x n ) for any i j, then p(x 1, x,, x n ) is called a symmetric polynomial.. The elementary symmetric polynomials of degree k, k = 0, 1,,, n, in n variables x 1, x,, x n is defined by σ k (x 1, x,, x n ) = x i1 x i x ik. 1 i 1 <i < <i k n
6 For example σ 0 = 1, σ 1 = x 1 + x + + x n, σ = x 1 x + x 1 x x n 1 x n,, σ n = x 1 x x n. 3. The kth power sum, k 0, of n variables x 1, x,, x n is defined by S k (x 1, x,, x n ) = x k i = x k 1 + x k + + x k n. 1 i n Fundamental Theorem of Symmetric Polynomials Any symmetric polynomial can be expressed as a polynomial in elementary symmetric polynomials in (or power sum of) those variables. NewtonGirard Formulae Let S k be the kth power sum and σ k be the elementary symmetric polynomials of degree k in x 1, x,, x n. Then for any positive integer m, m 1 ( 1) k σ k S m k + ( 1) m mσ m = 0. k=0 Here σ 0 = 1, σ k = 0 when k > n. Example: For m = 1,, 3,, n, we have S 1 σ 1 = 0 S σ 1 S 1 + σ = 0 S 3 σ 1 S + σ S 1 3σ 3 = 0. S n σ 1 S n ( 1) n 1 σ n 1 S 1 + ( 1) n nσ n = 0 6
7 For m > n, we have S m σ 1 S m 1 + σ S m + + ( 1) n σ n S m n = 0 Vieta s Formulae Let p(x) = a n x n + a n 1 x n a 1 x + a 0 C be a polynomial over C of degree n and α 1, α,, α n be the roots of p(x) = 0. Let σ k be the elementary symmetric polynomials of degree k in α 1, α,, α n. Then we have Application to recurrence sequences σ k = ( 1) k a n k a n. Combining Vieta s formula with NewtonGirard formula, we get an obvious relation a n S m + a n 1 S m 1 + a n S m + + a 0 S m n = 0. This means that the sequence S 0, S 1, S, satisfies the recurrence relation a n S k+n + a n 1 S k+n 1 + a n S k+n + + a 0 S k = 0, for k 0. More generally, fix any A 1, A,, A n C, let x k = A 1 α 1 k + A α k + + A n α n k. ( ) Then x 0, x 1, x, satisfies the recurrence relation a n x k+n + a n 1 x k+n 1 + a n x k+n + + a 0 x k = 0, for k 0. ( ) Equation p(x) = a n x n + a n 1 x n a 0 = 0 is called the characteristic equation. By solving it, we can find the general solution ( ) of the recurrence relation ( ). 7
8 Example: Fibonacci sequence The Fibonacci sequence 0, 1, 1,, 3,, 8, 13, is defined by the recurrence equation { F k = F k 1 + F k, for k > 1. F 0 = 0, F 1 = 1 The characteristic equation is x x 1 = 0 and its roots are 1±. Solving { A 1 + A = F 0 = 0, A 1 α 1 + A α = F 1 = 1 we have A 1 = 1, A = 1 and F k = 1 α 1 k 1 α k ( = 1 ( = ) k ( 1 + ) k ) k where the notation [x] means the largest integer not greater than x. Example 1 Find all rational polynomials p(x) = x 3 + ax + bx + c such that a, b, c are roots of the equation p(x) = 0. Solution By Vieta s formula a + b + c = a ab + bc + ca = b abc = c From the third equation (ab + a)c = 0. Thus ab = 1 or c = 0. If c = 0, then a + b = a and ab = b. Hence (a, b, c) = (0, 0, 0) or (1,, 0). 8
9 If ab = 1, then c = a b and 1 + b( a b) + ( a b)a = b a + b + b = 0 a 4 a + a b + a b = 0 a 4 a a + 1 = 0 Since a is rational, the only solution is a = 1 and (a, b, c) = (1, 1, 1). Hence the solution of the problem is (a, b, c) = (0, 0, 0), (1,, 0) or (1, 1, 1). Example Given that p(x) is a polynomial of degree n such that p(k) = k for k = 0, 1,, n. Find p(n + 1). Solution Take p(x) = ( ) x + 0 ( ) x + 1 ( ) x + + ( ) x, n then p(x) satisfies the condition of the problem and p(n + 1) = = ( ) ( ) ( ) ( ) n + 1 n + 1 n + 1 n n ( ) ( ) ( ) ( ) n + 1 n + 1 n + 1 n n = n+1 1 ( ) n n + 1 Example 3 x + y + z = 1 Given that x + y + z = 3 x 3 + y 3 + z 3 = 7. Find the value of x + y + z. 9
10 Solution Let S k and σ k be the kth power sum and symmetric sum of x, y, z. Then by Newton formula S 1 σ 1 = 0 S σ 1 S 1 + σ = 0 S 3 σ 1 S + σ S 1 3σ 3 = 0 σ 1 = 1 σ = 1 σ 3 = 1 Thus x, y, z are roots of the equation t 3 t t 1 = 0 and S k satisfies the recurrence relation S k+3 = S k+ + S k+1 + S k, k 0. Therefore S 4 = = 11 and S = = 1. Example 4 If x, y are nonzero numbers with x + xy + y = 0. Find ( x x+y )001 + ( y x+y )001. Solution Observe that x + y = 1 and x+y x+y x x+y y x+y = xy x +xy+y = xy xy. = 1. We know that x, x+y y x+y are roots of t t + 1 = 0. Thus S k = ( x x+y )k + ( y x+y )k satisfies the recurrence relation { S k+ = S k+1 S k, k 0 S 0 =, S 1 = 1 Then the sequence {S k }, k 0, is, 1, 1,, 1, 1,, 1, and S k = S l if k l(mod6). Therefore S 001 = S 3 =. Example (IMO 1999) Let n be a fixed integer. Find the least constant C such that the inequality ( ) 4 x i x j (x i + x j ) C x i 1 i<j n 1 i n holds for any x 1, x, x 3,, x n 0. For this constant C, characterize the instances of equality. 10.
11 Solution Since the inequality is homogeneous, we may assume that S 1 = σ 1 = 1. By Newton formula, S 4 σ 1 S 3 + σ S σ 3 S 1 + 4σ 4 = 0. Then x i x j (x i + x j ) = ( x ) 3 i x j 1 i<j n 1 i n j i = x 3 i (1 x i ) 1 i n = S 3 S 4 = σ S σ 3 S 1 + 4σ 4 = σ (1 σ ) + 4σ 4 σ 3 σ The last inequality holds since σ (1 σ ) = σ ( 1 σ ) 1 (σ + ( 1 σ )) = 1 8 by AMGM inequality and ( ) ( ) 3 ( ) n σ 4 σ 4σ 4 = 4 ( 4 4 n ( n 4) 4) ( n 4 4 ) σ3 ( n 3 ) σ 1 n = n 3 n σ 3σ 1 σ 3 σ 1 by symmetric mean inequality and the equality holds if and only if σ = 1 4 and σ 3 = σ 4 = 0. Therefore the least value of C is 1 8 and the equality holds for the original inequality if and only if two x i are equal and the rest are zero. Example 6 (IMO 006) Let P (x) be a polynomial of degree n > 1 with integer coefficients and let k be a positive integer. Consider the polynomial Q(x) = P (P (... P (P (x))... )), where 11
12 P occurs k times. Prove that there are at most n integers t such that Q(t) = t. Solution (by Tsoi Yun Pui) Denote P (P ( P (x)) ) by Q }{{} k (x). If there is at most one integer t satisfying k Q k (t) = t, then we are done. Otherwise, let s, t be integers such that Q k (s) = s, Q k (t) = t. As P (x) is a polynomial with integral coefficients, u v P (u) P (v) for any integers u, v. So s t P (s) P (t) Q (s) Q (t) Q k (s) Q k (t) = s t, and hence both s t P (s) P (t) and P (s) P (t) s t. This implies that i.e. P (s) P (t) = s t or P (s) P (t) = t s. P (s) s = P (t) t or P (s) + s = P (t) + t ( ) It is impossible to have P (s) P (t) = s t and P (u) P (t) = t u for distinct integral roots s, u, t of the equation Q k (x) = x. Otherwise P (s) P (u) = s t (t u) = s + u t. But P (s) P (u) = s u or u s. In either cases, it yields s = t or u = t. Contradiction. So only one equation in ( ) is true for all the integer roots of Q k (x) = x. In either cases, let us fix t. Then all integral roots of Q k (x) = x are also, at the same times, roots of the equation P (x) x = 0 or P (x) + x = 0. Note that P (x) x and P (x) + x are polynomials of degree n. So there is at most n such roots. Hence there are at most n integers t such that Q(t) = t. 1
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