Ecuaciones polinomiales

c 2007 The Author(s) and The IMO Compendium Group

Polynomial Equations
Duˇan Djuki´ s c

Contents
1 2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems with Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2

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Introduction

The title refers to determining polynomials in one or more variables (e.g. withreal or complex coefficients) which satisfy some given relation(s). The following example illustrates some basic methods: 1. Determine the polynomials P for which 16P (x2 ) = P (2x)2 . • First method: evaluating at certain points and reducing degree.

Plugging x = 0 in the given relation yields 16P (0) = P (0)2 , i.e. P (0) = 0 or 16. (i) Suppose that P (0) = 0. Then P (x) = xQ(x) for somepolynomial Q and 16x2 Q(x2 ) = 4x2 Q(2x)2 , which reduces to 4Q(x2 ) = Q(2x)2 . Now setting 4Q(x) = R(x) gives us 1 16R(x2 ) = R(2x)2 . Hence, P (x) = 4 xR(x), with R satifying the same relation as P . (ii) Suppose that P (0) = 16. Putting P (x) = xQ(x) + 16 in the given relation we obtain 4xQ(x2 ) = xQ(2x)2 + 16Q(2x); hence Q(0) = 0, i.e. Q(x) = xQ1 (x) for some polynomial Q1 . Furthermore, x2 Q1 (x2 ) =x2 Q1 (2x)2 + 8Q1 (2x), implying that Q1 (0) = 0, so Q1 too is divisible by x. Thus Q(x) = x2 Q1 (x). Now suppose that xn is the highest degree of x dividing Q, and Q(x) = xn R(x), where R(0) = 0. Then R satisfies 4xn+1 R(x2 ) = 22n xn+1 R(2x)2 + 2n+4 R(2x), which implies that R(0) = 0, a contradiction. It follows that Q ≡ 0 and P (x) ≡ 16.

We conclude that P (x) = 16

n 1 4x

for some n ∈N0 .

• Second method: investigating coefficients.

We start by proving the following lemma (to be used frequently): Lemma 1. If P (x)2 is a polynomial in x2 , then so is either P (x) or P (x)/x.

Proof. Let P (x) = an xn + an−1 xn−1 + · · · + a0 , an = 0. The coefficient at x2n−1 is 2an an−1 , from which we get an−1 = 0. Now the coefficient at x2n−3 equals 2an an−3 ; hence an−3 = 0, and so on.Continuing in this manner we conclude that an−2k−1 = 0 for k = 0, 1, 2, . . . , i.e. P (x) = an xn + an−2 xn−2 + an−4 xn−4 + · · · . △ Since P (x)2 = 16P (x2 /4) is a polynomial in x2 , we have P (x) = Q(x2 ) or P (x) = xQ(x2 ). In the former case we get 16Q(x4 ) = Q(4x2 )2 and therefore 16Q(x2 ) = Q(4x)2 ; in the latter

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Olympiad Training Materials, www.imo.org.yu, www.imocompendium.comcase we similarly get 4Q(x2 ) = Q(4x)2 . In either case, Q(x) = R(x2 ) or Q(x) = xR(x2 ) for some polynomial R, so P (x) = xi R(x4 ) for some i ∈ {0, 1, 2, 3}. Proceeding in this k way we find that P (x) = xi S(x2 ) for each k ∈ N and some i ∈ {0, 1, . . . , 2k }. Now it is enough to take k with 2k > deg P and to conclude that S must be constant. Thus P (x) = cxi n for some c ∈ R. A simpleverification gives us the general solution P (x) = 16 1 x for 4 n ∈ N0 .

Investigating zeroes of the unknown polynomial is also counted under the first method. A majority of problems of this type can be solved by one of the above two methods (although some cannot, making math more interesting!).

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Problems with Solutions
1 1 1. Find all polynomials P such that P (x)2 + P ( x )2 = P (x2 )P ( x2 ).Solution. By the introducing lemma there exists a polynomial Q such that P (x) = Q(x2 ) 1 1 or P (x) = xQ(x2 ). In the former case Q(x2 )2 + Q( x2 ) = Q(x4 )Q( x4 ) and therefore 1 1 2 2 Q(x) + Q( x ) = Q(x )Q( x2 ) (which is precisely the relation fulfilled by P ), whereas in the 1 1 1 latter case (similarly) xQ(x)2 + x Q( x )2 = Q(x2 )Q( x2 ) which is impossible for the left and right hand sidehave odd and even degrees, respectively. We conclude that P (x) = Q(x2 ), where Q is also a solution of the considered polynomial equation. Considering the solution of the least degree we find that P must be constant. 2. Do there exist non-linear polynomials P and Q such that P (Q(x)) = (x − 1)(x − 2) · · · (x − 15)? Solution. Suppose there exist such polynomials. Then deg P · deg Q = 15, so...