1987 AHSME Problems/Problem 24

Problem

How many polynomial functions $f$ of degree $\ge 1$ satisfy $f(x^2)=[f(x)]^2=f(f(x))$ ?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \text{finitely many but more than 2}\\ \qquad \textbf{(E)}\ \infty$

Solution

Let $f(x) = \sum_{k=0}^{n} a_{k} x^{k}$ be a polynomial satisfying the condition, so substituting it in, we find that the highest powers in each of the three expressions are, respectively, $a_{n}x^{2n}$ , $a_{n}^{2}x^{2n}$ , and $a_{n}^{n+1}x^{n^{2}}$ . If polynomials are identically equal, each term must be equal, so we get $2n = n^2$ and $a_{n} = a_{n}^{2} = a_{n}^{n+1}$ , so since $n \geq 1$ , we must have $n = 2$ , and since $a_{n} \neq 0$ , we have $a_{n} = 1$ . The given condition now becomes $x^4 + bx^2 + c \equiv (x^2 + bx + c)^2$ , so we must have $b = 0$ , or else the right-hand side would have a cubic term that the left-hand side does not. Thus we get $x^4 + c \equiv (x^2 + c)^2$ , so we must have $c = 0$ , or else the right-hand side would have an $x^2$ term that the left-hand side does not. Thus the only possibility is $f(x) = x^2$ , i.e. there is only $1$ solution, so the answer is $\boxed{\text{B}}$ .

1987 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1987 AHSME Problems/Problem 24

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