1987 AHSME Problems/Problem 24

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $\boxed{\text{B}}$.