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Difference between revisions of "1987 AHSME Problems/Problem 24"

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

Latest revision as of 15:46, 1 March 2018

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

See also

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
All AHSME Problems and Solutions

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