1987 AMC 12 Problem 24

Attempt Problem 24 of the 1987 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1987 AMC 12 solutions, or check the answer key.

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24.

How many polynomial functions ff of degree ≥1\ge1 satisfy f(x2)=[f(x)]2=f(f(x))? f(x^2)=[f(x)]^2=f(f(x))?
00
11
22
finitely many but more than 22
infinitely many
Answer: B
Concepts:polynomial degreeleading coefficientfunctional equation
Small hint:
Compare the degrees and leading coefficients of the three polynomials
Big hint:
After proving ff is a monic quadratic, compare coefficients in f(x2)=[f(x)]2f(x^2)=[f(x)]^2
Solution:
Let ff have degree n≥1n\ge1 and leading coefficient a.a. The three expressions have degrees 2n,2n,n2,2n,2n,n^2, so n2=2nn^2=2n and therefore n=2.n=2. Comparing leading coefficients in f(x2)=[f(x)]2f(x^2)=[f(x)]^2 gives a=a2,a=a^2, hence a=1.a=1.
Write f(x)=x2+bx+c.f(x)=x^2+bx+c. Then x4+bx2+c=(x2+bx+c)2. x^4+bx^2+c=(x^2+bx+c)^2. The cubic coefficient forces b=0,b=0, and then the quadratic coefficient forces c=0.c=0. Thus the only candidate is f(x)=x2,f(x)=x^2, which indeed satisfies all three expressions. There is exactly one function.
Thus the correct answer is B.
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