2007 AIME II Problems/Problem 14

Problem

Let $f(x)$ be a polynomial with real coefficients such that $\displaystyle f(0) = 1,$ $\displaystyle f(2)+f(3)=125,$ and for all $x$ , $\displaystyle f(x)f(2x^{2})=f(2x^{3}+x).$ Find $\displaystyle f(5).$

Solution

Note:The following solution(s) are non-rigorous.

Substitute the values $x = \pm i$ . We find that $\displaystyle f(i)f(-2) = f(-i)$ , and that $\displaystyle f(-i)f(-2) = f(i)$ . This means that $f(i)f(-i)f(-2)^2 = f(i)f(-i) \Longrightarrow f(i)f(-i)(f(-2) - 1) = 0$ . This suggests that $\pm i$ are roots of the polynomial, and so $\displaystyle (x - i)(x + i) = x^2 + 1$ will be a root of the polynomial.

The polynomial is likely in the form of $f(x) = (x^2 + 1)g(x)$ ; $g(x)$ appears to satisfy the same relation as $f(x)$ , so it also probably has the same roots. Thus, $f(x) = (x^2 + 1)^nh(x)$ is the solution. Guessing values for $h(x)$ , try $h(x) = 1$ . Checking a couple of values shows that $f(x) = (x^2 + 1)^2$ works, and so the solution is $f(5) = 676$ .

2007 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2007 AIME II Problems/Problem 14

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