2007 AIME II Problems/Problem 14

Problem

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

Solution

Let $r$ be a root of $f(x)$ . Then we have $f(r)f(2r^2)=f(2r^3+r)$ ; since $r$ is a root, we have $f(r)=0$ ; therefore $2r^3+r$ is also a root. Thus, if $r$ is real and non-zero, $|2r^3+r|>r$ , so $f(x)$ has infinitely many roots. Since $f(x)$ is a polynomial (thus of finite degree) and $f(0)$ is nonzero, $f(x)$ has no real roots.

Note that $f(x)$ is not constant. We then find two complex roots: $r = \pm i$ . We find that $f(i)f(-2) = f(-i)$ , and that $f(-i)f(-2) = f(i)$ . This means that $f(i)f(-i)f(-2)^2 = f(i)f(-i) \Longrightarrow f(i)f(-i)(f(-2)^2 - 1) = 0$ . Thus, $\pm i$ are roots of the polynomial, and so $(x - i)(x + i) = x^2 + 1$ will be a factor of the polynomial. (Note: This requires the assumption that $f(-2)\neq1$ . Clearly, $f(-2)\neq-1$ , because that would imply the existence of a real root.)

The polynomial is thus in the form of $f(x) = (x^2 + 1)g(x)$ . Substituting into the given expression, we have

$(x^2+1)g(x)(4x^4+1)g(2x^2)=((2x^3+x)^2+1)g(2x^3+x)$ $(4x^6+4x^4+x^2+1)g(x)g(2x^2)=(4x^6+4x^4+x^2+1)g(2x^3+x)$

Thus either $4x^6+4x^4+x^2+1=(4x^4+1)(x^2+1)$ is 0 for any $x$ , or $g(x)$ satisfies the same constraints as $f(x)$ . Continuing, by infinite descent, $f(x) = (x^2 + 1)^n$ for some $n$ .

Since $f(2)+f(3)=125=5^n+10^n$ for some $n$ , we have $n=2$ ; so $f(5) = \boxed{676}$ .

Comment: The answer is clearly correct, but the proof has a gap, i.e. there is no reason that $f(-2)\neq1$ . Since $f(x)$ has no real roots, the degree must be even. Consider $g(x)= f(x)/f(-x)$ . Then since $f$ is non-zero, $g(x)=g(2x^3+x)$ . Now the function $2x^3+x$ applied repeatedly from some real starting value of x becomes arbitrarily large, and the limit of $g(x)$ as $|x|$ approaches infinity is 1, so $g(x)$ =1 for all x, or $f(x)=f(-x)$ . Then $f(x)=h(x^2+1)$ for some polynomial $h(x)$ , and $h(x^2+1)h(4x^4+1)=h(4x^6+4x^4+x^2+1) = h((x^2+1)(4x^4+1))$ . Now suppose h has degree m. It is clearly monic. Assume that the next highest non-zero coefficient in h is k. Then, subtracting $((x^2+1)(4x^4+1))^m$ from both sides of the equation yields a polynomial equality with degree $4m+2k$ on the left and degree $6k$ on the right, a contradiction. So $h(x)=x^m$ , and $f(x)=(1+x^2)^m$ .

Solution 2

Let $x$ be $\frac{i}{\sqrt{2}}$

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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Solution

Solution 2

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