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Let $f(x)$ be a polynomial with real coefficients such that $f(0)=1$, $f(2)+f(3)=125$ and $f(x)f(2x^2)=f(2x^3+x)$. Find $f(5)$. (2007 AIME II, Problem 14)

I've tried many ways of solving but I never got to the answer. How do I do it?

Edit: I substituted $x=i$ and found that $f(2)=1$. But I don't know if a polynomial can have a complex domain. Can it?

Also found out that the function is odd (f(-x)=-f(x)) I think.

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$f(x)=1+2x^2+x^4$ seems to do the trick.

There is no systematic way to solve an arbitrary functional equation. Given that $f$ is a polynomial, you might want to derive some equations on its coefficients. Or you may just start playing with random polynomials right away. You'll discover pretty soon that $x^2+1$ fits the functional equation, but not the starting conditions. Wait, but then maybe we'll try a square of that?

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  • $\begingroup$ For this to be an answer, you should also prove that this particular polynomial is the only one that satisfies the conditions... otherwise, there very well could be another polynomial, with a different value at $5$. $\endgroup$
    – Clement C.
    Dec 27 '16 at 17:21
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    $\begingroup$ That's right. So we'll have to fiddle around with the coefficients, after all. $\endgroup$ Dec 27 '16 at 17:45

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