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If $f(x)$ is a polynomial satisfying $ f(x)f(\frac 1x) = f(x)+f(\frac 1x)$ and $f(3)=28$, then how could we find $f(4)$ ?

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    ... and then we know $f(x)-1 = x^m$ for some $m$ or $f(x)-1 = -x^m$ for some $m$ as the only ways to get it. Assuming INTEGER COEFFICIENTS (which is not stated), from $\pm 3^m = 27$ we have a good guess for $m$ and the sign.2011-11-22

1 Answers 1

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Solving the functional equation for $f\left(\frac{1}{x} \right) = \frac{f(x)}{f(x)-1}$. This means that $f(x)-1$ must be a monomial. Let $f(x) = 1 + c x^d$. Then $ c \left( \frac{1}{x} \right)^d +1 = \frac{1}{c} \left( \left( \frac{1}{x} \right)^d + c \right) $ This, implies $c^2 = 1$. Now use $f(3) = 28$ to determined $c$ and $d$. Since $28 = 1 + 1 \times 3^3$, we conclude $c=1$ and $d=3$.

Thus $f(4) = 1 + 4^3 = 65$.

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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/1839/discussion-between-sasha-and-max)2011-11-22