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I'm trying to prove that the entire functions such that

\begin{equation*} n^2f(1/n)^3+f(1/n)=0 \end{equation*}

for all $n\in\mathbb{N}$, are constant. I suppose I should prove that $f$ is bounded so that I can apply Liouville's theorem, but I don't see how.

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    $\begingroup$ You should aim for the identity theorem rather than Liouville's theorem. But the given relation doesn't imply that $f$ is constant, look for a non-constant $f$ satisfying it. $\endgroup$ – Daniel Fischer May 22 '15 at 14:20
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    $\begingroup$ The given equation implies $f(1/n)=0\lor n^2f(1/n)^2+1=0$, i.e. $f(1/n)=0\lor f(1/n)=\pm i/n$ ?! $\endgroup$ – Yves Daoust May 22 '15 at 14:29
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    $\begingroup$ Per @DanielFischer's comment, assume the equation $\frac{(f(z))^3}{z^2} + f(z)=0$ is satisfied for all nonzero values of $z$, and see what $f$ can be. $\endgroup$ – Dustan Levenstein May 22 '15 at 14:30
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$f(z)= iz$ will satisfie this equation and it is a non constant entire function

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