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This is a complex analysis problem. Let $f$ be an entire function and $f^2+2f+1$ be a polynomial. Prove that $f$ is a polynomial.

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an entire function with polynomial growth is a polynomial. –  Will Jagy Jan 14 '13 at 5:48
    
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$f+1$ is a polynomial if and only if $f$ is, so a good first step would be to notice that this is equivalent to showing that if $f^2$ is a polynomial and $f$ is entire, then $f$ is a polynomial. –  Jonas Meyer Jan 14 '13 at 6:15
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2 Answers

We have $(f(z)+1)^2=p(z)$ where p(z) is a polynomial. Then

$$\infty=\lim\limits_{z\to \infty} |p(z)|=\lim\limits_{z\to \infty}|f(z)+1|^2\implies \lim\limits_{z\to \infty}|f(z)|=\infty $$

Therefore $f$ is an entire function with a pole at $\infty$. So $f$ is a polynomial.

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Very Elegant Solution! –  user27456 Jan 14 '13 at 22:55
    
are you using the result: an entire function with pole of order $m$ at infinity must be a polynomial of degree atmost $m$? –  Bunuelian Trick May 3 '13 at 9:10
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Let

$$(f+1)^2=f^2+2f+1=a(z-z_1)^{n_1}..(z-z_k)^{n_k}$$

We claim that all $n_j$ are even, then the rest is simple (you still need to argue why you only get one root of a, not both).

Assume by contradiction that some $n_j=2l+1$. Then $$\frac{f(z)+1}{(z-z_{n_j})^l}$$

has a zero at $z_{n_j}$ thus $z_{n_j}$ is a zero of order at least $l+1$ for $f(z)+1$. This implies that $z_{n_j}$ is a zero of order at least $2l+2$ for $f(z)^2+2f(z)+1$, contradiction.

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