This problem is based on an question of Alibaba math contest: if $h(x)$ is an entire function on $\mathbb{C}$, and it is a solution of polynomial equation, i.e., for some non-vanishing $g(x,y)\in \mathbb{C}[x,y]$, $g(x,h(x))$ is always zero. Then $h(x)$ it self must be a polynomial.
My proof is $sup_{|x|=r}|h(x)|$ must be bounded by polynomial, then itself must be polynomial. But I'm uncomfortable with that proof. Are there any other ways to prove it? Say GAGA principle?
Writing $g(z,h(z))=0$ as $P_n(z)h^n(z)+P_{n-1}(z)h^{n-1}(z)+...P_0(z)=0, n \ge 1, P_n \ne 0$ and then on $|z|>R$ where $P_n$ doesn't vanish, getting:
$h^n(z)=-\frac{P_{n-1}(z)h^{n-1}(z)}{P_n(z)}-...-\frac{P_0(z)}{P_n(z)}$
we let $N= \max \deg P_k, 0 \le k \le n-1$.
Assuming $h$ is not a polynomial, it follows that $\infty$ is essential singularity so there is $|z_k| \to \infty, |h(z_k)| > |z_k|^{N+1}$.
Dividing the equation above by $h^{n-1}(z_k)|z|^{N+1}$ and using $|\frac{1}{h(z_k)}| < \frac{1}{|z_k|^{N+1}}$ and applying the triangle inequality we get
$1 < \frac{|h(z_k)|}{|z_k|^{N+1}} <|\frac{P_{n-1}(z_k)}{|z_k|^{N+1}P_n(z_k)|}|+|\frac{P_{n-2}(z_k)}{|z_k|^{2(N+1)}P_n(z_k)}|+...|\frac{P_{0}(z_k)}{|z_k|^{(n-1)(N+1)}P_n(z_k)}|$
Letting $|z_k| \to \infty$, RHS above goes to zero as each numerator has degree at most $N$ and $P_n \ne 0$ means $|P_n(z_k)| \ge c>0$ when $|z_k|>R$ as either $P_n$ is constant nonzero or $|P_n(z_k)| \to \infty$
This obviously leads to the contradiction $1<0$, so indeed $h$ must be a polynomial