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$f$ is entire and there exists $C>0$ and $M>0$ such that $|f(z)|leq C|z|^{5/2}$ for all $zinmathbb{C}$ where $|z|>M$. Prove that $f$ is polynomial of degree two.

I don't have a clear idea but given an entire function and we can bound it $dfrac{|f(z)|}{|z|^{5/2}}leq C$, and showing that it has removable singularity at $z=0$, using Riemann's theorem, I can show that $dfrac{f(z)}{z^{5/2}}$ is a constant by Liouville's theorem. However, I'm not sure what I should do to show that it is intact a polynomial.

Almost certainly, the author of this exercise expected you to use Cauchy's estimates:
$$frac{|f^{(n)}(a)|}{n!}lefrac{sup_t{|f(a+re^{it}|)}}{r^n}$$
for $r>0$. Here, then
$$frac{|f^{(n)}(0)|}{n!}lefrac{Cr^{5/2}}{r^n}$$
for $r>R$. If $nge3$ and we let $rtoinfty$ we get $f^{(n)}(0)=0$.
In the power series $f(z)=sum a_nz^n$ then $a_3=a_4=cdots=0$.

Let $f(z) = sum_{n=0}^infty a_n z^n$, and let $g(z) = sum_{k=0}^infty a_{k+3} z^k$, so that
$$
[f(z) - a_0 - a_1 z - a_2 z^2] = z^3 g(z).
$$
The function $g$ is entire and, by assumption, $lim_{|z| to +infty} g(z) = 0$:
$$
|g(z)| = frac{|f(z) - a_0 - a_1 z - a_2 z^2|}{|z|^{5/2}} cdot frac{1}{|z|^{1/2}}
leq left(C + frac{|a_0| + |a_1|, |z| + |a_2|, |z|^2}{|z|^{5/2}}right)frac{1}{|z|^{1/2}} to 0.
$$
Hence, by Liouville's theorem, we must have $g = 0$.

It may be interesting to note this follows from basic facts about Fourier series (Parseval) with more or less no complex analysis. Say $f(z)=sum c_nz^n$. Then $$C^2r^5gefrac1{2pi}int_0^{2pi}|f(re^{it})|^2,dt=sum_j|c_j|^2r^{2j}ge|c_n|^2r^{2n};$$hence $c_n=0$ for $nge3$.

Your approach can also be made to work, but you need to show that $h(z)=frac{f(z)}{z^3}$ has a removable singularity at $z=0$.

This can be done as follows:
Let $g(z)=frac{f(z)}{z^2}$. Then, as $lim_{z to 0}g(z)=0$, $g$ has a removable singularity at $z=0$, and if we remove it we have $g(0)=0$.

Now since, after removing the singularity, $g$ is entire and $g(0)=0$, then $h(z)=frac{g(z)}{z}$ also has a removable singularity at $0$.

Then $h$ becomes an entire function and
$$lim_{z to infty} h(z)=0$$
From here it is easy to conclude that $h$ is constant.