find all entire functions $f$ satisfying $f(\sqrt{n})=n^2$ for every positive integer $n$ and that are bound by $e^{3|z|}$ everywhere on $\Bbb C$ As the title suggests, I'm looking for to find all entire functions such that $$f(\sqrt{n})=n^2$$ for every positive integer $n$ and that  $$|f(z)|\leq e^{3|z|}$$ everywhere on $\Bbb C$.
I believe $f(z)=z^4$ is such a function, hence I would like to prove that this is the unique entire function satisfying these properties. My idea is  to use the Identity principle applied to something like $$ g(z)= \frac{1}{f(1/z)}-z^4$$ which has zeroes on the sequence $\{1/\sqrt{n}\}$ but there are some problems: is $g$ defined on $0$? is $g$ holomorphic near $0$? and I don't know how to deal with them.
EDIT Using Hadamard factorisation theorem we can write: $$f(z)=z^me^{az+b}\prod_{n=1}^{\infty}E_1\left(\frac{z}{a_n} \right)$$ since $f$ has order one by assumption. Now the condition $f(\sqrt{n})=n^2$ forces $az+b=0$ and I guess even all the factors $E_1$ are forced to be zero by it. Hence $m=4$ and we conclude.
 A: It is immediate that $f(z) = z^4$ has the desired values at $\sqrt{n}$, and it is not hard to check that it also satisfies the inequality $\lvert f(z)\rvert \leqslant e^{3\lvert z\rvert}$ for all $z\in \mathbb{C}$. In fact, locating the maximum of $r \mapsto r^4e^{-3r}$ on $[0,+\infty)$ shows $\lvert z^4\rvert \leqslant \bigl(\frac{4}{3e}\bigr)^4 e^{3\lvert z\rvert} < \frac{1}{16}e^{3\lvert z\rvert}$.
It is easy to find other entire functions that satisfy $g(\sqrt{n}) = n^2$ for all positive integers $n$ if we don't impose the restriction that $\lvert g(z)\rvert \leqslant e^{3\lvert z\rvert}$, e.g. $g(z) = f(z) + \sin (\pi z^2)$ or $g(z) = f(z) + \frac{1}{\Gamma(-z^2)}$. Thus the proof of uniqueness must make essential use of the inequality $\lvert f(z)\rvert \leqslant e^{3\lvert z\rvert}$.
The identity theorem would settle the matter by looking at
$$\frac{1}{f(1/z)} - z^4$$
if we knew that this function has a removable singularity at $0$. That is the case if and only if $f$ has a non-essential singularity at $\infty$, i.e. $f$ is a polynomial. But that isn't implied by the inequality, there are lots of transcendental entire functions satisfying that bound, like $z \mapsto e^{az}$ with $\lvert a\rvert \leqslant 3$ or, less obviously, $z \mapsto\cos \sqrt{z}$. So we need both, the growth restriction and the values at $\sqrt{n}$ to deduce uniqueness.
Hadamard's theory of the order and genus of entire functions is a powerful tool we can use to that effect. Since Hadamard's theory makes essential use of the zeros of entire functions, we need to rewrite the problem in a form that prescribes zeros of an entire function. Since we already have one solution, we can subtract that from an unknown solution $f$ and the problem becomes to determine all entire functions $h$ with $h(\sqrt{n}) = 0$ for all positive integers $n$ and $\lvert h(z)\rvert \leqslant 2e^{3\lvert z\rvert}$ for all $z \in \mathbb{C}$. [And then we must find those $h$ that additionally satisfy $\lvert z^4 + h(z)\rvert \leqslant e^{3\lvert z\rvert}$, but it's reasonable to expect that that problem will be easier to solve.]
If $h$ is such a function and $h \not\equiv 0$ (order and genus of the zero function aren't defined, so we exclude it from consideration here), then $\lvert h(z)\rvert \leqslant Ae^{B\lvert z\rvert}$ immediately implies that the order of $h$ is $\leqslant 1$. Since the genus of an entire function ($\not\equiv 0$) doesn't exceed its order, the genus $\gamma$ of $h$ is also at most $1$. By definition of the genus,
$$\sum \frac{1}{\lvert a_k\rvert^{\gamma + 1}} < +\infty$$
if $(a_k)$ are the zeros - with the exception of $0$, if $h(0) = 0$ - of $h$, listed according to their multiplicities. In particular,
$$\sum_{n = 1}^{\infty} \frac{1}{(\sqrt{n})^{\gamma+1}} < +\infty\,.$$
But this is the case if and only if $\frac{\gamma+1}{2} > 1$, i.e. $\gamma > 1$. That contradicts the inequality $\gamma \leqslant 1$ derived from the order, and hence there is no entire function vanishing an $\sqrt{n}$ for all positive integers $n$ and satisfying $\lvert h(z)\rvert \leqslant Ae^{B\lvert z\rvert}$ for some $A,B$ except the zero function.
Thus indeed $f(z) = z^4$ is the unique entire function of order $< 2$ with $f(\sqrt{n}) = n^2$ for all positive integers $n$.
