An entire function $g$ such that $|g(z^2)| \leq e^{|z|}$ and $g(m) = 0 \quad \forall m \in \mathbb{Z}$ is identically $0$ I have been trying to solve the following exercise from a collection of old complex analysis qualifier exams.

Suppose that $g$ is an entire function that satisfies the inequality $|g(z^2)| \leq e^{|z|}$. Also suppose that $g(m) = 0 \quad \forall m \in \mathbb{Z}$. Then prove that $g(z) \equiv 0$ (i. e. that $g$ is identically $0$).

So what I think is that the inequality by putting $z^{1/2}$ gives me $|g(z)| \leq e^{|z|^{1/2}}$ and this means that the entire function $g$ is of finite order and its order $\lambda = \lambda(g) \leq \frac{1}{2}$. Then I have been looking at the basic theorems for finite order entire functions but I don't really see if one of them would be helpful here.
So my question is how can I solve this problem? Is it really helpful to look at the theorems for finite order entire functions?
 A: Since $\mid g(z) \mid \leq exp(\mid z \mid^{\frac{1}{2}})$, $g$ has growth order $\leq \frac{1}{2}$ by definition, as you correctly remark.
If $g$ were not the zero function and if $(z_k)_k$ were some enumeration of its non-zero zeros, we would have for all $s\gt \frac{1}{2}$ (Stein-Shakarchi, Theorem 2.1, page 138)
$$  \sum_k    \frac{1}{\mid z_k \mid ^s }         \lt \infty   $$   In particular, taking $s=1$ and realizing that the positive integers are among the zeros $z_k$ according to the statement of the exercise, we would get 
$$\sum_{n=1} ^\infty    \frac{1}{n } \leq  \sum_k    \frac{1}{\mid z_k \mid  }         \lt \infty   $$
Since the harmonic series actually diverges ( $\sum_{n=1} ^\infty    \frac{1}{n }=\infty $ ) this is  false and we must have  $g=0$.  
N.B. The basic heuristic of growth order of an entire  function is that the faster the  function grows, the more zeros it has.
A baby case is that of a polynomial: if its degree is  $n$ it grows like $\mid z\mid^n $and has $n$ zeros.
