Given $f^2\left(\frac{1}{n}\right)g\left(\frac{1}{n}\right)= 0$ prove $f\equiv0$ or $g\equiv0$ Given the entire functions $f$,$g:\mathbb{C} \rightarrow \mathbb{C}$ for which : $$f^2\left(\frac{1}{n}\right)g\left(\frac{1}{n}\right)= 0 \ \ \ \ \ \ \ \forall\ n \in \mathbb{N}$$
I want to prove that $f\equiv0$ or $g\equiv0$
Here is my attempt.

Suppose that both $f$ and $g$ have a countable number of roots  inside the disk $D(0,1)$. Then $f^2g$ would also have a countable number of roots which leads to a contradiction because of the above relationship.
Therefore, either $f$ or $g$ has an uncountable number of roots in $D(0,1)$.
Let $g$ be that function.
This is where I'm stuck. From other examples I've seen, I think I need to make use of "accumulation points" somehow.
Is this approach correct? How should I procced or what other ways are there to solve this kind of problem?
 A: There is a countably infinite subset of $\{\frac{1}{n}\}_{n\in \Bbb N}$ on which either $f$ or $g$ is zero. By continuity, that function vanishes at zero.
Can you take it from here?
A: The Identity theorem (https://en.wikipedia.org/wiki/Identity_theorem) states that if we have two holomorphic functions $f$ and $g$ on some set $D$ and $f=g$ on $S \subseteq D$ where $S$ has an accumulation point, then $f = g$ on $D$.
Let $h(z) = f^2(z)g(z)$ and $p(z) = 0$. Since $f,g$ are entire, $h$ is entire, and so it's also holomorphic.
If we form the following sequence $(1/n)_{n\in\mathbb{N}}$, we can show that indeed $0$ is an accumulation point. Use the fact that an accumulation point of a set is the limit of some sequence of distinct elements of that set.
In this case our set $S = \{1/n : n \in \mathbb{N}\}$
Hence, the only point which could satisfy this is $0$ for our set $S$.
We also know that $h(1/n) = p(1/n)$ on our set $S$, but $S$ is just a subset of $D = \mathbb{C}$ which $h$ is analytic on.
Since all conditions are satisfied from the theorem, $h(z) = p(z) = 0$ on $D$.
But $h(z) = f^2(z)g(z) = 0$ on $D = \mathbb{C}$ which can only happen if $f \equiv 0$ or $g \equiv 0$ on $\mathbb{C}$.
