$f: \mathbb C \to \mathbb C$ is entire function such that $f(1/n)=1/n^2$ for all $n \in \mathbb N$, Then to show $f(z)=z^2$

Question

$f: \mathbb C \to \mathbb C$ is entire function such that

$f(1/n)=1/n^2$ for all $n \in \mathbb N$, Then to show $f(z)=z^2$

It seems that it is like use of Identity Theorem but I am not getting how to do it.

"Identity Theorem:if $f:D \to \mathbb C$" entire and limit point of zeros of $f$ contained in $D$, then $f=0$ in $C$" Please help me to connect.

Answer 1

Let $\phi(z) = f(z)-z^2$. Then $\phi({1 \over n}) = 0$ for all $n$ and ${ 1 \over n} \to 0$.

Answer 2

How about the function $g(z)=f(z)-z^2$?