Let $f_0$ be the function from $\mathbb{Z}^2$ to $\{0,1\}$ such that $f_0(0,0)=1$ and $f_0(x,y)=0$ otherwise. For each positive integer $m$, let $f_m(x,y)$ be the remainder when \[ f_{m-1}(x,y) + \sum_{j=-1}^{1} \sum_{k=-1}^{1} f_{m-1}(x+j,y+k) \] is divided by $2$.
Finally, for each nonnegative integer $n$, let $a_n$ denote the number of pairs $(x,y)$ such that $f_n(x,y) = 1$.
Find a closed form for $a_n$.

[i]Proposed by Bobby Shen[/i]
To find a closed form for \( a_n \), we need to analyze the behavior of the function \( f_m(x,y) \) and how it evolves over iterations. Let's break down the problem step by step.

1. **Initial Function Definition:**
   \[
   f_0(x,y) = 
   \begin{cases} 
   1 & \text{if } (x,y) = (0,0) \\
   0 & \text{otherwise}
   \end{cases}
   \]

2. **Recursive Definition:**
   For each positive integer \( m \), the function \( f_m(x,y) \) is defined as:
   \[
   f_m(x,y) = \left( f_{m-1}(x,y) + \sum_{j=-1}^{1} \sum_{k=-1}^{1} f_{m-1}(x+j, y+k) \right) \mod 2
   \]

3. **Counting Non-zero Terms:**
   We need to find \( a_n \), the number of pairs \((x,y)\) such that \( f_n(x,y) = 1 \).

4. **Pattern Observation:**
   We observe that the function \( f_m(x,y) \) evolves in a way that resembles cellular automata, specifically the 2D version of the Rule 90 cellular automaton. This automaton has a known pattern of growth.

5. **Binary Representation and Decomposition:**
   The problem can be simplified by considering the binary representation of \( n \). If \( n \) has a binary representation with a zero, we can decompose it into smaller parts. For example:
   \[
   a_{(101101110011)_2} = a_{(1011)_2} \cdot a_{(1110011)_2}
   \]

6. **Special Case Analysis:**
   We need to find \( a_{2^k-1} \) for all \( k \in \mathbb{N} \). This can be done by analyzing the recursive structure of the function.

7. **Recursive Formula:**
   By analyzing the recursive structure, we find that:
   \[
   a_{2^k-1} = \frac{5 \cdot 4^k + (-2)^{k+1}}{3}
   \]

8. **Combining Results:**
   Using the decomposition property and the special case analysis, we can combine the results to find \( a_n \) for any \( n \).

Conclusion:
The closed form for \( a_n \) is given by the recursive formula and the decomposition property. The final expression for \( a_n \) is:
\[
a_n = \frac{5 \cdot 4^k + (-2)^{k+1}}{3}
\]
where \( k \) is the largest integer such that \( 2^k - 1 \leq n \).

The final answer is \( \boxed{ \frac{5 \cdot 4^k + (-2)^{k+1}}{3} } \)