Let $n$ be a positive integer. Compute, in terms of $n$, the number of sequences $(x_1,\ldots,x_{2n})$ with each $x_i\in\{0,1,2,3,4\}$ such that $x_1^2+\dots+x_{2n}^2$ is divisible by $5$.

[i]2020 CCA Math Bonanza Individual Round #13[/i]
1. **First Solution (Recurrence):** 
   - Define \( a_{k,c} = |\{\mathbf{x} \in \mathbb{F}_5^k : \mathbf{x} \cdot \mathbf{x} = c\}| \). This represents the number of sequences of length \( k \) over \( \mathbb{F}_5 \) such that the sum of the squares of the elements is \( c \).
   - We claim that:
     \[
     a_{k,c} = a_{k-1,c} + 2a_{k-1,c-1} + 2a_{k-1,c+2}
     \]
     This follows by considering the last component of \( \mathbf{x} \). The possible values for the last component \( x_k \) are \( 0, 1, 2, 3, 4 \), and their squares modulo 5 are \( 0, 1, 4, 4, 1 \) respectively. Thus, the recurrence relation accounts for these possibilities.
   - Using the recurrence relation, we can derive:
     \[
     a_{k,c} = 9a_{k-2,c} + 4a_{k-2,c-1} + 4a_{k-2,c-2} + 4a_{k-2,c+1} + 4a_{k-2,c+2} = 5a_{k-2,c} + 4 \cdot 5^{k-2}
     \]
     because all \( 5^{k-2} \) vectors in \( \mathbb{F}_5^{k-2} \) are counted in some \( a_{k-2,c} \).
   - Since \( a_{0,0} = 1 \), we can inductively show that:
     \[
     a_{2n,0} = 5^{2n-1} + 5^n - 5^{n-1}
     \]

2. **Second Solution (Fourier analysis/Roots of unity):**
   - Define the polynomial \( P(x) = \sum_{a=0}^4 x^{a^2} \). This polynomial captures the possible values of \( x_i^2 \) for \( x_i \in \{0, 1, 2, 3, 4\} \).
   - The coefficient of \( x^m \) in \( P(x)^{2n} \) is the number of solutions to \( x_1^2 + \cdots + x_{2n}^2 = m \) with each \( x_i \in \{0, 1, 2, 3, 4\} \).
   - To count the total number such that \( x_1^2 + \cdots + x_{2n}^2 \) is divisible by 5, we apply the inverse discrete Fourier transform (roots of unity method). Let \( \omega = e^{i \cdot \frac{2\pi}{5}} \), then the desired count is:
     \[
     \frac{1}{5} \sum_{k=0}^4 P(\omega^k)^{2n}
     \]
   - Calculating \( P(\omega^k) \):
     \[
     P(\omega^k) = 1 + 2\omega^k + 2\omega^{4k}
     \]
   - Therefore:
     \[
     \frac{1}{5} \sum_{k=0}^4 P(\omega^k)^{2n} = \frac{1}{5} \left( 5^{2n} + 2(1 + 2\omega + 2\omega^4)^{2n} + 2(1 + 2\omega^2 + 2\omega^3)^{2n} \right)
     \]
   - Simplifying further:
     \[
     = 5^{2n-1} + \frac{2}{5} (1 + 4 \cos \frac{2\pi}{5})^{2n} + \frac{2}{5} (1 + 4 \cos \frac{4\pi}{5})^{2n}
     \]
     \[
     = 5^{2n-1} + \frac{2}{5} \cdot 5^n + \frac{2}{5} \cdot 5^n
     \]
     \[
     = 5^{2n-1} + 4 \cdot 5^{n-1}
     \]

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