$$
\text { Three. (50 points) Given the sequence }\left\{a_{n}\right\}: 1,3,5,7, \cdots \text {, }
$$

starting from the 5th term, $a_{n+4}$ is the unit digit of $a_{n}+a_{n+3}$. Find:
$$
a_{2008}^{2}+a_{2009}^{2}+a_{2010}^{2}+a_{2011}^{2}
$$

Can it be divisible by 4?
Consider the sequence $\left\{a_{n}\right\}$ and its modulo 2 residue sequence $\left\{b_{n}\right\}$.
Given that $a_{n+4}$ is the unit digit of $a_{n}+a_{n+3}$, we have $b_{n+4} \equiv b_{n}+b_{n+3}(\bmod 2)$.
By recursion, we get
$$
\begin{array}{l}
b_{n+10} \equiv b_{n+6}+b_{n+9} \\
\equiv\left(b_{n+2}+b_{n+5}\right)+\left(b_{n+5}+b_{n+8}\right) \\
\equiv b_{n+2}+b_{n+8} \\
\equiv b_{n+2}+\left(b_{n+4}+b_{n+7}\right) \\
\equiv b_{n+2}+\left(b_{n}+b_{n+3}\right)+\left(b_{n+3}+b_{n+6}\right) \\
\equiv b_{n+2}+b_{n}+b_{n+6} \\
\equiv b_{n+2}+b_{n}+\left(b_{n+2}+b_{n+5}\right) \\
\equiv b_{n}+b_{n+5}(\bmod 2) .
\end{array}
$$

Then, $b_{n+15} \equiv b_{(n+5)+10} \equiv b_{n+5}+b_{n+10}$
$$
\begin{array}{l}
\equiv b_{n+5}+\left(b_{n}+b_{n+5}\right) \\
\equiv b_{n}(\bmod 2) .
\end{array}
$$

It is evident that $\left\{b_{n}\right\}$ is a periodic sequence with a period of 15, hence
$$
\begin{array}{l}
a_{2008} \equiv b_{2008} \equiv b_{15 \times 133+13} \equiv b_{13} \equiv 0(\bmod 2), \\
a_{2009} \equiv b_{2009} \equiv b_{15 \times 133+14} \equiv b_{14} \equiv 0(\bmod 2), \\
a_{2010} \equiv b_{2010} \equiv b_{15 \times 134} \equiv b_{15} \equiv 0(\bmod 2), \\
a_{2011} \equiv b_{2011} \equiv b_{15 \times 134+1} \equiv b_{1} \equiv 1(\bmod 2) .
\end{array}
$$

Since the square of an even number is divisible by 4, and the square of an odd number is congruent to 1 modulo 4, we have
$$
S=a_{2008}^{2}+a_{2008}^{2}+a_{2010}^{2}+a_{2011}^{2} \equiv 1(\bmod 4) \text {. }
$$

Therefore, $S$ is not divisible by 4.