AHSME 1975 Problem 10

The sum of the digits in base ten of $(10^{4n^2+8}+1)^2$, where $n$ is a positive integer, is $\textbf{(A)}\ 4 \textbf{(B)}\ 4n \textbf{(C)}\ 2+2n \textbf{(D)}\ 4n^2 \textbf{(E)}\ n^2+n+2$