Question: If $n$ is odd then $n$ is congruent to $1 \pmod{8}$, but if $n$ is even the we have to do it by cases. 

Let $n=2k$

so $n^2=4k^2$.

When $k$ is even, then it's congruent to $0 \pmod{8}$, but when $k$ is odd, I don't see how it would be congruent to something in the options.

Answer: 
we have $$n\equiv 0,1,2,3,4,5,6,7 \mod 8$$ squaring we get $$n^2\equiv 0,1,4,1,0,1,4,1,\mod 8$$
**Hint** $\ $ Every integer has form $\, n = 4q+r\,\ $ for $\,\ r \in \{0, 1, 2, 3\}\,$ by the Division Algorithm.

Thus computing modulo $\,8\!:\ n^2\! = (4q+r)^2 \equiv r^2\!\in \{0, 1, 2, 3\}^2 \equiv \{0, 1, 4, 1\}$