Show that $n^2 \mod 5$ equals $0,1$, or $4$ for every integer $n$. Show that $n^2 \mod 5$ equals $0,1$, or $4$ for every integer $n$. Using divison in to cases.

Proof: let integer $n$ be given.
Case $1$: Suppose there exists an integer $k$ such that $n = 2k$
Case $2$: Suppose there exists an integer $k$ such that $n = 2k+1$

Do I have the right idea of having two cases for all integers, one that covers even numbers and one that covers odd, or am I not on the right track?

You want to know the outcome modulo $5$ so if we are to break things into cases it makes sense to break things into cases of modulo $5$.
If $n \equiv 0,1,2,3,4 \mod 5$ then 
$n^2 \equiv 0^2, 1^2, 2^2, 3^2, 4^2 \mod 5$ and ... the proof then practically writes itself.
It writes itself even better when you realize $3 \equiv -2 \mod 5$ and $4\equiv -1 \mod 5$ so
If $n \equiv 0,1,2,-2,-1 \mod 5$ then
$n^2 \equiv 0^2, 1^2, 2^2, (-2)^2, (-1)^2 \mod 5$.
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P.S.  If you want to do it the hard way we are trying to prove:
$n^2 = 5k + r$ and $0 \le r < 5$ then $r= 0,1$ or $4$ (and never $2$ or $3$).
We are dealing with remainders when divided by $5$ so if we let $n = 5j + m$ then
$n^2 = (5j + m)^2 = 25j^2 + 10jm + m^2 = 5(5j^2 + 2jm) + m^2 = 5k + r$ so  we must have $r = m^2 + 5(5j^2 + 2jm - 1)$.  And $m$ may be $0,1,2,3,4$ so $m^2$ may be $0,1,4,9,16$ and $r$ may be $0,1,2,3,4$
$r = 0 +  5(5j^2 + 2jm - 1)$ means $r= 0$
$r = 1 +  5(5j^2 + 2jm - 1)$ means $r = 1$.
$r = 4 + 5(5j^2 + 2jm - 1)$ means $r = 4$
$r = 9 + 5(5j^2 + 2jm - 1)= 4 + 5(5j^2 + 2jm)$ means $r =4$.
$r = 16 + 5(5j^2 + 2jm - 1)= 1 + 5(5j^2 + 2jm+2)$ means $r = 1$.
But that was the hard way. 
