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?
Best Answer
You should consider cases
$$n \equiv i \pmod 5$$
where $i \in \{0\,\ldots, 4\}$.