Show that $a^2$ congruent 0 (mod 4) or $a^2$ congruent 1 (mod 4). Proof: Suppose for the sake of contradiction, there is an integer a such that $a^2$ congruent $0 \pmod {4}$ and $a^2$ congruent $1 \pmod {4}$. Then $(a^2)-0=4k$ and $(a^2)-1=4l$ for some $k,l$ members of the integers. 
Now,
$$a^2-0=4k$$
$$a^2=4k$$
Thus,
$$a^2-1=4l$$
$$((4k)^2)-1=4l$$
$$(16k^2)-1=4l$$
$$(16k^2)-4l=1$$
$$2((8k^2)-2l)=1$$
$2m=1$ for some integer m where $m=(8k^2)-4l$
By the definition of an even integer $1$ is even, however we know $1$ is odd, therefore a contradiction.

I feel the answers provided aren't clear or complete enough, so I'm going to put my two cents in. The proof is done by cases.
Case $1$: $a = 2k$ is even
By the definition of divisibility and congruence, $$a^2 \equiv 0\pmod4\impliedby 4\mid(a^2-0)\impliedby 4\mid(2k)^2\impliedby4\mid4k^2$$ and since $\exists\, c\hspace{0.1cm}: 4k^2 =4\cdot c$ with $c=k^2$, the statement is true.
Case $2$: $a=2k+1$ is odd
The second case is done almost exactly, except you're proving that $a^2 \equiv 1\pmod4$ holds for odd numbers, so instead of inputting $n$ as $2k$, you input it as $2k+1$. You will find that the witness $c$ for this case is $c = k^2+k$. You have proven $a^2$ is congruent to either $0$ or $1\pmod4$ for all $a$, thus the proof is considered complete. $\square$
