For every integer $n$, the quantity $n^2 + 2n \equiv 0\pmod 4$ or $n^2 + 2n \equiv 3\pmod 4$ I'm trying to prove this question using induction
So far I have
Base Case
Let $n = 1$, $(1^2 + 2)\equiv 3 \pmod 4 $
Claim holds for base case
Induction
Assume $n = k$ holds, that is $k^2 + 2k \equiv 0\pmod 4$ or  $k^2 + 2k \equiv 3\pmod 4$
Let $n = k+1$ such that
$${(k+1)^2 + 2(k+1)}\equiv {k^2 + 2k + 1 + 2k + 2}\pmod 4$$
Then I substitute $k^2 + 2k$ with both $0$ or $3$ from the earlier assumption.
So we have
$$(0 + 2k + 3) \equiv (2k + 3) \pmod 4$$
Or,
$$(3 + 2k + 3) \equiv (2k + 6) \pmod 4$$
Where do I go from here?
 A: Separate cases for $n$ even and $n$ odd.
If $n$ is even, than it is $n=2m$. 
We get: $(2m)^2+2(2m)=4m^2+4m=4(m^2+m)\equiv 0\mod 4$
If $n$ is odd, then it is $n=2m+1$.
We get: $(2m+1)^2+2(2m+1)=4m^2+4m+1+4m+2=4(m^2+2m)+3\equiv 3\mod 4$
A: $$n^2+2n=(n+1)^2-1$$
Now $n+1\equiv0,\pm1,2\pmod4$
$\implies(n+1)^2\equiv0,1\pmod4$
$(n+1)^2-1\equiv?,?$
A: From where you are stuck, we can do the following case distinction:
Case 1: $k^2+2k \equiv 0 \mod 4$. Notice that in this case $k$ cannot be an odd number. And any even $k$ satisfies this (can be proven by induction). Therefore $k = 2m$ for some integer $m$. Then, 
$$2k+3 = 4m+3 \equiv3 \mod 4$$
Case 2: $k^2+2k \equiv 3 \mod 4$. Notice that in this case $k$ cannot be an even number. And any odd $k$ satisfies this (can be proven by induction). Therefore $k = 2m+1$ for some integer $m$. Then,
$$2k+6 = 4m+9 \equiv1 \mod 4$$
But instead of doing this, Cornman's answer is shorter and better since we are doing the case distinction anyway so why would we have to do an additional induction when a case distinction itself can be done for proving? I'm posting this just to complete your solution. So I refer to Cornman's answer.
A: You only need one equation.
$\begin{array}\\
(n+2)^2+2(n+2)
-(n^2+2n)
&=n^2+4n+4+2n+4
-(n^2+2n)\\
&=4n+8\\
\end{array}
$
which is divisible by $4$,
so whatever remainder $n$ has
mod 4, $n+2$ also has.
Since $n=0 \implies n^2+2n = 0$,
all even numbers have a
remainder of 0 mod 4.
Since $n=1 \implies n^2+2n = 3$,
all odd numbers have a
remainder of 3 mod 4.
