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?

$$n^2+2n=(n+1)^2-1$$
Now $n+1\equiv0,\pm1,2\pmod4$
$\implies(n+1)^2\equiv0,1\pmod4$
$(n+1)^2-1\equiv?,?$
