<s>[INST] 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?
 [/INST] 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.
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