Prove that if $k \in \mathbb{N}$, then $k^4+2k^3+k^2$ is divisble by $4$ I am trying to solve by induction and have established the base case (that the statement holds for $k=1$).
For the inductive step, I tried showing that the statement holds for $k+1$ by expanding $(k+1)^4+2\cdot(k+1)^3+(k+1)^2$, but this equals $16k^4+34k^3+3k^2+16k+4$, and $4$ cannot be factored out.

First note that $n^4+2n^3+k^2= n^2(n^2+2n+1)=n^2(n+1)^2$. Hence, the problem is to show that for every $n\geq 1$, 
$$
S_n : 4\mid n^2(n+1)^2
$$
is true. 

Without induction, the direct proof follows from the fact that among $n$ and $n+1$, one is even, and so one of $n^2$ and $(n+1)^2$ has a factor of $4$. But we will do it by induction anyway.

Base step: When $n=1, n^2(n+1)^2=4$, so $S_1$ holds.
Inductive step: Let $k\geq 1$ and let the induction hypothesis be that $S_k$ holds; in particular, suppose that $m$ is such that $k^2(k+1)^2=4m$. It remains to show that
$$
S_{k+1} : 4\mid (k+1)^2(k+2)^2
$$
follows. Indeed
\begin{align}
(k+1)^2(k+2)^2 &= (k^2+4k+4)(k+1)^2\\[0.5em]
&= k^2(k+1)^2+(4k+4)(k+1)^2\\[0.5em]
&= 4m+4(k+1)(k+1)^2\quad\text{(by $S_k$)}
\end{align}
shows that $(k+1)^2(k+2)^2$ is also divisible by $4$, concluding the proof of $S_{k+1}$ and hence the inductive step.
By mathematical induction, it is true that for any positive integer $n$, the number $n^2(n+1)^2$ is divisible by $4$. $\blacksquare$
