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.

$\underline{\text{Proof by induction:}}$
First, show that this is true for $k=1$:
$1^4+2\cdot1^3+1^2=4$
Second, assume that this is true for $k$:
$k^4+2k^3+k^2=4n$
Third, prove that this is true for $k+1$:
$(k+1)^4+2(k+1)^3+(k+1)^2=$
$4(k+1)^3+\color{red}{k^4+2k^3+k^2}=$
$4(k+1)^3+\color{red}{4n}=$
$4[(k+1)^3+n]$
Please note that the assumption is used only in the part marked red.

$\underline{\text{Proof by modular-arithmetic:}}$
Consider the following cases:


*

*$k\equiv0\pmod4 \implies k^4+2k^3+k^2\equiv0+0+0\equiv0\pmod4$

*$k\equiv1\pmod4 \implies k^4+2k^3+k^2\equiv1+2+1\equiv4\equiv0\pmod4$

*$k\equiv2\pmod4 \implies k^4+2k^3+k^2\equiv16+16+4\equiv36\equiv0\pmod4$

*$k\equiv3\pmod4 \implies k^4+2k^3+k^2\equiv81+54+9\equiv144\equiv0\pmod4$


Please note that this method is handy only when dealing with a relatively small divisor.
