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.
 A: I think you would want to make things more compact instead of expanding. We have $$k^4+2k^3+k^2 = k^2(k^2+2k+1) = k^2(k+1)^2 = [k(k+1)]^2$$ Now if we suppose that $[k(k+1)]^2$ is divisible by $4$ for all $k \in \{1,2,\ldots n\}$ then we can consider $$[(n+1)(n+2)]^2$$ Since $n$ is an integer, then either $n+1$ is even or $n+2$ is even. WLOG, suppose $n+2 = 2p$ is even. Then $$[(n+1)(n+2)]^2 = [(n+1)(2p)]^2 = 4[(n+1)p]^2 = 4q$$ for some integer $q$. I don't know if this really counts as induction since you don't use the induction hypothesis to show $P(n) \implies P(n+1)$. This argument can be completed simply by the fact that for any positive integer $n$, $n+1$ $n$ must be even. 
A: 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 $4$. $\blacksquare$
A: I would do it like this. As others have noted you have $$f(k)=k^4+2k^3+k^2=k^2(k+1)^2$$ and you are done if you can prove that $k(k+1)$ is even, because then its square is divisible by $4$ (an extra step you have to prove if you want to simplify the induction).
Now this is trivially true for $k=0$ and if true for $k=r$ we have $$(r+1)(r+2)=r(r+1)+2(r+1)$$$r(r+1)$ is even by inductive hypothesis, and $2(r+1)$ is even because it is divisible by $2$, so $(r+1)(r+2)$ is even.

If you want to keep the squares/fourth powers in, you can compute the difference between successive values as $$(n+1)^2(n+2)^2-n^2(n+1)^2=(n+1)^2\left((n+2)^2-n^2\right)$$
The second term is a difference of two squares equal to $2\cdot (2n+2)=4(n+1)$ so the difference is divisible by $4$. Again this avoids any full expansion of the expressions involved.
A: $\underline{\text{Proof by induction:}}$
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.
