Divisibility of $n^4 -n^2$ by 4 (induction proof) We have to show that $$ n^4 -n^2 $$  is divisible by 3 and 4 by mathematical induction  
Proving the first case is easy however I do not  know how what to do in the inductive step.  Thank you.
 A: Doing it by the book, though you were already told there are simpler/faster ways:
For $\;n=1:\;\;1^4-1^2=0\;\color{green}\checkmark\;$
Suppose it is true for $\;n\;$ and prove for $\;n+1\;$
$$(n+1)^4-(n+1)^2=(n+1)^2\left((n+1)^2-1\right)=(n+1)^2(n+1-1)(n+1+1)=$$
$$(n^2+2n+1)(n^2+2n)=n^4+4n^3+5n^2+2n=n^4-n^2+2n(2n^2+3n+1)=$$
$$=n^4-n^2+2n(n+1)(2n+1)$$
Now observe that the factor $\;2n(n+1)(2n+1)\;$ is always divisible by $\;4\;$ (as either $\;n\;$ or $\;n+1\;$ is even), and also by $\;3\;$ (since either $\;n\;$ is, or $\;n=1\pmod 3\;$ and then $\;2n+1=0\pmod 3\;$ , or $\;n=2\pmod3\;$ and then $\;n+1=0\pmod3\;$)
A: Since $a_n=n^4-n^2=n^2(n-1)(n+1)$, we can also write
$$
a_{n+1}=(n+1)^4-(n+1)^2=(n+1)^2n(n+2)
$$
Suppose $a_n=n^4-n^2$ is divisible by $3$; then $3$ divides one among $n$, $n-1$ and $n+1$. Since $n$ and $n+1$ appear in the decomposition of $a_{n+1}$. The only remaining case is when $3$ divides $n-1$; but in this case we can use $n+2=(n-1)+3$ and we are done.
Suppose $4$ divides $a_n$. Then either $2\mid n$ or $2\mid(n+1)$ (and also $n-1$). In the case $2\mid(n+1)$, we have $4\mid(n+1)^2$. In the case $2\mid n$, we have $2\mid(n+2)$, so $4\mid n(n+2)$.

Using Fermat's little theorem is of course easier; since $n^3\equiv n\pmod{3}$,
$$
n^4-n^2\equiv n\cdot n^3-n^2\equiv n^2-n^2\equiv0\pmod{3}
$$
Since $n^2\equiv n\pmod{2}$, we have
\begin{align}
n^2-n&\equiv n-n\equiv0\pmod{2}\\[4px]
n^2+n&\equiv n+n\equiv2n\equiv0\pmod{2}
\end{align}
Is this a proof by induction? Well, yes! One uses induction for proving Fermat's little theorem. ;-)
A: Let's prove that
if $n^4-n^2$ is a multiple of both $3$ and $4$,
then so is $(n+1)^4-(n+1)^2$.
Consider 
$
((n+1)^4-(n+1)^2)-(n^4-n^2)=4 n^3+6 n^2+2 n
$.
Ignoring $6n^2$, we have $4 n^3+2 n=3n^3+3n+n^3-n=3n^3+3n+6\binom{n+1}{3}$, which is always a multiple of $3$.
Ignoring $4n^3$, we have
$6 n^2+2 n=4n^2+2n(n+1)=4n^2+4\binom{n+1}{2}$, which is always a multiple of $4$.
We're done.
This problem is definitely one which is much easier not by induction but instead simply by arguing that $n^{4} - n^{2} = n^{2} (n + 1) (n - 1)$, as @eltonjohn suggested.
