<s>[INST] 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.
 [/INST] 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. ;-)
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