Prove by induction that $n^4-4n^2$ is divisible by 3 for all integers $n\geq1$. For the induction case, we should show that $(n+1)^4-4(n+1)^2$ is also divisible by 3 assuming that 3 divides $n^4-4n^2$. So,
$$ \begin{align} (n+1)^4-4(n+1)&=(n^4+4n^3+6n^2+4n+1)-4(n^2+2n+1)
\\ &=n^4+4n^3+2n^2-4n-3
\\ &=n^4+2n^2+(-6n^2+6n^2)+4n^3-4n-3
\\ &=(n^4-4n^2) + (4n^3+6n^2-4n)-3
\end{align}$$
Now $(n^4-4n^2)$ is divisible by 3, and $-3$ is divisible by 3. Now I am stuck on what to do to the remaining expression.
So, how to show that $4n^3+6n^2-4n$ should be divisible by 3? Or is there a better way to prove the statement in the title? Thank you!
 A: Well, obviously $6n^2$ is divisible by $3$.
And $4n^3-4n=4(n-1)n(n+1)$. Since $(n-1)n(n+1)$ is the product of three consecutive integers, one of them is divisible by $3$ and therefore their product has that property too.
A: Here is a way to prove it without induction. Notice that $$n^4-4n^2=n^2(n-2)(n+2)$$
If $3\not\mid n$ then $3\mid (n+1)-3$, that is $3\mid n-2$, or $3\mid (n+2)$.
A: $$n^4-4n^2=n^2(n^2-1-3)=n\underbrace{(n-1)n(n+1)}_{\text{product of three consecutive integers}}-3n^2$$
A: Because $$n^4-4n^2=n^4-n^2-3n^2=n(n-1)n(n+1)-3n^2$$
A: Wihout induction, it is simply a consequence of lil' Fermat:
As $3$ is prime, Fermat asserts that for any $n$, $\;n^3\equiv n\mod 3$, so
$$n^4-n^2=n^3\cdot n-n^2\equiv n^2-n^2\mod 3. $$
A: Alternatively, using modular arithmetic:
$$\begin{align}n&\equiv 0,1,2 \pmod 3 \\
n^2&\equiv 0,1 \pmod 3\\
4n^2&\equiv 0,1 \pmod 3\\
n^4&\equiv 0,1 \pmod 3\\
n^4-4n^2&\equiv 0 \pmod 3.\end{align}$$
A: Given that $6n^2$ is divisible by $3$, it's all a mater of showing that $4n^3-4n$ is divisible by $3$, and we can show that by showing that $n^3-n$ is divisible by $3$
And we know the later is true as follows: 
if $n$ itself is divisble by $3$, then obviously $n^3-n$ is divisible by $3$. So, the only other options are that $n=3k+1$ or that $n=3k+2$ for some $k$.
If $n=3k+1$, then $n^3-n$ works out to all multiples of $3$ except for the $+1$ at the end, and so subtracting $n$ will get rid of that $1$
And if $n=3k+2$, then $n^3-n$works out to all multiples of $3$ except for the $+8$ at the end, and so subtracting $n$ will leave $6$, so that's all  divisible by $3$ as well.
