Diophantine equation with perfect squares 
Find all the integer solutions of the equation:
  $$(n^2-4)n = 3b^2$$

Progress
I tried casework based on what $n$ is modulo $3$ but it didn't work. 
 A: $$n(n^2-4)=3b^2$$


*

*$n=\pm 2$,we have the solution $(n,b)=(\pm2,0)$

*$n^2>4$ ,then:
if $n$ is odd,then:


$$(n,n^2-4)=1$$
So,there are integers $k,l$, such that:
$$n=k^2 ,n^2-4=3l^2 $$
or
$$n=3k^2, n^2-4=l^2$$
At the second system,the only solution is $l=0$,so $n^2=4$(We have already seen this solution)
For the first system:
$$n=k^2 , n^2-4=3l^2 , n \text{ odd } $$
$$n^2 \equiv 1 \pmod 8$$
$$n^2-4 \equiv 5 \pmod8, \text{ so } 3l^2 \equiv 5 \pmod 8 \text{that is impossible}.$$
We have shown that if  $n$ is odd,we don't have solutions.
Now check the case,when $n$ is even.
A: If $p$ is an odd prime and $p \mid b$, then $p$ divides exactly one of $n, n-2, n+2$. Hence, in order for the equation to hold, we must have $p^2$ dividing one of those three. 
If $n$ is odd, that means we must have $3$ consecutive odd numbers, two of which are square, and one of the form $3k^2$. Since there are no square numbers that are $2$ or $4$ apart, this is impossible.
Can you show using similar reasoning that if $n$ is even, there are no solutions?
