Diophantine equation with perfect squares

Question

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.

Answer 1

$$n(n^2-4)=3b^2$$

• $n=\pm 2$,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.

Answer 2

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?