Let $n$ be a positive integer. Prove that $n(n + 1)(n + 2)(n + 3)$ cannot be a square or a cube. For the expression cannot be a square, there is an answer here. Prove that $ \sqrt{n(n+1)(n+2)(n+3)+1} $ is always a natural number
I am trying to prove it cannot be a cube.
There is a conclusion that every cube could be written as the difference of two squared, i.e.
$r^3$ = $p^2$ - $q^2$, where $p = r(r+1)/2$ and $q = r(r-1)/2$.
Therefore I think the conclusion holds if $n(n + 1)(n + 2)(n + 3)$ could not be expressed as the difference of two squared but I don't know how to continue.

1/ $n(n+1)(n+2)(n+3)$ can't be a square number
Denote $x = \sqrt{n(n+1)(n+2)(n+3)}$, we have
$$
\begin{align}
x^2 &= n(n+3)(n+1)(n+2) \\
&= (n^2+3n)(n^2+3n+2)\\
&= (n^2+3n+1)^2 -1
\end{align}
$$
It's easy to notice that
$$(n^2+3n)^2 <(n^2+3n+1)^2 -1 <(n^2+3n+1)^2$$
$$\implies n^2+3n < x <n^2+3n+1 \tag{1}$$
From $(1)$ we deduce that $x$ can't be a natural number so $n(n+1)(n+2)(n+3)$ can't be a square number.
2/ $n(n+1)(n+2)(n+3)$  can't be a cube number
Denote $x = \sqrt[3]{n(n+1)(n+2)(n+3)}$ and $y = n^2 +3n+1$ suppose $x \in \Bbb N+$. We have
$$x^3 = (n^2+3n+1)^2 -1 \implies x^3= y^2-1 \tag{2}$$
This diophantine equation $(2)$ is not trivial. An elementary proof of the problem can be found in the paper THE DIOPHANTINE EQUATIONS x3=Ny2±1 of (J. H. E. Cohn, 1991) (or this link) according to this answer and the unique solution of $(2)$ is $(x,y) =(2,3)$.
However, there doesn't exist $n \in \Bbb N^+$ such that $n^3+3n+1 = 3$.
Hence, $n(n+1)(n+2)(n+3)$  can't be a cube number.
