My assignment asks me to prove that there are only finitely many $ n\in\mathbb{N} $ such that the prime factorization of $n^3-n$ is of the form $p_1^{r_1} p_2^{r_2} p_3^{r_3}$ for $p_i$ primes and $r_n$ in $ \mathbb{N} $ including $0$.
I know that $n^3-n=n(n-1)(n+1)$ is always divisible by $2$ and $3$. I have noticed that there is the chance of $\le 3$ prime factors when $n$, $n-1$, or $n+1$ = $2^k$ or $3^k$ for some $k$ natural. The previous question was asked to prove that $\text{abs}(2^n-3^m) = 1$ has finitely many solutions but I'm not sure if this can be applied to this problem.
I'm not sure where to start for this problem because there are so many cases where it does work out to have $3$ unique prime factors. Any advice appreciated.
Case 1: $n$ is a power of $2$. Then, with finitely many exceptions (as proved in the previous question), neither $n-1$ nor $n+1$ will be a power of $3$. Hence, each of them have a factor that is neither $2$ nor $3$. Now prove that those factors can't be the same, and you're done.
Case 2: $n$ is a power of $3$. Similar to the previous.
Case 3: $n$ has a factor that is neither $2$ nor $3$. The only way this could work is if one of $\{n-1,n+1\}$ is a power of $2$, and the other a power of $3$. A variation of the previous question should prove that there are only finitely many such possibilities.