Do there exist $n,b\in \mathbb{N}, b<n$ such that $(n-b)(n^2-b)$ divides $n^3$

Question

My friend invented this problem in his spare time and asked me if I could solve it:

Do there exist $n,b\in \mathbb{N}, b<n$ such that $(n-b)(n^2-b)$ divides $n^3$ ?

I tried my best and all I can say is this:

He wrote a program to brute force $n$ and $b$ up to $10^5$ and there were no solutions so I assume there are none.

I also tried various ways to put a lower bound on the $(n-b)(n^2-b)$ in hope it will limit the size of $b$ or $n$, but none seemed to heve worked.

I have a hypothesis that the product here is just an additional , constraint meaning that the following claim also holds: if $(n-b) | n^3$, then $n^2-b$ does not. I also wrote a program to check it for $n$ and $b$ up to $10^5$ and it holds. However, I still didn't manage to prove it.

What seems to have some perspective is to consider the d=GCD(n,b) and write them in terms of it, but nothing I did made any real progress,except some numbers share all prime factors and $n-b$ divides $d^3$

I've seen similar problems get solved by making the statement into an equation(preferably quadratic) and then proving that discriminant cannont be a perfect square by something like quadratic residues, but I have no idea how to use the cubic formula here and it seems rather brute-forcish, not to mention I know nothing about residues of higher degree.

He already posted this question on Quora and got no sensible answer.

Answer 1

Indeed, a very good idea is to consider $d = \text{gcd}(n,b)$. Then $n=d\alpha$ and $b = d\beta$ with $\alpha>\beta\ge1$ and $\text{gcd}(\alpha, \beta)=1$. So $(n-b)(n^2-b) \mid n^3$ becomes: $$d(\alpha-\beta)(d^2\alpha^2-d\beta) \mid d^3\alpha^3$$ which is equivalent to $$(\alpha-\beta)(d\alpha^2-\beta) \mid d\alpha^3$$ In particular, $\alpha -\beta \mid d\alpha^3$. Since $\text{gcd}(\alpha, \beta)=1$ we obtain $\text{gcd}(\alpha-\beta, \alpha^3)=1$. Consequently, we get $\alpha -\beta \mid d$, so $d = (\alpha-\beta)\cdot d'$, for some $d' \ge 1$. Thus, replacing $d$ with $(\alpha-\beta)\cdot d'$ we get $$(\alpha-\beta)\left((\alpha-\beta)d'\alpha^2 - \beta \right) \mid (\alpha-\beta)d'\alpha^3$$ and thus, dividing both sides by $(\alpha-\beta)$ we obatin $$(\alpha-\beta)d'\alpha^2 -\beta \mid d'\alpha^3$$ Again, $\text{gcd}(\alpha, \beta)=1$ implies $\text{gcd}((\alpha-\beta)d'\alpha^2-\beta, \alpha^3)=1$ and hence $$(\alpha-\beta)d'\alpha^2 -\beta \mid d'$$ In particular, $(\alpha-\beta)d'\alpha^2 -\beta \le d'$. This inequality is equivalent to $$d\alpha^2 \le \frac{d}{\alpha-\beta}+\beta$$ But, since $\alpha \ge 2$, we have $d\alpha^2 \ge 2d\alpha=d\alpha+d\alpha> d + d\beta\ge \frac{d}{\alpha-\beta}+\beta$ yielding a contradiction.

Consequently, there are no such $n,b$.

Answer 2

In fact, a stronger claim holds: there are no positive integral $n, b$ with $b < n$ such that $n^2 - b$ divides $n^3$.

Indeed, assume for contradiction that such $n$ and $b$ exist. Note that $n^2 - b$ divides $n^3 - nb$. Hence $n^2 - b$ divides $n^3 - (n^3 - nb) = nb$. However, $0 < nb < n^2 - b$, and this leads us to a contradiction. Indeed, $b < n$, i.e., $b \le n - 1$. Hence $nb \le n(n - 1) = n^2 - n < n^2 - b$ (in the end we again use $b < n$).