If $\frac{a(m^n-1)}{n(m-1)}+1$ divides $m^n-1$, then $(n,m,a)=(2,5,1),(2,9,3),(2,13,1)$?

Question

Let $n,m,a$ be integers such that $n\gt 1, m\geq5$ and $a>0$.

Then, is the following true?

If $\frac{a(m^n-1)}{n(m-1)}+1$ divides $m^n-1$, then $(n,m,a)=(2,5,1),(2,9,3),(2,13,1)$.

With the help of computer, the answer is non-divisible in many cases. But I cannot prove it. Do you have any idea?

Answer 1

You've already got that $n=2$ or $3$, and that for $n=2$, $m\in\{5,9,13\}$.

For $n=3$, $$\frac{m^3-1}{\frac{a(m^3-1)}{3(m-1)}+1}=k\implies (3m-3-ak)(m^2+m+1)=3k$$ where $k$ is a positive integer.

So, writing $s=3m-3-ak$ where $s$ is a positive integer to have $$s=3m-3-a\cdot\frac{s(m^2+m+1)}{3}\implies asm^2+(as-9)m+as+3s+9=0$$ Considering the discriminant, we get $$\small (as-9)^2-4as(as+3s+9)\ge 0\implies 27\ge as(as+18+4s)\ge as(as+22)\implies a=s=1$$

Then, the solutions of $m^2-8m+13=0$ are not integers.