$\varphi(5^n - 1) = 5^m -1$ implies $\gcd(n, m) > 1$

Question

I'm thinking about a problem 2 from http://math.mosolymp.ru/upload/files/2020/khamovniki/11/S3_t4_final.pdf

Let $n$, $m$ be natural numbers such that $\varphi(5^n - 1) = 5^m - 1$. Prove that $n$ and $m$ are not coprime.

What I did so far: by this, we want to prove that $\gcd(5^n-1, 5^m-1)>4$. So, if $8 | 5^n - 1$, we are done, because each odd prime $p$ in the decomposition of $5^n-1$ spits out a $2$ in the decomposition of $\varphi(5^n - 1)$, so either $5^n-1=8$ (which it isn't) or $8|\varphi(5^n-1)$.

Also $5^n-1$ cannot be divided by a square of an odd prime.

So, the following is true: $5^n-1 = 4 p_1 \cdot ... \cdot p_t$ and $5^m - 1= 2 (p_1-1)...(p_t-1)$.

Some other remarks:

It is known and easy to prove that $k |\varphi(a^k-1) $ for every $k$, so $n |5^m-1$

Also, by LTE lemma, $m$ is even and so $n$ is odd. (And, I think, $m$ must be divisible by some huge power of 2, since ratio of $5^n-1$ to $5^m-1$ should be greater than $5$ and the product $\prod \frac{p}{p-1}$ diverges very slowly)

All that doesn't allow me to solve the problem. What am I missing here?

Answer 1

This is quite a tricky exercise in the famous Problems from the book by Titu Andreescu (Quadratic Reciprocity chapter). This was a previous Taiwanese TST problem and you can certainly expect it to be hard. Anyway i remember seeing the solution from the following link.

• https://www.imomath.com/index.php?options=325&lmm=0 $\textbf{Question 20.}$

Hope this helps.