Assume that $ \phi( 5^m - 1) = 5^{n} - 1$ for some relatively prime positive integers $ m$ and $ n$. We have $ \gcd(5^m - 1,5^n - 1) = 5^{\gcd(m,n)} - 1 = 4 = 2^2$. Therefore, all odd primes in prime factorization of $ 5^m - 1$ must come in the first degree. Suppose that $ 5^m - 1 = 2^s p_1 p_2 \dots p_k$ where $ s\geq 2$ and all $ p_i$ are distinct odd primes. Then $ \phi( 5^m - 1) = 2^{s - 1} (p_1 - 1)(p_2 - 1)\dots(p_k - 1)$ is divisible by $ 2^{s + k - 1}$. Hence, $ \min\{s,s + k - 1\}\leq 2$. Consider two cases: 1) If $ k = 0$ then $ s = 3$, implying that $ 5^m - 1 = 8$, a contradiction; 2) If $ k > 0$ then $ s = 2$ and $ k = 1$, implying that $ 5^m - 1 = 4 p$ and $ \phi( 5^m - 1) = 5^{n} - 1 = 2(p - 1)$. Hence, $ 4p = 5^m - 1 = 2\cdot 5^n + 2$, implying that $ - 1\equiv 2\pmod{5}$, a contradiction. The obtained contradiction proves that $ \phi( 5^m - 1) \ne 5^{n} - 1$ for all relatively prime positive integers $ m$ and $ n$.

In the case k>0, I understand s=2, but why k=1? I think ur wrong

is it true? I'm not sure $ \phi(5^{m} - 1) = 5^{n} - 1$ and $ (m,n) = 1$ $ p$ is prime; if $ p^{2}|5^{m} - 1$ $ \implies$ $ p|5^{n} - 1$ $ \implies$ $ p|5^{(m,n)} - 1 = 4$ $ \implies$ $ p = 2$ $ 5^{m} - 1 = 2^{a}.p_{2}....p_{k}$ $ \implies$ $ 5^{n} - 1 = 2^{a - 1}.(p_{2} - 1)...(p_{k} - 1)$ if $ k\geq2$ $ \implies$ $ 2^{a}|5^{m} - 1,2^{a}|5^{n} - 1$ $ \implies$ $ a = 2$ $ 5^{m} - 1 = 4.p_{2}...p_{k},5^{n} - 1 = 2.(p_{2} - 1)...(p_{k} - 1)$ $ 2^{k}|5^{n} - 1,2^{k}|5^{2^{k - 1}} - 1$ if $ (n,2) = 1$ $ \implies$ $ 2^{k}|4$ $ \implies$ $ k = 2$ $ 5^{m} = 4p_{2} + 1,5^{n} = 2.(p_{2} - 1) + 1 = 2p_{2} - 1$ $ \implies$ $ 5|4p_{2} + 1,5|2p_{2} - 1$ $ \implies$ $ 5|3$contradiction! $ \implies$ $ k > 2$ $ \implies$ $ (n,2) = 2$ $ (m,n) = 1$ $ \implies$ $ 2\nmid m$ $ \implies$ $ m = 2u + 1$ $ \implies$ $ 5^{m} = 5x^{2}$ $ \implies$ $ 5x^{2}\equiv 1\pmod{P_{i}}$ for $ i = 2,...,k$ $ c^{2}\equiv 5\pmod{P_{i}}$ $ \implies$ $ P_{i}\equiv \pm 1\pmod{10}$ if $ P_{s}\equiv 1\pmod{10}$ $ \implies$ $ 5|5^{n} - 1$ contradiction! $ \implies$ $ P_{i}\equiv - 1\pmod{10}$,$ P_{i}\equiv - 1\pmod{5}$ for $ i = 2,..,k$ $ 5^{m} - 1 = 4.p_{2}...p_{k}$ $ \implies$ $ ( - 1)^{k}\equiv - 1\pmod{5}$ $ \implies$ $ 2\nmid k$ $ 5^{n} - 1 = 2.(p_{2} - 1)...(p_{k} - 1)\equiv 2^{k}\equiv - 1\pmod{5}$ $ \implies$ $ 2|k$ Contradiction! if $ k = 1$ $ \implies$ $ 5^{m} - 1 = 2^{a},5^{n} - 1 = 2^{a - 1}$ $ \implies$ $ 2^{a - 1}|4$ ,$ 5^{n} - 1|4$ $ \implies$ $ n = 1$ $ \implies$ $ 5^{m} = 9$ contradiction! $ \Box$

We show the contrapositive. Suppose $\gcd(m,n)=1$, and write $5^m-1=2^e\prod_{i=1}^{k}p_i^{e_i}$ where $p_i$ are all odd primes. We have \[5^n-1=\varphi(5^m-1)=2^{e-1}\prod_{i=1}^{k}p_i^{e_i-1}\cdot\prod_{i=1}^{k}(p_i-1). \text{ } (*)\] Thus, $2^e|5^n-1$. Now, it is well known that $\gcd(5^m-1, 5^n-1)=5^{\gcd(m,n)}-1=4$, and it follows that $e_i=1$ for every $1\le i\le k$, and that $e=2$. Moreover, $2^3|5^j-1$ for all $2|j$, so $2\nmid m$. Thus, write $m=2m^{\prime}+1$. Now, since $p_i|5(5^{m^{\prime}})^2-1$ for $1\le i\le k$, $5$ must be a quadratic residue in $\mathbf{F}_{p^i}$, and it follows that $p_i \equiv\pm 1\pmod{5}$. However, the above work shows that $5\nmid p_i -1$ for any $1\le i\le k$, which means $p_i\equiv -1\pmod{5}$. Now, consider the prime factorization of $5^m-1$. Reducing this $\pmod{5}$ yields $(-1)^k = 1$, so $k$ is even. However, reducing $(*)$ mod $5$ yields $(-2)^{k+1}\equiv 1\pmod{5}$, which means $k\equiv 3\pmod{4}$, which contradicts $2|k$. We are done.