I found the excerpt here that said the second involves the p-adic analog of the above. How can these two statement related?
- The transcendence of $2^{\sqrt2}$ and $e^\pi$: Gelfand's proof. (Assuming some basic complex analysis).
- Nathanson's problem: show that $3^n\nmid5^n-2$ for $n>1$. (This involves the $p$-adic analogue of the above topic.)
I showed the second argument by the following, but try to find a way of the first prove and I don't know how they related,
Firstly, we assume that there exist an $n>1$ satisfying $3^n|(5^n-2)$. We have that $2,5$ are the primitive root modulo $3^n$. Thus, $$\left\{2^1,2^2,\dots , 2^{\varphi{(3^n)}}\right\}=\left\{5^1,5^2,\dots , 5^{\varphi(3^n)}\right\}.$$ Hence, we can choose $b\in[1,2\cdot 3^{n-1}]\cap\mathbb Z$ so that $2^n\equiv 5^b\pmod{3^n}$. Therefore, by the premise $5^n\equiv 2\pmod{3^n}$, $$5^{n^2}\equiv 2^n\equiv 5^b\pmod{3^n}\Longrightarrow 5^{|b-n^2|}\equiv 1\pmod {3^n}.$$ Since $5$ is a primitive root we have that $$\varphi(3^n)=2\cdot 3^{n-1}\Big||b-n^2|\Longrightarrow 2\cdot 3^{n-1}\le b-n^2<2\cdot 3^{n-1}.$$ As $|b-n^2|=b-n^2$ otherwise $n^2-b$ is still far less than $2\cdot 3^{n-1}.$ I hope it is valid and thank you in advance for my question.
The above is invalid as $b=n^2.$ I give a proof here,
Let $\alpha_i$ be the number such that $5^{\alpha_i}\equiv 2\pmod{3^i}$. We will show that $\alpha_i>i$ for all $i>1$. Notice that we have the proof that $\alpha_i\le\alpha_{i+1}$.
We assume otherwise that $\alpha_i>\alpha_{i+1}$ for some $i$. We have $$5^{\alpha_{i+1}}\equiv 2\pmod{3^{i+1}}\Longrightarrow 5^{\alpha_{i+1}}\equiv 2\equiv 5^{\alpha_i}\pmod{3^{i}}\Longrightarrow 5^{\alpha_i-\alpha_{i+1}}\equiv 1\pmod{3^{i}}.$$ Which means that $\varphi(3^i)\le \alpha_{i}-\alpha_{i+1}<\varphi(3^i)$, a contradiction.
By the definition of $\alpha_i$ we have that it must be in the form $6k+5$ for $i>1$ since $5^{\alpha_i}\equiv 2\pmod 9$ and $\alpha_i$ is not in the form $6k+3$ otherwise $5^{\alpha_i}\equiv 5^{6k+3}\equiv 8\not\equiv 2\pmod 9$. The same argument goes for $\alpha_i=6k+1$ that $5^{\alpha_i}\equiv 5\not\equiv 2\pmod 9$.
The problem is in the case $\alpha_i=\alpha_{i+1}$. Consider the set $$\left\{5^0,5^1,5^2,\dots,5^{\varphi(3^i)-1}\right\}.$$ Let $x$ be a smallest number that $5^x>3^i$. If $\alpha_i\le i$ then $5^{\alpha_i}=3^i+2$ or $5^{\alpha_i}=3^{i+1}+2$ since $5^{\alpha-1}<3^i\Longrightarrow 5^\alpha<5\cdot 3^i$.
Either case is impossible since if we assume $2=5^\alpha-3^\beta$ with $\alpha=6k+5$, as $\beta$ is odd otherwise $0\equiv 3^\beta+2\equiv (-2)^\beta+2\equiv 2^\beta+2 \pmod {5}\Longrightarrow 2^{\beta-3}\equiv 1\pmod 5$, which means that $4|(\beta-3)$ and it is impossible, then we have that $$2\equiv 5^{6k+5}-3^{\beta}\equiv 0,4,5 \pmod{7}$$ a contradiction occurs. Thus, $\alpha_i>i$ for any $i$ as desired. Hence, $3^n|(5^n-2)$ only when $n=1$.