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Let $a,binmathbb{N}$ be such that $ageq 2$, $anmid b$, and $a^n-1mid b^n-1$ for all $ninmathbb{N}$, then $b=1$.

PS: In fact, if we do not assume that $anmid b$, then the statement should be $b=a^k$ for some $kinmathbb{N}cup{0}$. The above statement can deduce the fact.

This is a solution here which was found in some paper in AMM. it's a very nice proof. but it's very constructive. For example, the construction of $Q_k(x)$, $p_k$ and $r_{k,n}$, it's not so easy to construct them. Maybe there is some direct method.

Here is my proof from AoPS.

$textbf{Problem 1.}$ Prove that if $b^n-1mid a^n-1$ for all $n$, then $a=b^k$ for some integer $k$.

$textbf{Lemma 1.}$ If $b^n-1mid a^n-1$ for all $n$, then $(b-1)(b^2-1)ldots (b^n-1)mid n!(a-1)(a-b)ldots (a-b^{n-1})$ for every $n$.

$textbf{Proof of Lemma 1.}$ First consider any prime $pmid (b-1)(b^2-1)ldots (b^n-1)$, so there exists minimal $d$ such that $pmid b^d-1$ and it's well-known fact that then $x^d-1equiv (x-1)(x-b)ldots (x-b^{d-1})pmod{p^{nu_p(b^d-1)}}$, so $P(x)=x^d-1$ has exactly $d$ consecutive roots $1, b, b^2,ldots , b^{d-1}$. From the condition $b^d-1mid a^d-1$ we get that $a$ is also the root of polynomial $x^d-1$. So easy to see that there exists only one $0leq d'leq d$, such that $a-b^{d'}equiv 0pmod{p^{nu_p(b^d-1)}}$. From this we get that for all $iequiv d'pmod{d}$, $a-b^iequiv 0pmod{p^{nu_p(b^d-1)}}$. Now from LTE lemma we get that $nu_p(b^{jd}-1)leqnu_p(b^d-1)+nu_p(j)$, for each $j$, so $nu_p(prod_{i=1}^n(b^i-1))leq[n/d]nu_p(b^d-1)+nu_p(n!)leqnu_p(n!(a-1)(a-b)ldots (a-b^{n-1}))$. Combining similar inequalities for every $p$ we get Lemma 1. $Box$

$textbf{Proof of Problem 1.}$ Firstly it's simple exercise to prove that if a prime $p$ is such that $pmid a$, then $pmid b$. So there exists a positive integer $t$, such that $anmid b^{t-1}$, but $amid b^t$. From Lemma 1 we get that $(2t)!frac{(a-1)(a-b)ldots (a-b^{t-1})(a-b^t)ldots (a-b^{2t-1})}{(b-1)(b^2-1)ldots (b^t-1)(b^{t+1}-1)ldots (b^{2t}-1)}inmathbb{Z}$. But for every $i$ we have that $gcd(a, b^i-1)$, so if we just look on the numerator of the fraction above then we get that $$gcd(a,1)gcd(a, b)ldots gcd(a, b^{t-1})a^tmid (2t)!frac{(a-1)(a-b)ldots (a-b^{t-1})(a-b^t)ldots (a-b^{2t-1})}{(b-1)(b^2-1)ldots (b^t-1)(b^{t+1}-1)ldots (b^{2t}-1)}$$ and if $anot= b^x$ for every natural $x$, then $$|(2t)!frac{(a-1)(a-b)ldots (a-b^{t-1})(a-b^t)ldots (a-b^{2t-1})}{(b-1)(b^2-1)ldots (b^t-1)(b^{t+1}-1)ldots (b^{2t}-1)}|geq gcd(a,1)gcd(a, b)ldots gcd(a, b^{t-1})a^t$$ But it's easy to check that $|(2t)!frac{(a-1)(a-b)ldots (a-b^{t-1})(a-b^t)ldots (a-b^{2t-1})}{(b-1)(b^2-1)ldots (b^t-1)(b^{t+1}-1)ldots (b^{2t}-1)}|leq (2t)!a^t$ and $gcd(a,1)gcd(a, b)ldots gcd(a, b^{t-1})geq 2^{frac{t(t-1)}{2}}$ (here we used the fact that $gcd(a, b^i)midgcd(a, b^{i+1})$ and $gcd(a, b^{i-1})not=gcd(a, b^i)$ for $ileq t-1$). So $2^{frac{t(t-1)}{2}}a^tleq (2t)!a^t$. Contradiction. $Box$

Reference for a generalization of this theorem and it's proof in French. See this article.

Langlois, Bruno, An arithmetical application of Hadamard’s quotient theorem, Ann. Univ. Sci. Budap. Rolando Eötvös, Sect. Comput. 44, 183-196 (2015). ZBL1389.11046.