Ytukyg

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. $Box$