Let $a$ and $b$ be integers, and denote by ${\rm ord}_p(a)$ the $\textbf{multiplicative order}$ of $a$ modulo $p$.
Assume that there exists a constant $K$ such that for all prime $p>K$, ${\rm ord}_p(b)$ divides ${\rm ord}_p(a)$. Is it that $b$ is a power of $a$?
I came up with this question, but I don't know if it's true or false, and I have no idea how to deal with it. Could you help me?
Thanks.
The (nontrivial) answer seems to be yes. Assume $a,b \geq 2$ to make matters simpler. The proof is a bit complex and it can probably be improved – I think that we can deal with sign issues by using a little bit more caution.
Let $k \geq 3$ be any large prime number (not dividing by any of the nonzero valuations at any prime of $a$ or $b$), and consider the number fields $K_a$ and $K_b$, where $K_x$ is the Galois closure of $R_x=\mathbb{Q}(x^{1/k})$. Now, $K_x$ is the tensor product of $R_x$ and $C=\mathbb{Q}(e^{2i\pi/k})$.
A large enough prime $p$ is totally split in $K_x$ iff it is split in $R_x$ and $C$, iff it is congruent to $1$ mod $k$ and $a$ is a $p$-th power mod $k$, iff $X^k-x$ splits simply in $\mathbb{F}_p$.
So, assume that $p$ is a large prime that splits completely in $K_a$. Then $k|p-1$ and $a$ is a $k$-th power mod $p$. Thus $\mathrm{ord}_p(a)$ divides $\frac{p-1}{k}$. Thus the same holds for $\mathrm{ord}_p(b)$, and thus (as $k|p-1$) $b$ is a $k$-th power in $\mathbb{F}_p$ and thus $p$ splits totally in $K_b$.
It follows by Cebotarev that $K_b \subset K_a$ and by dimension that $K_b=K_a$. In other words, $R_b$ is a subextension of $K_a$ (of degree $p$). Now, the Galois group of $K_a$ over $\mathbb{Q}$ can easily be shown to be the semi-direct product of $\mathbb{F}_k$ and $\mathbb{F}_k^{\times}$ (the latter acting on the former by multiplication), so it has cardinality $k(k-1)$.
How many subgroups of index $k$ does it have? Well, such a subgroup $H$ has order $k-1$, so its intersection with $Gal(K_a/C) \cong \mathbb{F}_k$ is trivial, so its projection to $Gal(C/\mathbb{Q}) \cong \mathbb{F}_k^{\times}$ is injective, hence an isomorphism and $H$ is cyclic, generated by its unique element above a fixed generator of $Gal(C/\mathbb{Q})$.
It follows that there are exactly $k$ extensions $H$ of index $k$, and thus $K_a$ has exactly $k$ subextensions of degree $k$ over $\mathbb{Q}$. Moreover, it’s easy to see that the $\mathbb{Q}(a^{1/k}e^{2il\pi/k})$, $0 \leq l < k$, all fit this description and are pairwise distinct (otherwise two of them are equal and it follows that one extension isomorphic to $R_a$ contains a $k$-th root of unity, which impossible by dimension since $k-1$ doesn’t divide $k$), and that the only one which is a subset of $\mathbb{R}$ is $R_a$. Thus $R_b=R_a$.
Let, for an automorphism $\sigma$ and $x=a,b$, $m(\sigma)$ be the number in $\mathbb{F}_k^{\times}$ such that $\sigma(e^{2i\pi/k})=e^{2i\pi m(\sigma)/k}$, and $n_x(\sigma)$ be the number in $\mathbb{F}_k$ such that $\sigma(x^{1/k})x^{-1/k}=e^{2i\pi n_x(\sigma)/k}$.
In $Gal(K_a/C)$, it’s easy to see that $\sigma \longmapsto n_x(\sigma)$ is an isomorphism, and thus $n_b=sn_a$ for an integer $k > s \geq 1$.
Therefore, if $x= b^{1/k}a^{-s/k} \in R_a=R_b$, the stabilizer in $Gal(K_a/\mathbb{Q})$ of $x$ contains $Gal(K_a/C)$ (so is divisible by $k$) and $Gal(K_a/R_a)$ (so is divisible by $k-1$) so is all of $Gal(K_a/\mathbb{Q})$, thus $b^{1/k} \in \mathbb{Q}a^{s/k}$. Thus $b/a^s$ is a $k$-th power in $\mathbb{Q}$.
In other words, consider the vectors $v_a=(v_p(a))_{p}$ of $p$-adic valuations of $a$ for all primes $p$, and the same vector $(v_p(b))_{p}$. We know that these vectors have finite support and are collinear mod every large prime number: it follows (using, say, determinants) that they are collinear in $\mathbb{Q}$.
In other words, we can find pairwise coprime positive integers $n,m$ and an integer $t$ such that $b=t^n, a=t^m$.
Then, if $p$ is a large prime, $\mathrm{ord}_p(b)=\frac{\mathrm{ord}_p(t)}{\mathrm{ord}_p(t) \wedge n}$. Using a similar argument for $a$, it follows that for almost every prime, $\mathrm{ord}_p(t) \wedge n$ is a multiple of $\mathrm{ord}_p(t) \wedge m$. But these two numbers are coprime, so there are finitely many primes $p$ such that the multiplicative order of $t$ mod $p$ is divisible by a prime dividing $m$.
But if $m>1$, this is in gross contradiction to Zsigmondy’s theorem (or, here, a LTE-based elementary statement saying that if $p$ is a prime divisor of $m$, there are infinitely many prime numbers dividing $t^{p^l}-1$ for some $l \geq 1$). So $t=a$, QED.
Edit: I know it’s been a year, but I realized that the argument could be made clearer. So let $a,b \geq 2$ be such that the multiplicative order of $b$ mod $p$ divides that of $a$ for almost all primes $p$. We want to show that $b$ is a power of $a$.
First, assume that $a,b$ are multiplicatively dependent, ie there are positive integers $r,s,t$ with $s,t$ coprime such that $a=r^s$, $b=r^t$. For $n \geq 10$, then $r^{sn}-1$ has a primitive prime divisor $p$ by Zsigmondy, so that $r^{tn}-1$ is also divisible by $p$. Thus $p$ divides the gcd of $r^{tn}-1$ and $r^{sn}-1$, which is $r^n-1$. But as $p$ is primitive, $n \geq sn$, so $s=1$ and $b$ is a power of $a$.
So we can assume that $a,b$ are multiplicatively independent. In other words, there is a large prime $k$ and primes $s,t$ such that $k$ doesn’t divide $v_s(a)v_t(b)-v_t(a)v_s(b)$ (in particular, $a,b$ are not $k$-th powers).
Let $C=\mathbb{Q}(\mu_k)$, $R_x=\mathbb{Q}(x^{1/k})$ and $K_x=R_xC$ for $x=a,b$. Note that $K_a,K_b$ are Galois over $\mathbb{Q}$.
As $C$ and $R_x$ have coprime degrees over $\mathbb{Q}$ ($k-1$ and $k$, respectively), $C \otimes R_x \rightarrow K_x$ is an isomorphism. Moreover, a prime number $p$ splits totally in $K_x$ iff it splits totally in $C$ and $R_x$, iff $k|p-1$ and $x$ is a $k$-th power mod $p$, iff $k|p-1$ and the order of $x$ mod $p$ divides $\frac{p-1}{k}$. In particular, for almost every $p$, if $p$ splits totally in $K_a$, then $p$ splits totally in $K_b$.
By Cebotarev, since $K_a$ and $K_b$ are Galois over $\mathbb{Q}$ with the same degree, this implies that $K_b \subset K_a$ hence $K_b=K_a$.
By Kummer theory (as $K_x/C$ is cyclic of Galois group $\mathbb{F}_k$), this implies that there are integers $0 \leq x,y < k$ not both zero such that $a^xb^y$ is a $k$-th power in $C$.
Taking norms, we see that $a^{(k-1)x}b^{y(k-1)}$ is the $k$-th power of a rational number. So $k|xv_s(a)+yv_s(b)$, $k|xv_t(a)+yv_t(b)$, which contradicts the assumption on $k,s,t$.