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$\DeclareMathOperator\ord{ord}$This is related to a question in the MO post, Does there exist a prime $p$ such that $\left|\frac{\mathrm{ord_{p}}(a)}{\mathrm{ord_{p}}(b)}-c\right|<\gamma$ for some small constant $\gamma$?

Let $p$ be a prime and $\ord_{p}(a)$ be the least positive integer $d$ such that $p\mid a^{d}-1$.

If $a$ and $b$ are two coprime natural numbers greater than 1, then does there exist a prime $p$ such that $ \frac{\ord_{p}(a)}{\ord_{p}(b)}>1$?

Edit: Some progress on this problem has been made recently here.

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    $\begingroup$ Only after reading the answer from @JoshuaZ did it even occur to me that $ord_p(a)$ might have a nonstandard meaning here and hence that the question might not be entirely trivial. $\endgroup$
    – Steven Landsburg
    Commented Nov 10, 2020 at 16:46
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    $\begingroup$ @StevenLandsburg $ord_p(a)$ is a standard notation for the multiplicative order of $a$ modulo $p$. $\endgroup$
    – Wojowu
    Commented Nov 10, 2020 at 16:58
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    $\begingroup$ Yeah, it can mean both the order of a in $Z/pZ$ but also the highest power of $p$ which divides $a$. This actually caused a small notational conflict in my thesis since I needed to talk about both. $\endgroup$
    – JoshuaZ
    Commented Nov 10, 2020 at 17:01
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    $\begingroup$ Please edit the question to include the definition of ord${}_p$. I would look at (non-tiny) prime factors of numbers of the form $b^n-1$, since that is a way of making ord${}_p(b)$ unusually small relative to the size of $p$, so that the desired inequality is much more likely to be true. $\endgroup$
    – Greg Martin
    Commented Nov 10, 2020 at 17:24
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    $\begingroup$ @JoshuaZ, I guess at least one almost never needs to talk about these two different notions for the same $a$ …. $\endgroup$
    – LSpice
    Commented Nov 10, 2020 at 22:18

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This question might be difficult, but heuristically it should be true. Essentially the same heuristic that yields the Artin primitive root conjecture, there should be for any such $a$ and $b$ be infinitely many primes $p$ where $a$ is a primitive root mod $p$, and $b$ is not.

Another heuristic approach is that if $a$ and $b$ are relatively prime, there should be infinitely many primes $p$ where $\frac{p-1}{2}$ is prime, $a$ is a quadratic non-residue mod $p$, and $b$ is a quadratic residue mod $p$. So if $p$ is large enough, one must have that the order of $a$ is $p-1$, and the order of $b$ is $\frac{p-1}{2}$.

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