# Does there exist a prime $p$ such that $\frac{\operatorname{ord}_{p}(a)}{\operatorname{ord}_{p}(b)}>1?$

$$\DeclareMathOperator\ord{ord}$$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.

• 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. Nov 10 '20 at 16:46
• @StevenLandsburg $ord_p(a)$ is a standard notation for the multiplicative order of $a$ modulo $p$. Nov 10 '20 at 16:58
• 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. Nov 10 '20 at 17:01
• 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. Nov 10 '20 at 17:24
• @JoshuaZ, I guess at least one almost never needs to talk about these two different notions for the same $a$ …. Nov 10 '20 at 22:18

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 $$\frac{p-1}{2}$$.