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Suppose $d_1, d_2$ are two fixed coprime integers, $\frac{d_1}{d_2} \neq \pm 1$. Given any $n > 0$, can we find a prime number $p$ such that the order of $d_1d^{-1}_2$ in the multiplicative group of the field $\mathbb{Z}/p\mathbb{Z}$ be greater than $n$?

edited Oct 3 at 9:47

Charles Matthews

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3 revisions, 2 users
Xiaolei Wu 100%

If $d_1/d_2 \ne \pm 1$, the order $\mod p$ is clearly at least $\log p/\log 2$, so yes. – Felipe Voloch Oct 3 at 2:48

Hi, Felipe, how do you see the order is at least logp/log2? I am not a number theorist, so I might be a little slow on this. – Xiaolei Wu Oct 3 at 2:56

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I'm not sure how Felipe got $\log p / \log 2$, but if $d_1^m \equiv d_2^m \mod p$ then $|d_1^m - d_2^m| = p k \ge p$. So, at least one of $|d_1|^m, |d_2|^m$ has to be at least $p$. Choosing $p$ larger than $\max(|d_1|,|d_2|)^m$ means the order has to be greater than $m$. – Douglas Zare Oct 3 at 3:08

Hi, Felipe. I actually think you are wrong now. There are always elements in Z/pZ has order 2 since the multiplicative group has order p-1 which is an even number if p is odd. – Xiaolei Wu Oct 3 at 3:20

Thank you, Douglas Zare. I see your proof now. I think I am going to close this question soon since it is actually very easy. – Xiaolei Wu Oct 3 at 3:29

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S. Carnahan · Accepted Answer · 2012-10-03 03:19:24Z UTC

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The answer to your question is "yes" (cf. Douglas Zare's comment). In fact, for all sufficiently large primes $p$, the order of $d_1 d_2^{-1}$ is greater than $n$. Here, "sufficiently large" means greater than $|d_1|^n$ and $|d_2|^n$.

answered Oct 3 at 3:19

S. Carnahan♦

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Can the order of a rational number in Z/pZ be as large as we want

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