MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $n$?

share|cite|improve this question
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 '12 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 '12 at 2:56
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 '12 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 '12 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 '12 at 3:29
up vote 1 down vote accepted

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 $|d_2|^n$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.