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S. Carnahan
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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 $|d_2|^n$.

S. Carnahan
  • 45.7k
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  • 114
  • 220