If we are looking for a value $n \equiv v \pmod p$ or $n \equiv v_r + v_i i \pmod p$, where $v_r+v_i\cdot i$ is a complex number modulo $p$, is it ever possible to have a situation where we can find both $n$ and $\log{(n)}$ modulo $p$? I know that, for instance, if $n=1$, then $\log{(n)} = 0$. However, I'm looking for a situation in which both values are nonzero.
Is it possible to get nontrivial $n$ such that we can find BOTH $n \pmod p$ and $\log{(n)} \pmod p$?
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Must we choose $n$ such that $\log(n)$ is an integer? (Depends on your definition of $\bmod$) – apnorton Dec 09 '14 at 00:21
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1@anorton: Not exactly. But I prefer to get something that is reasonably simple and close to an integer. I don't know what to expect from the answers yet, but I can help you get a feel for what I'm after. I'm trying to get a value for $n$ that I can approximate reasonably well using integers. If $n$ is too complicated somehow, I may not be able to use it. But if there's no alternative, I'd at least be curious to see what kind of values $n$ could be. To be more clear, you can use a fairly loose definition for $\bmod$, if you like. – Matt Groff Dec 09 '14 at 00:27