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Does there exist a prime $p \geq 7$ such that the order of $4$ in the multiplicative group of units in $\mathbb{Z}/p^n$ is odd for every positive integer $n$?

It would be nice if $7$ was already an example. I computed the order of $4$ modulo $7^n$ for $n=1,2,\ldots,12$ and it came out odd, but that's not particularly compelling evidence I guess.

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Hint: $4$ is always a square modulo $p^n$. What can you say about the order of its square root (whatever that is) modulo $7^n$? What does this imply about the order of $4$?

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    Ohhhhhhh I had it backward in my first comment. It's the order of $4$ which should divide the order of $2$! So yes, if the $2$ has odd order, then so does $4$. If $2$ has order $2k$ where $k$ is odd, then $4^k = 2^{2k} = 1$ modulo $7^n$, so the order of $4$ divides the odd number $k$ and is odd. Thanks, it makes sense now.2011-10-22
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Use Theorem 4.4 in Leveque, Fundamentals of Number Theory:

"Suppose $p$ is a prime and $p \nmid a$. Let $\mathrm{ord}_p a = t$ and let $p^z$ be [the exact power of $p$ dividing $a^t-1$]. Then if $p>2$ or $z>1$,

$ t_n = \mathrm{ord}_{p^n} a = t, \quad \text{for $n \leq z$}$

and

$ t_n = t p^{n-z}, \quad \text{for $n \geq z$}"$

So if the order of $4$ modulo $p$ is odd, then the order of $4$ is odd modulo every power of $p$.

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    Search "Leveque primitive root" on google. The google books link (first as I write this) links to the right chapter.2011-10-21