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Let $p$ be a prime number and let $d$ be a divisor of $(p-1)$. Look at $a^d \bmod p$ for every $a \in \{1,2,\dots, (p-1)\}$. What numbers do you get, and how frequently do they occur? Formulate a general rule, and prove that it holds for all prime numbers $p$, all $d \mid (p-1)$ and all integers $a \in \{1,2,\dots, (p-1)\}$.

I tried for a few different values of $p$, and found that we will get $\frac{p-1}{d}$ different solutions to $a^d \bmod p$, with a frequency of $d$ for each number. For example, for $p = 11$, $a \in \{1,2,\dots,9,10\}$ and $d = 1,2,5$ or $10$ we have

\begin{align*} &a^1 \bmod p = a \\ &a^2 \bmod p = \begin{cases} 1 \\ 4 \\ 9 \\ 5 \\ 3 \end{cases} \\ &a^{5} \bmod p = \begin{cases} 1 \\ 10 \\ \end{cases} \\ &a^{10} \bmod p = 1 \end{align*}

I'm just struggling to see how I can prove this in general. I know that the following is true for all $p$ and all $a \in \{1,2,\dots,(p-1)\}$: \begin{align*} &a^1 \bmod p = a \\ &a^{p-1} \bmod p = 1 \end{align*} But what about the values of $a^d \bmod p$ for all other $d$? How can I say something in general about those?

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Yes, there are exactly $\frac{p-1}{d}$ different solutions for $a^d \pmod{p}$. The idea is same as here.

Consider $a^d \equiv x \pmod{p}$ where $x$ is one of the solutions. We will find how many $a$ so that this congruence holds. Let $g$ be primitive root modulo $p$ then $\{ g,g^2, \cdots , g^{p-1} \}$ is the reduced residue system modulo $p$. Hence, it suffices to find $1 \le l \le p-1$ so that $g^{ld} \equiv x \pmod{p}$. Note that $\text{ord}_p(g)=p-1$ so it suffices to find $l$ so that $ld \equiv k \pmod{p-1}$ where $g^k \equiv x \pmod{p}$. We see that there are $d$ possible value for $l$. This follows there are $d$ solutions for $a$ so that $a^d \equiv x \pmod{p}$ for a fixed $x$.

There are $p-1$ values for $a$, and each solution $a^d \pmod{p}$ needs $d$ values for $a$, so that must means there are $\dfrac{p-1}{d}$ solutions for $a^d \pmod{p}$.

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