Order of an Element in $\Bbb Z_p$

Question

I have a very simple question & would appreciate it very much if you could assist.

Let $n$ be an odd positive composite integer and let $p$ be a prime divisor of $n$, where $p$ is not of the form $p= 2^m +1, m \geq 1$. Let $a \in \mathbb Z_{p} -\{0\}$. Further, let $r$ be the order of $a$ (i.e. $a^r \equiv 1 \bmod{p}$).

Show that if $a^n \equiv a \pmod{n}$, then

1. the integer $r$ divides $n-1$.
2. $r$ is not a power of two (i.e. $r\not= 2^j$ for some $j\geq 1$). Note that $n$ is a composite integer. For prime $n$, it easy to find examples in which the order is a power of two: $2^{16} \equiv 1 \pmod{257},~2^{64} \equiv 1 \pmod{641}.$

I tried the following.

Since $a^n\equiv a \pmod{n} \Longrightarrow a^{n}\equiv a\pmod{p}$, it follows that \begin{eqnarray*} a^{n}& \equiv & (a^{p})^\frac{n}{p} \pmod{p},\\ &\equiv & a^\frac{n}{p} \pmod{p},\\ &\equiv & a^{(k_1(p-1)+k_2)} \pmod{p},\\ &\equiv & a^{k_2} \equiv a \pmod{p}, \end{eqnarray*} for some positive integers $k_1$ and $k_2$.

Further, \begin{eqnarray*} a^{n}&\equiv & a^{k_2} \equiv a \pmod{p},\\ a^{n-1}&\equiv & a^{k_2-1} \equiv 1 \pmod{p}, \end{eqnarray*} Now, I know that $r|(k_2-1)$. Does this imply that $r|(n-1)$?

Answer 1

Hint:

1. You can assume that $n-1>r$, otherwise there is nothing to prove. Now write $$n-1=qr+s$$ and consider $a^{n-1}\equiv_p 1$. What can you say about $s$?