(a) Let $p$ be prime and $\gcd (a, p) = 1$. Show that if the order of $a \pmod p$ is $k$ and $k$ is even then
$$ a^{\frac{k}{2}} \equiv (p− 1) \pmod p$$
(b) Let $n \ge 2$ and $\gcd (a, n) = 1$.
Disprove that if the order of $a \pmod n$ is $k$ and $k$ is even then
$$ a^{\frac{k}{2}} \equiv \pm 1 \pmod n $$
(a) The order of $a \pmod p$ being $k$ means
$$ a^k \equiv 1 \pmod p $$
We're told $k$ is even, so $k=2j$ for some integer $j$,
$$ (a^{j})^2 \equiv 1 \pmod p $$
By Proposition (3.14), this means $a^j \equiv 1 \pmod p$ or $a^j \equiv -1 \pmod p$.
But $a^j \equiv 1 \pmod p$ is not possible as the order is $k$, the least positive integer such that $a^k \equiv 1 \pmod p$, and $j<k$. This leaves
$$a^j \equiv -1 \equiv ( p-1 ) \pmod p $$
That is, $a^{\frac{k}{2}} \equiv (p− 1) \pmod p$.
(b) We disprove the statement with a counter-example.
Let's set $n=8, a=3$.
We confirm that $n=8 \ge 2$, and $\gcd(a,n)=\gcd(3,8)=1$.
Because $3^1 \equiv 3 \pmod 8$ and $3^2 \equiv 1 \pmod 8$, the order $k$ of $a \pmod n$ is 2, which is even.
And so
$$ 3^{\frac{2}{2}} \equiv 3 \not \equiv \pm 1 \pmod 8 $$
This counter-example disproves the given statement.