Prove that $a^\text{1/2 ord(a)}\equiv - 1\pmod n$ Given is that $a∈ℤ_n^*$, with $n>2$.
We assume that there exists a positive integer $\text{e}$ such that $a^e\equiv - 1\pmod n$
We need to show that $1/2$ $\text{ord(a)}$ is the smallest such number.
I know that $\text{ord(a)}$ is the smallest positive integer such that $a^{\text{ord}(a)}\equiv 1\pmod{n}$
But how do I prove that $a^\text{1/2 ord(a)}$ is congruent to $-1 \text{(mod n)}$?  

$a^e\equiv -1\pmod n\Rightarrow a^{2e}\equiv 1\pmod n$. This implies $\text{ord}(a)\mid 2e$. If $\text{ord}(a)$ was odd, we would have $\text{ord}(a)\mid e$ and so $a^e\equiv 1\pmod n$, impossible because $n>2$. So $\text{ord}(a)$ is even, say $\text{ord}(a)=2m$, and we know that $m\mid e$. Subtracting multiples of $2m$ from $e$ we can assume $0<e<2m$, hence $e=m$, that is $a^m\equiv -1\pmod n$.
