$m>2$ and $n > 2$ are relatively prime $\Rightarrow$ no primitive root of $mn$ 
Show that if $m>2$ and $n > 2$ are relatively prime, there is no primitive root of $mn$

I know that $mn > 4$, and thus $\varphi(mn)$ is an even number so that I might write $\varphi(mn) = 2x$ for an integer $x$. If I could prove that $x = \frac{1}{2} \varphi(mn)$ is the order of some integer $a$ modulo $mn$, then I've proven that there is no primitive root of $mn$. 
Since $m$ and $n$ are relatively prime, I can write the equation
\begin{align}
a^{\frac{1}{2} \varphi(mn)} \equiv 1 \pmod{mn}
\end{align}
as a set of congruences
\begin{align}
\begin{cases}
a^{\frac{1}{2} \varphi(mn)} \equiv 1 \pmod{m} \\
a^{\frac{1}{2} \varphi(mn)} \equiv 1 \pmod{n}
\end{cases}
\end{align}
This is where I get stuck. Am I on the right track? Or is there a better way to prove this?

$A=\Bbb Z/mn\Bbb Z$ has always divisors of zero in particular $mn=0$. If $a$ were a primitive root then for all element $x$ distinct of $0$ in $A$ we have $x=a^k$ for some integer $k$. It follows $m=a^s$ and $n=a^t$ for some integers $s$ and $t$. But then $mn=a^{s+t}=0$ absurde.
