The argument below was used by Gauss to rule out primitive roots when in fact they don't exist.
Suppose that $g$ is a primitive root of $m \gt 2$. Then the congruence $x^2\equiv 1\pmod{m}$ has precisely $2$ solutions.
Now let $m=pq$ where $p$ and $q$ are distinct odd primes. By the Chinese Remainder Theorem, the congruence $x^2\equiv 1\pmod{pq}$ has $4$ solution. (Just combine in all possible ways the solutions mod $p$ and mod $q$.)
Added: Alternately, we show the stronger result that if $a\gt 2$ and $b\gt 2$ are relatively prime, then $ab$ does not have a primitive root. So we need to show there is no element of order $\varphi(ab)$ modulo $ab$. Suppose to the contrary that $g$ is such an element.
We have $\varphi(ab)=\varphi(a)\varphi(b)$. Both $\varphi(a)$ and $\varphi(b)$ are even. It follows that $\varphi(a)$ and $\varphi(b)$ each divide $m$, where $m=\frac{\varphi(a)\varphi(b)}{2}$.
From this we can see that $g^m\equiv 1\pmod{a}$, and $g^m\equiv 1\pmod{b}$, and hence $g^m \equiv 1\pmod{ab}$, contradicting the assumption that $g$ has order $\varphi(ab)$.