TITLE: Proving a number has no primitive roots
QUESTION [5 upvotes]: How do you prove an arbitrary number $n$ has no primitive roots without finding all numbers less than $n$ which are also coprime to $n$ and exhausting that none of the order of these numbers modulo $n$ are equal to $\phi(n)$?

REPLY [2 votes]: The key observation, which leads to half of the main theorem, is that

If $n=ab$, where $a,b >1$ are odd and coprime, then $n$ does not have a primitive root because for $m=lcm(\phi(a),\phi(b))$ we have $x^m \equiv 1 \bmod n$ for all $x$ coprime with $n$. Note that $m < \phi(n)$ because $\phi(a)$ and $\phi(b)$ are both even.

The harder part of the main theorem is the existence of primitive roots for primes and powers of primes.