Prove that there exists no primitive character mod $k$ if $k=2m$, where $m$ is odd. This is from Apostol Chapter 8 Exercise 5 extended for all characters, real or not.
(Corrected using solution of Bruno Joyal)
Using Apostol Thm 8.16: $m|k$ so take $(a,k)=(b,k)=1$ (so that $a,b$ are odd) and take $a\equiv b \pmod m$.
Then $a\equiv b \pmod k$ as either
$a=b+rm\equiv b \pmod k$ if $r$ even or
$a\equiv b+m \pmod k$ if $r$ odd. But $b,m$ odd so $(b+m,k)\not=(a,k)=1$. A contradiction.
Hence whenever $(a,k)=(b,k)=1$ with $a\equiv b \pmod m$ then $a\equiv b \pmod k$ so $\chi(a)=\chi(b)$ for every character mod $k$.
Thus $m$ is an induced modulus for $\chi$ so there can be no primitive characters mod $k$.