Do two identical primitive Dirichlet characters have the same modulus

Question

If $\chi_1\ (\mathrm{mod}\ q_1),\chi_2\ (\mathrm{mod}\ q_2)$ are primitive Dirichlet characters with $\chi_1(n)=\chi_2(n)$ for all $n\in\mathbb{Z}$, can we deduce, that $q_1=q_2$? So far I just got that $(q_1,q_2)>1$: Assume $\mathrm{gcd}(q_1,q_2)=1$ with $q_1,q_2\neq1$. Thus $1=q_1x+q_2y$ for some integers $x,y$. Therefore \begin{equation} 1=\left|\chi_1(-1)\right|=\left|\chi_1(-1+q_1x)\right|=\left|\chi_2(-q_2y)\right|=0, \end{equation} a contradiction.

Answer 1

Since $\chi_1=\chi_2$ is periodic with period $q_1$ and also periodic with period $q_2$, it is periodic with period $g=\gcd(q_1,q_2)$. Then it must in fact be a Dirichlet character (mod $g$). (A little checking here is required to show that the zero values are all correct.) The primitivity of $\chi_1$ and $\chi_2$ then implies that $q_1=g=q_2$.