Solving $m^3 \equiv n^6 \pmod{19}$

Question

I'm studying for a first year Discrete Mathematics course, I found this question on a previous paper and am lost on how to solve:

Let $n$ be a fixed arbitrary integer, prove that there are infinitely many integers $m$ s.t.: $m^3 \equiv n^6 \pmod{19}$

Thank you

Answer 1

Since $n$ is given, you take $m=n^2$. This satisfies the congruence $m^3 \equiv n^6 \pmod{19}$. Now to generate infinitely many integer solutions, set $m_k=n^2+19k$, where $k \in \Bbb{Z}$.

Answer 2

Use Discrete Logarithm wrt primitive root $g$

$3$ind$_gm\equiv6$ind$_gn\pmod{\phi(19)}$

$\iff$ind$_gm\equiv2$ind$_gn\pmod6$

$\implies m=n^2g^{6k}$ where $k\equiv0,1,2\pmod3$

As $2^4\equiv-3,2^9\equiv2(-3)^2\equiv-1\pmod{19},g$ can be chosen to be $2$