(exam study help) Prove that there are infinitely many integers $m$ such that: $m^3 \equiv n^6 \pmod{19}$

Question

I'm studying for a first year discrete math final exam. I was wondering if anyone could help me with this proof. I started by writing down what I know is true, but I can't seem to bring what I have so far to some meaningful conclusion.

Let $n$ be a fixed but arbitrary integer. Prove that there are infinitely many integers m such that: $$m^3\equiv n^6 \pmod {19}$$

$$m^3\equiv n^6 \pmod {19} \Rightarrow 19|(n^6-m^3)$$ then:

$n=19k+r $ for $ k\in \mathbb{R}$

$m=19l+r $ for $ l\in \mathbb{R}$

$n^6=(19k+r)^6 $

$m^3=(19l+r)^3 $

now:

$$(n^6-m^3)=(19k+r)^6-(19l+r)^3$$

$$=19(k^6-l^3)+r$$

Answer 1

First notice m = $n^2$ is a solution
for the equivalence relation.
Assume there are only finite many solutions.
Let m be the largest. m + 19 is a larger solution.
Thus there are infinitely many solutions.

Answer 2

$(m + 19k)^3 \equiv m^3 \mod 19$ so whatever is true about about $m$ and $m^3$ $\mod 19$ will be true about all infinite $m + 19k$ $\mod 19$

So if there exist any $m^3 \equiv n^6 \mod 19$ then there exist an infinite number of $(m + 19k)^3 \equiv n^6\mod 19$.

And for $m = n^2$ we have $(n^2)^3 \equiv n^6 \mod 19$.