Find all $n \in \mathbb{N}$ such that $30^n \equiv 20 \pmod {38}$

Question

The full problem was to actually find all $n \in \mathbb{N}$ such that $w^{{30}^n} = -w^{4^5+21}$, with $w$ being a primitive 38th root of unity.

I've proven that the right side of the equation equals 1, so now I'm left with proving that $30^{n}+208 \equiv 0 \pmod {38}$ which leads to $30^n \equiv 20 \pmod{38}$ And here I got stuck, I've got no clue how to continue. Any help would be appreciated!

Answer 1

Just write out the period

$30\equiv -8\pmod{38}$

$30^2\equiv 64\equiv-12\pmod{38}$

$30^3\equiv96\equiv20 \pmod{38}$

$30^4\equiv-160\equiv30\pmod{38}$

The period is short and you can see every $3k$ is a solution. Sometimes the dumb way is the best way.

Answer 2

Since $30^n\equiv 20\pmod2$ for all $n>0$, we only need to consider $11^n\equiv30^n\equiv 20\equiv1\pmod{19}$. Since the order of $11\pmod{19}$ is $3$, this happens for all positive multiples of $3$.