Find $x$ so $x^{31}\equiv 2 \pmod{81}$

Question

$81$ isn't a prime so I can't use Fermat's little theorem.

$81=3^{4} \Rightarrow \varphi(81)=54$ but about the $gcd(x,81)$ , I don't know if it's equals to $1$ or not

and euler theorem won't help because $x^{54}\equiv 1 \pmod{81}$ and I need to find $x^{31}$

How do I approach this problem?

Answer 1

As $(31,54)=1$ we find by trial $1=31\cdot7-4\cdot54$

$$x^{31\cdot7-4\cdot54}=(x^{31})^7\cdot(x^{54})^{-4}\equiv2^7\cdot1^{4}\pmod{81}$$

Answer 2

This question is screaming for Hensel's Lemma. Let $f(x) = x^{31}-2$. Solve $f(x) \equiv 0 \pmod{3}$ and get $x\equiv 2 \pmod{3}$. Then lift to 9 to get $x\equiv 2 \pmod{9}$. Then lift to 27 to get $x\equiv 20 \pmod{27}.$ And finally lift to 81 to get $x\equiv 47 \pmod{81}$.