Solving a congruence, tricky implication

Question

We want to prove that

$$243x \equiv 1 \mod 2018 \implies x^{403} \equiv 3 \mod 2018$$

My try :

Assume that $243x \equiv 1 \mod 2018$

We have $x^{2016} \equiv 1 \mod 2018$ (by Fermat ($1009$ is prime) and oddness of $x$) so $x^{2015} \equiv 243 \mod 2018$

but $403 \times 5 = 2015 $

hence $(x^{403})^5 \equiv 243 \mod 2018$ or equivalently $(x^{403})^5 \equiv 3^5 \mod 2018$. I wonder how to get from this last congruence to the desired $x^{403} \equiv 3 \mod 2018$ ?

Thanks for any suggestions.

Answer 1

Hint:

$3^5x\equiv1 \implies x\equiv 3^{-5} \implies x^{403} \equiv 3^{-2015} $

and $3^{2016}\equiv1\pmod{2018}$