Prove an implication in modular arithmetic

Question

Prove for any integer $x$:

if $x^{13} = 1 \mod \ 17$ then $x = 1 \mod 17$

Answer 1

$$x=(x^{13})^5(x^{16})^{-4}$$

Now apply Fermat little theorem

This will hold true for any odd integer $m$

All we need to find is Integers $a,b$ such that $$am+16b=1$$ which is always available using

https://artofproblemsolving.com/wiki/index.php/Bezout%27s_Lemma

Answer 2

Hint: Take $x^{13}\equiv1\bmod17$ and raise both sides to the fifth power.

Answer 3

Since $17$ is a prime, all members of the multiplicative group mod $17$ have order dividing $17-1 = 16$. Since $13$ and $16$ are coprime, the order must be $1$, i.e. $x \equiv 1$.