Why is $xy^{-1}$ equal to 1?

Question

$x^3≡y^3 \mod5 \iff (xy^{−1})^3≡1 \mod 5$ all I have to do now is to prove that $xy^{-1}$ is equal to $1$.

Can someone tell me how that can be done?

Answer 1

If $x$ and $y$ are divided by $5$ it's wrong.

If no, it's true because $$(xy^{-1})^3\equiv x^3y^{-3}\equiv y^3y^{-3}=1$$

Now, prove that $$a^3\equiv1\Leftrightarrow a\equiv1,$$ for which use $$a^4\equiv1$$ by the Fermat's little theorem.

I'll prove the Fermat's little theorem for $p=29$.

Let $\gcd(a,29)=1$ and $ia\equiv r_i,$ where $1\leq r_i\leq28.$

Thus, $$\prod_{i=1}^{28}(ia)\equiv\prod_{i=1}^{28}r_i$$ or $$a^{28}28!\equiv28!$$ or $$a^{28}\equiv1.$$ I used that for any integers $\{i,j\}\subset[1,28]$ we have: $$ia\equiv ja\Leftrightarrow (i-j)a\equiv0\Leftrightarrow i=j.$$ Now, if $a^3\equiv1,$ so $a^{27}\equiv1$ or $a^{28}\equiv a,$ which gives $a\equiv1.$

Answer 2

$x^3\equiv y^3\implies (x^3)^3\equiv(y^3)^3\iff x^9\equiv y^9 $.

Now apply lil' Fermat.