Saudi Arabian Mathematical Competitions

Let $u = x / z$ and $v = y / z$ with $x$, $y$, $z$ positive integers, and $(x, y, z) = 1$. The statement is equivalent to saying that $x^{n} \equiv y^{n} \pmod{z^{n}}$.

If $z > 1$, assume that there is an odd prime divisor $p \mid z$. Let $r$ be the least positive integer such that $x^{r} \equiv y^{r} \pmod{p}$, so $r \mid n \Rightarrow n = r k$. Let $a = v_{p}(n)$ and $b = v_{p}(x^{r} - y^{r})$. By LTE, we have
$$v_{p}(x^{n} - y^{n}) = v_{p}((x^{r})^{k} - (y^{r})^{k}) = v_{p}(x^{r} - y^{r}) + v_{p}(k) \leq b + v_{p}(n) = a + b.$$
But $v_{p}(x^{n} - y^{n}) \geq v_{p}(z^{n}) \leq n \Rightarrow n \leq a + b \Rightarrow p^{n} \leq p^{a} \cdots p^{b} = n p^{b}$. This cannot be true for infinitely many positive integers $n$.

If $z$ has no odd prime divisor, then $z$ is a power of $2$. From this we have $x$, $y$ are both odd.

If $n$ is odd then
$$2^{n} \mid x^{n} - y^{n} = (x - y)(x^{n-1} + x^{n-2} y + \cdots + x y^{n-2} + y^{n-1}).$$
But the second factor is odd since $n$ is odd, we get $2^{n} \mid x - y$ for $x \neq y$. It is clear that there are only finitely many such $n$.

If $n$ is even, let $s = v_{2}(x^{2} - y^{2})$ and $c = v_{2}(n)$. By LTE we have
$$v_{2}(x^{n} - y^{n}) = c + s - 1 \Rightarrow n \leq c + s - 1 \Rightarrow n \leq \log_{2} n + s - 1.$$
Which does not hold for sufficiently large even values of $n$.