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 $x^n\equiv y^n(\mathrm{mod}{z}^n)$.

If $z>1$, assume that there is an odd prime divisor $p\mid z$. Let $r$ be the least positive integer such that $x^r\equiv y^r(\mathrm{mod}p)$, so $r\mid n\Rightarrow n=rk$. Let $a=v_p(n)$ and $b=v_p(x^r-y^r)$. By LTE, we have $$v_p(x^n-y^n)=v_p((x^r{)}^k-(y^r{)}^k)=v_p(x^r-y^r)+v_p(k)\le b+v_p(n)=a+b.$$ But $v_p(x^n-y^n)\ge v_p(z^n)\le n\Rightarrow n\le a+b\Rightarrow p^n\le p^a\cdots p^b=n{p}^b$. 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 $x\ne y$. 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 $$v_2(x^n-y^n)=c+s-1\Rightarrow n\le c+s-1\Rightarrow n\le {\log }_{2}n+s-1.$$ Which does not hold for sufficiently large even values of $n$.