Solve the equation $n^{x}+n^{y}=n^{z}$ in natural numbers.
Obviously, $n \geq 2$. Considering $x \leq y \leq z$, we rewrite our equation as $n^{z-x}-n^{y-x}=1$.

If $y>x$, then 1 is divisible by $n$. But this is not true ( $n \geq 2$ ), so $y=x$. From this, $n^{z-x}=2$, which means $n=2, z-x=1$.

## Answer

$n=2, y=x, z=x+1, x$ - any natural number.