4. find all pairs $(u, v)$ of natural numbers such that

$$
\frac{u v^{3}}{u^{2}+v^{2}}
$$

is a prime power.
## Solution:

Let $p^{n}$ be this prime power, transforming yields the equivalent equation $u v^{3}=p^{n}\left(u^{2}+v^{2}\right)$. Set $d=\operatorname{ggT}(u, v)$ and write $u=d x, v=d y$, then $x$ and $y$ are divisors. The equation becomes

$$
d^{2} x y^{3}=p^{n}\left(x^{2}+y^{2}\right)
$$

Now $x y^{3}$ is divisor-foreign to $x^{2}+y^{2}$, because every common prime divisor of, for example, $x$ and $x^{2}+y^{2}$ also divides $y^{2}$ and thus $y$, in contradiction to the divisor-foreignness of $x$ and $y$. Consequently, $x y^{3}$ is a divisor of $p^{n}$, so $x$ and $y$ are both $p$ powers. In addition, at least one of these two numbers must be equal to 1 due to divisor strangeness. We now distinguish between three cases:

- Let $x=y=1$. Then $d^{2}=2 p^{n}$, so the left-hand side is divisible by 4 and therefore $p=2$. In addition, $n=2 k-1$ is odd. This yields $u=v=d=2^{k}$ with $k \geq 1$. These are actually solutions.
- Let $x=p^{r}$ with $1 \leq r \leq n$ and $y=1$. Then $d^{2}=p^{n-r}\left(p^{2 r}+1\right)$. Here, the two factors on the right are divisors because of $r \geq 1$ and therefore both are squares. A contradiction, because the second factor lies between two consecutive squares due to $\left(p^{r}\right)^{2}<p^{2 r}+1<\left(p^{r}+1\right)^{2}$, so it cannot be one itself.
- The case $x=1$ and $y=p^{s}$ with $1 \leq s \leq n / 3$ also leads to a contradiction.

The pairs you are looking for are therefore

$$
(u, v)=\left(2^{k}, 2^{k}\right) \quad \text { with } \quad r \geq 1
$$