Determine all solutions of the equation

$$
p^{n}=x^{3}+y^{3}
$$

with $p$ prime and $x, y, n \in \mathbb{N}^{*}$.
By dividing $x$ and $y$ by an appropriate power of $p$, we can assume that $x$ or $y$ is not divisible by $p$. Since $p$ must divide $x^{3} + y^{3}$, if $p$ does not divide one of them, it does not divide the other. We start by treating the case $p=2$ separately: we have $2^{n} = (x+y)(x^{2} - xy + y^{2})$, but since $x$ and $y$ are odd, the second factor is also odd, so it must be 1, i.e., $(x-y)^{2} + xy = 1$, which implies $x = y = 1$ and $n = 1$. By multiplying $x$ and $y$ by the same power of two, we get $x = y = 2^{k}$ and $n = 3k + 1$ with $k \geq 0$.

If $p$ is odd, we use the LTE: we have $n = v_{p}(x^{3} + y^{3}) = v_{p}(x + y) + v_{p}(3)$. If $p \neq 3$, then $n = v_{p}(x + y)$, so $x + y = p^{n}$ and $x^{2} - xy + y^{2} = 1$. We again obtain $x = y = 1$ and $p = 2$, which is absurd. If $p = 3$, then $n = v_{3}(x + y) + 1$, so $3(x + y) = p^{n}$ and $(x - y)^{2} + xy = 3$, which gives $x = 1$, $y = 2$ or vice versa. We thus obtain the solutions $x = 3^{k}$, $y = 2 \times 3^{k}$ and $n = 3k + 2$ or $x = 2 \times 3^{k}$, $y = 3^{k}$ and $n = 3k + 2$.