(BxMO 2010, P4)

Find all quadruplets $(a, b, p, n)$ of strictly positive integers with $p$ prime such that $a^{3}+b^{3}=p^{n}$
The solution presented below is deliberately written exclusively with simple ideas. After noting its tedious nature, I recommend looking for the solution using Zsigmondy.

We rewrite $(a+b)\left(a^{2}-a b+b^{2}\right)=p^{n}$. Since $a+b \geq 2$, we know that $p \mid a+b$. We have $a^{2}-a b+b^{2}=(a-b)^{2}+a b$, so $a=b=1$ or $a^{2}-a b+b^{2} \geq 2$ and thus $p \mid a^{2}-a b+b^{2}$. Let's consider the second case. Then $p$ also divides $(a+b)^{2}-\left(a^{2}-a b+b^{2}\right)=3 a b$. We recall that $p \mid a+b$, so $p \mid a$ if and only if $p \mid b$. Thus, either $p=3$, or $p \mid a$ and $p \mid b$. In this case, we rewrite $a=p a^{\prime}$ and $b=p b^{\prime}$, and we get $a^{\prime 3}+b^{\prime 3}=p^{n-3}$, noting that $n-3>0$ because $a^{\prime} \geq 1$ and $b^{\prime} \geq 1$. We continue this process until we reach a solution $\left(a_{0}, b_{0}, p, n_{0}\right)$ for which $p$ does not divide $a_{0}$. From the above, we end up in one of the special cases $a_{0}=b_{0}=1$, which gives the solution $(1,1,2,1)$, or $p=3$. If $p=3$, then $3 \mid a_{0}+b_{0}$, so $9 \mid\left(a_{0}+b_{0}\right)^{2}$. If $9 \mid a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}$, then $9 \mid 3 a_{0} b_{0}$, a contradiction. Thus, $a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}=3$, so $\left(a_{0}-b_{0}\right)^{2}+a_{0} b_{0}=3$, from which we get $a_{0}=1$ and $b_{0}=2$ or vice versa.

Thus, there are 3 possible solutions when $p$ does not divide $a$, which are $(1,1,2,1),(1,2,3,1),(2,1,3,1)$. Finally, there are 3 families of solutions to the initial equation: $\left(2^{k}, 2^{k}, 2,3 k+1\right),\left(3^{k}, 2 \cdot 3^{k}, 3,3 k+2\right)$, and $\left(2 \cdot 3^{k}, 3^{k}, 3,3 k+2\right)$ with $k \in \mathbb{N}$, which are easily verified to be valid.