Problem 1. Find all positive integers $a, b, n$ and prime numbers $p$ that satisfy

$$
a^{2013}+b^{2013}=p^{n} \text {. }
$$

(Matija Bucić)
First solution. Let's denote $d=D(a, b), x=\frac{a}{d}, y=\frac{b}{d}$. With this we get

$$
d^{2013}\left(a^{2013}+b^{2013}\right)=p^{n}
$$

So $d$ must be a power of $p$, so let $d=p^{k}, k \in \mathbb{N}_{0}$. We can divide the equality by $p^{2013 k}$. Now let's denote $m=$ $n-2013 k, A=x^{671}, B=y^{671}$. So we get

$$
A^{3}+B^{3}=p^{m}
$$

and after factorisation

$$
(A+B)\left(A^{2}-A B+B^{2}\right)=p^{m}
$$

(From the definition, $A$ and $B$ are coprime.)

Let's observe the case when some factor is $1: A+B=1$ is impossible as both $A$ and $B$ are positive integers. And $A^{2}-A B+B^{2}=1 \Leftrightarrow(A-B)^{2}+A B=1 \Leftrightarrow A=B=1$, so we get a solution $a=b=2^{k}, n=2013 k+1, p=2, \forall k \in \mathbb{N}_{0}$. If both factors are larger than 1 we have

$$
\begin{gathered}
p \mid A+B \\
p \mid A^{2}-A B+B^{2}=(A+B)^{2}-3 A B \\
\Longrightarrow p \mid 3 A B
\end{gathered}
$$

If $p \mid A B$, in accordance with $p \mid A+B$ we get $p \mid A$ and $p \mid B$, which is in contradiction with $A$ and $B$ being coprime. So, $p \mid 3 \Longrightarrow p=3$.

Now we are left with 2 cases:

- First case: $A^{2}-A B+B^{2}=3 \Leftrightarrow(A-B)^{2}+A B=3$ - so the only possible solutions are $A=2, B=1$ i $A=1, B=2$, but this turns out not to be a solution as $2=x^{671}$ does not have a solution in positive integers.
- Second case: $3^{2} \mid A^{2}-A B+B^{2}$ - then we have:

$$
\begin{gathered}
3\left|A+B \Longrightarrow 3^{2}\right|(A+B)^{2} \\
3^{2} \mid A^{2}-A B+B^{2}=(A+B)^{2}-3 A B \\
\Longrightarrow 3^{2} \mid 3 A B \\
\Longrightarrow 3 \mid A B
\end{gathered}
$$

And as we have already commented the case $p \nmid A B \Longrightarrow$ doesn't have any solutions.

So all the solutions are given by

$$
a=b=2^{k}, n=2013 k+1, p=2, \forall k \in \mathbb{N}_{0}
$$