## Exercise 3

Let $p$ be a prime number. Find all integers $a, b, c \geq 1$ such that:

$$
a^{p} + b^{p} = p^{c} .
$$
Solution. The answer is:

(i) $p=2$ and $(a, b, c)=\left(2^{u}, 2^{u}, 2 u+1\right)$ for an integer $u \geq 0$

(ii) $p=3$ and $(a, b, c)=\left(2 \cdot 3^{u}, 3^{u}, 2+3 u\right)$ or $(a, b, c)=\left(3^{u}, 2 \cdot 3^{u}, 2+3 u\right)$ for an integer $u \geq 0$.

For an integer $n \geq 1$ and a prime number $p$ we denote $v_{p}(n)$ the greatest integer $k \geq 0$ such that $p^{k}$ divides $n$. We first prove three lemmas.

## Lemma 1

Let $p$ be a prime number. Then, for $1 \leq j \leq p-1, \binom{p}{j}$ is divisible by $p$.