Find all integers $x, y$ such that $2 x^{3}+x y-7=0$.
First, it is clear that $x \neq 0$. The integer $x$ divides 7, so we have $x = -7, -1, 1$ or 7. We obtain:

- if $x = -7, -2 \cdot 7^3 - 7y - 7 = 0$, thus $y = -2 \cdot 7^2 - 1 = -99$,
- if $x = -1, y = -9$,
- if $x = 1, y = 5$,
- if $x = 7, y = -97$.

The solutions are therefore $(-7, -99), (-1, -9), (1, 5)$ and $(7, -97)$.

## 3 Monday, 24th afternoon: Vincent Bouis

Definition 124. Let $p$ be a prime number and $n$ a non-zero integer. The $p$-adic valuation, denoted $v_p(n)$, is the greatest positive integer such that $\frac{n}{p^{v_p(n)}}$ is an integer.

Theorem 125. Every number $n \in \mathbb{N}$ greater than or equal to 2 can be uniquely written (up to the order of the $p_i$) as:

$$
n = \prod_{i=1}^{k} p_i^{\alpha_i}
$$

for some $k \in \mathbb{N}_{>0}$, $\alpha_1, \alpha_2, \ldots, \alpha_k$ strictly positive integers, and $p_1, p_2, \ldots, p_k$ distinct prime numbers.

Remark 126. With the above notation, $\alpha_i$ is the $p_i$-adic valuation of $n$.

## - Exercises -