Find all solutions $(x, y) \in \mathbb{N}^{2}$ of the equation:

$$
x \wedge y + x \vee y = x + y
$$
Let $(x, y) \in \mathbb{N}^{2}$ be a solution pair. Let $d = x \wedge y$. There exist $x^{\prime}, y^{\prime} \in \mathbb{N}$ such that:

$$
\left\{\begin{array}{l}
x = d x^{\prime} \\
y = d y^{\prime} \\
x^{\prime} \wedge y^{\prime} = 1
\end{array}\right.
$$

The equation becomes:

$$
\begin{aligned}
1 + x^{\prime} y^{\prime} = x^{\prime} + y^{\prime} & \Longleftrightarrow \left(x^{\prime} - 1\right)\left(y^{\prime} - 1\right) = 0 \\
& \Longleftrightarrow x^{\prime} = 1 \text{ or } y^{\prime} = 1
\end{aligned}
$$

Thus $(x, y)$ is of the form $(d, d k)$ or $(d k, d)$ with $d, k \in \mathbb{N}$.

Conversely, these pairs are indeed solutions.