Find the $x, y$ integers such that $x^{3}(y+1)+y^{3}(x+1)=19$.
Again, nothing very pretty. First, note that the problem is symmetric in $x, y$. If $d$ is the gcd of $x$ and $y, d^{3}$ divides 19, so $d=1: x$ and $y$ are coprime.

Let's try to handle simple cases: if $x=0$, then $y^{3}=19$ which is impossible, if $x=-1$, we get $-(y+1)=19$, i.e., $y=-20$. We will assume henceforth that $x, y$ are different from 0 and -1. Conversely, the pairs $(-1,-20)$ and $(-20,-1)$ are solutions.

If $x$ and $y$ are positive, they cannot be too large, as all quantities present are positive. Indeed, if $x \geq 3, 19 \geq x^{3}(y+1) \geq 27$ which is impossible. Therefore, $x$ and $y$ are 1 or 2. Since they are coprime, we have $(x, y)=(1,2),(2,1),(1,1)$. For $(x, y)=(1,1)$, we get $4=19$, a contradiction. For $(x, y)=(2,1)$, we get $82+13=19$. Thus $(2,1)$ and $(1,2)$ are solutions.

If $x$ and $y$ are negative, we have $x \leq-2$ and $y \leq-2$. If $x \leq-3$, we have $x^{3}(y+1)+y^{3}(x+1) \geq 27>19$ which is impossible. Therefore, $y>-3$ and $x>-3$ by symmetry, so $x=-2=y$. However, in this case, $x$ and $y$ are not coprime, a contradiction.

The remaining case is where $x$ is strictly positive and $y$ is strictly negative (the other case is identical by symmetry). In this case, $x^{3}(y+1) \leq 0$ and $y^{3}(x+1) \leq 0$ so $(x, y)$ is not a solution.

Thus, the solutions to the equation are the pairs $(2,1),(1,2),(-20,-1),(-1,-20)$.

## 2 Mid-term Training

## Statements