Find all real numbers $x$ and $y$ such that

$$
\left(x-\sqrt{x^{2}+1}\right)\left(y-\sqrt{y^{2}+1}\right)=1 \text { and }\left(x^{2}+y+2\right)\left(y^{2}+x+2\right)=8 .
$$
Let $(x, y)$ be a potential solution. The equalities

$$
\left\{\begin{array}{l}
\left(x-\sqrt{x^{2}+1}\right)\left(x+\sqrt{x^{2}+1}\right)=x^{2}-\left(x^{2}+1\right)=-1 \\
\left(y-\sqrt{y^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right)=y^{2}-\left(y^{2}+1\right)=-1
\end{array}\right.
$$

indicate that

$$
\left\{\begin{array}{l}
x+\sqrt{x^{2}+1}=\sqrt{y^{2}+1}-y \\
\sqrt{x^{2}+1}-x=y+\sqrt{y^{2}+1}
\end{array}\right.
$$

Subtracting equation (2) from equation (1) indicates that $x=-y$, and the equation $\left(x^{2}+y+2\right)\left(y^{2}+x+2\right)=8$ can then be rewritten as

$$
8=\left(x^{2}-x+2\right)\left(x^{2}+x+2\right)=\left(x^{2}+2\right)^{2}-x^{2}
$$

If we let $z=x^{2}$, it then becomes necessary to solve the equation
$0=(z+2)^{2}-z-8=z^{2}+3z-4=(z+3/2)^{2}-25/4=(z+3/2)^{2}-(5/2)^{2}=(z+4)(z-1)$.
Since we know that $z=x^{2} \geqslant 0$, we deduce that $z=1$, hence $x= \pm 1$ and $y=-x= \pm 1$.
Conversely, when $(x, y)=( \pm 1, \mp 1)$, it is easily verified that both equalities in the problem statement are satisfied. In conclusion, the two solutions are

$$
(x, y)=(1,-1) \text { and }(x, y)=(-1,1)
$$

**Graders' Comments** This very challenging problem was tackled by only a few students, and the graders were pleased to note that five students managed to solve it.
The key idea for this problem was to recognize what are called the conjugates of a difference of the form $a-b$ where $a$ and $b$ are terms whose squares $a^{2}$ and $b^{2}$ are very close. Indeed, in such cases, it is often useful to also study the term $a+b$, which is called the conjugate of the term $a-b$, and this is due to the remarkable identity $a^{2}-b^{2}=(a+b)(a-b)$. This is exactly how the solution above begins, and the students who solved this problem are those who had the idea to introduce these conjugates.
Moreover, several students mistakenly believed that $x$ and $y$ had to be integers, or even that the terms $x-\sqrt{x^{2}+1}$ and $y-\sqrt{y^{2}+1}$ had to be integers as well. It is important to read the problem statement carefully!
Similarly, several students had the very good reflex to look for particular solutions, but many of them stopped at the cases where $x=0$ (or $y=0$), $x=y=1$, or more generally at the case where $x=y$, without considering the analogous case where $x=-y$. What a pity! In such an equation, it is always important to not overlook the case where one of the two unknowns (or both) could be negative.