Exercise 4. Find all real numbers $x$ and $y$ such that

$$
\left(x-\sqrt{x^{2}+1}\right)\left(y-\sqrt{y^{2}+1}\right)=1 \text { and }\left(x^{2}+y+2\right)\left(y^{2}+x+2\right)=8
$$
Solution to Exercise 4 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

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

Equation (1) - (2) indicates that $x=-y$, and the equality $\left(x^{2}+y+2\right)\left(y^{2}+x+2\right)=8$ 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 is then necessary to solve the equation

$0=(z+2)^{2}-z-8=z^{2}+3 z-4=(z+3 / 2)^{2}-25 / 4=(z+3 / 2)^{2}-(5 / 2)^{2}=(z+4)(z-1)$.

Since we know that $z=x^{2} \geqslant 0$, we deduce that $z=1$, so $x= \pm 1$ and $y=-x= \pm 1$.

Conversely, when $(x, y)=( \pm 1, \mp 1)$, it is easily verified that both equalities in the statement are satisfied. In conclusion, the two solutions are

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

Graders' Comments This very difficult problem was tackled by very 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 two terms whose squares $a^{2}$ and $b^{2}$ are very close. Indeed, in such cases, it is often relevant to also study the term $a+b$, which is called the conjugate of the term $a-b$, and this in virtue of the remarkable identity $a^{2}-b^{2}=(a+b)(a-b)$. This is exactly how the solution above begins, and the students who succeeded in this problem are more or less those who had the idea of introducing 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 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 not to forget the case where one of the two unknowns (or both) could be negative.

## Senior Statements