Exercise 3. Find all real numbers $x, y$ and $z$ such that

$$
\left\{\begin{array}{r}
x^{2}-4 y+7=0 \\
y^{2}-6 z+14=0 \\
z^{2}-2 x-7=0
\end{array}\right.
$$
Solution to Exercise 3 It is not about deploying a non-conventional calculation technique to solve a system of polynomial equations, but rather to cleverly combine the different equations.

Let $(x, y, z)$ be a solution triplet of the system. A simple way to combine the equations is to add them, in order to group the terms in $x$, the terms in $y$, and the terms in $z$, and complete the squares. Here, this gives:

$$
\begin{aligned}
0 & =x^{2}-4 y+7+y^{2}-6 z+14+z^{2}-2 x-7 \\
& =x^{2}-2 x+1+y^{2}-4 y+4+z^{2}-6 z+9 \\
& =(x-1)^{2}+(y-2)^{2}+(z-3)^{2}
\end{aligned}
$$

Each of the squares is non-negative, so each of the squares must be zero. If $(x, y, z)$ is a solution to the system, we therefore have $x=1, y=2$, and $z=3$.

Conversely, we can verify that:

$$
\left\{\begin{aligned}
1^{2}-4 \times 2+7 & =0 \\
2^{2}-6 \times 3+14 & =0 \\
3^{2}-2 \times 1-7 & =0
\end{aligned}\right.
$$

Thus, the unique solution to the system is the triplet $(1,2,3)$.

Graders' Comment: The exercise is very well done, but few students think to verify their solution: in every problem where solutions are obtained without equivalence, it is essential to check the solutions at the end!