Exercise 2. Find all real numbers $x, y$ such that

$$
x(x-2 y)+y(2 y-4)+4=0
$$
## Solution to Exercise 2

Let's develop the equation to obtain

$$
x^{2}-2 x y+2 y^{2}-4 y+4=0
$$

The terms $x^{2}-2 x y$ remind us of the remarkable identity $x^{2}-2 x y+y^{2}=(x-y)^{2}$. We then separate the $2 y^{2}$ into $y^{2}+y^{2}$ to write

$$
(x-y)^{2}+y^{2}-4 y+4=0
$$

But we then recognize a second remarkable identity $y^{2}-4 y+4=(y-2)^{2}$. Factoring this, we get

$$
(x-y)^{2}+(y-2)^{2}=0
$$

However, the sum of squares of real numbers is always non-negative, and can only be zero if both squares are zero. Thus, if $x$ and $y$ are solutions to the equation, we must have

$$
x-y=0 \quad \text{and} \quad y-2=0
$$

Therefore, the only possible solution to the equation is $x=y=2$. We can verify conversely that this is indeed a solution, so it is the only one.

Graders' Comment: Students generally found the solution using the method presented in the solution. Unfortunately, most of them forget to verify that the solution they found is indeed a solution to the problem. It is indeed sufficient that at one step, the equivalence is not justified for the verification to be necessary. Some considered the expression as a polynomial in $x$, whose roots need to be found, which only exist for certain values of $y$.