Exercise 10. Find all functions $f$ from $\mathbb{R}$ to $\mathbb{R}$ such that, for all real numbers $x, y$,

$$
f(x-f(y))=1-x-y
$$
Solution to Exercise 10: First, let's test some particular values: for $y=0$, we have $f(x-f(0))=1-x$. We almost have $f(x)$, but we need to make the $f(0)$ "disappear". For this, let $z$ be any real number. By setting $x=z+f(0)$, we get $f(z)=1-(z+f(0))=1-f(0)-z$. For $z=0$, we have $f(0)=1-f(0)$ and thus $f(0)=\frac{1}{2}$. Therefore, for any real $z$, $f(z)=\frac{1}{2}-z$.

Let's verify that this function is indeed a solution:

$$
f(x-f(y))=f\left(x-\frac{1}{2}+y\right)=\frac{1}{2}-x+\frac{1}{2}-y=1-x-y
$$

Thus, the function that associates $\frac{1}{2}-z$ to any real $z$ is indeed the unique solution to this equation.

Graders' Comments: The exercise is very well done. Some students forgot to verify that the function they obtained after their analysis indeed satisfies the equation and were penalized by one point. A few others, because they were too hasty in this verification, claimed that no function is a solution or that all functions of the form $x \mapsto -x+c$ are solutions.