Exercise 7. We denote by $\mathbb{R}_{>0}$ the set of strictly positive real numbers. Find all functions $f: \mathbb{R}_{>0} \mapsto \mathbb{R}_{>0}$ such that

$$
f(x+f(x y))+y=f(x) f(y)+1
$$

for all strictly positive real numbers $x$ and $y$.
Solution to Exercise 7 Let $f$ be a potential solution to the problem. In the following, we will denote $\mathbf{E}_{x, y}$ the equality from the statement. Let $a$ and $b$ be two strictly positive real numbers such that $f(a)=f(b)$. The equalities $\mathbf{E}_{1, a}$ and $\mathbf{E}_{1, b}$ indicate that

$$
a=f(1) f(a)+1-f(1+f(a))=f(1) f(b)+1-f(1+f(b))=b
$$

which means that $f$ is injective.

Next, let $c$ and $d$ be two real numbers such that $0 < c < d$. From the equality $\mathbf{E}_{c, d}$, we deduce that $f(d) \geqslant f(c)$, which means that $f$ is increasing.

We then set $p=\inf \{f(x): x>0\}$ and $q=\inf \{f(x): x>p\}$. Since $f$ is increasing, we know that $f(x) \rightarrow p$ as $x \rightarrow 0$ and $x>0$, and that $f(y) \rightarrow q$ as $y \rightarrow p$ and $y>p$. For a fixed positive real number $y$, if we let $x$ tend to 0, we deduce that $f(x) \rightarrow p$ and that $f(x y) \rightarrow p$, so that $x+f(x y) \rightarrow p$ and that $f(x+f(x y)) \rightarrow q$. Consequently, the equality $\mathbf{E}_{x, y}$ indicates that

$$
y-1=f(x) f(y)-f(x+f(x y)) \rightarrow p f(y)-q
$$

which means that $f(y)=y+(q-1) / p$.

The function $f$ is therefore an affine function, of the form $f: t \mapsto u t+v$. Conversely, once the coefficients $u$ and $v$ are fixed, the equality $\mathbf{E}_{x, y}$ indicates that

$$
u^{2} x y+u x+u v+v+y=f(x+f(x y))+y=f(x) f(y)+1=u^{2} x y+u v x+u v y+v^{2}+1
$$

or equivalently that $u(v-1) x+(u v-1) y=u v+v-v^{2}-1$. This means that

$$
u v-u=u v-1=u v+v-v^{2}-1=0
$$

We deduce that $u-1=(u v-1)-(u v-u)=0$, that is, $u=1$, and then that $v-1=u v-u=0$, that is, $v=1$. But then the triple equality

$$
u v-u=u v-1=u v+v-v^{2}-1=0
$$

is indeed satisfied.

In conclusion, the function $f: t \mapsto t+1$ is the unique solution to the problem.

Graders' Comments This exercise was extremely difficult, and no student (nor any POFM supervisor) managed to pass the crucial step of proving that $f$ was increasing. Several students had excellent reflexes: looking for a particular solution (the function $f: t \mapsto t+1$), examining particular values of $x$ and $y$ (here, $x=1$), investigating the injectivity of $f$, conjecturing that $f$ was increasing (by noticing an infinity of values $x < y$ such that $f(x) < f(y)$).

$\triangleright$ Many students looked at what happened when $y=1$, but not when $x=1$. Others looked at the case where $x=1$, but did not think to deduce the injectivity of $f$, even though they had the equality $f(1+f(x))+x=1+f(1) f(x)$, which made the injectivity of $f$ immediate.

$\triangleright$ Several students checked if the identity or constant functions could be solutions. It is a pity that they did not directly treat the case of affine functions, which was just as easy to handle, and which would have allowed them to identify the only solution to the problem.