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 $y$.
Let $f$ be a potential solution to the problem. In the following, we will denote $\mathbf{E}_{x, y}$ the given equality. 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 $0 < c < d$. Since $f$ is injective, we have $f(c) \neq f(d)$. If $f(c) > f(d)$, then there would exist a $z$ such that $c < z < d$ and $f(z) = f(c)$, which contradicts the injectivity of $f$. Therefore, $f(d) \geq f(c)$, 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 real number $y > 0$, if we let $x$ tend to 0, we deduce that $f(x) \rightarrow p$ and that $f(xy) \rightarrow p$, so that $x + f(xy) \rightarrow p$ and that $f(x + f(xy)) \rightarrow q$. Consequently, the equality $\mathbf{E}_{x, y}$ indicates that

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

which means that $f(y)=y+(q-1)/p$.
The function $f$ is therefore an affine function, of the form $f: t \mapsto ut + v$. Conversely, once the coefficients $u$ and $v$ are fixed, the equality $\mathbf{E}_{x, y}$ indicates that

$$
u^2 xy + ux + uv + v + y = f(x + f(xy)) + y = f(x) f(y) + 1 = u^2 xy + uvx + uvy + v^2 + 1
$$

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

$$
u(v-1) = uv - 1 = uv + v - v^2 - 1 = 0
$$

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

$$
u(v-1) = uv - 1 = uv + v - v^2 - 1 = 0
$$

is indeed satisfied.
In conclusion, the function $f: t \mapsto t + 1$ is the unique solution to the problem.
Comment from the graders This problem was extremely difficult, and no student (nor any mentor from the POFM) managed to pass the crucial step of proving that $f$ is 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 $f$ is 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.