Problem (Algebra):
Let $\mathbb{R}^{+}$be the set of positive real numbers. Determine all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that, for all positive real numbers $x$ and $y$,

$$
f(x+f(x y))+y=f(x) f(y)+1
\tag{*}
$$

Solution:
A straightforward check shows that $f(x)=x+1$ satisfies (*). We divide the proof of the converse statement into a sequence of steps.

Step 1: $f$ is injective.

Put $x=1$ in (*) and rearrange the terms to get

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

Therefore, if $f\left(y_{1}\right)=f\left(y_{2}\right)$, then $y_{1}=y_{2}$.

Step 2: $f$ is (strictly) monotone increasing.

For any fixed $y \in \mathbb{R}^{+}$, the function

$$
g(x):=f(x+f(x y))=f(x) f(y)+1-y
$$

is injective by Step 1. Therefore, $x_{1}+f\left(x_{1} y\right) \neq x_{2}+f\left(x_{2} y\right)$ for all $y, x_{1}, x_{2} \in \mathbb{R}^{+}$with $x_{1} \neq x_{2}$. Plugging in $z_{i}=x_{i} y$, we arrive at

$$
\frac{z_{1}-z_{2}}{y} \neq f\left(z_{2}\right)-f\left(z_{1}\right), \quad \text { or } \quad \frac{1}{y} \neq \frac{f\left(z_{2}\right)-f\left(z_{1}\right)}{z_{1}-z_{2}}
$$

for all $y, z_{1}, z_{2} \in \mathbb{R}^{+}$with $z_{1} \neq z_{2}$. This means that the right-hand side of the rightmost relation is always non-positive, i.e., $f$ is monotone non-decreasing. Since $f$ is injective, it is strictly monotone.

Step 3: There exist constants $a$ and $b$ such that $f(y)=a y+b$ for all $y \in \mathbb{R}^{+}$.

Since $f$ is monotone and bounded from below by 0 , for each $x_{0} \geqslant 0$, there exists a right limit $\lim _{x \searrow x_{0}} f(x) \geqslant 0$. Put $p=\lim _{x \searrow 0} f(x)$ and $q=\lim _{x \searrow p} f(x)$.

Fix an arbitrary $y$ and take the limit of $(*)$ as $x \searrow 0$. We have $f(x y) \searrow p$ and hence $f(x+f(x y)) \searrow q$; therefore, we obtain

$$
q+y=p f(y)+1, \quad \text { or } \quad f(y)=\frac{q+y-1}{p}
$$

(Notice that $p \neq 0$, otherwise $q+y=1$ for all $y$, which is absurd.) The claim is proved.

Step 4: $f(x)=x+1$ for all $x \in \mathbb{R}^{+}$.

Based on the previous step, write $f(x)=a x+b$. Putting this relation into (*) we get

$$
a(x+a x y+b)+b+y=(a x+b)(a y+b)+1,
$$

which can be rewritten as

$$
(a-a b) x+(1-a b) y+a b+b-b^{2}-1=0 \quad \text { for all } x, y \in \mathbb{R}^{+}
$$

This identity may hold only if all the coefficients are 0 , i.e.,

$$
a-a b=1-a b=a b+b-b^{2}-1=0 .
$$

Hence, $a=b=1$.

We provide another proof that $f(x)=x+1$ is the only function satisfying $(*)$.

Put $a=f(1)$. Define the function $\phi: \mathbb{R}^{+} \rightarrow \mathbb{R}$ by

$$
\phi(x)=f(x)-x-1
$$

Then equation $(*)$ reads as

$$
\phi(x+f(x y))=f(x) f(y)-f(x y)-x-y .
\tag{1}
$$

Since the right-hand side of (1) is symmetric under swapping $x$ and $y$, we obtain

$$
\phi(x+f(x y))=\phi(y+f(x y))
$$

In particular, substituting $(x, y)=(t, 1 / t)$ we get

$$
\phi(a+t)=\phi\left(a+\frac{1}{t}\right), \quad t \in \mathbb{R}^{+}
\tag{2}
$$

Notice that the function $f$ is bounded from below by a positive constant. Indeed, for each $y \in \mathbb{R}^{+}$, the relation $(*)$ yields $f(x) f(y)>y-1$, hence

$$
f(x)>\frac{y-1}{f(y)} \quad \text { for all } x \in \mathbb{R}^{+}
$$

If $y>1$, this provides a desired positive lower bound for $f(x)$.

Now, let $M=\inf _{x \in \mathbb{R}^{+}} f(x)>0$. Then, for all $y \in \mathbb{R}^{+}$,

$$
M \geqslant \frac{y-1}{f(y)}, \quad \text { or } \quad f(y) \geqslant \frac{y-1}{M}
\tag{3}
$$

Lemma 1. The function $f(x)$ (and hence $\phi(x)$ ) is bounded on any segment $[p, q]$, where $0<p<q<+\infty$.

Proof. $f$ is bounded from below by $M$. It remains to show that $f$ is bounded from above on $[p, q]$. Substituting $y=1$ into $(*)$, we get

$$
f(x+f(x))=a f(x)
\tag{4}
$$

Take $z \in[p, q]$ and put $s=f(z)$. By (4), we have

$$
f(z+s)=a s \quad \text { and } \quad f(z+s+a s)=f(z+s+f(z+s))=a^{2} s
$$

Plugging in $(x, y)=\left(z, 1+\frac{s}{z}\right)$ to $(*)$ and using (3), we obtain

$$
f(z+a s)=f(z+f(z+s))=s f\left(1+\frac{s}{z}\right)-\frac{s}{z} \geqslant \frac{s^{2}}{M z}-\frac{s}{z}
$$

Now, substituting $(x, y)=\left(z+a s, \frac{z}{z+a s}\right)$ to $(*)$ and applying the above estimate and the estimate $f(y) \geqslant M$, we obtain

$$
\begin{aligned}
& a^{2} s=f(z+s+a s)=f(z+a s+f(z))=f(z+a s) f\left(\frac{z}{z+a s}\right)+1-\frac{z}{z+a s} \\
& \geqslant M f(z+a s) \geqslant \frac{s^{2}}{z}-\frac{M s}{z} \geqslant \frac{s^{2}}{q}-\frac{M s}{p}
\end{aligned}
$$

This yields $s \leqslant q\left(\frac{M}{p}+a^{2}\right)=: L$, and $f$ is bounded from above by $L$ on $[p, q]$.



Applying Lemma 1 to the segment $[a, a+1]$, we see that $\phi$ is bounded on it. By $(2)$ we get that $\phi$ is also bounded on $[a+1,+\infty)$, and hence on $[a,+\infty)$. Put $C=\max \{a, 3\}$.

Lemma 2. For all $x \geqslant C$, we have $\phi(x)=0$ (and hence $f(x)=x+1$ ).

Proof. Substituting $y=x$ to (1), we obtain

$$
\phi\left(x+f\left(x^{2}\right)\right)=f(x)^{2}-f\left(x^{2}\right)-2 x
$$

hence,

$$
\phi\left(x+f\left(x^{2}\right)\right)+\phi\left(x^{2}\right)=f(x)^{2}-(x+1)^{2}=\phi(x)(f(x)+x+1) .
\tag{5}
$$

Since $f(x)+x+1 \geqslant C+1 \geqslant 4$, we obtain that

$$
|\phi(x)| \leqslant \frac{1}{4}\left(\left|\phi\left(x+f\left(x^{2}\right)\right)\right|+\left|\phi\left(x^{2}\right)\right|\right)
\tag{6}
$$

Since $C \geqslant a$, there exists a finite supremum $S=\sup _{x \geqslant C}|\phi(x)|$. For each $x \in[C,+\infty)$, both $x+f\left(x^{2}\right)$ and $x^{2}$ are greater than $x$; hence they also lie in $[C,+\infty)$. Therefore, taking the supremum of the left-hand side of (6) over $x \in[C,+\infty)$, we obtain $S \leqslant S / 2$ and hence $S=0$. Thus, $\phi(x)=0$ for all $x \geqslant C$.

It remains to show that $f(y)=y+1$ when $0<y<C$. For each $y$, choose $x>\max \left\{C, \frac{C}{y}\right\}$. Then all three numbers $x, x y$, and $x+f(x y)$ are greater than $C$, so $(*)$ reads as

$$
(x+x y+1)+1+y=(x+1) f(y)+1, \text { hence } f(y)=y+1
$$

Final answer: $f(x)=x+1$