Problem 3. Let $\mathbb{R}^{*}$ be the set of non-zero real numbers. Find all functions $f: \mathbb{R}^{*} \rightarrow \mathbb{R}^{*}$ such that

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

for all $x, y \in \mathbb{R}^{*}, y \neq-x^{2}$.

Alexander Ivanov
3. Set $\alpha=f(1)$. Then setting $y=1$ and $x=1$ in

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

gives

$$
f\left(x^{2}+1\right)=f^{2}(x)+1
$$

and

$$
f(y+1)=\alpha^{2}+\frac{f(y)}{\alpha}
$$

respectively. Using (3), we consecutively get

$$
\begin{gathered}
f(2)=\alpha^{2}+1, f(3)=\frac{\alpha^{3}+\alpha^{2}+1}{\alpha} \\
f(4)=\frac{\alpha^{4}+\alpha^{3}+\alpha^{2}+1}{\alpha^{2}}, f(5)=\frac{\alpha^{5}+\alpha^{4}+\alpha^{3}+\alpha^{2}+1}{\alpha^{3}}
\end{gathered}
$$

On the other hand, setting $x=2$ in (2) gives $f(5)=\alpha^{4}+2 \alpha^{2}+2$. Therefore $\frac{\alpha^{5}+\alpha^{4}+\alpha^{3}+\alpha^{2}+1}{\alpha^{3}}=\alpha^{4}+2 \alpha^{2}+2 \Longleftrightarrow \alpha^{7}+\alpha^{5}-\alpha^{4}+\alpha^{3}-\alpha^{2}-1=0$, whence

$$
(\alpha-1)\left[\alpha^{4}\left(\alpha^{2}+\alpha+1\right)+(\alpha+1)^{2}\left(\alpha^{2}-\alpha+1\right)+2 \alpha^{2}\right]=0
$$

Since the expression in the square brackets is positive, we have $\alpha=1$. Now (3) implies that

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

and therefore $f(n)=n$ for every positive integer $n$.

Now take an arbitrary positive rational number $\frac{a}{b}$ ( $a, b$ are positive integers). Since (4) gives $f(y)=y \Longleftrightarrow f(y+m)=y+m, m$ is a positive integer, the equality $f\left(\frac{a}{b}\right)=\frac{a}{b}$ is equivalent to

$$
f\left(b^{2}+\frac{a}{b}\right)=b^{2}+\frac{a}{b}
$$

Since the last equality follows from (1) for $x=b$ and $y=\frac{a}{b}$ we conclude that $f\left(\frac{a}{b}\right)=\frac{a}{b}$.

Setting $y=x^{2}$ in (4), we obtain $f\left(x^{2}+1\right)=f\left(x^{2}\right)+1$. Hence using (2) we conclude that $f\left(x^{2}\right)=f^{2}(x)>0$. Thus $f(x)>0$ for every $x>0$. Now (1), the inequality $f(x)>0$ for $x>0$ and the identity $f\left(x^{2}\right)=f^{2}(x)$ imply that $f(x)>f(y)$ for $x>y>0$. Since $f(x)=x$ for every rational number $x>0$, it easily follows that $f(x)=x$ for every real number $x>0$.

Finally, given an $x0$. Then $x y>0$ and (1) gives

$$
\begin{aligned}
x^{2}+y & =f\left(x^{2}+y\right)=f^{2}(x)+\frac{f(x y)}{f(x)} \\
& =f\left(x^{2}\right)+\frac{x y}{f(x)}=x^{2}+\frac{x y}{f(x)}
\end{aligned}
$$

i.e. $f(x)=x$. Therefore $f(x)=x$ for every $x \in \mathbb{R}^{*}$. It is clear that this function satisfies (1).