## T-2 A

Determine all functions $f: \mathbb{R} \backslash\{0\} \rightarrow \mathbb{R} \backslash\{0\}$ such that

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

holds for all nonzero real numbers $x$ and $y$.

Answer. $f(x)=\frac{1}{x^{2}}$ and $f(x)=-\frac{1}{x^{2}}$


Solution 1. Let $f$ be any function with the desired property and set $\alpha=f(1)$.

Lemma. Let $x \in \mathbb{R}^{\neq 0}$ be arbitrary and put $z=x^{2} f(x)$. Then $f(z)=z, f\left(z^{2}\right)=2 z-\alpha$ and $z^{2}=3 z-2 \alpha$.

Proof. Substituting $y=1$ and $y=z$ into the given functional equation we obtain $f(z)=z$ and $f\left(z^{2}\right)+\alpha=z+f(z)$, whereby the first two parts of the claim are proved. Applying the first part to $z$ in place of $x$ we infer that $z^{2} f(z)=z^{3}$ is a fixed point of $f$ as well, i.e., $f\left(z^{3}\right)=z^{3}$. On the other hand we may plug $y=z^{2}$ into the given equation, thus getting $f\left(z^{3}\right)+\alpha=$ $z+f\left(z^{2}\right)=3 z-\alpha$. Comparing the two previous results we learn indeed $z^{3}=3 z-2 \alpha$.

In the particular case $x=1$ we have $z=\alpha$ and the third part of the lemma tells us $\alpha^{3}=\alpha$. Since the number $\alpha$ is a value attained by $f$, it cannot vanish, so $\alpha= \pm 1$.

Let us now return to the situation of the above lemma. The third equation may now be rewritten as $(z-\alpha)^{2}(z+2 \alpha)=z^{3}-3 z+2 \alpha^{3}=0$. It follows that either $z=\alpha$ or $z=-2 \alpha$.

Assume there were a nonzero real number $x$ such that $z=x^{2} f(x)$ has the property $z=-2 \alpha$. Then our lemma yields $f(z)=-2 \alpha$, whence $z^{2} f(z)=-8 \alpha^{3}=-8 \alpha \notin\{\alpha, 2 \alpha\}$, which means that $z$ in place of $x$ violates the result from the previous paragraph. This proves that $z=\alpha$ holds for all real $x \neq 0$.

In other words we have $f(x)=\frac{\alpha}{x^{2}}$ for all nonzero real numbers $x$. Due to $\alpha= \pm 1$ this shows that $f$ is one of the two functions mentioned in the answer.

It is easy to verify that they do indeed solve the functional equation under consideration both of its sides being equal to $\alpha+\frac{\alpha}{y^{2}}$.
