
Problem T-1. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

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

for all $x, y \in \mathbb{R}$.


Solution. Putting $x=y=0$, we get $f(0)=1$.

Putting $x=0, y=z$, we get

$$
f(2 z)=f(z)+z
$$

Putting $x=z, y=-z f(z)$, we get

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

Replacing $z$ by $2 z$ in (2) and using (1), we obtain

$$
f\left(4 z^{2}\right)=2 z f(2 z)-2 z+1=2 z(f(z)+z)-2 z+1=2 z f(z)+2 z^{2}-2 z+1
$$

Using (1) for $2 z^{2}$ and then for $z^{2}$ in place of $z$, and afterwards using (2), we obtain

$$
f\left(4 z^{2}\right)=f\left(2 z^{2}\right)+2 z^{2}=f\left(z^{2}\right)+z^{2}+2 z^{2}=z f(z)-z+1+3 z^{2}
$$

Comparing (3) and (4) we have

$$
\begin{aligned}
2 z f(z)+2 z^{2}-2 z+1 & =z f(z)-z+1+3 z^{2} \\
z f(z)-z^{2}-z & =0 \\
z(f(z)-z-1) & =0
\end{aligned}
$$

For $z \neq 0$ we obtain $f(z)=z+1$. For $z=0$, we have $f(0)=1$. Thus, for all $z \in \mathbb{R}$, we have

$$
f(z)=z+1
$$

This function indeed satisfies the functional equation.
