Can you tell me if my solution is so far correct and help me complete the second case?
The statement of the problem :
Find all functions $f: \mathbb R \to \mathbb R$ with the following property:
$$
f(xf(y) + f(x+y)) = y(f(x)+1) + f(x) , \forall x,y \in \mathbb R.
$$
My approach :
For $x = y = 0$ we get
$$
f(f(0)) = f(0).
$$
For $x = 0$ we get
$$
f(f(y))=y(f(0)+1)+f(0) , \forall y \in \mathbb R.
$$
Now the function $y \mapsto ay+c$ is bijective , where $a$ and $c$ real constants and $a \neq 0$ . So , for $f(0) \neq -1$ we get
$$
\begin{align}
&y \mapsto y(f(0)+1)+f(0) \text{ is bijective,}\\
\implies &y \mapsto f(f(y)) \text{ is bijective,}\\
\implies &f \text{ is injective and $f$ is surjective,}\\
\implies &f \text{ is bijective}.
\end{align}
$$
Making $y = 0$ in the main equation we get
$$f(f(x))=f(x) ,\forall x \in \mathbb R.
$$
And since $f$ is bijective we obtain that
$$f(x)=x ,\forall x \in \mathbb R,
$$
which satisfies the initial equation.
Now we have left the case where $y \mapsto y(f(0)+1)+f(0)$ is not bijective , more precisely when
$$
\begin{align}
f(0) &= -1 \\
\implies f(f(y)) &= f(0) = -1 ,\forall y \in \mathbb R.
\end{align}
$$
Making $y=0$ we get
$$f(-x+f(x))=f(x),\forall x \in \mathbb R.
$$
From here I didn't know what to do. I think that we either have to show that $f \equiv -1$ or reach a contradiction. I hope that at least the first part is correct.
I am open to all your suggestions and solutions. Thank you!
Best Answer
Note that $f(x) \equiv -1$ is also a solution of the functional equation
$$ f(xf(y) + f(x+y)) = y(f(x)+1) + f(x) , \qquad \forall x,y \in \mathbb{R} \tag{1}\label{fe}$$
To examine other possibilities, assume that $f$ is not identically $-1$.
Let $a \in \mathbb{R}$ be an arbitrary element satsifying $f(a) \neq -1$. Then plugging $x = a$ to $\eqref{fe}$ gives $$ f(af(y) + f(a+y)) = y(f(a) + 1) + f(a). $$ This shows that $f$ is surjective.
Now we plug $x = 0$ to $\eqref{fe}$. Then $$ f(f(y)) = y(f(0) + 1) + f(0). $$ Since $y \mapsto f(f(y))$ is surjective, the same is true for the right-hand side. In particular, $f(0) \neq -1$ and this in fact shows that $f$ is bijective.
Finally, we plug $y = 0$ to $\eqref{fe}$. Then $$ f(xf(0) + f(x)) = f(x), $$ hence by the injectivity of $f$ we get $$ x f(0) + f(x) = x, \qquad\text{i.e.,}\qquad f(x) = (1 - f(0)) x. $$ This then shows that $f(0) = 0$ and therefore $f(x) = x$.
Conclusion. The functional equation $\eqref{fe}$ has two solutions: $f(x) \equiv -1$ or $f(x) = x$.