Determining all functions satisfying the functional equation $f \big( x + y f ( x ) \big) + f ( x y ) = f ( x ) + f ( 2019 y )$

I came up with the following functional equations problem, which is from Romanian Master of Mathematics 2019 Day 2, Problem 5 (still lack of solution by 2019 February 28th). I found this problem interesting, but I do not know how to do it. I want to know in general, how can one deal with such a functional problem. Are there any recommend books, lecture notes and etc.

Determine all functions $$f : \mathbb R \to \mathbb R$$ satisfying $$f \big( x + y f ( x ) \big) + f ( x y ) = f ( x ) + f ( 2019 y )$$ for all real numbers $$x$$ and $$y$$.

Through some investigation, I guess, $$f ( x ) = \text {some constant}$$ or $$f ( x ) = 2019 - x$$ assuming it is a linear function. I do not know whether those are the only solutions.

• All constant functions are clearly solutions, so if $f:\ \Bbb{R}\ \longrightarrow\ \Bbb{R}$ is a nonconstant solution then plugging in $x=2019$ shows that $$f(2019+yf(2019))=f(2019),$$ and hence $f(2019)=0$ as $f$ is nonconstant. Plugging in $y=1$ shows $$f(x+f(x))+f(x)=f(x)+f(2019),$$ and hence $f(x+f(x))=f(2019)=0$. Plugging in $x=z+f(z)$ yields $$f((z+f(z))+yf(z+f(z))))+f((z+f(z))y)=f(z+f(z))+f(2019y),$$ and because $f(z+f(z))=0$ for all $z\in\Bbb{R}$ this implies $$f((z+f(z))y)=f(2019y),$$ for all $y,z\in\Bbb{R}$. So $z+f(z)\neq0$ for all $z\in\Bbb{R}$ as $f$ is nonconstant. Mar 1 '19 at 15:32
• There is an answer here: google.com/… Jun 13 '20 at 16:20
• I edited the answer from the above link and added it as an answer here, so that there would be access to the answer on this site, especially in case that the link breaks or something. Also, I made some parts clearer in the edited version. Jun 14 '20 at 20:50

Consider $$N \in \mathbb R \setminus \{ 0 \}$$ (in the case of the above problem, $$N = 2019$$). There are three types of functions $$f : \mathbb R \to \mathbb R$$ satisfying $$f \big( x + y f ( x ) \big) + f ( x y ) = f ( x ) + f ( N y ) \tag 0 \label 0$$ for all $$x , y \in \mathbb R$$: $$f ( x ) = N - x \text ; \tag i \label i$$ $$f ( x ) = c \text { for an arbitrary constant } c \text ; \tag {ii} \label {ii}$$ $$f ( x ) = 0 \text { for } x \ne 0 \text {, and } f ( 0 ) \text { is arbitrary.} \tag {iii} \label {iii}$$ A straightforward check shows that all three types satisfy the equation, hence we just need to show that they are the only ones.
First of all, substituting $$N x$$ for $$x$$ in \eqref{0}, we arrive at the equation $$f \big( N x + y f ( N x ) \big) + f ( N x y ) = f ( N x ) + f ( N y ) \text .$$ After a change $$g ( x ) = \frac { f ( N x ) } N$$ this eqation reads $$g \big( x + y g ( x ) \big) + g ( x y ) = g ( x ) + g ( y ) \tag 1 \label 1$$ for all $$x , y \in \mathbb R$$, which does not depend on $$N$$. Now we investigate the corresponding functions $$g$$.
Setting $$x = 1$$ in \eqref{1} we get $$g ( 1 + y g ( 1 ) ) = g ( 1 )$$. If $$g ( 1 ) \ne 0$$, then $$1 + y g ( 1 )$$ attains all real values, so we arrive at the answer \eqref{ii}. Otherwise, $$g ( 1 ) = 0$$, and by setting $$y = 1$$ in \eqref{1} we get $$g \big( x + g ( x ) \big) = 0$$. If $$a = 1$$ is the unique real number with $$g ( a ) = 0$$, then we obtain $$x + g ( x ) = 1$$, whence $$g ( x ) = 1 - x$$, which falls into \eqref{i}. Also, if $$g ( x ) = 0$$ for all $$x \ne 0$$, we get the remaining answer \eqref{iii}. Hence in the sequel we assume that $$g ( 1 ) = 0$$, $$g ( a ) = 0$$ for some $$a \ne 1$$, and also there exists $$s \ne 0$$ with $$g ( s ) \ne 0$$, and arrive at a contradiction.
If $$b$$ is an arbitrary zero of $$g$$, then by plugging $$x = b$$ in \eqref{1} we get $$g ( b y ) = g ( y )$$. Recalling that $$g ( g ( 0 ) ) = g \big( 0 + g ( 0 ) \big) = 0$$, we obtain $$g \big( g ( 0 ) y \big) = g ( a y ) = g ( y ) \text . \tag 2 \label 2$$
Now, we show that $$g$$ is $$p$$-periodic, where $$p = ( a - 1 ) s$$. Indeed, setting $$x = a s$$ in \eqref{1} and using \eqref{2}, we get $$g \big( a s + y g ( s ) \big) = g \big( a s + y g ( a s ) \big) = g ( a s ) + g ( y ) - g ( a s y ) = g ( s ) + g ( y ) - g ( s y ) = g \big( s + y g ( s ) \big) \text ,$$ which proves the required periodicity, since $$y g ( s )$$ attains all real values. Letting $$x = p$$ in \eqref{1} and using periodicity we get $$g \big( y g ( 0 ) \big) + g ( p y ) = g ( 0 ) + g ( y )$$. By \eqref{2}, we arrive at $$g ( p y ) = g ( 0 )$$, which leads to the contradiction $$0 = g ( 1 ) = g \big( p \cdot p ^ { - 1 } \big) = g ( 0 ) = g \big( p \cdot \frac s p \big) = g ( s ) \ne 0$$.