-Knock knock! -Who's there? -Iran 2016

rmtf1111 wrote: -Knock knock! -Who's there? -Iran 2016

Actually $f\equiv c$ for any constant $c$ works. Just put it in the equation and check.

rmtf1111 wrote: We distinguish three cases - either $A=\{0\}$ or $B=\{1\}$ or $f$ is trivially constant. How do we get this line from the previous argument?

stroller wrote: rmtf1111 wrote: We distinguish three cases - either $A=\{0\}$ or $B=\{1\}$ or $f$ is trivially constant. How do we get this line from the previous argument? If neither of $A$ or $B$ is $\{0\}$ or $\{1\}$ respectively, then there exists a $x\in A$ and $m \in B$ such that $x(m-1)\neq 0$ and so by varying $z$ the expression $y+x(m-1)z$ will span the whole real axis, making $f$ constant

I claim that the set of solutions are $f(x)\equiv c$, $f(x)=\begin{cases}0& x\neq 0\\ c& x=0\end{cases}$, and $f(x)=2019-x$. If we denote the assertion as $P(x,y)$, note that $P(2019,y)$ yields $f(2019+yf(2019))=f(2019)$. If $f(2019)\neq 0$, we can modify $y$ such that $2019+yf(2019)$ ranges over all reals. Hence, $f\equiv c$ for some constant $c$. If $f(2019)=0$, note that $P(x,1)$ gives $f(x+f(x))=0$. Now, suppose we have a $z\neq 0$ such that $f(z)=0$. (note that if $z=0$, then $P(0,y)$ would yield $f$ is constant). $P(z,x)$ yields that $f(zy)=f(2019y)$, or equivalently $f(x)=f\left(\frac{2019}{z} x\right)$. Now, consider $P\left(x_0,\frac{c}{f(x_0)}\right)$ and $P\left(\frac{2019x_0}{z},\frac{c}{f(x_0)}\right)$ for some $x_0$ such that $x_0,f(x_0)\neq 0$ (If such an $x_0$ does not exist, then we will get our second set of solutions, which do indeed work.) They have the same RHS, and comparing LHS gives that $f\left(\frac{2019x_0}{z}+c\right)=f(x_0+c)$. As $c$ is arbitrary, we get that $f$ has period $\frac{2019x_0}{z}-x_0=p$. Now, we consider $P(x,y)$ and $P(x+p,y)$, which gives that $f(xy)=f(xy+xp)$. Varying $xy$ and $x$, we see that if $p\neq 0$, we get that $f$ is constant. So, we must have $p=0\implies z=2019$. So, $2019$ is our only root. However, looking back at $P(x,1)$, we see that $x+f(x)=2019$. So, we must have $f(x)=2019-x$, which indeed satisfies the assertion. Thus, we see that the three aforementioned solution forms are the only ones.

We claim that the answers are $f(x) = \begin{cases} 0 & \mbox{if } x \neq 0 \\ c & \mbox{if } x = 0 \end{cases}$, $f(x) = c$, $f(x) = 2019 - x$, where $c \in \mathbb{R}$ is a constant. It is easy to check that these functions work. We now show these are all. Let $P(x, y)$ denote the given assertion. From $P(2019, y)$, we get \begin{align*} f(2019 + yf(2019)) + f(2019y) &= f(2019) + f(2019y) \\ f(2019 + yf(2019)) &= f(2019). \end{align*}If $f(2019)$ is nonzero, as $y$ varies, $2019 + yf(2019)$ achieves all real values, implying that $f(x) = f(2019)$ for all $x$. Thus, in this case, $f$ is constant. Henceforth, suppose $f(2019) = 0$, and that $f$ is not identically $0$. Now, suppose there exists $u \neq 2019$ for which $f(u) = 0$. From $P(u, y)$, we have \begin{align*} f(u + yf(u)) + f(uy) &= f(u) + f(2019y) \\ f(uy) &= f(2019y). \end{align*}Thus, $f(uy) = f(2019y)$ for all $y$. Now, from $P(ux, y)$ and $P(2019x, y)$, we respectively have \begin{align*} f(ux + yf(ux)) + f(uxy) &= f(ux) + f(2019y) \\ f(2019x + yf(2019x)) + f(2019xy) &= f(2019x) + f(2019y). \end{align*}Since $f(uy) = f(2019y)$ for all $y$, we find that $f(ux + yf(ux)) = f(2019x + yf(2019x))$. Additionally, since $f$ is not identically $0$, there exists $t$ for which $f(2019t) \neq 0$. Now, we have \begin{align*} f(ut + yf(ut)) &= f(2019t + yf(2019t)) \\ f(ut + yf(ut)) &= f(ut + yf(ut) + (2019 - u)t). \end{align*}Since $f(ut) = f(2019t) \neq 0$, $ut + yf(ut)$ achieves all real values as $y$ varies across $\mathbb{R}$. Thus, $f(x) = f(x + (2019 - u)t)$ for all $x$, implying that $f$ is periodic with period $(2019 - u)t$. Now, suppose $t \neq 0$. Let $k = (2019 - u)t$ (which is nonzero because $u \neq 2019$). From $P(x + k, y)$, we have \begin{align*} f(x + k + yf(x + k)) + f(xy + ky) &= f(x + k) + f(2019y). \end{align*}Noting that $f(x + k) = f(x)$, we have \begin{align*} f(x + yf(x)) + f(xy + ky) &= f(x) + f(2019y). \end{align*}Comparing with $P(x, y)$ yields \begin{align*} f(xy + ky) &= f(xy). \end{align*}Now, fix $y \neq 0$. As $x$ varies across the real numbers, $xy$ achieves all real values; thus, we have $f(z + ky) = f(z)$ for all $z \in \mathbb{R}$, implying that $ky$ is also a period of $f$. However, since $k \neq 0$, as $y$ varies across the real numbers, $ky$ achieves all real numbers. This is enough to imply that every real number is a period of $f$, which is enough to imply that $f$ is constant. Thus, $f(x) = f(2019) = 0$ for all $x$; however, we had assumed that $f$ is not identically zero, contradiction. Thus, we must have $t = 0$, meaning that $f(x) = \begin{cases} 0 & \mbox{if } x \neq 0 \\ c & \mbox{if } x = 0 \end{cases}$, where $c \neq 0$ is a real constant. Otherwise, $u$ does not exist, meaning that $f(r) = 0$ only for $r = 2019$. Then, from $P(x, 1)$, we have \begin{align*} f(x + f(x)) + f(x) &= f(x) + f(2019) \\ f(x + f(x)) &= 0. \end{align*}This yields $x + f(x) = 2019$, implying that $f(x) = 2019 - x$. Thus, the only solutions to the given functional equation are $f(x) = c$ (where $c \in \mathbb{R}$ is a constant) and $f(x) = 2019 - x$, as claimed. $\Box$

Let $P(x,y)$ denote the assertion. The answer is $f$ constant; $f(x)=0$ for $x\ne0$; and $f(x)=2019-x$ for all $x$. These can be easily verified. We now check these are the only answers. Assume that $f$ is nonconstant. If $f(0)=0$, then $P(0,y)$ gives $f(2019y)=0$, so $f\equiv0$; thus we may assume $f(0)\ne0$. First, $P(2019,y)$ gives $f(2019+yf(2019))=f(2019)$, so if $f(2019)\ne0$, then $2019+yf(2019)$ is surjective and $f$ is constant. Henceforth $f(2019)=0$. Now $P(x,1)$ and $P(0,1)$ give \[f(x+f(x))=f(f(0))=f(2019).\]If there is no $z\ne2019$ with $f(z)=2019$, then $x+f(x)=2019$ for all $x$, so $f(x)=2019-x$. Henceforth assume such $z$ exists, and let $a=z/2019\ne1$. Assertions $P(2019x,y)$ and $P(zx,y)$ give \[f(2019x+yf(2019x))=f(zx+yf(zx))=f(zx+yf(2019x)).\]If $x_0=2019x$, then we have $f(x_0+yf(x_0))=f(ax_0+yf(x_0))$. Assume we can select $x_0$ such that $x_0\ne0$ and $f(x_0)\ne0$; otherwise we are done. From this, $x_0+yf(x_0)$ attains all real numbers, so we have $f(x)=f(x+(a-1)x_0)$ for all $x$. Denote $p=(a-1)x_0$, so that $f$ is periodic modulo $p$. To finish, note that $P(x,y)$ and $P(x+p,y)$ yield $f(xy+py)=f(xy)$. As we fix $xy$ and vary $y$, we have $f$ is constant, so we are done.

TheUltimate123 wrote: Let $P(x,y)$ denote the If there is no $z\ne2019$ with $f(z)=2019$, then $x+f(x)=2019$ for all $x$, so $f(x)=2019-x$. Isn't it $f(z)=0$?

Solved with nprime06 We claim the solutions are $f\equiv c$, $f(x)\equiv 2019-x$, and $f(x)=c \text{ if }x= 0 \text{, else }f(x)=0, $ and it's easy to check that they work. First by $P(2019,y)$ we see that $f(2019+yf(2019))=f(2019)$. If $f(2019)\ne 0$, then $2019+yf(2019)$ is surjective and hence $f$ is constant. Henceforth assume $f(2019)=0$ and that $f$ is not constant. Claim: $f$ is injective at $2019$. Proof. FTSOC suppose $\exists\text{ } a\ne 2019\in\mathbb R$ st. $f(a)=0$. \[P(a,y)\implies f(a+yf(a))+f(ay)=f(a)+f(2019y)\iff f(ay)=f(2019y)\quad (\spadesuit). \]Then, \begin{align*}f(ax+yf(ax))\stackrel{P(ax,y)}=&f(ax)+f(2019y)-f(axy)\stackrel\spadesuit=f(2019x)+f(2019y)-f(2019xy)\\\stackrel{P(2019x,y)}=&f(2019x+yf(2019x))\quad (\clubsuit). \end{align*}Since $f$ is non constant, $\exists \text{ }b\ne 1$ st. $f(ab)=f(2019b)\ne 0$. Then by replacing $x$ with $b$ in $\clubsuit$ and varying $y$ we see that $yf(ab)$ is surjective; hence $f$ is periodic with period $(2019-a)b$. Let $0\ne c=(2019-a)b\implies f(x+c)=f(x)\quad (\heartsuit)$. Next, \begin{align*}f((x+c)y)\stackrel{P(x+c,y)}=&f(x+c)+f(2019y)-f(x+c+yf(x+c))\\ \stackrel\heartsuit=&f(x)+f(2019y)-f(x+yf(x))\stackrel{P(x,y)}=f(xy)\iff f(z+cy)=f(z). \end{align*}However, varying $y$ over the reals we see that $f$ is constant, our desired contradiction. Hence, $a=2019,b=1,$ and $c=0$ which gives us the solution $f=\begin{cases} c\ne 0 & x=0 \\ 0 & x\ne 0 \end{cases}$. $\square$ By the Claim, we have $P(x,1)\implies f(x+f(x))=f(2019)\implies x+f(x)=2019\implies f(x)=2019-x$ is our last solution to the FE. Having found all the solutions, we are done. $\blacksquare$

Does this work? This uses the same idea as APMO 2019/5. The only solutions are $f\equiv 2019-x,$ $f\equiv c,$ and \[f(x)\equiv\begin{cases}0&x\ne 0\\ c&x=0\end{cases}.\]It is easy to check that these work, so assume $f$ is not one of these. Let $P(x,y)$ denote the assertion. $\textbf{Claim: }$ $f(2019)=0$ $P(2019,y)$ gives $f(2019+yf(2019))=f(2019).$ If $f(2019)\ne 0,$ then $2019+yf(2019)$ takes all real values as $y$ varies, so $f$ is constant. This contradicts our original assumption, so $f(2019)=0.$ $\blacksquare$ Now let $S$ be the set of reals $c$ satisfying $f(cx)=f(x)$ for all real $x.$ $\textbf{Claim: }$ $S$ is nonempty $\emph{Proof: }$ $P(x,0)$ gives $f(x+f(x))=0$ for all $x.$ Therefore, if the only root of $f$ is $2019,$ then $f\equiv 2019-x,$ contradiction. Hence there exists $a\ne 2019$ such that $f(a)=0.$ From $P(a,y)$ and $P(2019,y),$ we find that $f(ay)=f(2019y)$ for all $y.$ Thus, $$f\left(2019y\cdot\frac{a}{2019}\right)=f(ay)=f(2019y).$$Since $2019y$ can take any real value, we are done. $\blacksquare$ Fix $c\in S.$ Since $c\in S\iff\frac{1}{c}\in S,$ we may assume $|c|>1.$ The assertions $P(cx,y)$ and $P(x,y)$ yield \begin{align*} f(cx+yf(x)) &= f(cx)+f(2019y)-f(cxy)\\ &= f(x)+f(2019y)-f(xy)\\ &= f(x+yf(x)). \end{align*}Therefore, $M=\frac{cx+yf(x)}{x+yf(x)}\in S.$ Choose $x\ne 0$ such that $f(x)\ne 0,$ and let $y=\frac{1}{f(x)}.$ Then, $M=\frac{cx+1}{x+1},$ so as $x$ varies, $M$ takes all values in the interval $(1,c).$ This implies that $f(a)=f(b)$ for all $b\in (a,ca),$ so $f$ must be constant, contradiction.

Really nice problem! I have a weak spot for entirely or nearly-entirely wrapped FEs on $\mathbb{R}$ For storage: Clearly $$\boxed{f(x)=c \ \ \forall x \in \mathbb{R}}$$is a solution for all values of $c$. Henceforth assume that $f$ is non-constant. Let $P(x,y)$ denote the original assumption. If $f(2019) \neq 0$, then $P \left (2019,\dfrac{y-2019}{f(2019)} \right )$ implies $$f(2019)=f(y)$$for all $y$, which contradicts that $f$ is non-constant. Hence $f(2019)=0$, and so $P(x,1)$ $\implies$ $f(x+f(x))=0$ for all $x$. If $x+f(x)=2019$ for all $x$, we get $$\boxed{f(x)=2019-x \ \ \forall x \in \mathbb{R}}$$which is indeed a solution. Else suppose $x+f(x)=u \neq 2019$ for some $x$. Then $P \left (u, \dfrac{y}{2019} \right )$ $\implies$ $f(y)=f(ky)$ for all $y$, where $k =\dfrac{u}{2019} \neq 1$. If $f(x)=0$ for all $x \neq 0$, then we get $$\boxed{f(x)=0 \ \text{for} \ x \neq 0 \ \text{and} \ f(0)=c}$$which is a solution for all values of $c$. Else $f(t) \neq 0$ for some $t \neq 0$. Comparing $P \left ( t, \dfrac{y}{f(t)} \right )$ and $P \left ( kt, \dfrac{y}{f(t)} \right )$ we get $$f(kt+y)=f(t+y)$$for all $y$ $\implies$ $f$ is periodic with period $p=(k-1)t \neq 0$. Comparing $P \left ( 0, \dfrac{y}{p} \right )$ and $P \left ( p, \dfrac{y}{p} \right )$ we get $f(y)=f(0)$ for all $y$, which contradicts that $f$ is non-constant. $\blacksquare$

How is this a killer problem? The only $f$'s are constant $f$, $f(x)=2019-x,$ and $$ f(x)=0\forall x\ne 0$$They clearly work. Let $P(x,y)$ denote the assertion $f(x + yf(x)) + f(xy) = f(x) + f(2019y).$ We prove a lemma: Lemma: If $f(y)=f(Cy)\forall y$ for some constant $C\ne 1,$ then $f(x)=0\forall x\ne 0$, or $f$ is constant. Proof. Suppose $f(x)\ne 0,$ for some $ x\ne 0$ $P(x,y)$ gives $f(x+yf(x))=f(x)+f(2019y)-f(xy)$ $P(x,Cy), P(Cx,y)$ gives $f(x+Cyf(x))=f(x)+f(2019Cy)-f(Cxy)=f(x)+f(2019y)-f(xy)=f(x+yf(x)).$ Similarly $f(Cx+yf(x))=f(x+yf(x))$ Let $a=yf(x).$ Then we can see $f(x+a)=f(x+Ca),$ and the range of $a$ is $\mathbb{R}.$ Hence, $f(x+a)=f(\frac xC+a)$ for all $a.$ This implies $f$ is periodic with period $|x-\frac xC|=T$ Now, $P(z+T,y)-P(z,y)$ suggest $f((z+T)y)=f(zy)$ for all $y,$ which implies $f$ is constant because there are no restrictions on $z,y$ other than them being reals. If $f(y)=0\forall y\ne 0,$ then we can see $f(0)$ can be anything. Suppose $f(a)=0.$ Then $P(a,y)$ yields $f(a+0)+f(ay)=f(a)+f(2019y),$ or $f(ay)=f(2019y).$ If $a\ne 2019,$ we are done by our lemma, so $f(2019)=0.$ Also, $P(x,1)$ gives $f(x+f(x))=0.$ If $f(a)=0,$ we can see $a=2019$ is forced, so $x+f(x)=2019,$ or $f(x)\equiv 2019-x.$

The only solutions are $f(x) = c$ for a constant $c$ and $f(x) = 2019 - x$. These both obviously work. Let $P(x,y)$ denote the assertion. Plug in $P(2019, y)$ to get $f(2019 + yf(2019)) = f(2019)$. If $f(2019)\neq 0$, then $2019+yf(2019)$ ranges over all real numbers, so $f(x)$ is constant. Therefore, assume $f(2019) = 0$. Then, $P(x,1)$ gives $f(x+f(x)) = f(2019) = 0$. Consider if there existed some $a\neq 2019$ such that $f(a) = 0$, and $f$ was not a constant function. By $P(a,y)$, it gives \[f(a+yf(a)) + f(ay) = f(a) + f(2019y) \Rightarrow f(ay) = f(2019y)\]This means $f(x) = f(\frac{a}{2019}x)$. Now, by $P(\frac{a}{2019}x, y)$, it gives \[f\left(\frac{a}{2019}x + yf\left(\frac{a}{2019}x\right)\right) + f\left(\frac{a}{2019}xy\right) = f\left(\frac{a}{2019}x\right) + f(2019y)\]\[\Rightarrow f\left(\frac{a}{2019}x + yf(x)\right) + f(xy) = f(x) + f(2019y)\]By $P(x,y)$, this means $f(\frac{a}{2019}x + yf(x)) = f(x + yf(x))$. Consider some fixed $x$ such that $f(x) \neq 0$ (exists by non-constant function). This means that, since we can vary $y$, we must have $f(r) = f\left(r + \frac{2019-a}{2019}x\right)$, so $f$ is a periodic function. Let the period be $k$, so $f(r) = f(r+k)$. By $P(k, y)$, it gives \[f(k+yf(k)) + f(ky) = f(k) + f(2019y) \Rightarrow f(yf(0)) + f(ky) = f(0) + f(2019y)\]By $P(0, y)$, we get $f(yf(0)) = f(2019y)$. Substituting into the above equation,w e get $f(ky) = f(0)$, so $f$ is a constant function, a contradiction. Therefore, assume the only value of $a$ such that $f(a) = 0$ is $a = 2019$. Since $f(x+f(x)) = 0$, this means $x + f(x) = 2019$, so $f(x) = 2019-x$. These are our only solutions.

CantonMathGuy wrote: Determine all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying \[f(x + yf(x)) + f(xy) = f(x) + f(2019y),\]for all real numbers $x$ and $y$. To begin with, we would try to extract the constant solutions. Clearly, any constant \(f\equiv c\) works, now assume that \(f\) is non-constant. We shall now prove a claim. Claim.\(f(z)=0\iff z=2019\) or \(f(x)= 0\iff x\neq 0\). Proof. Note that \(P(z,y)\) gives us that \[f(z+yf(z))+f(zy)=f(z)+f(2019y)\]implying that \(f(zy)=f(2019y)\). This implies that \(f(kx)=f(x)\) for all \(x\in\mathbb{R}\) where \(k=\frac{z}{2019}\). Then, \(P(kx,y)\) gives us that \[f(kx+yf(x))+f(xy)=f(x)+f(2019y)\]This implies that \[f(kx+yf(x))=f(x+yf(x))\]implying that \(f\) is periodic with period \(T=(k-1)x\). Now, let \(a\) be such that \(f(a)\neq0\) and put \(T=(k-1)a\) which is fixed. Then, \(P(x+T,y)\) and \(P(x,y)\) together give us that \[f(xy+yT)=f(xy)\]and this holds for every \(x\) and \(y\). This forces \(T=0\) otherwise \(f\) is constant. Therefore, \(k=1\), i.e. \(z=2019\) or \(a=0\), i.e. \(f(x)=0\) for all \(x\neq0\), because if \(z\neq2019\), then no other \(a\) shall work, implying that the only such \(a\) is \(0\). Therefore, we have another "almost" constant solution, that is \(f(x)=0\) for all \(x\neq0\). Now assume \(f(2019)=0\). Then, \(P(x,1)\) yields \[f(x+f(x))=0\]implying that \(f(x)+x=2019\) or \(f(x)=2019-x\). Therefore, the only solutions are \(f\equiv c\) for any real constant \(c\), \(f(x)=0\) for all \(x\neq0\) and \(f(x)=2019-x\) for all \(x\in\mathbb{R}\).

Let $P(x,y)$ be the given assertion. Clearly any constant function works so assume otherwise. Note that $f(2019)=0$ from $P(2019,x).$ $P(x,1)$ yields $f(x+f(x))=0$ which means $f(x)=2019-x$ if injectivity at zero, this works. Assume $f(c)=0$ for some $c\neq 2019.$ $P(c,x/2019)$ gives $f(x)=f(cx/2019).$ Moreover $P(0,x)$ implies $f(xf(0))=f(2019x).$ We may have $f(x)=0$ for all non zero $x$ and $f(0)=k$, this works. If not then, comparing $P(x,y)$ with $P(cx/2019,y)$ gives $f(cx/2019+yf(x))=f(x+yf(x)).$ This implies $f$ is periodic with some period $p.$ But then $P(p,y)$ gives $f$ is constant, contradiction.

We claim the solutions are $f\equiv c$ where $c$ is some constant or $f\equiv -x+2019$. They clearly work. Let $P(x,y)$ be the assertion. Consider $P(2019,y)$: \begin{equation} f(2019+ y \cdot f(2019) = f(2019) \end{equation} Either $f(2019)\neq 0$, in which case $f$ is constant, or $f(2019)=0$. Henceforth, we assume $f(2019)=0$ and that $f$ is not constant. Next, note that if $f(z)=0$, then $P(z,y)$ tells us that $f(zy) = f(2019y)$. Thus, if we let $r=\frac{z}{2019}$, then $f(rx) = f(x)$ for all $x$. Note that multiplying $x$ by $r$ only changes the $f(x+yf(x))$ term in $P(x,y)$. Thus, since $f$ is non-constant, we may select $\alpha$ such that $f(\alpha)=C\neq 0$. Thus, we have $f(\alpha r + y\cdot C) = f(r+ y\cdot C)$, and $f$ is periodic with period $T = \alpha r - r$. This finally derives a contradiction since if we increase $x$ by $T$, the only value that changes in $P(x,y)$ is $f(xy)$. Then, by selecting $y = \frac{\delta }{T}$, $f$ is periodic with all periods, so $f$ must be constant, contradiction. Finally, consider if the only such $z$ is 2019. Then, note that $P(x,1)$ gives $f(x+f(x)) = 0$, so this implies that $x+f(x) = 2019$ for all $x$, recovering $f(x)\equiv -x+2019$, and we have shown that the stated functions are the only possible ones. $\blacksquare$.

The answer is $f \equiv c$, $f(x)=2019-x$, and $f(x)=0$ if $x \neq 0$ with $f(0)\neq 0$ arbitrary. The first two clearly work. For the third solution, note that if $x$ is nonzero, then the LHS is $f(xy)$ and the RHS is $f(2019y)$, which are equal. If $x$ is zero, then the LHS is $f(yf(0))+f(0)$ and the RHS is $f(0)+f(2019y)$. Since $f(0) \neq 0$ these two sides are also equal. We now prove that these are the only solutions; let $P(x,y)$ denote the given assertion. Clearly every constant solution works, so assume that $f$ is nonconstant. From $P(2019,y)$, we find that $f(2019+yf(2019))=f(2019)$ for all $y \in \mathbb{R}$. If $f(2019)\neq 0$, then this means that $f(x)=f(2019)$ for all $x$, so $f$ is constant—contradiction. Thus $f(2019)=0$. Suppose that $f(a)=0 \implies a=2019$, i.e. $f$ has a single root. From $P(x,1)$ we obtain $f(x+f(x))=0$ for all $x$, hence $x+f(x)=2019 \implies f(x)=2019-x$ for all $x$, which is one of our solutions. Thus suppose that $f$ has some other root $r$. From $P(r,x)$ we obtain $f(rx)=f(2019x)$, so $f(x)=f(kx)$ where $k=\tfrac{r}{2019}$. If there does not exist $x \neq 0$ with $f(x) \neq 0$, then we get the third solution as described. Thus suppose that such an $x$ exists. From $P(x,y)$ and $P(kx,y)$ we find that $$f(kx+yf(x))=f(kx+yf(kx))=f(kx)+f(2019y)-f(kxy)=f(x)+f(2019y)-f(xy)=f(x+yf(x)).$$Since $f(x) \neq 0$, $yf(x)$ spans the real numbers. Furthermore, $kx \neq x$ as $x \neq 0$, so $f$ is in fact periodic with period $(k-1)x:=T \neq 0$. Now consider some $x \in \mathbb{R}$ (not related to $\tfrac{T}{k-1}$). From $P(T,x)$, we obtain $$f(T+xf(T))+f(xT)=f(T)+f(2019x) \implies f(xf(0))+f(xT)=f(0)+f(2019x).$$On the other hand, $P(0,x)$ yields $$f(xf(0))+f(0)=f(0)+f(2019x),$$hence $f(xT)=f(0)$. Since $T \neq 0$, this means that $f$ is constant for all reals: contradiction. Thus we are done. $\blacksquare$

We change the $2019$ to $2023$ which still works. The solutions are $f\equiv c$ and $f \equiv 2023 - x$. Assume $f\not\equiv c$. So there exists a $p$ such that $f(p) \neq 2023$. $P(2023,y) \implies f(2023 + yf(2023)) = f(2023)$. If $f(2023) \neq 0$, then choosing $y \rightarrow\dfrac{y}{f(2023)} - 2023$ gives $f\equiv c$, contradiction. Otherwise $f(2023) = 0$. Now $P(0,y) \implies f(yf(0)) = f(2023y) \implies f(f(0)) =f(2023) = 0$. If $f(0) \neq 2023$, then $f(y) = f\left(\frac{f(0)y}{2023}\right) = f(ry); \; r\neq 1$. Now $P(x,1) \implies f(x + f(x)) = 0$. Furthermore, $P(x + f(x),y) \implies f(y(x+f(x))) = f(2023y). \qquad\ (\clubsuit)$ Now in the main equation, comparing $P(x,y)$ and $P(rx,y) \implies f(rx+yf(x)) = f(x+yf(x))$. Thus $k = \dfrac{rx + yf(x)}{x + yf(x)} \implies y = \dfrac{x(k-1)}{f(x)(r-k)}$ for some $f(x) \not\equiv 0$. Then we can similarly get $f\equiv c$, contradiction. Thus we must have that $f$ is injective or $f(0) = 2023$. If $f$ is injective, then $(\clubsuit)$ gives that $f(x) = 2023 - x$. Otherwise $f(0) = 2023$. Now putting $y=1$ in $(\clubsuit)$, we get $f(x+f(x)) = f(2023) = 0$. Now comparing $P(x+f(x),y)$ and $P(2023,y)$, we get $f((x+f(x))y) = f(2023y)$ and we can finish similarly by assuming $r = \dfrac{2023}{x+f(x)} \neq 1$ and then using APMO trick as earlier we are done.

$$f(x+yf(x))+f(xy)=f(x)+f(2019y)$$ The answer is $f(x)=c$ and $f(x)=2019-x$, and $f(x)=0$ on $x\neq 0$ for any $f(0)$, which clearly work. Plugging $x=2019$, we have $$f(2019+yf(2019))=f(2019).$$If $f(2019)\neq 0$, then $2019+yf(2019)$ can be any real number, which implies that $f$ is constant. Therefore, if $f$ is nonconstant, then $f(2019)=0$. Assume this from now on. Plugging $y=1$, we have $$f(x+f(x))=0.$$If $2019$ is the only value of $x$ for which $f(x)=0$, we would be done as we must have $x+f(x)=2019$. Thus, from now on assume there exists $r\neq 2019$ with $f(r)=0$. Seeing the abundance of $f(x)$, assert $x=r$ to get $$f(ry)=f(2019y).$$Thus, $$f(y)=f(\frac{r}{2019}y).$$This means that $f$ is "exponentially periodic". For periodic functions, a common strategy is to replace $x$ with its next cycle, so $x\rightarrow \frac{r}{2019}x$ (not changing $f(x)$ or $f(xy)$) gives $$f(\frac{r}{2019}x+yf(x))+f(xy)=f(x)+f(2019y).$$Thus, $$f(x+yf(x))=f(\frac{r}{2019}x+yf(x)).$$Assuming $f$ is not always zero on nonzero inputs (we already found that as a solution), plugging in any nonzero non-root of $f$ gives us that $f$ is periodic with a nonzero period (additively), since $yf(x)$ can be any real. Doing the same periodic trick, $x\rightarrow x+P$ where $P$ is the period, $f(x+yf(x))$ does not change since $x+yf(x)$ increases by $P$, and $f(x)$ and $f(2019y)$ also do not change. Thus, $$f(xy)=f(xy+Py).$$Plugging $x=\frac{1}{y}$, we have $$f(1)=f(1+Py),$$so $f$ is constant as $Py$ can be any nonzero real and we are done. remark: the $f(0)$ not having to equal 0 is quite tricky, and I initially fakesolved this. The error was that when claiming that $$f(x+yf(x))=f(\frac{r}{2019}x+yf(x))$$implies that $f$ is periodic for $f(x)\neq 0$, one has to make sure that the period is not zero, but when $x=0$, the period is 0. So this only shows that when $f$ is nonzero on a nonzero input, $f$ is periodic.

$f\left(x+yf\left(x\right)\right)+f\left(xy\right)=f\left(x\right)+f\left(cy\right)$ $f\left(x+f\left(x\right)\right)=f\left(c\right)$ $f\left(c+yf\left(c\right)\right)=f\left(c\right)$ So, if $f$ is injective $f\left(c\right)=0$ and $f\left(x\right)+x=c$ which means $f\left(x\right)=c-x$. So, all we need is to prove that function is injective. Suppose $f\left(c\right)\neq0$. Then, $f\left(c+y\right)=f\left(c\right)$ and the function is a constant. (Any constant is fine.) So, we have $f\left(c\right)=0$. Hence, $f\left(x+f\left(x\right)\right)=0$. We need injectivity at 0 to finish. Suppose $f\left(a\right)=f\left(b\right)=0$ for distinct $a,b$. Then, plugging $P\left(a,y\right)$ we get $f\left(ay\right)=f\left(cy\right)=f\left(by\right)$. Let $r=\frac{a}{b}$. Let $S$ be the set of all $r$ for which $f\left(x\right)=f\left(rx\right)$. Then, plug $P\left(x,r\right): f\left(x+rf\left(x\right)\right)=0$. Note, plugging $P\left(rx,y\right)$ gives $f\left(rx+yf\left(x\right)\right)=f\left(x+yf\left(x\right)\right)=f\left(x+rf\left(x\right)\right)$ This means that $f(rx+yf(x))=f(x+yf(x))$. Take $f(x)\ne0$, and note that $f$ is a periodic function with period $(r-1)x$ as $yf(x)$ can be arbitrary. Let $T=(r-1)x$. Then, $f(x)=f(x+T)$. Plug $P(x+T,y)$ to get that $f((x+T)y)=f(xy)$. Hence, if $T$ is the period then so is $yT$. Since $y$ can be arbitrarily small, we get that $f$ is a constant for all numbers other than 0. If $f(x)=0$ for every $x\ne0$ then we can have $f(0)$ and still original FE works. So, solutions are: f(x)=c-x; f(x) = constant f(x)=0 for $x\ne0$ and $f(0)$ is arbitrary.

Answers: $f \equiv c$, $f(x)=2019-x$, and $f(0)=c$ with $f(k)=0$ for $k\neq 0$. Let $P(x,y)$ denote the assertion. First, $x=2019$ gives $f(2019+yf(2019))=f(2019)$ so $f$ is constant or $f(2019)=0.$ If $f(0)=0$, then $P(0,y)$ gives $f \equiv 0$, so assume that $f(0)\neq 0$. Now, $y=1$ gives $f(x+f(x))=0$. If no nonzero real number is sent to a nonzero number, we obtain the third solution set. If $2019$ is the only number sent to $0$ then $f(x+f(x))=0$ gives our $f(x)=2019-x$ solution Thus suppose $f(a)=0$ for $a \neq 0, 2019$; we will show no more solutions exist. $P(a,y)$ now gives that $f(ay)=f(2019y)$, or $f(x)=f(\frac{2019}{a}x)$. Claim: Here, $f$ is periodic with period $d$. Proof. There exists a nonzero $d$ with $f\left (\frac{d}{\frac{2019}{a}-1}\right)\neq 0$. Then, let $m-n=d$. $P(\frac{2019}{a}x,y)$ gives $f(\frac{2019}{a}x+yf(x))=f(x+yf(x))$. Letting $x=d$ there exists a $y$ with the LHS as $m$ and RHS as $n$ as desired. $\blacksquare$ Now, let $a\neq b$. Let $y, x_1, x_2$ be reals with $d \mid x_1-x_2$ and $a=yx_1$, $b=yx_2$. $P(x_1,y)$ and $P(x_2,y)$ give that $f(a)=f(b)$ since all terms except the $f(xy)$ are the same.

The answer is \[f(x) = \boxed{\begin{cases} 2019-x \\ c, \ c \in \mathbb{R} \\ 0, \ x \neq 0 \ \text{ and } \ f(0) = c \end{cases}} \] which all work. Denote the given assertion as $P(x,y)$. Notice that $f$ can be constant, so henceforth assume otherwise. Observe that $P(2019,y)$ gives \[f(2019+yf(2019)) = f(2019).\] If $f(2019) = k \neq 0$, then the RHS is always equal to $k$, making $f$ constant. Hence, $f(2019) = 0$. Then, $P(x,1)$ yields \[f(x+f(x))=0.\] Suppose that $x+f(x) = 2019$. We easily see that $f(x) = 2019-x$ is a solution, so assume otherwise from now on. Let $x+f(x) = r$ such that $r$ is another root of $f$. Plug in $P(r,y)$ to get \[f(ry) = f(2019y) \implies f(y) = f\left( \frac{r}{2019} y\right).\] Obviously, if $f(x) = 0$ for all values except $x = 0$, this is a solution. Else, assume there is a value $s \neq 0$ such that $f(s) \neq 0$. Comparing $P(s,\tfrac{y}{f(s)})$ and $P(\tfrac{r}{2019}s,\tfrac{y}{f(s)})$ shows that \[f(s+y) = f\left(\frac{r}{2019} s + y \right).\] This shows the periodicity of $f$ so we let the period be $p$. Comparing $P(0,\tfrac{y}{p})$ and $P(p,\tfrac{y}{p})$ yields $f(y) = f(0)$ for all $y$, contradicting the nonconstant condition. We have exhausted all of the cases, so we are done. Remark: We apply the $x \neq 0$ condition while proving periodicity, because otherwise the period would equal $0$.

Clearly $f(x)\equiv k$ for some constant $k$ satisfies our equation, so let's assume that $f$ is non-constant. Let $P(x,y)$ denote the given equation. $P(0,x)$ implies that $f(xf(0))=f(2019x)$. Next we have, $$P(f(0),x)\implies f(f(0)+xf(f(0))) = f(f(0))$$this means that $f(f(0))=0$ since otherwise $f$ is constant. It also means $f(2019)=0$. So $P(x,1)$ gives us $f(x+f(x))=0$ and so $$P(x+f(x),y)\implies f((x+f(x))y)=f(2019y) \qquad (1)$$Now we will prove a result which will help us "unwrap" $f$ and also finish the problem. Claim: If $f(tx)=f(x)$ for all $x$, then $t=1$. Proof. Then, we can check that $P(t,y)$ and $P(1,y)$ gives us $$f(t+yf(1))=f(2019y)=f(1+yf(1)).$$Note that this equation itself implies $f(1)\neq 0$ (otherwise $f(2019y)=0$, a contradiction). Thus we get $f(x)=f(x+t-1)$ for all real $x$. Now $$P(t-1,x)\implies f(t-1+xf(0))+f(x(t-1))=f(0)+f(2019x) \implies f(xf(0))+f(x(t-1))=f(0)+f(2019x)$$while $P(0,x)$ implies $f(xf(0))+f(0)=f(0)+f(2019x)$ and so we get $f(x(t-1))=f(0)$ for all $x$. This is a contradiction unless $t=1$. So we're done. //// Applying the claim for $(1)$ we get that $x+f(x)=2019$ for all $x$, thus $f(x)=2019-x$. Conversely it's easy to check this function works. Edit: oops, seems like I missed the solution $f(x)=\begin{cases} 0 & x\neq 0 \\ k & x=0\end{cases}$

The solutions are constant $f$, $f(x)=2019-x$ and \[f(x)\equiv\begin{cases}0&x\ne 0\\ c&x=0\end{cases}\]these clearly work. We will suppose that $f$ isn't of the first and third from, hence there exists $x\neq 0$ such that $f(x)\neq 0$. Let $P(x,y)$ denote the given assertion. $\bullet$ $P(2019,x)\Rightarrow f(2019+yf(2019))=f(2019)$, hence $f(2019)=0$ since $f$ isn't constant. $\bullet$ $P(x,1)\Rightarrow f(x+f(x))=0$ Now suppose that there exists $a\neq 2019$ such that $f(a)=0$. $P(a,y)$ gives $f(ay)=f(2019y)$. Now subtracting $P(ax,y)$ from $P(2019x,y)$ gives $$f(ax+yf(2019x))=f(2019x+yf(2019x))$$ Therefore, fixing $x\neq 0$ such that $f(2019x)\neq 0$ (notice that $x\neq 0$ from our assumption at the beginning) and $y\mapsto \frac{y}{f(2019x)}$ gives $f(y+2019x)=f(y+ax)$, $\forall y\in \mathbb R$ and since $a\neq 2019$ we get $f(x)=f(x+c)$, $\forall x\in\mathbb R$, where $c=|x(2019-a)| \neq 0$. Now subtracting $P(x+c,y)$ from $P(x,y)$ gives $f(xy)=f(xy+yc)$ so $x\mapsto \frac{a}{y}$ gives $f(a+yc)=0$, $\forall y\in\mathbb R\setminus \{0\}$ so $f\equiv 0$. Finally, $f$ is injective at $0$ so from $f(x+f(x))=0=f(2019)$ we get $f(x)=2019-x$. $\blacksquare$

Let $P(x, y)$ denote assertion. $P(2019, y)$ \[f(2019+yf(2019))=f(2019)\]Thus we get either $f(2019)=0$ or $f(x)=c$ for some fixed $c$. Suppose $f(2019)=0$. $P(x, 1)$ \[f(x+f(x))=0\]Suppose $f(k)=0$ and $k\neq 2019$. $P(k, y)$ \[f(ky)=f(2019y)\]Thus if we let $m=\frac{k}{2019}$ we get $f(x)=f(mx)$. $P(mx, y)$ \[f(mx+yf(x))=f(x)+f(2019y)-f(xy)=f(x+yf(x))\]Thus this implies that $f$ is periodic, now let the period be $t$. $P(x+t, y)-P(x, y)$ \[f((x+t)y)=f(y)\]Which implies that for all $x\neq 0$ $f(x)=0$. Thus we get either $f$ is constant, $f$ is $0$ for everything that is not $0$ or $f(x)=2019-x$.

\[f(x + yf(x)) + f(xy) = f(x) + f(2019y)\]Answers are $f\equiv c$ and $f(x)=2019-x$ and $f(x)=0$ for all $x\neq 0$ and $f(0)$ is any real number. Constant functions clearly satisfy the equation hence assume that $f$ is nonconstant. Case $I$: $f$ is periodic. Let $T$ be the period. Comparing $P(x,y)$ with $P(x+T,y)$ implies $f(xy)=f(xy+Ty)$.Plug $\frac{x}{y},\frac{y}{T}$ to get $f(x)=f(x+y)$ for $y\neq 0$ which is obvious for $y=0$. So $f$ is constant which contradicts with our initial assumption.$\square$ Case $II$: $f$ is not periodic. Claim: $f(2019)=0$. Proof: If $f(2019)\neq 0,$ then plugging $P(2019,\frac{y-2019}{f(2019)})$ gives $f(y)=f(2019)$ but this is impossible since $f$ is non-constant.$\square$ Claim: For $x\neq 0,$ we have $f(x)\in \{0,2019,2019-x\}$. Proof: Suppose that $f(x)\neq 0,2019$. \[P(x,\frac{x}{2019-f(x)}): \ f(x+\frac{xf(x)}{2019-f(x)})+f(\frac{x^2}{2019-f(x)})=f(x)+f(\frac{2019x}{2019-f(x)})\]Since $x+\frac{xf(x)}{2019-f(x)}=f(\frac{2019x}{2019-f(x)}),$ we conclude that \[f(x)=f(\frac{x^2}{2019-f(x)})\]We compare $P(x,y)$ with $P(\frac{x^2}{2019-f(x)},y)$ in order to verify \[f(x+yf(x))=f(\frac{x^2}{2019-f(x)}+yf(x))\]Now plug $x,\frac{y-x}{f(x)}$ which gives us the equation \[f(y)=f(y+\frac{x^2}{2019-f(x)}-x) \ \ \text{for} \ f(x)\neq 0,2019\]Since $f$ is not periodic, we must have $\frac{x^2}{2019-f(x)}=x$. We have assumed that $x\neq 0$ so $x=2019-f(x)\iff f(x)=2019-x$ which proves our claim.$\square$ If there exists $f(a)=0$ where $a\neq 2019,$ then choose $P(a,\frac{y}{2019})$ which yields $f(y.(\frac{a}{2019}))=f(y)$. Since $f$ is nonconstant, we can choose an $x$ such that $f(x)\neq 0$. Comparing $P(x,y)$ with $P(\frac{xa}{2019},y)$ gives us \[f(x+yf(x))=f(\frac{xa}{2019}+yf(x))\]By plugging $y\rightarrow \frac{y-x}{f(x)},$ \[f(y)=f(y+\frac{xa}{2019}-x)\]Since $f$ is not periodic, we get that $xa=2019x\iff x(a-2019)=0$. So $x=0$ and this means if $x\neq 0,$ then $f(x)=0$ which is a solution. If $f(b)=2019$ and $b\neq 0,$ then plug $P(x,1)$ to get $f(x+f(x))=0$. Choosing $x=b$ gives $f(b+2019)=0$. But by looking at the previous case, if $f(b+2019)=0$ and $b\neq 0,$ then we get the function $f(x)=0$ for all $x\neq 0$. Now, we have worked on the cases $f(a)\in \{0,2019\}$ for $a\neq 2019,0$ respectively. We have $f(x)=2019-x$ for $x\neq 0$. \[P(0,1): \ f(yf(0))=f(2019y)=2019-2019y\]Note that $f(0)\neq 0$. So put $y=1$ to verify $2019-f(0)=f(f(0))=0$ thus, $f(0)=2019$. We get the function $f(x)=2019-x$ for all reals $x$ as desired.$\blacksquare$

Answers are $f(x)=c,f(x)=2019-x,f(x)=0\forall x\ne 0$, which we can check work. First $P(2019,y)$ gives $f(2019+yf(2019))=f(2019)$, so either $f$ constant or $f(2019)=0$. If $f$ is injective at $0$ then $P(x,1)$ gives $f(x+f(x))=0$ implies $f(x)=2019-x$. Otherwise suppose $f(2019s)=0$ for $s\ne 1$. Now $P(2019s,\tfrac x{2019})$ gives $f(x)=f(sx)$ for all $x$. Next $P(sx,y)$ gives \[f(sx+yf(x))=f(sx+yf(sx))=f(sx)+f(2019y)-f(sxy)=f(x)+f(2019y)-f(xy)=f(x+yf(x)).\]If $f(x)\ne 0$, then $f$ has period $p=(s-1)x$. If $x\ne 0$ then $p\ne 0$ and $P(x+p,y)$ gives \[f(xy+py)=f(x+p)+f(2019y)-f(x+p+yf(x+p))=f(x)+f(2019y)-f(x+yf(x))=f(xy).\]This implies $f$ constant. Otherwise $f(x)\ne 0\implies x=0$. This gives the answers.

Did this a while back but never posted solution The only solutions are $f \equiv c$ for some constant $c$, $f(x) = 2019 - x \forall x$, and $f(x) =0 \forall x\ne 0$ where $f(0)$ is any real not equal to $0$. Let $P(x,y)$ be the given assertion. Clearly all constants work, so assume $f$ isn't constant. $P(2019. y): f(2019 + y f(2019)) = f(2019)$. If $f(2019) \ne 0$, then $f$ is constant so $f(2019) = 0$. Note that $f(x) = 0 \forall x\ne 0$ always works, so assume that there is some nonzero real number that isn't a zero of $f$. Claim: $f$ is injective at $0$. Proof: Suppose not and $f(c) = 0$ for some $c\ne 2019$. Firstly, if $c = 0$, then $P(0,y)$ gives $f(2019y) = 0$, so $f$ is constant, absurd. Therefore $c\ne 0$. $P(c, x): f(cx) = f(2019x)$. Let $r = \frac{2019}{c}\ne 1$. We have $f(x) = f(rx)$ for all reals $x$. $P(rx, y): f(rx + y f(x)) + f(rxy) = f(rx) + f(2019y)$. Comparing with the original FE gives \[ f(x + y f(x)) = f(rx + y f(x))\] Now let $t\ne 0$ satisfy $f(t) \ne 0$. Note that $y f(t)$ is surjective over reals as $y$ varies, so \[ f(t + y) = f(rt + y) \implies f(x) = f(x + t(r-1)) \forall x\in \mathbb R\]Now let $d = t(r-1) \ne 0$. $P(x+d, y) - P(x,y): f((x+d) y) = f(xy) \implies f(xy) = f(xy + dy)$. Setting $x =0 $ gives $f(dy) = f(0)$, so $f$ is constant, a contradiction. $\square$ $P(x,1): f(x + f(x)) = 0\implies x + f(x) = 2019 \implies f(x) = 2019 - x$.

The answers are $f(x)=c$ for some $c\in \mathbb R$ $f(x)=2019-x$ $f(x)=0$ for all nonzero values $x$ and $f(0)=c$ for some $c\in\mathbb R$ all of which verifiably work. From now on, assume $f$ is neither the first nor the third, and we'll prove that $f$ must be identically $2019-x$. Let $P(x,y)$ denote the assertion $f(x + yf(x)) + f(xy) = f(x) + f(2019y)$. Claim: $f(2019)=0$. Proof. If we have $f(2019)\neq0$ then $P(2019, \tfrac{z-2019}{f(2019)})$ gives $f(z)=f(2019)$ for all reals $z$, which is a contradiction to $f$ being nonconstant. Claim: $f(x)=0\implies x=2019$ Proof. Suppose there exists $r\neq 2019$ such that $f(r)=0$. Then taking $P(r,x)$ over $x\in \mathbb R\setminus 0$ gives $f(rx)=f(2019x)=c$ such that $c\neq 0$. If such a value $x$ does not exist then $f$ is constantly zero over all values except at zero, contradiction. We have \begin{align*} P(rx,y)&\implies f(rx+cy)+f(rxy)=c+f(2019y) \\ P(2019x,y)&\implies f(2019x+cy)+f(2019xy)=c+f(2019y) \\ \end{align*}Note that $f(rxy)=f(2019xy)$ so $f(rx+cy)=f(2019x+cy)$. Fixing $x$ and varying $y$ tells us that $f$ is periodic with some period $k$. Then comparing $P(0,y)$ with $P(k,y)$ gives $f(yk)=f(0)$ for all $y$. Note that from this we can easily deduce that $f$ is constant, contradiction. Now, we have $P(x,1)$ giving $f(x+f(x))=f(2019)=0$ which implies $x+f(x)=2019$ for all $x$ giving $f(x)=2019-x$ as desired.

New year solve ! RMM 2019 P5 wrote: Determine all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying \[f(x + yf(x)) + f(xy) = f(x) + f(2019y),\]for all real numbers $x$ and $y$. The solutions to this functional equation is, \[f(x) = \boxed{\begin{cases} 2019-x \\ c, \ c \in \mathbb{R} \\ 0, \ x \neq 0 \ , \ f(0) = c \end{cases}} \]It can be checked that these work. First we plug in $y=1$ to get $f(x+f(x))=f(2019)$ and then, $x=2019,y=\frac{x-2019}{f(2019)}$ to get $f(x)=f(2019)=c \forall x \in \mathbb{R}$ in accordance with the condition that $f(2019) \not = 0$. Now consider the case when $f(2019)=0$ .Here we give a claim which forms the crux of the problem. Claim : $f(x)$ is injective at $2019$ or $f$ is constant at all points with $f(2019=0$. Proof : Let $k$ be a real such that $f(k)=0$ where $k\ge 0$ and $k \not = 2019)$ .The assertion $P(k,y)$ gives that $$f(ky)=f(2019y)$$Making plausible substitution gives , $$f(x)=f(rx)=f(r^2x)=\cdots=f(r^nx), r \in \mathbb{R}^+,r \ge 1$$Now the assertion $P(rx,y)$ gives, $$f(rx+yf(rx))+f(rxy)=f(rx)+f(2019y) \implies f(rx+yf(x))+f(xy)=f(x)+f(2019y)=f(xy)+f(x+yf(x))=f(xy)+f(rx+ryf(x))$$Plugging in $y=\frac{1}{f(x)}$ in the above equation gives, $$f(rx+1)=f(rx+r) \implies f\text{ is periodic with period r-1 }$$But for $R$ abruptly large we get $f(R+r-1)=f(R) \implies f(x)=f(\frac{R+r-1}{R}\cdot x)$.Which gives for $s \in (1,1+e)$ for $e\ge 0$ and small, $f(sx)=f(x) \forall s$.Which implies $f$ is constant other than at $f(0)$ which gives rise to the second solution .Otherwise any such $k$ does not exist i.e. the function is injective at $2019$ $\square$ Since the function is injective at $2019$ and we have $f(x+f(x))=f(2019)$ we get that , $$x+f(x)=2019 \implies f(x)=2019 -x$$Thus we have obtained all the solutions for the problem and hence WE DUN $\blacksquare$

The only solutions are $f(x) = c$, $f(x)=\begin{cases} 0 & x \neq 0 \\ c &x=0 \end{cases}$, and $f(x)=2019-x$. We have $P(2019, y) \implies f(2019+yf(2019)) = f(2019)$. If $f(2019) \neq 0,$ then $2019+yf(2019)$ can attain any real value, implying $f$ is constant. Hence, assume $f(2019)=0$ and $f \not\equiv 0$. Now, $P(x, 1) \implies f(x+f(x))=f(2019)=0$. If $2019$ is the only root of $f$, then we must must have $x+f(x)=2019 \implies f(x)=-x+2019$. Now, assume there exists another $z \neq 0 $ that is a root of $f$. Then $P(z, y) \implies f(yz)=f(2019y) \implies f\left( \frac{yz}{2019}\right)=f(y)$. Now, setting $P\left( \frac{xz}{2019}, y \right)$ and $P(x, y)$ equal gives us $f (\frac{xz}{2019}+yf(x))=f(x+yf(x))$. We will now prove $f$ is periodic. Let $T=x\left(\frac{z}{2019}-1 \right)$ be the period of $f$. Then $yf(x)$ can attain any real value, except for when $x=0$ for $f \equiv 0$. Now, $P(x+T, y) \implies f(xy+Ty)-f(xy) \implies f(c+Ty)-f(c)$. It follows easily that $f \equiv a$ for $x \neq 0$. Clearly, we must have $a=0$, yielding $f(x)=\begin{cases} 0 & x \neq 0 \\ c &x=0 \end{cases}$ as the last solution.