$P(x,0): f(f(x)+f(0))=xf(x)$- hence notice that if $f(x)=f(t) \neq 0 => x=t$. $P(0,0): f(2f(0))=0$. $P(0,2f(0)): f(f(0)+f(2f(0)))=2f(0)f(2f(0))=>f(f(0))=0$. $P(f(0),f(0)): f(0)=2f(0) \cdot f(2f(0)) =0 =>f(0)=0$. $P(x,0): f(f(x))=xf(x)$. $P(x+y,0): f(f(x+y))=(x+y)f(x+y)=f(f(x)+f(y))$. Now consider all $x,y$ such that $f(x) $ and $f(y) \neq 0$ $=>$ By the first statement, $f(x+y)=f(x)+f(y)$. Putting in $(f(x),f(y)) (\neq 0)$ into this gives us $f(f(x)+f(y))=f(f(x))+f(f(y))=xf(x)+yf(y)$. But, by problem statement, $f(f(x)+f(y))=(x+y)(f(x)+f(y))=xf(x)+xf(y)+yf(x)+yf(y) => xf(y)+yf(x)=0=>f(y)=- \frac{yf(x)}{x}$. Putting $x=y => f(x)=-f(x)=>$ Contradiction to $f(x) \neq 0$. Hence $f \equiv 0$ is the only solution.

Note

ubermensch wrote: ... $P(x+y,0): f(f(x+y))=(x+y)f(x+y)=f(f(x)+f(y))$. Now consider all $x,y$ such that $f(x) $ and $f(y) \neq 0$ $=>$ By the first statement, $f(x+y)=f(x)+f(y)$ No, in order to conclude $f(x+y)=f(x)+f(y)$, the constaint is $f(f(x+y))$ (or $f(f(x)+f(y))$) different from zero, and not, as you wrote, $f(x)\ne 0$ and $f(y)\ne 0$

As above, plugging in $P(0, 0)$, $P(0, 2f(0))$, and $P(f(0), f(0))$ one could obtain $f(0) = 0$, and $P(x, 0)$ yields: $$ f(f(x)) = x f(x) \; \forall x\in \mathbb{R} \ldots(1) $$ By the equation $(1)$, if there exists real numbers $x$ and $y$ such that $f(x) = f(y) \neq 0$, then $xf(x) = yf(y)$ and thus $x = y \neq 0$ (otherwise $f(x) = f(y) = 0$). Also, we know that $f(f(x)) = 0$ if and only if $f(x) = 0$. From this on, first suppose that there exists a real number $t \neq 0$ such that $f(t) = 0$. Then, $P(x, t)$ with $(1)$ yields: $$ f(f(x)) = f(f(x+t)) \; \forall x\in \mathbb{R} $$ Suppose that $f(f(x)) \neq 0$ for some $x\in \mathbb{R}$. Then, applying the above property we just obtained twice, we get $x = x + t$; a contradiction. Hence, $f(f(x)) = 0$ for all $x\in \mathbb{R}$, and this implies $f\equiv 0$. Now, we claim that $t = 2$ works; i.e. $f(2) = 0$. $P(1, 0)$ yields $f(f(1)) = f(1)$, so either $f(f(1)) = f(1) = 0$ or $f(1) = 1$. Meanwhile, $P(1, 1)$ yields $$ f(2f(1)) = 2f(2) $$The former implies $2f(2) = f(0) = 0$, while the latter implies $f(2) = 2f(2)$. Both yields $f(2) = 0$, as desired.

We have that the only answer is $f \equiv 0$. Let $P(x,y)$ denote the given assertion.From $P(x,-x)$ we have that $f(f(x)+f(-x))=0$.This implies that $f(2f(0))=0$. From $P(x,0)$ we have that $f(f(x)+f(0))=xf(x)$. So let $S$ be the set of all given reals so that $f(x)=0$, i.e. $S=\{ x|f(x)=0 \}$, and assume that the $card(S)\geq 2$. If we set $x \in S$ in $f(f(x)+f(0))$ we get that $f(f(0))=0$. Now from $P(x,f(0))$ we have that $f(f(x))=(x+f(0))f(x+f(0))$, and setting $x=f(0)$ we get that $f(0)=0$. From this we have that $f(f(x))=xf(x)$.Thus our assertion is transformed into $f(f(x)+f(y))=f(f(x+y))$. Now define set $A$ to be $A=\mathbb{R} \backslash S$. If $x \in A$ from $f(f(x))$ we get that $f$ is injective whenever $x \in A$. Now set $x+y \in A$, we get that $f(x)+f(y)=f(x+y)$. If $x,y \in S$, we have that $f(x+y)=0$, which leads to a contradiction. If $x \in S$, we have that $f(x+y)=f(y)$, by injectivity we must have that $x+y=y \implies x=0$, but we have that $card(S) \geq 2$, so we get a contradiction. Thus we have that if $x+y \in A$, we must have that $x,y \in A$. But since we have that $x$ and $y$ are arbitrary we have that $A= \mathbb{R}$, but that is impossible since $card(S)=2$. Thus we have that $S=\mathbb{R}$. Thus we have that $ f \equiv 0$. Now assume that $card(S) = 1$, meaning that $f$ is injective at $x=0$. Notice, again that $f$ must be injective because of $f(f(x))=xf(x)$. Meaning that, by injectivity, we must have that $f(x)+f(y)=f(x+y)$, i.e. $f$ is additive on $\mathbb{R}$. This implies that: $$f(f(x)+f(y))=f(f(x))+f(f(y))=xf(x)+yf(y)=(x+y)f(x+y)$$By setting $x=y$ we get that $2xf(x)=2xf(2x)$, which implies that $x=2x$, but this only holds when $x=0$, thus we get a contradiction.

First observe that f is clearly bijective. Now assume that $f(x)$ is not$=0.$ Let $P(x, y)$ be the assertion $P(0,0)$ gives $f(2f(0))=0$ and by Surjectivity, there exist some $a$ for which $f(a)=0.$ $P(0,a)$ gives $f(f(0))=0$ and then using injectivity we have: $f(0)=0.$ $P(x,0)$ gives $f(f(x))=xf(x),$ $P(x,-x)$ gives $f(f(x)+f(-x))=f(0)$ which gives: $f(x)=-f(-x)$ $P(0,-x)$ gives, $f(f(-x))=-xf(-x)$ which is equivalent to: $f(-f(x))=f(f(x))$ and then using injectivity we have, $f(x)=-f(x)$ a contradiction. Hence The only solution to this equation if $f(x)=0.$

Let $P(x,y)$ denote the assertion $f(f(x)+f(y))=(x+y)f(x+y)$. $P(0,0)\Rightarrow f(2f(0))=0$ $P(2f(0),0)\Rightarrow f(f(0))=0$ $P(f(0),f(0))\Rightarrow f(0)=0$ $P(x,0)\Rightarrow f(f(x))=xf(x)$ If $f(a)=f(b)\ne0$, then we have $a=b$. Assume some nonzero $k$ exists with $f(k)=0$, then: $P(x,k)\Rightarrow f(f(x))=f(f(x+k))$ If $f(f(x))\ne0$ for some $x$, then $x=x+k$, contradiction. Thus we assume $f(f(x))=0$ for all $x$, or $xf(x)=0$. This gives the solution $\boxed{f(x)=0}$. Otherwise, assume $f(k)=0\Rightarrow k=0$. Then our statement of partial injectivity becomes: If $f(a)=f(b)$ and $a\ne0$, then $a=b$. Motivated by this, set $x\ne0$: $P(y,x-y)\Rightarrow f(f(y)+f(x-y))=f(f(x))\Rightarrow Q(x,y):f(x-y)=f(x)-f(y)$ $Q(0,x)\Rightarrow f(-x)=-f(x)$ $Q(x,-y)\Rightarrow f(x+y)=f(x)+f(y)$ $P(1,1)\Rightarrow f(1)=0$ $P(x,1)\Rightarrow f(x)=0$, contradiction with the assumption that $f(x)\ne0\forall x\ne0$.

OGGY_666 wrote: First observe that f is clearly bijective. Now assume that f(x) is not=0. Let P(x, y)be the assertion P(0,0) gives f(2f(0))=0 and by Surjectivity, there exist some a for which f(a)=0. P(0,a) gives f(f(0))=0 and then using injectivity we have: f(0)=0. ... Hence The only solution to this equation if f(x)=0. Uhhh? If $f$ is bijective, how can the only solution be $f(x)=0$ which is obviously not bijective?

@ above See my second statement I have assumed that f(x) is not =0. Infact, Subtracting P(b,0) from P(a,0) gives us two condition that either f is 0 or f is injective. And for f being surjective, P(x,-x) gives that f(f(x)+f(-x))=0, which is comes out from the fact that f(0)=0 which has come from f being injective.

OGGY_666 wrote: @ above See my second statement I have assumed that f(x) is not =0. Infact, Subtracting P(b,0) from P(a,0) gives us two condition that either f is 0 or f is injective. And for f being surjective, P(x,-x) gives that f(f(x)+f(-x))=0, which is comes out from the fact that f(0)=0 which has come from f being injective. No. You are "begging the question." A function whose solution is only $f(x)=0$ can never be "clearly surjective." Or maybe you mean "surjective at zero" (?)

Denote the assertion by $P(x,y).$ $P(0,0)$ gives $f(2f(0))=0.$ Then $P(2f(0),0)$ gives $f(f(0))=0.$ Finally $P(f(0),f(0))$ yields $f(0)=0.$ $P(x,0)$ implies $f(f(x))=xf(x)$ so if $f(a)=f(b)\neq 0$ then $a=b.$ Also note that $f(f(x))=0\implies f(x)=0$ this works. Otherwise, suppose $f(a)=0$ then $P(x,a)$ yields $a=0.$ Now $P(x,-x)$ gives $f(-x)=-f(x).$ And $P(-x,0)$ gives $f(f(x))=f(-f(x))$ implying $f\equiv 0.$

Solved with some basic subtitutions

Vlados021 wrote: Find all functions $f:\mathbb R\to\mathbb R$ satisfying the equality $$ f(f(x)+f(y))=(x+y)f(x+y) $$for all real $x$ and $y$. (B. Serankou) $\color{blue}\boxed{\textbf{Answer: }f\equiv 0}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$f(f(x)+f(y))=(x+y)f(x+y)...(\alpha)$$In $(\alpha) y=0:$ $$\Rightarrow f(f(x)+f(0))=xf(x)...(\beta)$$$$\Rightarrow f \text{ is inyective}...(I)$$In $(\alpha) x=y=0:$ $$\Rightarrow f(2f(0))=0...(II)$$In $(\alpha) x=2f(0), y=0:$ By $(I):$ $$\Rightarrow f(f(0))=f(0)...(III)$$By $(I),(II)$ and $(III):$ $$\Rightarrow f(0)=2f(0)$$$$\Rightarrow f(0)=0$$In $(\alpha) y=-x:$ $$\Rightarrow f(f(x)+f(-x))=0$$By $(I):$ $$\Rightarrow -f(x)=f(-x)$$$$\Rightarrow f \text{ is odd}...(IV)$$In $(\beta) x=-x:$ By $(IV):$ $$\Rightarrow f(f(x))=-xf(x)$$By $(\beta):$ $$\Rightarrow xf(x)=0$$$$\Rightarrow \boxed{f\equiv 0}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

We uploaded our solution https://calimath.org/pdf/BelarusMO2019-7.pdf on youtube https://youtu.be/3gd-E43sAbc.

Posting for storage, sorry if this is a repeat solution \begin{align*} P(0,0)&: f(2f(0))=0 \\ P(2f(0),0)&: f(f(0))=0 \\ P(f(0),f(0))&: \boxed{f(0)=0} \\ P(x,0)&: f(f(x)=xf(x) \implies f \text{ is injective or zero everywhere} \\ P(x,-x)&: f(x)+f(-x)=0 \\ P(-x,0)&: f(f(-x))=-xf(-x)=xf(x)=f(f(x)) \implies f(x)=f(-x) \\ \end{align*}Thus, $f(x)=0$.

navier3072 wrote: Posting for storage, sorry if this is a repeat solution This is not a repeat solution. Just a wrong new one. navier3072 wrote: $ f(f(x)=xf(x) \implies f \text{ is injective or zero everywhere} $ No. This means that $f(a)=f(b)$ implies either $f(a)=f(b)=0$, either $a=b$, which is quite different. Example, the function $f(x)=0$ $\forall x\le 0$ and $f(x)=x^{\frac{1+\sqrt 5}2}$ $\forall x>0$ is neither allzero, neither injective and matches $f(f(x))=xf(x)$ $\forall x$

pco wrote: navier3072 wrote: Posting for storage, sorry if this is a repeat solution This is not a repeat solution. Just a wrong new one. navier3072 wrote: $ f(f(x)=xf(x) \implies f \text{ is injective or zero everywhere} $ No. This means that $f(a)=f(b)$ implies either $f(a)=f(b)=0$, either $a=b$, which is quite different. Example, the function $f(x)=0$ $\forall x\le 0$ and $f(x)=x^{\frac{1+\sqrt 5}2}$ $\forall x>0$ is neither allzero, neither injective and matches $f(f(x))=xf(x)$ $\forall x$ as you mentioned there are a lot of solutions with lack of proof