Let $P(x,y)$ denote the given assertion. $P(x,x) \implies f(x+f(x))=0$ $P(0,y) \implies f(f(0))=0$ $P(f(0),y) \implies f(0)f(f(0)+f(y))=(y-f(0))f(0)$ So $f(0)=0$ or $ f(f(0)+f(y))=(y-f(0))$ If $ f(f(0)+f(y))=(y-f(0))$, setting $y \rightarrow f(y)+y \implies f(y)=-y+f(0)$ and checking we see that $f(x)=-x+k$ truly satisfies the condition Suppose now $f(0)=0$ $P(x,0) \implies f(f(x))=-f(x)$ So $P(x,y)$ becomes $xf(x+f(y))=(x-y)f(x)$ So $P(x,f(y)+y)$ gives us $f(x)=0$ or $f(x)=-x$ Finally $f(x)=0, f(x)=-x+k$ where $k$ is a constant

Suppose firstly that $f(f(x)) = 0$ for all $x$. Then $f(x+f(y)) = 0$ for all $x\neq 0$. If $f(0) = 0$, then $y=0$ gives $f(x) = 0$ for all $x$ and this is a solution; if $f(0) \neq 0$, then $x=-f(y))$ gives a contradiction. So we may assume that $f(f(x)) \neq 0$ for some $x$. But then $f$ is injective, as if $f(y) = f(z)$, then $(y-x)f(f(x)) = (z-x)f(f(x))$ and so $y=z$. Now $x=0$, $y=1$ gives $f(f(0)) = 0$ while $y=x \neq 0$ gives $f(x+f(x)) = 0$ and hence $f(x) = f(0) - x$ for $x\neq 0$ (but this also trivially holds for $x=0$). Conversely, direct substitution shows that $f(x) = k-x$ for any constant $k$ is a solution.

Let $P(x,y)$ denotes the given assertion. Then $P(x,x)$ yields $f(x+f(x))=0$ for all $x\ne 0$. Further, $P(0,x)$ gives with $x\ne 0$ that $f(f(0))=0$. Combining, $f(x+f(x))=0$ holds for all $x$. Now, let there is an $x_0$ such that $f(f(x_0))\ne 0$. If $f(y_1)=f(y_2)$ for some $y_1,y_2$, then considering $P(x_0,y_1)$ and $P(x_0,y_2)$ we obtain $y_1=y_2$: $f$ is injective. Using $f(x+f(x))=0$, we get $x+f(x)=c$ for some constant $c$, which yields $f(x)=-x+c$. Lastly, suppose $f(f(x))=0$ for all $x$, so that $xf(x+f(y)) = 0$ for all $x,y$. Taking $y=f(r)$, we get $xf(x) =0$ for all $x$, so $f(x)=0$ for $x\ne 0$. Lastly, taking $x\ne 0$ and using $f(f(x)) = 0$, we get $f(0)=0$, too. So $f\equiv 0$ identically.

cretanman wrote: Find all functions $f\colon \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$, \[xf(x+f(y))=(y-x)f(f(x)).\] North Macedonia Let $P(x,y)$ denote the given assertion. 1)Let for all real $x$ $f(f(x))=0$ then from $P(x,y)$ we get $f(x+f(y))=0$ $x=a-f(y)$ $\rightarrow$ for all a $f(a)=0$ $f \equiv 0$ 2)For same $x$ $f(f(x))$$\neq$$0$ Claim:Function is injective Prove: Assume that for some $a$$\neq$$b$ $f(a)=f(b)$ then $( f(f(0))$$\neq$$0)$ from $P(x,a)$ and $P(x,b)$ we get $a=b$ contradiction $P(y,0)$ $\rightarrow$ $f(f(0))=0$ $P(f(0)-f(x),x)$ $\rightarrow$ $(x+f(x)-f(0))×f(f(f(0)-f(x)))=0$ for $x$$\neq$$0$ $f(f(f(0)-f(x)))$$\neq$$0$ because from injectivety we get $f(0)=f(y)$ but it is impossible then for all $x$ $x+f(x)-f(0)=0$ $\rightarrow$ $f(x)=-x+c$ (there $c=f(0)$)

My 100-th post ! The answers are $ f(x)=c-x$ for every x and $f(x)=0$ for every x which clearly work (c is an arbitrary constant) Let's prove these are the only ones. We have two cases Case 1: $f(f(x))=0$ for every x. Then our $P(x,y)$ becomes $f(x+f(y))=0 $ for every x not 0 and y real,but is easy to see it works also for x=0. But then let a be any real. Put $P(a-f(y),y)$ to get $f(a)=0$ so f is constant 0 function Case 2: There exists a real such that $f(f(a))$ is not 0. If we put x=0 we get that $f(f(0))=0$ so a is not 0 Then put $P(a,y)$ to get that f is bijective To finish put $P(x,x)$ to get that $f(x+f(x))=0$ for every x real( because for x=0 we proved already) and by injectivity we get f(x)=c-x and we are done !

Both $f(x) =-x+c$ and $f =0$ are solutions. Let's prove they are the only ones. Let $f(0)=c$, by $P(0,y)$ we get $f(c)=0$ , and by $P(x,x)$ we get $f(x+f(x))=0$. If there is a number $a \neq0$ such that $f(a)\neq0$ : applying $P(a,c)$ give us $f(f(a)) \neq 0$. then $P(a,y)$ give us that $f$ is injective so $x+f(x)=c$ thus $f(x)=-x+c.$ Else : for any $x\neq0$ we have $f(x)=0$, so if we suppose $c\neq0$ we can notice that $P(c,0)$ : $cf(2c)=-c^2$ then $c=0$ absurde, so $f(x) =0$ for any $x$.

Proposed by Nikola Velov

Put $y=x\neq0$ gives $f(x+f(x))=0$. Let $f(c)=0$. Case 1: $f(f(x))=0\forall x\in\mathbb{R}$. Then $xf(x+f(y))=0$ so $f(x)=0\forall x\neq -f(y)$. If f is nonconstant then $f(x)=0\forall x$, if constant then $f(x)=f(c)=0$ Case 2: $\exists x: f(f(x))\neq0$. $f(a)=f(b)$ by $P(x,a),P(x,b)$ imply $a=b$ so $f$ is injective. Assume $\exists t:f(t)\neq c-t$ then $$P(c-f(t),t)\Rightarrow(f(t)+t-c)f(f(c-f(t)))=0$$so there's at most one such $t=t_0$. $$P(\frac{t_0}2,t_0)\Rightarrow f(\frac{t_0}2+f(t_0))=f(f(\frac{t_0}{2}))\Rightarrow f(t_0)=c-t_0$$So either $f(x)=c-x$ or $f(x)=0$ for all $x\in \mathbb{R}$, both solutions fit.

Let $P(x,y)=xf(x+f(y))=(y-x)f(f(x))$ $P(0,x)\Longrightarrow xf(f(0))=0 \Longleftrightarrow f(f(0))=0 \text{, for } x\neq0$ (1) $P(x,x)\Longrightarrow xf(x+f(x))=0 \Longleftrightarrow f(x+f(x))=0 \text{, for } x\neq0$ (2) Claim: The function is injective iff $f(f(x))\neq0$ Proof: Let $f(a)=f(b)$ $P(x,a)$ yields: $xf(x+f(a))+xf(f(x))=af(f(x))$ $P(x,b)$ yields: $xf(x+f(b))+xf(f(x))=bf(f(x))$ Thus $af(f(x))=bf(f(x))$, which implies that the function is injective when $f(f(x))\neq0$ $\square$. Case 1: $f(f(x))\neq0$ From (1) and (2) we have that: $f(x+f(x))=f(f(0))$, however since the function is injective when $f(f(x))\neq0$, thus $x+f(x)=f(0)\Longleftrightarrow f(x)=-x+f(0)\Longrightarrow f(x)=-x+c$ where $c$ is a constant.

Case 2: $f(f(x))=0$ Since $f(f(x))=0$, the functional equation becomes $xf(x+f(y))=0$. Let $Q(x,y)=xf(x+f(y))$ $Q(x,f(x))$ yields: $xf(x)=0\Longleftrightarrow f(x)=0$ for $x\neq0$ $Q(-f(0),0)$ yields: $-f(0)^2=0\Longleftrightarrow f(0)=0$, thus $f(x)=f(0)=0$ which implies that $f(x)=0, \forall x \in \mathbb{R}$ So to sum up, $\boxed{f(x)=-x+c} \text{ and } \boxed{f(x)=0}, \forall x \in \mathbb{R}$ $\blacksquare$

Case 1. $f$ isn't injective: $\exists a\neq b\in\mathbb{R}$ such that $f(a) = f(b)$. Comparing $(x,a)$ and $(x,b)$ yields that $f(f(x)) = 0$ $\forall x\in\mathbb{R}$. Plugging in $(x,f(y))$ now gives $xf(x) = 0$ for all $x$, hence $f(x) = 0$ $\forall x\neq 0$. However, $(x,y) = (-f(0),0)$ implies that $-f(0)^2 = 0$ so $f\equiv 0$, which is indeed a solution. Case 2. $f$ is injective: Plugging in $(x,x)$ for $x\neq 0$ yields $f(x+f(x)) = 0$ and $(x,y) = (0,1)$ gives $f(f(0)) = 0$, so $f(x+f(x)) = 0$ for all $x$, hence $x+f(x) = y+f(y)$ $\forall x,y\in\mathbb{R}$ (due to the injectivity of $f$), whence $f(x) = c-x$ which works for all constants $c$.

First if F(a)=F(b) then a=b prove P(x,a) and P(x,b) ==> a=b or F(F(x))=0 if F(F(x))=0 ==> xF(x+F(y))=0 P(x-F(y) , y)==> F(x)=0 ∀ x∈R if a=b ==> P(0,y) and P(x,x) ==> F(x+F(x))=F(F(0))=0 ==> x+F(x)=F(0)=K F(x)=K-x and it's easy to verify that F(x)=0 orF(x)=K-x ∀ K∈R is the Solution.

why am i so scared of fes lmao... Anyways, $P(0,0)$ gives $f(f(0)) = 0$, now if $f(f(x)) = 0$ for all $x$, we have $f(x) = 0$, otherwise we can take $P(x,a),P(x,b)$ with $f(f(x)) \neq 0, f(a) = f(b)$ to get $(a-x)f(f(x)) = (b-x)f(f(x))$, so $a=b$. Then $P(x,x)$ gives $f(x + f(x) ) = 0 = f(f(0))$, so $x + f(x) = f(0)$, so $f(x) = k - x$. All functions of this form work, so we are done.

I denote by $P(x, y)$ plugging some $x$ and $y$ into the given equation. if $f(f(x)) = 0$ for all $x$ then: $P(x, f(y)): xf(x) = 0 \implies f(x) = 0$ for all $x$. now Suppose that there exists $a$ such that $f(f(a)) \neq 0$ $P(a, \frac{a+y}{f(f(a))}): af(...) = y$, in other words $f$ is surjective. $P(0, y): yf(f(0)) = 0 \implies f(f(0))=0$ $P(f(0)-f(y), y): 0=(y+f(y)-f(0))f(f(f(0)-f(y)))$ Now suppose that $f(f(b)) = 0$ for some $b$. $P(b, y): bf(b + f(y)) = 0$ Case 1: $b \neq 0$ $bf(b + f(y)) = 0 \implies f(x) = 0$ since $f$ is surjective. Case 2: $b = 0$ $bf(b + f(y)) = 0 \implies 0=0$ Now let's focus on nonzero solutions, we know that for some $y$: $y + f(y) - f(0) = 0 \Leftrightarrow f(y) = -y + f(0)$ or $f(0) - f(y) = 0 \Leftrightarrow f(y) = f(0)$ suppose that there exists some $c \neq 0$ for which $f(c) = f(0)$ $P(c, y): cf(c + f(y)) = 0 \implies f(x) = 0$ since c is not zero and $f$ is surjective. But since we are looking for nonzero solutions, such c does not exist. so $f(y) = -y + f(0)$ or $f(y) = 0$ for all $y$

Let $P(x,y)$ be the assertion $xf(x+f(y))=(y-x)f(f(x))$. We first make some observations: - $P(x,x)$ yields $xf(x + f(x)) = 0 \ \forall x$, so $f(x+f(x)) = 0 \ \forall x \neq 0$. - $P(x,x + f(x))$ yields \[ x f\big(x + f(x+f(x))\big) = \big((f(x) + x) -x) f(f(x)) \iff xf(x) = f(x) f(f(x)) \]if $x \neq 0$. Then either $f(x) = 0$ or $x = f(f(x))$. Note that $f(x) = 0$ is a solution. We continue from $f(f(x)) = x$. If $x\neq 0$, the original equation becomes $xf(x + f(y)) = (y-x)x \implies f(x + f(y)) = y-x. \quad (\star)$. Claim. $f(f(0)) = 0$. Proof. Pick $a\neq 0$ such that $a + f(a) \neq 0$. Then $f(a+f(a)) = 0$ by the first observation. Thus $f(f(0)) = f(f(f(a+f(a)))) = f(a + f(a)) = 0$ by observation two. Otherwise, $f(x) = -x$ and the result is direct. By the claim, $(\star)$ and $y = f(0)$ gives \[ f(x) = f(0) - x. \]Hence $f(x) = c-x$ and $f(x) = 0$ are the only possible solutions. A direct substitution checks that they are indeed valid solutions.

Solution: The answers are $f \equiv -x+c$ for some real constant $c$ and $f \equiv 0$. Clearly both of them work and we now proceed to show the other direction. Denote $P(x,y)$ as the assertion to the functional equation. It is easy to see $f \equiv 0$ is the only constant solution. Henceforth, assume that $f$ is non-constant. If $f(a) = f(b)$ for $a,b \in \mathbb{R}$, then comparing $P(x,a)$ and $P(x,b)$ would give $a=b$(note that we used $f$ non-constant here) proving injectivity. Surjectivity follows immediately after fixing $x$ and varying $y$ on the LHS. This means $f^{-1}$ exists. $P(x,x)$ would give $f(x+f(x)) = 0$ for $x \ne 0$. Due to injectivity, we get that $f(x) = -x + c$ for all $x \ne 0$. Finally, it remains to prove $f(0) = c = f^{-1}(0) \iff f(f(0)) = 0$. This can be shown by $P(0,x)$ easily and we're done. $\blacksquare$

Assume that $f(a)=f(b)$. $P(x,a)$ and $P(x,b)$ we have: $(a-x)f(f(x)) = (b-x)f(f(x))$ either $f(x) = 0 $ for all $x$. or $f$ is injective. $P(x,x) \implies f(x) =c-x .$

Call the assertion $p(x,y)$. $f\equiv0$ satisfies. let $f$ be a non-constant function. $p(0,1)$ gives $f(f(0))=0$. $p(x,x)$ ($x\ne0$) gives $f(x+f(x))=0$. We show that the function is injective. Take $x$ such that $f(f(x))\ne0$. {Or else $f\equiv0$.) Then, $p(x,y_1)$ and $p(x,y_2)$ gives $y_1=y_2$. Since $f(x+f(x))=0=f(f(0))$, we have $f(0)=x+f(x)\Longrightarrow f(x)=f(0)-x$. Re-checking in the original equation, we see $xf(x+f(y))=xf(x+f(0)-y)=x(y-x)=(y-x)(f(0)-(f(0)-x))=(y-x)f(f(x))$. So, the solution set is $\boxed{f\equiv0}$ and $\boxed{f(x)=c-x\forall x\in\mathbb{R}}$, where $c\in\mathbb{R}$ is a constant.

y-1. We find : 1st solution f(x)=0 x-0. _____f(f(0))=0_____. X-0 yf(f(x))=0 x-› f(0)-f(y) (f(0)-f(y))(f(f0))=(f(y)-f(0)+y)(f(f(f(0)-y))) || 0. f(0)-y≠f(0) y≠0 —> f(f(0)-y)≠0 f(y)=f(0)-y Checking:true

The only solutions are $f(x)=0$ and $f(x)=c-x$ for all reals $c$. These can be directly verified. Let $P(x,y)$ be the given assertion. Case 1: $f(0)=0$ If $f(f(x))=0$ for all $x$ then $xf(x+f(y))$ is $0$ for all $x$ and $y$. By varying $x\neq 0$ and from the fact $f(f(y))=0$ we get that $f$ is the zero function. Assume otherwise that $f(f(x))$ attains a non zero value. Then varying $y$ gives that $f$ is surjective. If there exists a $x\neq 0$ such that $f(x)=0$ then varying $f(y)$ gives that $f$ is the zero function. Then $P(x,x)$ gives $x+f(x)=0$ for all $x\neq 0$ implying that $f(x)=-x$. Case 2: $f(0)\neq 0$ By taking $P(x,x)$ we get $f(x+f(x))=0$ for all $x\neq 0$, thus $f(x)$ must be zero for some $x$. Let $a$ and $b$ be such that $f(a)=f(b)=0$ then $P(a,b)$ gives $a=b$. Let $c$ be the unique root of $f$. Then from $P(x,x)$ for all $x\neq 0$ we must have $f(x)=c-x$. $P(0,y)$ gives that $f(0)=c$.

We claim the only functions are $\boxed{f(x)\equiv 0}$ and $\boxed{f(x)\equiv c-x}$ for $c\in\mathbb{R}$. It is easy to check that these work. Let $P(x,y)$ denote the given assertion. $P(0,1)$ gives $f(f(0))=0$. If $f(f(x))\equiv 0$, $P(x-f(0),0)$ yields $f(x)=0$ if $x\neq f(0)$. The case $x=f(0)$ was covered already. Hence $f(x)\equiv 0$. So assume there exists $k$ such that $f(f(k))\neq 0$. Then $P(k,x)$ implies that $f$ is injective. Now $P(x,x)$ for $x\neq 0$ gives $f(x+f(x))=0=f(f(0))$. By injectivity, we have $f(x)=f(0)-x$. This also agrees with the $x=0$ case so we are done. $\square$

Let $P(x,y)$ denote the assertion given in the question. $P(x,x)$ gives us that $f(x+f(x)) = 0$ for all $x \ne 0$ $-------(*)$ $P(0,y)$, where $y$ is positive, gives us $f(f(0))=0$ $-------(**)$ Now assume, $f(x_1) = f(x_2)$ for some $x_1,x_2$, then for some $x \ne 0$, $\Longrightarrow xf(x+f(x_1))=xf(x+f(x_2))$ $\Longrightarrow f(f(x))(x_1-x) = f(f(x))(x_2-x)$ which means that $f(f(x)) = 0 \ \ \forall x$ or $x_1 = x_2$ First lets consider the case when $f(f(x)) = 0$ for all $x$, then from $P(x,y)$, we have $xf(x+f(y)) = 0$ $\Longrightarrow x=0$ or $f(x+f(y)) = 0$. We can vary $x$ above to get all real values except $f(f(y))$ for a fixed $y$ and if $x=0$, then we also get $f(f(y)) = 0$ by our assumption. Hence, we get the solution $f(x) = 0 \ \ \forall x \in \mathbb{R}$ Considering the second case, $f$ becomes injective in this case. From $(*)$ and $(**)$, we get that $f(x) = c-x, c \in \mathbb{R}$ which satisfies our original equation. Hence, the only solutions we get are $f(x)=0$ or $f(x) = c-x \ \ \forall x \in \mathbb{R}$.

My first comment claim) $f$ is injective proof) assume $a,b$ such that $a\neq b$ and $f(a)=f(b)$ $P(x,a)-P(x,b)$ will give that $a=b$. $P(x,x) \rightarrow xf(x+f(x))=0$…. (1) $P(0,x) \rightarrow xf(f(0)=0$…. (2) From (1,2) we get $xf(x+f(x))=xf(f(0) ) \rightarrow x+f(x)=f(0)$ $\rightarrow f(x)= c-x$ by plugging $P(x-f(x),x)$ in (1) we get $f(x)=0$ so we deduce that $f(x)=0$ and $f(x)=c-x$ are only solutions,

Cute FE. $P(x,y):xf(x+f(y))=(y-x)f(f(x))$ Case 1. $f(f(x)) = 0\ \forall x\in \mathbb R$. We have that $xf(x+f(y)) = 0$. If $f$ is constant, then $f \equiv 0$. If not, there exists $a\neq b \in \text{Im}f$. Take $y$ such that $f(y) = a$ and vary $x$ such that $x+a$ covers $\mathbb R$ and for the case where $x=0$ just swap $a$ with $b$ and take $x=a-b \neq 0$. Hence we get $f \equiv 0$. Case 2. there exists $a$ such that $f(f(a)) \neq 0$. Suppose that there exists $y_{1}\neq y_{2}$ with $f(y_{1})=f(y_{2})$. From $P(a,y_{1})$ and $P(a,y_{2})$ we get a contradiction. So $f$ is injective. $P(0,1) \Rightarrow f(f(0)) = 0$. And from $P(x,x)$ with $x\neq 0$ we get that $f(x+f(x)) = 0$ so from injectivity we have $x+f(x)=f(0)$. Notice that $0$ also satisfies this relation so we get that $f(x)=c-x$ which does satisfy the FE.

Let $P(x,y)$ denote the given assertion. Now note that: $P(x,x)$ gives $xf(x+f(x))=0$ $P(0,x)$ gives $f(f(0))=0$ Moreover, $P(f(0),x)$ and $P(0,x)$ give $f(0)f(f(0)+f(x))=(x-f(0))f(0)$, therefore we check the two cases $f(0)=0$ and $f(0)\neq0$. Case 1: $f(0)=0$ $P(x,0)$ gives $f(x)=-f(f(x))$ $P(-f(x),x)$ and $P(x,0)$ give $0=(x+f(x))(f(-f(x))$ Case 1.1: $x+f(x)=0$, hence $f(x)=-x, \forall x\in\mathbb{R}$ Case 1.2: $f(-f(x))=0$. $P(x,0)$ and $P(x,-f(x))$ give $xf(x)=f(x)^2+xf(x)$ so $f(x)\equiv 0$. Case 2: $f(0)\neq 0$ and $f(f(0)+f(x))=x-f(0)$ (which in fact also shows surjectivity). Let $f(0)=a$, then from $f(f(0))=0$ we know $f(a)=0$. Claim: $f$ is injective. Proof: Suppose there exist $b, c \in\mathbb{R}$ s.t. $f(b)=f(c)$. Now look at $P(a,b): a(b-a)=af(a+f(b))=af(a+f(c))=a(c-a)$, therefore $b=c$ and $f$ is injective. Looking back at $P(x,x)$ and using injectivity finally shows $x+f(x)=a$ or $f(x)=a-x$. We conclude that $\boxed{f(x)=c-x}$ for some constant $c$ and $\boxed{f(x)\equiv 0}$ are the only solutions. $\square$

Lunch break!

Let $P(x,y)$ be the assertion of $xf(x+f(y))=(y-x)f(f(x))$ \(Our\) \(claim\) \(is\) $f \equiv 0$ or $f(x)=u-x$ for some constant $u$ Now $f \equiv 0$ is obvious We assume $f \not\equiv 0$ Now $P(0,y)$ gives us $y f(f(0))=0$ so $f(f(0))=0$ Let $f(0)$ \(be\) $u$ then $f(u)=0$ Now $P(x,x)$ gives us $xf(x+f(x))=0$ then $f(x+f(x))=0$ Now as $f \not\equiv 0$ there exists some $c \ne 0$ such $f(f(c)) \ne 0$ Now $P(c,y)$ gives us $cf(c+f(y))=(y-c)f(f(c))$ proving its bijectivity Now then $f(x+f(x))=0=f(u)$ Then $x+f(x)=u$ then $f(x)=u-x$ for some constant $u$

Let $f$ be a solution to this equation. Letting $x = y$ gives $xf(x + f(x)) = 0$. Letting $x = 0$ and $y \neq 0$ gives $f(f(0)) = 0$. Let's then take $y$ such that $f(y) = 0$ : we get $xf(x) = (y-x)f(f(x))$. We are going to distinguish two cases : - if there exists $x$ such that $f(f(x)) \neq 0$, then $y = \frac{xf(x)}{f(f(x))} + x$. So if there are two different $y$ such that $f(y) = 0$ then from this equation these two $y$ should be equal, absurd. Therefore there exists at most one $y$ such that $f(y) = 0$ : we have already established that $f(f(0)) = 0$, so this $y$ must be $f(0)$. For $x \neq 0$, we have $f(x+f(x)) = 0$ from the $x=y$ equation. So $x+f(x) = f(0) \Longrightarrow f(x) = -x + f(0)$. Combining with $f(0) = -0 + f(0)$, we can affirm that $f$ is of the form $- x + k$ where $k$ is a constant. - if not, this means that $f(f(x)) = 0$ for all $x$. Therefore, $xf(x) = 0$. Clearly for $x \neq 0$, we necessarily have $f(x) = 0$. By the way, with $y = 0$, we have $xf(x+f(0)) = 0$ since $f(f(x)) = 0$. Letting $x = -f(0)$ we have $-f(0)^2 = 0$ so $f(0) = 0$. In summary, $f \equiv 0$. To conclude there are two potential solutions : $f(x) = 0$ and $f(x) = -x + k$ where $k$ is a constant. We easily check that there are indeed solutions.

$P(x,x)$ and $P(0,x)$ gives that $x+f(x)$ is always annihilated by $f$, then $P(x,x+f(x))$ yields that $$ xf(x)=f(x)f(f(x)) $$giving either $f(x)=0$ or $f(f(x))=x$ for each $x$. If there is $x \neq 0$ such that the latter hold, $f$ is injective, giving $f(x) \equiv c-x$, otherwise we have the constant $0$ function everywhere except $0$ and a quick check reveals at $0$ too, both of which works.

Let the problem statement be denoted by $P(x,y)$, we have that \[P(x,x): xf(x+f(x))=0 \implies f(x+f(x))=0, \ \ \ \forall x \in \mathbb{R*}\]wherein $\mathbb{R*}$ is reals without $0$. Now we have \[P(-f(y),y): -f(y)f(0)=(y+f(y))f(f(-f(y)))\]where letting $y \ne 0$ we would have, $f(y)f(0)=0$ forcing either $f(0)=0$ otherwise $f(y) \equiv 0 \ \ \forall y \in \mathbb{R*}$ Now if we have $f \equiv 0$ that acts as a solution, assuming there are more we must have some $f(f(x)) \neq 0$ otherwise we have, \[xf(x+f(y))=0 \ \forall (x,y) \in \mathbb{R} \implies xf(x+f(f(0)))=0 \implies xf(x)=0 \implies f(x)=0 \ \forall x \in \mathbb{R*}\]however for some $y \ne 0$, we must have that $0=f(f(y))=f(0)$ which indicates identically $0$ a direct contradiction to our assumption. Now we have established our claim so let $f(y)=f(y'), \ y \ne y'$ then we compare $P(x,y)$ and $P(x,y')$ to obtain that $f$ is injective. However since \[P(0,1): f(f(0))=0 \ \land \ f(x+f(x))=0 \ \ \forall x \in \mathbb{R*}\]so we must have $f(x)+x=f(0)$ for some constant $f(0)$ and thus $f(x)=f(0)-x \ \forall x \in \mathbb{R*}$ however we must also have that, $f(0)=f(0)-0$ and so we have $f \equiv f(0)-x \ \ \forall x$. Thus our solutions are $\boxed{f \equiv 0; f(x)=f(0)-x \ \ \forall x}$ which are trivial to check that they work.

Denote by $P(x, y)$ the given assertion. For an $y \neq 0$ $P(0, y): f(f(0)) = 0$ For an $x \neq 0$ $P(x, x): f(x + f(x))=0$ Case 1: $f$ is injective $\Rightarrow x + f(x) = f(0)$ so $f(x) = f(0) - x, \forall x \in \mathbb{R} $ Case 2: $f$ is not injective Then we can find $a, b \in \mathbb{R}$ such that $a \neq b$ and $f(a) = f(b)$ Then for any $x$ such that $x \neq a, x \neq b, x \neq 0$ we have: $P(x, a) : xf(x + f(a)) = (a-x)f(f(x)) $ $P(x, b): xf(x + f(b)) = (b-x)f(f(x)) $ Assume LHS $\neq 0$ then RHS $\neq 0$. It follows that $a-x = b-x$ which is false, by the way we picked $x$. So LHS $= 0$ therefore $f(x + f(a)) = 0$. $\forall x$ such that $x \neq a, x \neq b, x \neq 0$. We are only left to check if $f(a + f(a)), f(b+ f(b)), f(f(a))$ are $0$. By $P(a,a)$ and $P(b,b)$ we get $af(a + f(a)) = bf(b + f(b))= 0$ If $a, b \neq 0$ we are done. So assume one of them is $0$. If $a = 0$: $\Rightarrow f(b + f(0)) = 0$ and by the first observation $f(f(0)) =0$. We also have $f(x + f(0)) = 0$ for all $x \neq 0, x\neq b$. So basically, $f \equiv 0$. $\blacksquare$

nice problem for P1 level of BMO! Clearly $\boxed{f(x)=0\quad\forall x\in\mathbb{R}}$ is a solution, so we'll now look for all nonzero solutions. Let $P(x,y)$ be the assertion of the functional equation given. Take $x=y$, so $xf(x+f(x))=0$. For $x\ne 0$, the equation yields $f(x+f(x))=0\dots(1)$. Now taking $P(0,x)$ easily gives us that $xf(f(0))=0$, which is $f(f(0))=0\dots(2)$ for $x\ne 0$. Suppose there exists $a,b\in\mathbb{R}$ so that $f(a)=f(b)$. Hence, by comparing $P(1,a)$ and $P(1,b)$ yields $a-1=b-1$ $\Longleftrightarrow$ $a=b$ $\Longleftrightarrow$ $f$ is injective. So by applying injectivity at $(1)$ and $(2)$ will implies $x+f(x)=f(0)$ $\Longleftrightarrow$ $f(x)=f(0)-x$. Now by letting $f(0)=c$ for some $c\in\mathbb{R}$, we easily get $\boxed{f(x)=c-x\quad\forall x\in\mathbb{R}}$, which is indeed a solution. Therefore, we are done. $\blacksquare$

P(1,0) implies that f(f(0))=0.if f(f(x)) isn't constant then f is injective .And P(x,x) implies f(x+f(x))=0=f(f(0)) and we get f(x)=c-x from injectivity.If f(f(x)) is constant f(f(x))=0 and y>>f(0) implies f(x)=0 Answers: f(x)=0 and f(x)=c-x.

Suppose $f(f(x))=0$ holds for all $x$. Then if $x\neq 0$, we have $f(x+f(y))=0$. Fix $y=y_0$. Then $f(f(y_0))=0$, and $f(f(y_0)+x)=0$ for $x\neq 0$ but $x+f(y_0)$ along with $f(y_0)$ produce entire set of reals. Hence $f$ is $0$. We check easily that this works. Now, subbing $x=0$ gives $0=yf(f(0))$. Pick $y$ nonzero, then $f(f(0))=0$. Hence, assume that $f(f(x_0))\neq 0$ for some $x_0\neq 0$. Substituting that for $x$, we have \[f(c+f(y))=c'(y-x_0)\]for constants $c',c,x_0$ all non-zero. $f$ is injective as there is an isolated $f(y)$ on the LHS, noting $c'\neq 0$. $f$ is also surjective because shifting and scaling the reals still gives the reals on the RHS. Hence, there is an unique $k$ such $f(k)=0$. Letting $x=y$ initially, we have $xf(x+f(x))=0$, so when $x\neq 0$, then $f(x+f(x))=0=f(k)$, injectiveness yields $f(x)=k-x$. Of course, when $x=0$, then $f(0)=k=k-0$ too. One checks this works easily.