oVlad wrote: Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that \[f(x(x + f(y))) = (x + y)f(x),\]for all $x, y \in\mathbb{R}$. $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ is a solution. So let us look from now for non allzero solutions. Let $P(x,y)$ be the assertion $f(x(x+f(y)))=(x+y)f(x)$ $P(0,2)$ $\implies$ $f(0)=0$ If $f(u)=0$ for some $ u$ : subtracting $P(x,0)$ from $P(x,u)$, we get $uf(x)=0$ $\forall x$ and so $u=0$ $P(-x,x)$ $\implies$ $f(-x(-x+f(x)))=0$ $\implies$ $x(f(x)-x)=0$ $\forall x$ $\implies$ $\boxed{\text{S2 : }f(x)=x\quad\forall x}$, which indeed fits.

oVlad wrote: Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that \[f(x(x + f(y))) = (x + y)f(x),\]for all $x, y \in\mathbb{R}$. Let $P(x,y)$ be the assertion 1)İf for all $x$ $f(x)=0$ then $\boxed{f(x)=0\quad\forall x}$ 2) Assume that here exits such $x$ that $f(0) \neq 0$ Claim:$f(x)$ is injective. Proof: Assume that for $a \neq b$ $f(a)=f(b)$ then from $P(x,a)$ and $P(x,b)$ (there $f(x) \neq 0$) we get $a=b$ contradiction. From $P(0,y)$ $\rightarrow$ $f(0)=0$ P(-f(x),x) $\rightarrow$ $(f(x)-x)f(-f(x))=0$ from injectivety for $x \neq 0$ $f(-f(x)) \neq 0$ then we get for all $x$ $f(x)=x$

$x = y = 0$ yields $f(0) = 0$ Suppose that there exist number $c$ such that $f(c) \neq 0$ We will prove that $f$ is one to one. Suppose $f(a) = f(b) \Rightarrow c(c + f(a)) = c(c + f(b)) \Rightarrow f(c(c + a)) = f(c(c + b)) \Rightarrow (c + a)f(c) = (c + b)f(c) \Rightarrow a = b$ Let $x \rightarrow -x , y \rightarrow x$ Then $f(x^2 - xf(x)) = 0 = f(0) \Rightarrow f(x) = x $ for all $x \neq 0$ But $f(0) = 0$ So $f(x) = x$ for all $x$ Also $f(x) = 0$ satisfies the equation and along with $f(x) = x$ , these are the only solutions !

Very cute problem The only solutions are the zero and the identity function, both of which are easily verified to work. We structure the proof in some Claims: Claim 1: $f(0)=0$. Proof: Take $x=0$ and $y=1$ in the given equation $\blacksquare$ Claim 2: $f(x^2)=xf(x)$ for all real $x$. Proof: Take $y=0$ in the given equation $\blacksquare$ Claim 3: If $f(1)=0$, then $f$ is the zero function. Proof: Suppose that $f(1)=0$. Then, for all real $x$, $f(x)=(x+1)f(x)-xf(x)=f(x(x+f(1)))-f(x^2)=f(x^2)-f(x^2)=0,$ as desired $\blacksquare$ Claim 4: If $f(1) \neq 0$, then $f$ is the identity function. Proof: Suppose that $f(1) \neq 0$. We initially prove that $f(u)=0$ is equivalent to $u=0$. Indeed, take $x=1$ and $y=u$ in the given equation to obtain $f(1)=f(1(1+f(u)))=(1+u)f(1),$ and so $f(1)u=0$, which implies that $u=0$, as desired. Now, take $x=-f(y)$ in the given equation to obtain $(y-f(y))f(-f(y))=0,$ and since $f(x)=0$ implies $x=0$, we conclude that $f(x)=x$ for all nonzero inputs $x$. Since $f(0)=0$, this implies that $f(x)=x$ for all $x$, as desired $\blacksquare$

Clearly $f\equiv 0$ works. Now assume that $f(z) \neq 0$ for some $z \in \mathbb{R}$. Plugging $x = z$ and varying $y$ shows that $f$ is surjective. Plugging $x = 1 - f(y)$ (which covers all reals as $f$ is surjective) implies that $(y-f(y))f(1-f(y)) = 0$, so for every $y\in\mathbb{R}$ either $f(y) = y$ or $f(1-f(y)) = 0$. However, plugging $x = z$ in the original FE shows that \[f(z^2+zf(y)) = (z+y)f(z)\]Therefore $f$ is injective as otherwise if we assume that $f(a) = f(b)$ for some $a\neq b$, comparing $(x,y) = (z,a)$ and $(x,y) = (z,b)$ yields a contradiction. Therefore if for two different $y_{1}, y_{2}$ we have \[f(1-f(y_{1}))) = 0 = f(1-f(y_{2}))\Longrightarrow 1-f(y_{1}) = 1-f(y_{2})\Longrightarrow y_{1}=y_{2}\]implying that $f(y) = y$ for all real $y$ except for at most one. However, $f$ is surjective, so the latter case is impossible and therefore $f(x) = x$ for all $x\in\mathbb{R}$ which can also be checked to be a working solution.

Let $P(x,y)=f(x(x + f(y))) = (x + y)f(x)$ $P(0,0)$ yields $f(0)=0$ $P(x,0)\Longrightarrow f(x(x+f(0)))=xf(x)\Longleftrightarrow f(x^2)=xf(x)$ (!) $P(-x,0)\Longrightarrow f(x^2)=-xf(-x)$ (1) $P(x,-x)$ yields $f(x(x+f(-x)))=0\Longleftrightarrow f(x(x+f(-x)))=f(0)$ (2) Claim:The function is injective iff $f(x)\neq0$ Proof: Let $f(a)=f(b)$ $P(x,a)$ yields $f(x^2+xf(a))=xf(x)+af(x)\Longleftrightarrow f(x^2+xf(a))-xf(x)=af(x)$ $P(x,b)$ yields $f(x^2+xf(b))=xf(x)+bf(x)\Longleftrightarrow f(x^2+xf(b))-xf(x)=bf(x)$ Thus $af(x)=bf(x)$, which implies that $a=b$ if and only if $f(x)\neq0$ $\square$ Case 1: $f(x)\neq0$ Since $f$ is injective when $f(x)\neq0$, we get that from (2), $x^2+xf(-x)=0\Longleftrightarrow -xf(-x)=x^2$ furhtermore from (1), $f(x^2)=x^2$ thus from (!) $xf(x)=x^2\Longrightarrow f(x)=x, \forall x \neq0$. Since $f(0)=0$, this implies that $f(x)=x, \forall x \in \mathbb{R}$ Case 2: $f(x)=0$ This clearly works for all $x\in \mathbb{R}$ So to sum up: $\boxed{f(x)=x \text{, and } f(x)=0, \forall x \in \mathbb{R}}$ $\blacksquare$

claim1 let f(a)=f(b) then a=b or f(x)=0 proof note P(x,a) and P(x,b) ==> a=b or f(x)=0 claim2 f(x)=x P(0,y) f(0)=0 P(1-f(x),x) wher f(x)≠1 f(1-f(x))=(1-f(x)+x)f(1-f(x)) ==> f(x)=x or f(c)=1 but by claim1 and P(c,y) f(c(c+y))=c+y=f(c+y)==> c=1

Rewrite as $f(x^2 + xf(y)) = xf(x) + yf(x)$. $P(0,y)$ forces $f(0) = 0$. We can see by varying $y$ either $f$ is zero, or it is bijective. Assume the second case. Now consider the unique value of $y$ such that $f(y) = -x$. We can then see that we are forced to have $(x + y)f(x) = 0$, so we have $f(-x) = -x$, or $f(x) = x$. Both solutions obviously work, so we are done.

Clearly $\color{red} f\equiv 0$ works, so we suppose $f$ is not zero everywhere. Setting $x=0$ we find $f(0)=0$ and setting $y=0$ we find $f(x^2)=xf(x)$. Now let $f(x_0)=0$, then setting $y=x_0$ we get $f(x^2)=xf(x)+x_0f(x)$ i.e. $x_0=0$, since $f\not \equiv 0$. Now $-xf(-x)=f(x^2)=xf(x)$ which implies $f$ is odd. Setting $y=-x$ implies $\color{red} f(x)=x$, which works.

$f(x)\equiv0$ holds. Otherwise, $f$ is clearly bijective. Putting $y=1-x$ yields $$f(x(x+f(1-x))=f(x)$$Since $f$ is injective, $f(x(x+f(1-x)))=f(x)$. Thus, for $x\neq0$, $f(1-x)=1-x$. For all $x\neq1$, $f(x)=x$. Let's show that $f(1)=1$. Putting $y=1$ and $x\neq1$, $x^2+xf(1)\neq1$ yields $$x^2+xf(1)=x^2+x$$thus $f(1)=1$ and $f(x)\equiv x$. The only solutions are $f(x)\equiv0$ and $f(x)\equiv x$.

oVlad wrote: Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that \[f(x(x + f(y))) = (x + y)f(x),\]for all $x, y \in\mathbb{R}$. $$f(x(x + f(y))) = (x + y)f(x)...(\alpha)$$In $(\alpha) x=0:$ $$\Rightarrow f(0)=yf(0)$$$$\Rightarrow f(0)=0$$In $(\alpha) y=0:$ $$\Rightarrow f(x^2)=xf(x)$$$$\Rightarrow f \text{ is odd}$$If $f(x)=c, \forall x \in \mathbb{R}$ Replacing in $(\alpha):$ $$f(x)=0, \forall x \in \mathbb{R}$$If $f \text{ is not constant}:$ $$\Rightarrow \exists \text{ }a/ f(a)\neq 0$$In $(\alpha) x=a:$ $$\Rightarrow f(a(a+f(y)))=(a+y)f(a)$$$$\Rightarrow f \text{ is surjective}$$In $(\alpha):$ $$f(x^2+xf(y))=(x+y)f(x)=xf(x)+yf(x)...(\theta)$$In $(\theta) x=-f(y)$: $$\Rightarrow f(f(y)^2-f(y)^2)=f(y)f(f(y))-y(f(f(y)))$$$$\Rightarrow f(0)=(f(y)-y)f(f(y))$$$$\Rightarrow (f(y)-y)f(f(y))=0$$$$\Rightarrow f(y)=y \text{ or } f(f(y))=0$$If $\exists \text{ }b\neq 0/f(b)=0:$ In $(\alpha) x=b:$ $$f(b^2+bf(y))=0$$Note that $b^2+bf(y)$ passes through all real $$\Rightarrow f(x)=0, \forall x\in\mathbb{R}(\Rightarrow \Leftarrow)$$ $$\Rightarrow f(y)=y,\forall y\in\mathbb{R}$$ $$\Rightarrow f(x)=0 \text{ and } f(x)=x \text{ are the only solutions}_\blacksquare$$

Let $P(x,y)$ be the assertion. $P(0,0) \rightarrow f(0)=0$ $P(-x,x) \rightarrow f(-x(-x+f(x)))=0$ $P(-f(x), x) \rightarrow f(-f(x))=0 \text{ OR } f(x) = x$ Suppose that $f(-f(x))=0$. Then, there exists $n$ such that $f(n)=0$. $P(x, n) \rightarrow f(x^2)=(x+n)f(x)$ $P(x, 0) \rightarrow f(x^2)=xf(x)$ Subtracting gives $nf(x)=0$, so either $f(x)=0$ or $n=0$. If $n=0$, then that means $f(k)=0$ iff $k=0$, so by $P(-x,x)$ above, we must have $-x(-x+f(x))=0 \rightarrow f(x)=x$, which was already established above, so there are no new solutions here. In summary, the solutions are $f(x)=x, 0$, which both clearly work.

We claim that $f\equiv0$ and $f(x)=x$ for all $x\in\mathbb{R}$ are the only solutions. Suppose the function is constant. Then, $f\equiv0$ is the only solution. Now we assume that the function is non-constant. Call the original equation $p(x,y)$. $p(0,0): \ \ f(0)=0$. Suppose $f(x_1)=f(x_2)$.Choose $x$ such that $f(x)=0$ (this is possible since the function is non-constnant. Then, comparing $p(x,x_1)$ and $p(x,x_2)$ gives $x_1=x_2$. So, the function is injective. $$p(-f(y),y): \ \ f(-f(y))(y-f(y))=0$$However, if $f(-f(y))=0\Longrightarrow f(y)=0\Longrightarrow y=0$. So, for every $y\ne0$, we have $f(y)=y$. Combining with $f(0)=0$, we get that $f(x)=x$ for all $x\in\mathbb{R}$, as claimed. $\square$

a relatively easy problem Solution:

$P(0,x)$ gives $f(0)=xf(0)$ $P(0,-x($ gives $f(0)=-xf(0)$ Dividing by x on both sides yields $f(0)=-f(0)$ hence $2f(0)=0$ $$\Rightarrow f(0)=0$$$P(x,0)$ gives $f(x^2)=xf(x)$ $P(-x,0)$ gives $f(x^2)=-x(fx)$ $$\Rightarrow$$$f$ is odd $P(-f(x),x)$ gives $0=(-f(x)+x)(f(-f(x))$ $$\Rightarrow f(x)=x$$

One solution is $f(x) = 0,$ so assume that $f(x)$ is not all $0.$ Claim: $f(0) = 0.$ Proof: Plug in $x = y = 0$ into the given assertion. Claim: If $f(x) = 0,$ then $x = 0.$ Proof: Plug in $y = 0$ to get $f(x^2) = x f(x).$ This means that if $f(y) = 0$ and $f(x)$ is nonzero (exists since $f(x)$ is not always $0$) then $$f(x(x+0)) = f(x^2) = x f(x) = (x+y)f(x).$$This means that $y = 0.$ Now plug in $x = -y$ to get $$f(x(x+f(-x)) = 0.$$Consequently, $x(x+f(-x)) = 0,$ so for nonzero $x,$ we have $f(-x) = -x.$ This means that $f(x) = x$ for all reals $x$ since $f(0) = 0.$ Hence the only solutions are $f(x) = 0$ and $f(x) = x.$

Let $P(x,y)$ be the assertation $f(x(x+f(y)))=(x+y)f(x)$ We see that $\boxed{f(0)=0}$ works, so let $f$ be a non allzero function. $P(0,y) \Rightarrow f(0)=yf(0) \Leftrightarrow f(0)=0$ $P(x,0) \Rightarrow f(x^2)=xf(x)$ $P(-f(y),y) \Rightarrow 0=f(0)=(y-f(y)f(-f(y))$ We see that $\boxed{f(x)=x}$ also works, so let there $\exists$ $t \neq 0$, such that $f(-f(t))=0$ $P(x,-f(t)) \Rightarrow xf(x)=f(x^2)=(x-f(t))f(x) \Leftrightarrow f(t)=0$ $P(x,t) \Rightarrow xf(x)=f(x^2)=(x+t))f(x) \Leftrightarrow t=0$, which is a contradiciton.

$f(x(x + f(y))) = (x + y)f(x), \forall x, y \in \mathbb{R}$ Let $x = y = 0,$ we have $f(0) = 0$ From this, if we let $y = 0,$ we have $f(x^2) = xf(x), \forall x \in \mathbb{R}$ Let $y = - x,$ we have $f(x(x + f(- x))) = 0$ Suppose that there exists $a \ne 0$ satisfies $f(a) = 0$ Let $y = a,$ we have $f(x^2) = (x + a)f(x), \forall x \in \mathbb{R}$ So $af(x) = 0, \forall x \in \mathbb{R}$ or $f(x) = 0, \forall x \in \mathbb{R}$ If $f(x) = 0 \Longleftrightarrow x = 0$ then $x(x + f(- x)) = 0,$ or $f(x) = x, \forall x \ne 0$ Combine with $f(0) = 0,$ we have $f(x) = x, \forall x \in \mathbb{R}$

$$P(0; y) \Longrightarrow f(0) = 0$$$$P(-f(y); y) \Longrightarrow (y-f(y))f(-f(y)) = 0$$$$f(y) = y \hspace{2mm} \forall y$$ $\newline$ Case 2: $f(-f(y)) = 0$ $$P(x; 0) \Longrightarrow f(x^2) = xf(x)$$$$P(x; -f(y)) \Longrightarrow xf(x) = f(x^2) = (x-f(y)) f(x)$$$$ x = x - f(y)$$$$ f(y) = 0 \hspace{2mm} \forall y$$

Case 1: $f(0)=0$ $P(-f(x); x)$ $\Longrightarrow$ $0=(x-f(x))f(-f(x))$ $\rightarrow$ $f(-f(x))=0$ or $f(x)=x$. $\boxed{f(x)=x}$ is a solution. Now, let's look at case where $f(-f(x))=0$. Let's assume that $f(x)$ $\ne$ $0$ $\rightarrow$ $\exists$$d$, $f(d)$ $\ne$ $0$. Claim: if $f(a)=f(b)=0$ $\rightarrow$ $a=b$ Proof: $P(d; a)$, $P(d; a)$ $\rightarrow$ $a=b$. $f(-f(x))=0$, $f(0)=0$. So, by the claim $\rightarrow$ $\boxed{f(x)=0}$ $\forall$$x$ $\in$ $\mathbb{R}$ (as we assumed above that $f(x)$ $\ne$ $0$) Now, let's prove that there is no composition of solutions. Assume that $\exists$$a$, $f(a)=a$, $\exists$$b$, $f(b)=0$, $a$, $b$ $\ne$ 0 We can say that Claim above is true as there exists a number "a" that $f(a)$ $\ne$ 0. $P(b, b)$ $\Longrightarrow$ $f(b^2)=0$, by the claim $b^2=0$ $\rightarrow$ $b=0$. We get a contradiction. Case 2: $f(0)$ $\ne$ $0$ $P(x; 0)$ $\Longrightarrow$ $f(f(x))=xf(0)$ $\rightarrow$ $f(x)$ $is$ $injective$. $P(0; 0)$ $\Longrightarrow$ $f(f(0))=0$ $P(f(0); x)$ $\Longrightarrow$ $f(x)=1-f(0)$ $\rightarrow$ $f(x)=k$. $k$ is a constant number. From equation, we get $k=1$ $\rightarrow$ $f(x)=0$ $\forall$$x$ $\in$ $\mathbb{R}$. But $f(0)$ $\ne$ $0$. So, we get a contradiction. Solutions: $\boxed{f(x)=0}$, $\boxed{f(x)=x}$ $\forall$$x$ $\in$ $\mathbb{R}$.

$f(x)=0,f(x)=x$ are the solutions. Take $P(0,y)$ giving $f(0)=0.$ Take $P(x,0)$ giving $f(x^2)=xf(x).$ Take $P(x,k)$ for $f(k)=0$ to get $xf(x)=f(x^2)=(x+k)f(x),$ so either $f(x)=0$ for all $x$ or $k=0.$ Take $P(x,a)$ and $P(x,b)$ for $f(a)=f(b)$ and $f(x)\ne 0$ to get $f$ injective. Take $P(x,1-x)$ to get $x+f(1-x)=1$ for $x\ne 0,$ so $f(y)=y$ for $y\ne 1$ by setting $y=1-x.$ Take $P(1-f(1),1)$ to get $1-f(1)=0$ or $2-f(1)=1,$ both giving $f(1)=1.$ Thus $f(x)=0,f(x)=x$ are the only possible solutions and it is easy to check both/ @above i think you got pointwise trapped .

OronSH wrote: $f(x)=0,f(x)=x$ are the solutions. Take $P(0,y)$ giving $f(0)=0.$ Take $P(x,0)$ giving $f(x^2)=xf(x).$ Take $P(x,k)$ for $f(k)=0$ to get $xf(x)=f(x^2)=(x+k)f(x),$ so either $f(x)=0$ for all $x$ or $k=0.$ Take $P(x,a)$ and $P(x,b)$ for $f(a)=f(b)$ and $f(x)\ne 0$ to get $f$ injective. Take $P(x,1-x)$ to get $x+f(1-x)=1$ for $x\ne 0,$ so $f(y)=y$ for $y\ne 1$ by setting $y=1-x.$ Take $P(1-f(1),1)$ to get $1-f(1)=0$ or $2-f(1)=1,$ both giving $f(1)=1.$ Thus $f(x)=0,f(x)=x$ are the only possible solutions and it is easy to check both/ @above i think you got pointwise trapped . Thanks, I think I corrected it. Would be happy if you mention if there is still a mistake

actually it might be fine You should reword it into showing that $f(x) = 0$ for all $x$ with $f(-f(x)) = 0$ and then proving pointwise

$P(0,0) \implies f(0)=0$. Clearly $f(x) \equiv 0$ is a solution, henceforth assume that $f$ is nonconstant. Choose $x$ such that $f(x) \neq 0$, then we see that $f$ is bijective. If $x \neq 0$, $P(x,-x) \implies f(x(x+f(-x)))=0 \implies x(x+f(-x))=0 \implies f(-x)=-x$. It follows that $f(x)=x$ for all $x$. $\therefore f(x)=0 \text{ } \forall x$ or $f(x)=x \text{ } \forall x$.

We claim that the only solutions are $f(x) \equiv 0$ and $f(x)=x$, which work. We prove that they are the only solutions. Let $P(x,y)$ be the given assertion. First, $P(0,0)$ yields $f(0 \cdot (0 + f(0))) = 0$, or $f(0)=0$. Thus, if $f(x) \equiv c$ where $c$ is a constant, we must have $f(x) \equiv 0$. Now suppose that $f(x) \not\equiv c$. We claim that $f(x)$ is injective at $x=0$. Since we already have $f(0)=0$, we just need to show that if $f(x)=0$ then $x$ must be equal to $0$. For the sake of contradiction, suppose there exists $k \neq 0$ such that $f(k)=0$. Then, $P(x,k)$ gives $f(x^2) = (x+k)f(x)$. However, we know that $P(x,0)$ yields $f(x^2) = xf(x)$, so if $f(x) \not\equiv 0$, we must have $x+k=x$, or $k=0$, contradiction. Next, $P(-f(y), y)$ yields $f(0) = f(-f(y))(y-f(y))$ or $f(-f(y))(y-f(y)) = 0$. This implies that $f(-f(y)) = 0$ or $y-f(y)=0$. For the former case, we must have $-f(y)=0$, or $y=0$. If $y \neq 0$, then we must have $y-f(y)=0$, or $f(y)=y$. But we also know that $f(0)=0$, so we have $f(y)=y$ for all $y \in \mathbb{R}$. $\blacksquare$

The solutions are $\boxed{f\equiv0}$ and $\boxed{f\equiv \text{id}}$, which clearly work. We now prove that they are the only ones. Let $P(x,y)$ denote the assertion. By $P(0,y)$, we find that $f(0) = yf(0)$. Set $y\ne 1$; then we have $f(0)=0$. Since rearranging the given yields $$\frac{f(x(x+f(y)))-xf(x)}{f(x)}=y,$$note that either $f\equiv0$ or it is injective. We now consider the latter case. By $P(-y,y)$, we find that $f(y^2-yf(y)) = 0 = f(0)$. Since $f$ is injective, $y^2 - yf(y) = 0$, and hence $f(y)=y$ on all nonzero reals. Combining this with the fact that $f(0)=0$, we can see that $f$ is just the identity function in this case, which finishes. $\blacksquare$

Prove injectivity and try $P(x,1-x)$

$P(x,y) : f(x(x+f(y)))=(x+y)f(x)$ $f \equiv 0$ is a solution so suppose that there exists $a$ such that $f(a) \neq 0$. $P(0,0) \Rightarrow f(0) = 0.$ Claim: $f$ is injective. Proof: assume FTSOC that there are $x_{1} \neq x_{2}$ such that $f(x_{1}) = f(x_{2})$. From $P(a,x_{1})$ and $P(a,x_{2})$ we get that $$(a+x_{1})f(a) = (a+x_{2})f(a)$$which is a contradiction. Finally, from $P(x,-x)$ we get $f(x^{2}+xf(-x)) = 0$ so from injectivity we have that $x^{2} + xf(-x) = 0$ so $f(x)=x$ which does satisfy the given equation.

oVlad wrote: Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that \[f(x(x + f(y))) = (x + y)f(x),\]for all $x, y \in\mathbb{R}$. Morale booster If $f$ is constant, then $f \equiv 0$, so assume not. $P(0, 0)$ gives $f(0) = 0$. $P(x, 0)$ gives $f(x^2) = xf(x)$. If there exists $\alpha$ such that $f(\alpha) = 0$ (which we know exists), then $P(x, \alpha)$ gives $f(x^2) = (x + \alpha)f(x) = xf(x) \implies \alpha = 0$, so $f$ is injective on $0$. $P(-y, y)$ gives $f(y^2 - yf(y)) = 0 \implies f(y) = y$, done.

A0?? It's obvious that $\boxed{f(x)=0\quad\forall x\in\mathbb{R}}$ is a solution, so assume that $f$ is not zero everywhere. Take $x=-y$, so easily we get $f(x^2-xf(x))=0$. Take $x=0$, so we have $f(0)=xf(0)$ $\Longleftrightarrow$ $f(0)=0$ for $x\ne 0$. Now take $x=1$ and then we can get $f(1+f(x))=(x+1)f(1)$ which trivially implies that $f$ is injective. So, applying injectivity at $f\equiv 0$ implies $x^2-xf(x)=0$ $\Longleftrightarrow$ $\boxed{f(x)=x\quad\forall x\in\mathbb{R}}$, which is indeed a solution. Done. $\blacksquare$

We will prove the $f \equiv 0$ or $f(x)=x$ $f \equiv 0$ works so we assume its not the case Let $P(x,y)$ be the assertion $P(0,y)$ implies $f(0)=yf(0)$ Or $f(0)=0$ Now there must be some $c$ such $f(c)$ is not $0$.(or it would be $f \equiv 0$) Now $P(c,y)$ $f(c(c+f(y)))=(c+y)f(c)$ We will prove its injectivity We assume there is some $a,b$ such $f(a)=f(b)$ $c(c+f(a))=c(c+f(b))$ $f(c(c+f(a)))=f(c(c+f(b)))$ $(c+a)f(c)=(c+b)f(c)$ $a=b$ Now $P(x,-x)$ implies $f(x(x+f(-x)))=0=f(0)$ $x(x+f(-x))=0$ $f(-x)=-x$ or $f(x)=x$ $\square$

straightforward soln We claim that $f(x) =0 $ and $f(x)=x \forall x \in \mathbb{R}$ are the only such functions. Let $P(x,y)$ denote the assertion. $P(0,0) \implies f(0)=0…. (1)$ $P(-f(y),y) \implies (y-f(y))f(-f(y))=0$, so for each $y \in \mathbb{R}, f(y)=y$ or $f(-f(y))=0$

I claim that the only functions that work are $f(x) = x$ and $f \equiv 0$. Now first of all notice that if $f$ is constant, then $f \equiv 0$ is the only solution so assume that $f$ is non constant. Let $P(x, y)$ be the assertion, $P(0, y) \rightarrow f(0) = 0$ and $P(-x, x) \rightarrow f(x^2-xf(x)) = 0$. Now if there is some other $z \neq 0$ such that $f(z) = 0$, $P(x, z) \rightarrow f(x^2) = (x+z)f(x)$ and thus $xf(x) = (x+z)f(x)$ and from here we can conclude that $z$ must be $0$ and thus $f(x) = x$ $\blacksquare$