IstekOlympiadTeam wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ satisfying the equation \[|x|f(y)+yf(x)=f(xy)+f(x^2)+f(f(y))\]for all real numbers $x$ and $y$. Let $P(x,y)$ be the assertion $|x|f(y)+yf(x)=f(xy)+f(x^2)+f(f(y))$ Let $a=f(1)$ and $b=-f(-1)$ $P(1,1)$ $\implies$ $f(a)=0$ $P(0,1)$ $\implies$ $f(0)=0$ Let $x\ge 0$ : $P(\sqrt x,0)$ $\implies$ $f(x)=0$ $\forall x\ge 0$ $P(0,y)$ $\implies$ $f(f(y))=0$ $\forall y$ Let $x<0$ : $P(x,-1)$ $\implies$ $f(x)=-bx$ $\forall x<0$ Plugging this back in original equation, we get $\boxed{f(x)=c(x-|x|)\text{ }\forall x}$ which indeed is a solution, whatever is $c\le 0$ *edit : too late *

The solutions are $f(x) = 0 \ | \ x \geq 0$ and $f(x) = -xf(-1) \ | \ x < 0$. It is easy to verify that these work by checking combinations of $\pm x, \pm y$. We now prove that these are the only solutions. Let $P(x, y)$ be the given assertion. $P(1, 1) \implies f(f(1)) = 0$ and $P(0, 1) \implies f(0) = 0$. $P(-1, 0) \implies f(1) = 0$ using the fact that $f(0) = 0$. $P(x, 0) \implies f(x^2) = 0 \implies f(x) = 0 \ | \ x \geq 0$, which proves the first part of the claim. $P(-1, x) \ | \ x > 0 \implies f(x) + xf(-1) = f(-x) + f(f(x))$. Since $x > 0$, we know that $f(x) = 0 = f(f(x))$. Substituting these yields $f(-x) = xf(-1) \ | \ x > 0 \implies f(x) = -xf(-1) \ | \ x < 0$, and the proof is complete.

Nice Problem.

IstekOlympiadTeam wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ satisfying the equation \[|x|f(y)+yf(x)=f(xy)+f(x^2)+f(f(y))\]for all real numbers $x$ and $y$. A magical problem Claim 1.$f(f(x))=0$ Proof: $P(0,x)\implies f(f(x))=xf(0)-2f(0)$ $P(1,x)\implies f(f(x))=xf(1)-f(1)$ Then $xf(0)-2f(0)=xf(1)-f(1)\implies f(0)=f(1),2f(0)=f(1)\implies f(0)=f(1)=0$ $f(f(x))=xf(1)-f(1)=0$ $|x|f(y)+yf(x)=f(xy)+f(x^2)$ Claim 2.$f(x^2)=0$ Proof: $x\leq 0,P(x,x)\implies f(x^2)=0$ $|x|f(y)+yf(x)=f(xy)$ $f(xy)=|x|f(y)+yf(x)=|y|f(x)+xf(y)\implies (x-|x|)f(y)=(y-|y|)f(x),y=-1\implies f(x)=c(x-|x|),c\leq 0$ $\boxed{f(x)=c(x-|x|)\text{ }\forall x},c\leq 0$

Functional_equation wrote: IstekOlympiadTeam wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ satisfying the equation \[|x|f(y)+yf(x)=f(xy)+f(x^2)+f(f(y))\]for all real numbers $x$ and $y$. A magical problem Claim 1.$f(f(x))=0$ Proof: $P(0,x)\implies f(f(x))=xf(0)-2f(0)$ $P(1,x)\implies f(f(x))=xf(1)-f(1)$ Then $xf(0)-2f(0)=xf(1)-f(1)\implies f(0)=f(1),2f(0)=f(1)\implies f(0)=f(1)=0$ $f(f(x))=xf(1)-f(1)=0$ $|x|f(y)+yf(x)=f(xy)+f(x^2)$ Claim 2.$f(x^2)=0$ Proof: $x\leq 0,P(x,x)\implies f(x^2)=0$ $|x|f(y)+yf(x)=f(xy)$ $f(xy)=|x|f(y)+yf(x)=|y|f(x)+xf(y)\implies (x-|x|)f(y)=(y-|y|)f(x),y=-1\implies f(x)=c(x-|x|),c\leq 0$ $\boxed{f(x)=c(x-|x|)\text{ }\forall x},c\leq 0$ Why $c\leq 0$? i dont understand.

pco wrote: IstekOlympiadTeam wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ satisfying the equation \[|x|f(y)+yf(x)=f(xy)+f(x^2)+f(f(y))\]for all real numbers $x$ and $y$. Let $P(x,y)$ be the assertion $|x|f(y)+yf(x)=f(xy)+f(x^2)+f(f(y))$ Let $a=f(1)$ and $b=-f(-1)$ $P(1,1)$ $\implies$ $f(a)=0$ $P(0,1)$ $\implies$ $f(0)=0$ Let $x\ge 0$ : $P(\sqrt x,0)$ $\implies$ $f(x)=0$ $\forall x\ge 0$ $P(0,y)$ $\implies$ $f(f(y))=0$ $\forall y$ Let $x<0$ : $P(x,-1)$ $\implies$ $f(x)=-bx$ $\forall x<0$ Plugging this back in original equation, we get $\boxed{f(x)=c(x-|x|)\text{ }\forall x}$ which indeed is a solution, whatever is $c\le 0$ *edit : too late * why $c\le 0$?

LoloVN wrote: why $c\le 0$? As suggested, just plug $f(x)=c(x-|x|)$ in original equation and you get $f(f(y))=0$ and so $f(y)\ge 0$ and so $c\le 0$

$P(1,1)\Rightarrow f(f(1))=0$ $P(0,1)\Rightarrow f(0)=0$ $P(x,0)\Rightarrow f(x^2)=0$ $P(0,-1)\Rightarrow f(f(-1))=0$ $P(-x^2,-1)\Rightarrow f(-x^2)=x^2f(-1)$ So $\boxed{f(x)=\begin{cases}0&\text{if }x\ge0\\cx&\text{if }x<0\end{cases}}$ which works for any constant $c\in\mathbb R$.

Let $P(x,y)$ be the assertion of $\vert x\vert f(y)+yf(x)=f(xy)+f(x^2)+f(f(y))$. We have $$P(0,x)\implies f(f(x))=(x-2)f(0)\quad (1)$$and $$P(x,0)\implies f(x^2)=\vert x\vert f(0)-f(0)-f(f(0))\quad (2)$$Claim. $f(0)=0$. Proof. Suppose not, i.e. $f(0)\neq 0$. By $(1)$, we get that $f$ is injective. Hence by $(1)$, $x=2\implies f(f(2))=0$ and $x=3\implies f(f(3))=f(0)\implies f(3)=0$ and thus $f(f(2))=f(3)\implies f(2)=3$. By $(1)$, $x=0\implies f(f(0))=-2f(0)$. By $(2)$, $x=2\implies f(4)=f(0)-f(f(0))=3f(0)$. Now, $P(2,2)\implies f(4)=2f(2)=6$, thus $f(0)=2$. Thus, $3=f(2)=f(f(0))=-2f(0)=-4$, contradiction. \end{proof} Hence, $(1)$ and $(2)$ give $f(f(x))=0$ and $f(x^2)=0$ for all $x\in\mathbb R$. Thus, $P(x,y)$ simplifies to $\vert x\vert f(y)+yf(x)=f(xy)$. Now $P(x>0,-1)\implies f(-x)=xf(-1)$, hence $f(x)=-xf(-1)\forall x<0$. Suppose there is $a$ so that $f(a)<0$, thus taking $x=f(a)$, we get that $0=f(f(a))=-f(a)f(-1)\implies f(-1)=0\implies f\equiv 0$, contradiction. Hence, $f(x)\geq 0\forall x$, this indeed implies that $f(f(x))=0$ for all $x\in\mathbb R$. Now, we verify that $P(x,y)$ is true for any $x,y$. For $x>0,y>0$, this is obvious. If $x>0,y<0$, then $-xy f(-1)=\vert x \vert f(y)=f(xy)=-xyf(-1)$, which is true. If $x<0,y>0$, then $-xy f(-1)=y f(x)=f(xy)=-xyf(-1)$, which is true. If $x<0,y<0$, then $-y\vert x\vert f(-1)-xyf(-1)=\vert x\vert f(y)+y f(x)=f(xy)=0$, which is true. We conclude that our solutions are $$f(x)=\begin{cases} 0\forall x\geq 0\\ -xf(-1)\forall x<0, \end{cases}$$where $f(-1)\geq 0$.

Let $P(x,y)$ the assertion of the given F.E. Case 1: $f(0)=0$ Then by $P(x,0)$ $$f(x^2)=0 \implies f(t)=0 \; \forall t \in \mathbb R^+$$$P(0,x)$ $$f(f(x))=0$$Now do $P(-x,-y)$ where $x,y \in \mathbb R^+$ $$f(t)=ct \; \forall t \in \mathbb R^-$$Now by $P(-1,x)$ where $x \in \mathbb R^+$ $$c=f(-1) \implies f(c)=0 \implies c \ge 0$$Hence the sol in this case is $f(x)=c(x-|x|)$ for $c \ge 0$ Case 2: $f(0) \ne 0$ Note that by $P(1,1)$ $$f(f(1))=0 \implies f(1)=3$$And then by $P(0,1)$ $$f(0)=0 \; \text{contradiction!!}$$Hence the solution is $\boxed{f(x)=c(x-|x|) \; \forall x \in \mathbb R \; \text{and} \; c \ge 0}$ thus we are done