mathwiz0803 wrote: Plugging in $x=0$ we get that $f(0) = f(0)+y(f(f(0)+y))\implies y(f(f(0)+y)) = 0$. Since $y$ is not necessarily 0, we have that $f(f(0)+y) = 0$. Wrong,this sentence didn't told us anything about $f(f(0))$.Indeed $y$ is not necessarily be $0$,but if $y=0$,then there's nothing more that this statement can told.

@Below:Oh yes,thank.

TLP.39 wrote: mathwiz0803 wrote: Plugging in $x=0$ we get that $f(0) = f(0)+y(f(f(0)+y))\implies y(f(f(0)+y)) = 0$. Since $y$ is not necessarily 0, we have that $f(f(0)+y) = 0$. Wrong,this sentence didn't told us anything about $f(f(0))$.Indeed $y$ is not necessarily be $0$,but if $y=0$,then there's nothing more that this statement can told.

Ah I thought wrong so deleted one time well I don't think he stated wrong but just omitted explanation substitution of $y=-f(0)$ yields $-f(0)^2 =0$ so $f(0)=0$ then $f(y)=0$

Wow. What a nice waste of a seemingly pleasant morning. MF163 wrote: Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$f(x^2y)=f(xy)+yf(f(x)+y)$$for all real numbers $x$ and $y$. $P(x,0)$ and $P(0,y) \implies y \cdot f(f(0)+y) = 0 \forall y \in \mathbb{R} \implies f(k) = 0 \forall \ k \in \mathbb{R} - \{0\}$. Now for $f(0)$, consider the expression $P(0, -f(0))$. This yields the desired result so $f(x) \equiv 0 \forall x \in \mathbb{R}$.

Why $y\cdot f(f(0)+y)=0 $ implies $f(k)=0$ for all $k\in \mathbb{R}-{0}$? It should be $0$ for a single $k$

duongbgbg32 wrote:

We dont need to use (1) because, if $f(0) \ne 0$, let $ y=-f(0) \implies f(0)=0$ . so $f(y)=0, \forall y $

MF163 wrote: Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$f(x^2y)=f(xy)+yf(f(x)+y)$$for all real numbers $x$ and $y$. Case 1: $f$ is a constant function, $f(x)=c$ where $c \in \mathbb R$. We will replace this to get: $$y \cdot c=0 \implies c=0 \implies f(x)=0 \; \forall x \in \mathbb R$$Case 2: $f$ is non-constant, let $P(x,y)$ the assertion. $P(0,y)$ $$yf(f(0)+y)=0 \implies f(t)=0 \; \forall t \in \mathbb R$$This is a contradiction since $f$ is non-constant and the expresion $f(0)+y$ is surjective, and second $f(0)=0$ by plugging $y=-f(0)$. Thus the only sol is: $\boxed{f(x)=0 \; \forall x \in \mathbb R}$ Thus we are done

Isn't this trivial? $P(0,-f(0))\Rightarrow f(0)=0$ $P(0,x)\Rightarrow f(x)=0\forall x\ne0\Rightarrow\boxed{f(x)=0}$.

MF163 wrote: Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$f(x^2y)=f(xy)+yf(f(x)+y)$$for all real numbers $x$ and $y$. Too trivial for a Baltic Way P5. \(P(0,-f(0))\) and \(P(0,x)\) kill the problem.