parmenides51 wrote: Find all functions $f : R \to R$ such that $$f(xf(y))= (1 - y)f(xy) + x^2y^2f(y)$$for all reals $x, y$. If $f$ is constant then $f(x)\equiv 0$.Let $f$ non-constant. Let $P(x,y) :f(xf(y))= (1 - y)f(xy) + x^2y^2f(y)$ $P(0,x):f(0)=(1-x)f(0)\Leftrightarrow f(0)=0$ $P(x,1):f(xf(1))=x^2f(1)\,\,(**)$ $P(1,1):f(f(1))=f(1)$ $P(1,x):f(f(x))=(1-x)f(x)+x^2f(x)\,\,(***)$ $P(1,f(1)): f(f(f(1)))=(1-f(1))f(f(1))+f(1)^2f(f(1))\,\,(*)$. Set $f(1)=a$.We have $f(a)=a$ So $(*):f(f(a))=(1-a)f(a)+a^2f(a)\Leftrightarrow a=(1-a)a+a^3\Leftrightarrow a=0\,\,or\,\,a=1$ If $a=1$ then $(**)\Rightarrow f(x)=x^2$ But $f(xf(y))=x^2y^4\neq (1-y)x^2y^2+x^2y^4=(1-y)f(xy)+x^2y^2f(y)$ So $a=0$ $f(x_1)=f(x_2)\overset{(***)}{\Rightarrow}x_1^2-x_1+1=x_2^2-x_2+1\Leftrightarrow x_1=x_2\,\,or\,\,x_1+x_2=1\,\,(****)$ $P(\dfrac{1}{x},x):f(\dfrac{1}{x}f(x))=(1-x)f(1)+f(x)\Leftrightarrow f(\dfrac{1}{x}f(x))=f(x),x\neq 0$ So $\dfrac{1}{x}f(x)=x\,\,or\,\,\dfrac{1}{x}f(x)+x=1\Leftrightarrow f(x)=x^2\,\,or\,\,f(x)=x-x^2$ Suppoce that there exist $x_1\neq0$ such that $f(x_1)=x_1^2$ Then $f(x_1^2)=x_1^2(x_1^2-x_1+1)$ If $f(x_1^2)=x_1^4$ then $x_1^4=x_1^2(x_1^2-x_1+1)\Leftrightarrow x_1=1$,contradiction since $f(1)=0\neq1^2$ So $f(x_1^2)=x_1^2-x_1^4$ and $x_1^2-x_1^4=x_1^2(x_1^2-x_1+1)\Leftrightarrow x_1=\dfrac{1}{2}$ Therefore $f(x)=x^2\Leftrightarrow x=\dfrac{1}{2}$.But $f(\dfrac{1}{2})=\dfrac{1}{4}=\dfrac{1}{2}-\dfrac{1}{2^2}$ So $f(x)=x-x^2,\forall x\in \mathbb{R}$ which is accepted.

parmenides51 wrote: Find all functions $f : R \to R$ such that $$f(xf(y))= (1 - y)f(xy) + x^2y^2f(y)$$for all reals $x, y$. Let $P(x,y)$ be the assertion $f(xf(y))=(1-y)f(xy)+x^2y^2f(y)$ $P(x,1)\implies f(xf(1))=x^2f(1)$ $\ast$ Case (1): $f(1)\not=0$ $\implies$ put $x=\frac{x}{f(1)}$ in $\ast$ $\implies f(x)=x^2\frac{1}{f(1)}=cx^2$ Substitute it in the equation and we will get that $c=0$ $\implies f(x)=0$ Substitute it in the equation and you will get that it is a solution!! Case (2): $f(1)=0$ and $f(x)\not=0$ we get from $\ast$ that : $f(0)=0$ $P(\frac{1}{x},x) : x\not=0 \implies$ $f(\frac{f(x)}{x})=f(x)$ $P(1,x)\implies f(f(x))=(1-x)f(x)+x^2f(x)$ (1) $P(1,\frac{f(x)}{x}) :x\not=0 \implies$ $f(f(x))=(1-\frac{f(x)}{x})f(x)+\frac{f(x)^3}{x^2}$ (2) (1)$-$(2) $\implies x^4f(x)-x^3f(x)=f(x)^3-xf(x)^2$ Since $f(x)\not=0$ so we can divide by $f(x)$ $\implies f(x)^2-xf(x)-(x^4-x^3)=0$ It is a quadratic equation $\smiley$ $\implies$ either: $f(x)=\frac{x+\sqrt{x^2+4x^4-4x^3}}{2}=\frac{x+\sqrt{x^2(2x-1)^2}}{2}=\frac{x+2x^2-x}{2}=x^2$ But $f(1)=0$ $\implies contradiction !!$ or: $f(x)=\frac{x-\sqrt{x^2+4x^4-4x^3}}{2}=\frac{x-\sqrt{x^2(2x-1)^2}}{2}=\frac{x-2x^2+x}{2}=\frac{2x-2x^2}{2}=x-x^2$ Substitute it in the equation and you will get that it is solution !! And the solution work when $x=0$ because $f(0)=0=0-0^2$ And we won a pizza

Let $P(x,y)$ be the assertion of $f(xf(y))= (1 - y)f(xy) + x^2y^2f(y)$. $P(x,1)\implies f(xf(1))=x^2f(1)$ $P(1,1)\implies f(f(1))=f(1)$ $P(0,y)\implies f(0)=(1-y)f(0)\implies f(0)=0$ $P(x,f(1))\implies f(xf(1))=(1-f(1))f(xf(1))+xf(1)^3\implies f(1)f(xf(1))=xf(1)^3$. If $f(1)=c\neq 0$, meaning that $f(cx)=cx^2$ and because $cx$ takes all real values, then $f(x)=cx$. Plugging this back into our original, $$c^2xy=(1-y)cxy+cx^2y^3 \implies cxy(c+y-1-xy^2)\implies c=0\implies \text{Contradiction.}$$ If $f(1)=0$: $P(1,y)\implies f(f(y))=(1-y)f(y)+y^2f(y)=f(y)(y^2-y+1)$ $P(x,\frac{1}{x})\implies f(xf(\frac{1}{x}))=f(\frac{1}{x})$ $P(\frac{1}{x},x)\implies f(\frac{1}{x}f(x))=f(x)$ Suppose now there exist $f(d)=0$ so that $d\neq 0,1$, thus $P(x,d)\implies (1-d)f(xd)=0\implies f\equiv 0$. Now, suppose only $d$ values satisfying $f(d)=0$ are $d=\{1 ,2\}$: Let $f(a)=f(b)\neq 0$ (then $a,b\neq 0,1$), thus $a^2-a=b^2-b\implies a=b \text{ or } a+b=1$. Case one. $f$ is only injective, thus $f(\frac{1}{a}f(a))=f(a)\implies f(a)=a^2$ for all $a\in \mathbb{R}-1$ and $f(1)=0$. Let $q,w, qw,qw^2\neq 0,1$: $P(q,w)\implies q^2w^4=(1-w)q^2w^2+q^2w^4\implies \text{Contradiction.}$ Case two. There is $f(k)=f(l)$ and $f(g)=f(h)$ so that $k=l$ and $g+h=1$. Thus, $f(k)=k^2$ and $\frac{1}{g}f(g)+g=1\implies f(g)=g(1-g)$. $P(\frac{k}{g},g)\implies f(k(1-g))=(1-g)k^2+k^2g^3(1-g)=k^2(1-g)(g^3+1)\implies \text{Contradiction.}$ Contradiction, since $f(k(1-g))= k^2(1-g)^2\text{ or } k(1-g)(1-k+kg)$ is not equal to the $\mathcal{RHS}$. Case three. if for any $a,b\neq 0,1$, we have $f(a)=f(b)$, then $a+b=1$. $\frac{1}{x}f(x)+x=1\implies f(x)=x(1-x)$ for all $x\in \mathbb{R}$ (notice that $x=0,1$ also satisfy this so we can use this also for $x=0,1$). Checking if it works: $f(xf(y))= (1 - y)f(xy) + x^2y^2f(y)\implies f(xy(1-y))=xy(1-y)(1-xy)+x^2y^3(1-y)$. $\mathcal{LHS}= f(xy(1-y))=xy(1-y)(1-xy(1-y))=xy(1-y)(1-xy+xy^2)$ $\mathcal{RHS}=xy(1-y)(1-xy)+x^2y^3(1-y)=xy(1-y)(1-xy+xy^2)$. Thus, $\mathcal{LHS}=\mathcal{RHS}$. Answer. $f\equiv 0\, \forall x \in \mathbb{R}$ and $f=x(1-x)\, \forall x \in \mathbb{R}$.