assuming this is correct, did it in 6:59 Let $P(x,y)$ denote the given assertion. $P(0,0): f(f(0))=0$. $P(x,0): f(f(x))=xf(0)$. $P(f(0),0): f(0)=f(0)^2\implies f(0)\in \{0,1\}$. Case 1: $f(0)=0$. Then $f(f(x))=0$. If exist $k\ne 0$ with $f(k)=0$ then $P(k,x): kf(x)=0\implies \boxed{f\equiv 0}$. If not, then $f(x)=0\implies f\equiv 0$, contradiction. Case 2: $f(0)=1$. Then $f(f(x))=x$. Clearly must exist $k$ with $f(k)=0$ because of $P(0,x)$. Then $P(k,x): f(f(k-x))=kf(x)\implies k-x=kf(x)\implies f(x)=\frac{k-x}{k}$ so $f$ is linear. Clearly if $ax+b$, then by involution $a=1$ or $-1$. But $f(0)=1$ so $b=1$. Thus, $f(x)=x+1$, not possible or $\boxed{f(x)=1-x}$, which works.

dangerousliri wrote: Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all real numbers $x$ and $y$, $$f(f(x-y)-yf(x))=xf(y).$$ $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ is a solution.So let us from now look only for non allzero solutions. Let $P(x,y)$ be the assertion $f(f(x-y)-yf(x))=xf(y)$ Let $a=f(0)$ Since $f$ is not allzero, $P(x,y)$ implies $f(x)$ surjective, and so $f(f(x))$ surjective too. $P(x,0)$ $\implies$ $f(f(x))=ax$ and so $a\ne 0$ (else $f(f(x))$ would not be surjective). $P(0,-x)$ $\implies$ $f(f(x)+ax)=0$ $P(f(x)+ax,0)$ $\implies$ $a=(f(x)+ax)a$ and so, since $a\ne 0$, $f(x)=1-ax$ Setting there $x=0$, we get $a=1$ and so $\boxed{\text{S2 : }f(x)=1-x\quad\forall x}$ xhich indeed fits.

dangerousliri wrote: Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all real numbers $x$ and $y$, $$f(f(x-y)-yf(x))=xf(y).$$ $P(0,0):f(f(0))=0$ $P(f(0),0):f(f(f(0)))=f(0)^2\Rightarrow f(0)=f(0)^2\Rightarrow f(0)=0 or1$ Case1If $f(0)=0$ then: if there is $t$ sucj that $f(t)\neq 0$ then from the first we have $f$ surjective. $P(x,0):f(f(x))=0\xrightarrow[]{f,surjective}f(x)=0$ contradiction Otherwise $f(x)=0$ Case2If $f(0)=1$ then: $P(x,0):f(f(x))=x\Rightarrow f:1-1$ $P(x,f(y)):f(f(x-f(y))-f(x)f(y))=xy=f(f(y-f(x))-f(x)f(y))\xrightarrow[]{f:1-1}f(x-f(y))-f(x)f(y)=f(y-f(x))-f(x)f(y)\Rightarrow f(x-f(y))=f(y-f(x))\xrightarrow[]{f:1-1}x-f(y)=y-f(x)\Rightarrow f(x)=c-x$ Move to the first to take $c=1$.

Case 1: $f$ is constant function so we have $f(x)=0$ Case 2: $f$ not constant Claim $f$ is injective. Proof: $x_1=x_2 \rightarrow f(x_1-y)=f(x_2-y) \rightarrow f(x_1-y)-yf(x_1)=f(x_2-y)-yf(x_2) \rightarrow f(f(x_1-y)-yf(x_1))=f(f(x_2-y)-yf(x_2)) \rightarrow x_1f(y)=x_2f(y)$ Proven. Subs $x=1$ we have $f(f(1-y)-yf(1))=f(y) \rightarrow f(1-y)-yf(1)=y \dots (1)$ Subs $y=0$ to $(1)$ we have $f(1)=0$ From $(1)$ Subs $f(1)=0$ we have $f(1-y)=y \rightarrow f(y)=1-y$

Jufri wrote: Case 1: $f$ is constant function so we have $f(x)=0$ Case 2: $f$ not constant Claim $f$ is injective. Proof: $x_1=x_2 \rightarrow f(x_1-y)=f(x_2-y) \rightarrow f(x_1-y)-yf(x_1)=f(x_2-y)-yf(x_2) \rightarrow f(f(x_1-y)-yf(x_1))=f(f(x_2-y)-yf(x_2)) \rightarrow x_1f(y)=x_2f(y)$ Proven. Subs $x=1$ we have $f(f(1-y)-yf(1))=f(y) \rightarrow f(1-y)-yf(1)=y \dots (1)$ Subs $y=0$ to $(1)$ we have $f(1)=0$ From $(1)$ Subs $f(1)=0$ we have $f(1-y)=y \rightarrow f(y)=1-y$ You have the injectivity totally wrong.

Did this in three minutes. $P(0,0)\Rightarrow f(f(0))=0$ $P(f(0),0)\Rightarrow f(0)\in\{0,1\}$ If $f(0)=0$: $P(x,0)\Rightarrow f(f(x))=0$ $P(f(x),x)\Rightarrow f(x)=0$ If $f(0)=1$: $P(x,0)\Rightarrow f(f(x))=x\Rightarrow f$ injective $P(0,-x)\Rightarrow f(f(x)+x)=0\Rightarrow f(x)=c-x$, testing gives $f(x)=1-x$. Both of these are solutions.

The only constant that works is $\boxed{ f \equiv 0} $, which is our first solution. Let $f$ be non-constant. Let $(x,y) $ be the assertion in given equation. As $f$ is nonzero, we directly get $f$ is surjective from given equation. $P(x,0)$: $f(f(x))=xf(0)$ Hence, $f(0)\neq 0$ (otherwise combining with $f$ surjective, we get $f$ is constant. Contradiction.) Hence $f$ is injective. $P(0,0)$ : $f(f(0))=0$ $P(0,-y)$: $ f(f(y)+yf(0))=0$ Hence, $f(0)=f(y)+yf(0)$ (as $f$ is injective.) Hence, $f(y)=f(0)(1-y)...(1)$ $y=f(0)$ in $(1)$ and $f(f(0))=0$ gives: $f(0)=0$ or $1$ . As, $f(0)\neq0$, $f(0)=1$. Hence from $(1)$ we get that : $\boxed{f(x)=1-x}$ which indeed works and is our second solution.

Denote the assertion by $P(x,y).$ We have $f(f(0))=0$ by $P(0,0).$ And moreover $f(0)\in \{0,1\}$ by $P(f(0),0).$ If $f(0)=0$ then $P(x,0)$ implies $f(f(x))=0$ and so $P(f(x),x)$ implies $f(x)=0$ which fits. Otherwise $P(x,0)$ implies $f(f(x))=x$ and in particular $f(1)=0$ and $f$ is injective. Finally $P(1,1-x)$ implies $f(x)=1-x$ which fits.

Let $P(x, y)$ denote the given assertion. $P(x,0)$ gives us that $f(f(x)) = x.f(0)$ Also $P(\frac{x}{f(y)},y)$ gives us that f is surjective. Now there are two cases: Case 1: $f(0) = 0$. So $f(f(x)) = 0$ and by subjectivity, we have $f(x) = 0$ which is a solution. Case 2: $f(0) \neq 0$. So $f(f(x)) = x.f(0)$ results in f being injective. Now, $P(1,1)$ gives us that $f(f(0) - f(1)) = f(1)$ and by injectivity we have $f(0) - f(1) = 1$ so $f(0) = f(1) + 1$ Also, $P(1,x)$ gives $f(f(1-x) - x.f(1)) = f(y)$ which by injectivity means $f(1-x) - x f(1) = x$ which means $f(1-x) = x(f(1) + 1) = x.f(0)$. Now if we put $x:= 1-x$ we have $f(x) = (1-x)f(0)$. This means f(1) = 0 so f(0) = 1 and so the equation becomes $f(x) = 1-x$, which is indeed a solution. In conclusion, we proved that the only possible functions are $f(x) = 0$ and $f(x) = 1-x$, in which the equality holds. P. S: In the third line, I supposed that there exists an f(y) such that f(y) is not equal to 0. I forgot to clarify about it. This is kinda like proving by contradiction which helps us prove f(x) = 0 in the first case, and it's perfectly fine in the second case.