Tintarn wrote:

What about $f(x)=1-x$ ?

Let $y=0$, we obtain: $f(f(x)) = xf(0)\cdots\cdots\cdots (1)$. We will now consider two cases. Case 1. $f(0)\neq 0$: It follows that $f$ is bijective from $(1)$ as the $RHS$ is bijective. Since $f$ is surjective, there exists $a\in\mathbb{R}$ such that $f(a) = 0$. Let $y=a$ and $x=1$. We obtain $$f(a+f(1)) = f(a).$$As $f$ is injective, we have $a+f(1) = a\iff a = 1$, again, because $f$ is injective. Finally, $y=1$ implies $$f(x+f(x)) = xf(1) = 0 = f(1)\implies f(x) = 1-x,$$which is indeed a solution. Case 2. $f(0) = 0$: It follows from $(1)$ that $f(f(x)) = 0, \forall x$. $f(x) = 0$ is clearly a solution. Assume there exists another function with $f(x_0)\neq 0$ for some $x_0\neq 0$. Plug in $y=x_0$ to obtain $$f(xx_0 + f(x)) = xf(x_0).$$Again, the $RHS$ is bijective, so $f$ is surjective, therefore there exists $a\in\mathbb{R}$ such that $f(a) = x_0$. However, $f(f(a)) = 0\iff f(x_0) = 0$, a contradiction.

dangerousliri wrote: What about $f(x)=1-x$ ? Thanks for pointing that out. There was a little computational error in the case $f(0)=1$ leading to $f(x)=-x$ (which is of course not a solution) instead of the immediate $f(x)=1-x$. It is now fixed and everything should be fine. Btw, nice problem, did you propose it?

dangerousliri wrote: Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that: $$f(xy+f(x))=xf(y)$$for all $x,y\in\mathbb{R}$. Solution. Denote by $P(x,y)$ the assertion $f(xy+f(x))=xf(y)$. Consider the two cases below. (1). Assume $f(0)\ne0$. Since that $P(x,0)\Longrightarrow f\left(f(x)\right)=xf(0)$ for all $x\in\mathbb{R}$, $f$ is injective, which implies that $$P(1,y)\Longrightarrow f\left(y+f(1)\right)=f(y)\Longrightarrow y+f(1)=y\Longrightarrow f(1)=0,$$thereby $P(x,1)\Longrightarrow f\left(x+f(x)\right)=0\Longrightarrow x+f(x)=1\Longrightarrow f(x)=1-x$. (2). If $f(0)=0$, then $P(x,0)\Longrightarrow f\left(f(x)\right)=0$ for all $x\in\mathbb{R}$, whence \begin{align*}&f(xy+f(x))=xf(y)\,\forall x,y\in\mathbb{R}\\ \Longrightarrow&f\left(f(xy+f(x))\right)=f\left(xf(y)\right)\,\forall x,y\in\mathbb{R}\\ \Longrightarrow&0=f\left(xf(y)\right)\,\forall x,y\in\mathbb{R},\\ \end{align*}which implies $0=f(z)\,\forall z\in\mathbb{R}$, since that $f(z_0)\ne0$ gives $0=f\left(\frac{z_0}{f(z_0)}\cdot f(z_0)\right)=f(z_0)$. $\blacksquare$

Tintarn wrote: dangerousliri wrote: What about $f(x)=1-x$ ? Thanks for pointing that out. There was a little computational error in the case $f(0)=1$ leading to $f(x)=-x$ (which is of course not a solution) instead of the immediate $f(x)=1-x$. It is now fixed and everything should be fine. Btw, nice problem, did you propose it? Yes.

dangerousliri wrote: Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that: $$f(xy+f(x))=xf(y)$$for all $x,y\in\mathbb{R}$. Solution. We have \[f(f(x)) = xf(0).\]Suppose $f(0)\neq 0$. This gives, $f$ is bijective. We have \[f(x)\cdot f(0)=f(f(f(x))) = f(xf(0))\implies f(1)f(0) = f(f(0)) = 0\cdot f(0) = 0,\]but $f(0)\neq 0$, so $f(1) = 0$. Therefore $f(x+f(x)) = 0 = f(1)$. As $f$ is injective, $\boxed{f(x) = 1-x}$ for all $x$ which indeed is a solution. Now suppose $f(0) = 0$. We have $f(f(x)) = 0$. If there exists a real $a$ such that $f(a)\neq 0$, then $f(xa+f(x))=xf(a)$ which gives $f$ is surjective, so there exists a real $b$ such that $f(b) = a$ giving $0=f(f(b)) = f(a)$ which is a contradiction. Therefore there is no such $a$ giving $\boxed{f(x)=0}$ for all $x$ which indeed is a solution. $\blacksquare$

Setting $x=0$ results in $f(f(0)) = 0$. Setting $y=0$ results in $f(f(x)) = x f(0)$. Since the range of $f$ is a real number, $x$ can be set to equal $f(0)$, and doing so results in $f(f(f(0))) = f(0) = f(0)^2$. Thus, $f(0) ( f(0) - 1 ) = 0$. If $f(0) = 1$, then $f(1) = 0$ and $f(f(x)) = x$. Setting $y=1$ results in $f(x + f(x)) = 0$. Thus, from definition of function, $f(f(x + f(x)) = f(0)$, so $x + f(x) = 1$. Then one of the working functions is $\boxed{f(x) = 1-x}$. If $f(0) = 0$, then setting $y=0$ results in $f(f(x)) = 0$. From definition of function, $f(f(xy + f(x))) = f(x f(y)) = 0$. Assume that there exists $y$ such that $f(y) = a, a \neq 0$. Since $\tfrac{y}{a}$ is a real number, setting that for $x$ results in $f(y) = 0$, a contradiction. Thus, $f(y) = 0$ for all $y$, so the other function that works is $\boxed{f(x) = 0}$.

Let $P(x,y)$ be the assertion. $P(0,0)\implies f(f(0))=0$. $P(f(0),0)\implies f(0)=f(0)^2$, thus $f(0)=0$ or $f(0)=1$. Case 1. $f(0)=0$. $P(x,-\frac{f(x)}{x})\implies f\left(-\frac{f(x)}{x}\right)=0$ for all $x\neq 0$. $P(-\frac{f(x)}{x},x)\implies f(-f(x))=-\frac{f(x)^2}{x}$. $P(f(x),-1)\implies -\frac{f(x)^2}{x}=f(-f(x))=f(x)f(-1)$, thus $f(x)=0$ either $f(x)=-xf(-1)$. Suppose we have $a\neq 0$ so that $f(a)=-af(-1)$: $P(a,a)\implies f(a^2-af(-1))=-a^2f(-1)$, thus $-a^2f(-1)=0$ or $-a^2f(-1)+af(-1)^2=-a^2f(-1)\implies af(-1)^2=0$, we conclude that $f(-1)=0$, thus $\boxed{f(x)=0}$ for all $x$. Case 2. $f(0)=1$. Since, $f(f(0))=0$, then $f(1)=0$. $P(x,0)\implies f(f(x))=x$, thus $f$ is bijective. $P(x,1)\implies f(x+f(x))=0=f(1)$, by injectivity, we have $\boxed{f(x)=1-x}$. This obviously satisfies given.

dangerousliri wrote: Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that: $$f(xy+f(x))=xf(y)$$for all $x,y\in\mathbb{R}$. Kiyoot problem! Let $P(x, y)$ be the assertion. We claim that the two solutions to the given functional equation are $\boxed{f(x) = 1-x}$ for all reals $x$ and $\boxed{f(x) = 0}$ for all reals $x$, these functions do work and therefore it suffices to prove that they are the only such functions. $P(x, 0) \implies f(f(x)) = xf(0)$. If $f(0) \neq 0$ then we immediately get that $f$ is bijective. Now, $P(1, 0) \implies f(f(1)) = f(0)$ or due to bijectivity we get that $f(1) = 0$ and therefore $P(x, 1) \implies f(x + f(x)) = 0 = f(1) \implies \boxed{f(x) = 1 - x}$. If $f(0) = 0$, then $f(f(x)) = 0$ and so $f(xy + f(x)) = xf(y) \implies f(f(xy + f(x)) = 0 = f(xf(y)) \dots (1)$, so if there exists a real number $z$ such that $f(z) \neq 0$, then in $(1)$ let $x = \dfrac{z}{f(z)}$ and $y = z$, this implies that $f(z) = 0$ which is the desired contradiction, which means that $\boxed{f(z) = 0}$ is the only solution when $f(0) = 0$ and $\boxed{f(z) = 1-z}$ is the only solution when $f(0) \neq 0$ which is the desired contradiction.

classic, I love it

One word for this FE: kawaii! We claim the solutions are $f(x) = 1-x$, $f(x) = 0$. Note that $P(x, 0) \implies f(f(x)) = xf(0)$. We have two cases: Case 1: $f(0) \ne 0$. Then $f$ is injective, so $P(1, x) \implies f(x+f(1)) = f(x) \implies f(1) = 0.$ And now $P(x, 1) \implies f(x+f(x)) = 0 \implies x+f(x) = 1$, yielding the first solution. Case 2: $f(0) = 0$. Assume $\exists$ $x_0 \ne 0$ with $f(x_0) \ne 0$. Then $P(x, x_0) \implies f$ is surjective. But $f(f(x)) = 0 \forall x \implies f(x) = 0 \forall x$, contradiction. So $f(x) = x \forall x$. And we're done, as both work.

Let $P(x,y)$ denote the given assertion. By $P(x,0)$ we have $f(f(x)) = x.f(0)$ Now there are two cases. Case 1: $f(0) = 0$. Then we have $f(f(x)) = 0$. We are going to prove by contradiction that $f(x) = 0 \quad \forall x \in \mathbb{R}$, so we suppose that $\exists a \in \mathbb{R} \quad f(a) \neq 0$. By $P(x,f(a))$ we get that $f$ is surjective. So $f(f(x)) = 0$ means $f(x) = 0 \quad \forall x \in \mathbb{R}$, contradiction. So $f(x) = 0 \quad \forall x \in \mathbb{R}$ is the only solution in this case, in which the given equality holds. Case 2: $f(0) \neq 0$. So $f(f(x)) = x.f(0)$ results in $f$ being injective. By putting $x := 1$ in the same equality, we have that $f(f(1)) = f(0)$, and $f$ is one to one so $f(1) = 0$. Then $P(x,1): f(x + f(x)) = 0 = f(1)$ and because $f$ is injective we have $x + f(x) = 1$. So the only solution in this case is $f(x) = 1 - x \quad \forall x \in \mathbb{R}$

This is the solution