socrates wrote: Find all functions $f$ defined on all real numbers and taking real values such that \[f(f(y)) + f(x - y) = f(xf(y) - x),\] for all real numbers $x, y.$ Let $P(x,y)$ be the assertion $f(f(y))+f(x-y)=f(xf(y)-x)$. We get: $P(0,0)\rightarrow f(f(0))=0$ $P(0,f(0)),P(f(0),f(0))\rightarrow f(0)=0$ $P(x,0)\rightarrow f(x)=f(-x)\Rightarrow P(x,-x),P(x,x)\rightarrow f(x)\equiv 0$ Hence $ f(x)\equiv 0 \,\,\, \forall x\in \mathbb{R}$.

Nice problem!! Let $P(x,y)$ the given relation. Let $y\in\mathbb{R}$ such that $f(y)\neq 2.$ Then $P(-\frac{y}{f(y)-2},y)$ gives $f(f(y))=0.$ So, if $f(y)\neq 2$ the initial gives: $f(x-y)=f(xf(y)-x)$ and for $x=0$ the last one gives $f(-y)=f(0)$ We conclude that either $f(y)=2$ or $f(-y)=f(0).$ But if $f(y)=2$ then $f(-y)=f(0)-f(2)$ So $f(y)=c_1$ for some $y$ and $f(y)=c_2$ for all the other and $c_1=2\neq 0.$ Take $y_0$ such that $f(y_0)=2.$ Then the initial gives: $f(2)+f(x-y_0)=f(x)$ Taking $x=2$ to the last one we get $f(2-y_0)=0$ so $c_2=0$ and $f(0)=f(2).$ So $f(y)=2$ for some $y$ and $f(y)=0$ for all the other. Now take $x=y=y_0,$ then $f(2)+f(0)=f(y_0)=2.$ But $f(0)=f(2)$ so $f(0)=f(1)=1,$ this is a contradiction. So $f(x)=0$ for all $x.$

Problem (Baltic Way 2014): Find all functions $f$ defined on all real numbers and taking real values such that \[f(f(y)) + f(x - y) = f(xf(y) - x),\]for all real numbers $x, y.$ Solution : Let $P(x,y)$ be the assertion $f(f(y)) + f(x - y) = f(xf(y) - x)$ $f \equiv 1$ isn't obviously a solution. Thus there exists some $a \in \mathbb R$ such that $f(a) \neq 1$ $P(\frac{f(a)}{f(a)-1},a) \implies$ $$ f(\frac{f(a)}{f(a)-1}-a)=0$$ $\implies$ there exists a $k$ such that $f(k)=0$ $P(x,k) \implies f(0)+f(x-k)=f(-x)$ $P(0.k) \implies f(-k)=0$ $P(k,k) \implies f(0)=0$ $\implies f(0)=0$ $$P(0,y) \implies f(f(y)) + f(-y) = 0$$ $$P(\frac{-y}{f(y)-2},y) \implies f(y)=0$$for all $f(y) \neq 2$ Now let there exist some $t \in R$ such that $f(t)=2$ $f(f(t))= - f(t) = -2$ But we know that that for all $f(x) \neq 2$, we have $f(x)=0$ A contradiction. $\implies \boxed{f \equiv 0}$ which indeed is a solution.

$x=y=0 f(f(0))=0$ $y=f(0) f(0)+f(x-f(0))=f(-x)$ $x=0 f(0)+f(-f(0))=f(0)$ $ f(-f(0))=0$ $x=f(0) 2f(0)=f(-f(0))=0$ $ f(0)=0$ $y=0 f(x)=f(-x)$ $x=y f(f(x))=f(xf(x)-x)$ $y=-x f(f(x))+f(2x)=f(xf(x)-x)$ $ f(x)=0$

Sweet and Simple . Answer: We show that the only function satisfying the assertion is $\boxed{f(x) \equiv 0}$. Proof: Let (as usual) $P(x,y)$ denote the assertion $f(f(y)) + f(x-y) = f(xf(y) - x)$. Claim 1. $f(0) =0$. $\longrightarrow$ $P(0,0) \rightarrow f^2(0) + f(0) = f(0) \implies f^2(0) =0$ Now let $f(0) =c $ for some $c \in \mathbb{R}$. Consider $P(0,c) \rightarrow f^2(c) + f(-c) = c \iff f(-c)=0$. Also consider $P(c,c) \rightarrow f^2(c) + c = f( c f(c) -c) \implies 2c = 0 \iff c= 0$. Claim 2. $f$ is even. $\longrightarrow$ Consider $P(x,0) \rightarrow f(x) = f(-x)$ Claim 3. $f \equiv 0$. $\longrightarrow$ Consider the two equations : $P(x,x)$ and $P(x,-x)$. (1) $\rightarrow f^2(x) + f(2x) =f(xf(x) -x).$ (2) $\rightarrow f^2(x) = f(xf(x) -x).$ Compairing the two equations yields $ f(2x) = 0 \forall x\in \mathbb{R}$.

Let $P(x,y)$ denote the assertion $f(f(y))+f(x-y)=f(xf(y)-x).$ We claim that the only solution is $\boxed{f(x)=0}.$ Clearly, this solution does work and we aim to prove that this is the only solution. Claim 1: $f(f(0))=0.$ Note that $P(0,0)$ gives $f(f(0))+f(0)=f(0)$ implying the desired. Let $f(0)=a$ so $f(a)=0$ so $f(f(0))=0$. Claim 2: The only value of $a$ that works is $a=0$ so $f(0)=0.$ Let $P(0,a)$ so $a+f(-a)=f(0)=a\implies f(-a)=0.$ Let $P(a,a)$ so $2a=f(-a)$. Hence, $2a=f(-a)=0\implies a=0.$ Thus, $f(0)=0.$ Claim 3: $f$ is even. Let $P(x,0)$ so $f(x)=f(-x)$ with the fact that $f(0)=0.$ Claim 4: $f(x+y)=f(x-y)$ for all real $x,y\in \mathbb{R}$. Consider $P(x,y)$ and $P(x,-y)$ which give, respectively, \begin{align*} f(f(y))+f(x-y)=f(xf(y)-x)\\f(f(-y))+f(x+y)=f(xf(-y)-x). \end{align*}Rearranging and noting that $f$ is even gives $$f(x-y)=f(xf(y)-x)-f(f(y))=f(xf(-y)-x)-f(f(-y))=f(x+y).$$ Hence, we can set $x=y=\frac{x}{2}$ in $f(x-y)=f(x+y)$ so $f(x)=f(0)=0.$ $\blacksquare$

Taking $x=y=0$ gives $f(f(0))=0$. Taking $y=f(0)$ implies $f(0)+f(x-f(0))=f(-x)$. Letting $x=f(0)/2$ in the previous equation yields $f(0)=0$. Putting this back into the equation it was derived from, we see $f(x)=f(-x)$ or that $f$ is even. From the original FE, we have $f(x-y)=f(xf(y)-x)-f(f(y)).$ Swapping the sign on $y$ and noting evenness, $f(x+y)=f(xf(y)-x)-f(f(y))$. Hence $f(x-y)=f(x+y)$ for all $x,y$, which shows the function is constant. Hence, there is only one solution: $\boxed{f(x)=0}$.

Let $P(x,y)$ denote the assertion $f(f(y))+f(x-y)=f(xf(y)-x).$ $P(0,0)$: $f(f(0))+f(0)=f(0)\implies f(f(0))=0$. $P(0,y)$: $f(f(y))+f(-y)=f(0)$. $P(x,f(0))$: $f(f(f(0)))+f(x-f(0))=f(xf(f(0))-x) \implies f(0)+f(x-f(0))=f(-x)$. Last two together give us that $f(f(x))=-f(x-f(0))$, which means that when $x=f(0)$, thus $f(f(f(0)))=-f(f(0)-f(0)) \implies f(0)=-f(0)\implies f(0)=0$ Hence, $-f(-x)=f(f(x))=-f(x)\implies f(x)=f(-x)$. $P(x,y)$: $f(f(y))+f(x-y)=f(xf(y)-x)$ $P(y,x)$: $f(f(x))+f(y-x)=f(yf(x)-y)$ Since $f(x-y)=f(y-x)$, thus $f(y)-f(x)=f(f(x))-f(f(y))=f(xf(y)-x)-f(yf(x)-y)$, where the first equality comes from knowing that $-f(x)=f(f(x))$. Now, set $x$ to be $0$ and we get that $f(y)-f(0)=f(0\cdot f(y)-0)-f(yf(0)-y)\implies f(y)=-f(-y)=-f(y) \implies f(y)=0\forall y \in \mathbb{R}$. We claim $\boxed{f(y)=0\forall y \in \mathbb{R}}$ to be solution to this problem.

Nice Problem.

Nice problem.

The only solution is that $f \equiv 0$. By setting $x=y=0$, we have that $f(f(0))=0$. For simplicity let's say that $c=f(0)$. Now let's plug into the assertion $y \rightarrow c$, then we have that $c+f(x-c)=f(-x)$. Now let's plug into the assertion $y = 0$, then we must have that $f(x)=f(xc-x)$ Plug in $x=0$, then we must have that $f(f(y))=-f(-y)=-c-f(y-c)$ Set $y=0$ into the upper relation to get that $f(-c)=-c$ Into the original assertion plug in $y=-c$, then we must have that $-c+f(x+c)=f(-xc-x)$. Into the upper relation plug in $x=0$ to get that $c=0$. This implies that $f(x)=f(-x)$. This easily implies that $f(x+y)=f(x-y)$ and then we just set $x=y$, to get that $f(2x)=0$. Thus this implies that $f \equiv 0$.

Let $P(x,y)$ be the assertion $f(f(y)) + f(x - y) = f(xf(y) - x)$. $P(0,0)\Rightarrow f(f(0))=0$ $P(0,f(0))\Rightarrow f(-f(0))=0$ $P(f(0),f(0))\Rightarrow 2f(0)=f(-f(0))\Rightarrow f(0)=0$ $P(x,0)\Rightarrow f(x)=f(-x)$ $P(x,x)\Rightarrow f(f(x))=f(xf(x)-x)$ $P(x,-x)\Rightarrow f(f(x))+f(2x)=f(xf(x)-x)$ Comparing these, $\boxed{f(x)=0}$, which works.

Same solution but anyways. Answer: $f(x) = 0 \ \forall \ x$. Proof: Let $P(x,y)$ denote the given assertion. Claim: $f(0) = 0.$ Proof: $P(0,0) \to f(f(0)) + f(0) = f(0) \implies \boxed{f(f(0)) = 0}$. $P(0 , f(0)) \to f(0) + f(-f(0)) = f(0) \implies \boxed{f(-f(0)) = 0}$. $P(f(0) , f(0)) \to f(0) + f(0) = f(-f(0)) = 0 \implies \boxed{f(0) = 0}$. Claim: $f$ is even. Proof: $P(x,0) \to f(f(0)) + f(x) = f(xf(0) - x) \implies \boxed{f(x) = f(-x)}$. $P \left(\frac x2 , \frac x2 \right) \to f \left( f \left (\frac x2 \right) \right) + f(0) = f \left( \frac x2 f \left( \frac x2 \right) - \frac x2 \right) \implies f \left( f \left (\frac x2 \right) \right) = f \left( \frac x2 f \left( \frac x2 \right) - \frac x2 \right)$ $P \left(\frac x2 , -\frac x2 \right) \to f \left( f \left ( - \frac x2 \right) \right) + f(x) = f \left( \frac x2 f \left( - \frac x2 \right) - \frac x2 \right)$ $\implies f \left( f \left ( \frac x2 \right) \right) + f(x) = f \left( \frac x2 f \left( \frac x2 \right) - \frac x2 \right)$ (Using $f$ is even.) $\implies f \left( f \left ( \frac x2 \right) \right) + f(x) = f \left( f \left ( \frac x2 \right) \right)$ (Equating $P \left(\frac x2 , \frac x2 \right)$ and $P \left(\frac x2 , -\frac x2 \right)$ .) $\implies f(x) = 0$.

Let $P(x,y)$ be the assertion. $P(0,0)\implies f(f(0))=0$. $P(x,0)\implies f(x)=f(x(f(0)-1))$. If $c=f(0)-1 \neq 1$, then comparing $P(x,y)$ with $P(x,yc)$ gives $f(x)=f(x+y(1-c))$, which implies that $f$ is constant since $c\neq 1$. So only solution is $f(x)=0\ \forall x\in \mathbb{R}$ in this case. If $f(0)=2$, then we have $f(2)=0$. From $P(x,2)$ we get $2+f(x-2)=f(-x)$, but plugging $x=4$ and $x=-2$ gives a contradiction. Thus our only solution is $\boxed{f(x)=0 \ \forall x\in \mathbb{R}}$, which clearly fits.

Let $P(x,y)$ be the assertion of $f(f(y)) + f(x - y) = f(xf(y) - x)$. We have $P(0,0)\implies f(f(0))=0$, thus $P(0,f(0))\implies f(-f(0))=0$ and therefore $P(f(0),f(0))\implies f(0)=0$. Now, $P(x,0)\implies f(x)=f(-x)$ and $P(0,x)\implies f(f(x))=-f(-x)=-f(x)$. This means that $$-f(x)=f(f(x))=f(-f(x))=f(f(f(x)))=-f(f(x))=f(x)\implies f\equiv 0,$$this obviously works.

Let $P(x,y)$ denote the assertion. $P(0,0)\implies f(f(0))=0$ $P(1,0) \implies f(1)=0$ $P(0,1) \implies f(-1)=0$ $P(1,1) \implies f(0)=0$,so by induction $\boxed{f(x)=0 \forall x\in \mathbb{R}}$

Sprites wrote: Let $P(x,y)$ denote the assertion. $P(0,0)\implies f(f(0))=0$ $P(1,0) \implies f(1)=0$ $P(0,1) \implies f(-1)=0$ $P(1,1) \implies f(0)=0$,so by induction $\boxed{f(x)=0 \forall x\in \mathbb{R}}$ Will you tell me how does $P(1,0) \implies f(1)=0 ?$

Sprites wrote: Let $P(x,y)$ denote the assertion. $P(0,0)\implies f(f(0))=0$ $P(1,0) \implies f(1)=0$ $P(0,1) \implies f(-1)=0$ $P(1,1) \implies f(0)=0$,so by induction $\boxed{f(x)=0 \forall x\in \mathbb{R}}$ How do you induct to prove over $\mathbb R$?

hakN wrote: Let $P(x,y)$ be the assertion. $P(0,0)\implies f(f(0))=0$. $P(x,0)\implies f(x)=f(xf(0))$. Plugging $x=1$ in above, we get $f(1)=0$. $P(0,x)\implies f(f(x))=f(0) - f(-x)$. Plugging $x=1$ in above, we get $f(-1)=0$. $P(x,1)\implies f(0) + f(x-1) = f(-x)$. Plugging $x=1$ in above, we get $f(0)=0$. So we get $\boxed{f(x)=0 \forall x\in \mathbb{R}}$ which clearly fits. Can u explain how u received line 3 How f(x)=f(xf(0))$

Let $P(x,y)$ be the assertion. $P(0,0): f(f(0))=0.$ Let $f(0)=a \Rightarrow f(a)=0$ $P(a,a):f(-a)=2a$ $P(0,a): 3a=f(0)=a\Rightarrow f(0)=0$ $P(x,0):f(-x)=f(x) \forall x \in \mathbb{R}$ $P(x,-y):f(f(y))+f(x+y)=f(xf(y)-x)\Rightarrow f(x-y)=f(x+y) \forall x,y (1)$ In (1), let $x=y$, we get $f(x)=a \text{(a=const)}$ It is easy to show that $f(x)=0$ is the only function which works.

$$x=0 \implies f(0)=f(f(y))+f(-y)\rightarrow (1),x=y \implies f(0)=f(xf(x)-x)-f(f(x)) \implies f(f(0))=0$$$$x=y=f(0)\implies 2f(0)=f(-f(0)),y=f(0),x=0 \implies f(0)=f(f(f(0)))+f(-f(0))=f(0)+2f(0)=3f(0)\implies f(0)=0$$$$y=0 \implies f(x)=f(-x),f(0)=0 \implies f(f(x))=-f(x)\text{from (1)},f(x)=-(-f(x))=-f(f(x))=f(f(f(x)))=f(-f(x))\implies f(-f(x))=f(x)$$Finally, $$x=f(x),y=0 \implies f(f(x))=f(-f(x)) \implies f(x)=-f(x) \implies f(x)=0$$

Let $P(x,y)$ denote the given assertion. $P(0,0)$ gives $f(f(0))=0.$ Further, $P(0,f(0))$ gives $f(-f(0))=0$ and finally $P(f(0),f(0))$ gives $f(0)=0.$ Then $P(x,0)$ implies $f$ is even. Subtracting $P(x,x)$ from $P(x,-x)$ implies $f(2x)=0.$ And it's easy to verify the identity zero function works.

Rahmanhabibi wrote: Can u explain how u received line 3 How f(x)=f(xf(0))$ It's wrong.

The only solution is $\boxed{f\equiv 0}$, which works. Let $P(x,y)$ denote the given assertion. Let $f(0) = a$. $P(0,0): f(a) = 0$. $P(0,a): a + f(-a) = a$, so $f(-a) = 0$. $P(a,a): 2a = f(-a)$, so $a=0$. $P(x,0): f(x) = f(-x)$. $P(0,x): f(f(x)) + f(x) = 0$. $P(x+y,y): f(f(y)) + f(x) = f(xf(y) - x)$. $P(x+y,-y): f(f(y)) + f(x+2y) = f(xf(y) - x)$. So $f(x) = f(x+2y)$, which implies $f$ is constant, so $f\equiv 0$.

I claim that the only solution is $f(x) \equiv 0$, which works. Consider the following: \begin{align*} P(0, 0): &\rightarrow f(f(0)) = 0 \\ P(0, x): &\rightarrow f(f(x)) + f(-x) = f(0) \\ P(x, x): &\rightarrow f(f(x)) + f(0) = f(xf(x)-x) \\ P(x, 0): &\rightarrow f(x) = f(-x) \\ \\ P(0, f(0)): &\rightarrow f(-f(0)) = 0 \\ P(f(0), f(0)): &\rightarrow f(0) = 0 \\ \\ P(-x, x) \wedge P(x, 0): &\rightarrow f(f(x)) + f(-2x) = f(xf(x) - x) \\ P(-x, x) \wedge P(x, x): &\rightarrow f(-2x) = 0. \end{align*} Thus we have that as $-2x$ is surjective, $f(x) \equiv 0$ for all $x \in \mathbb{R}$. $\blacksquare$

We claim that the only solution is $f\equiv0$, which clearly fits. We refer the assertion $f(f(y))+f(x-y)=f(xf(y)-x)$ as $p(x,y)$. Note that if $f(x)$ is a constant function then $f\equiv0$. Assume that $f(x)$ is a non-constant function. Note that $p(0,-y)$ gives $f(y)=f(f(-y))$. Since the function is non-constant, there exists $y$ such that $f(y)\ne2$. Then, $p\left(\frac{y}{2-f(y)},y\right)$ gives $f(f(y))=0$. But $f(f(y))=f(-y)$. So, $f(-y)=0$. Now, $p\left(-\frac{y}{2},-y\right)$ gives $f(f(-y))=0=f(y)$. We conclude that either $f(y)\in\{0,2\}$. Note that $p(0,0)$ gives $f(f(0))=0=f(0)$. But $p(x,0)$ gives $f(x)=f(-x)$. Take $y$ such that $f(y)=2$. We have $2=f(y)=f(-y)=f(f(y))=f(2)$. Then, $p(x+2,2)$ gives $2+f(x)=f(x+2)$. It follows that $f(x)=0$ and $f(x+2)=2$. There is no restriction on $x$, so we can take any real and $f(x)=0$. So, $f\equiv0$, contradicting $f(2)=2$. So, there are no solutions to the given equation other than $f\equiv0$. $\square$

Let $P(x,y)$ denote the assertion. Note that if $f(y)\neq 2$, $f(f(y))=0$. Proof: Choose $x=-\frac{y}{f(y)-2}$. Now clearly exists a number $l$ such that $f(l)\neq 2$, since $f(x)=2$ for all $x$ is not a solution. Hence $f(f(l))=0, f(f(f(l)))=f(0)=0$ and so $f(0)=0$. Now suppose exists $k$ such that $f(k)=2$, $P(k,k): f(2)+f(0)=f(k)=2$ so $0=f(0)=f(2)-2$. So $f(2)=2$. Now $P(x,0): 0+f(x)=f(-x)$, so $f(x)=f(-x)$. $P(0,k): 2+f(-k)=0$. So $f(-k)=-2$ at the same time $f(k)=2$ and since $f(k)=f(-k)$, $2=-2$ which is not okay So there is no $k$ such that $f(k)=2$. So $f(f(x))=0$ for all $x$. $P(0,y): f(-y)=f(0)$ so $f$ is constant. Now checking only $f(x)=0$ works.

My solution is similar to #$6,7$ but I'm gonna post it anyway

Let $P(x, y)$ denote the assertion. $P(0,0)$ gives $f(f(0))+f(0)=f(0)$ so $f(f(0))=0$. $P(x, 0)$ gives $f(x)=f(x(f(0)-1))$. $P(1, y)$ gives $f(f(y))+f(1-y)=f(f(y)-1)$ so plugging $y=f(0)$ gives $f(0)+f(1-f(0))=f(-1)$ but $f(x)=f(x(f(0)-1))$ so $f(0)+f(-1)=f(-1)$ so $f(0)=0$. Plugging this back into $f(x)=f(x(f(0)-1))$ gives $f(x)=f(-x)$. $P(0, y)$ gives $f(f(y))+f(-y)=0$ so $f(f(y))=-f(y)$. We claim that the only solution is $\boxed{f(x)=0}$. Assume FTSOC that there exists some $c\ne0$ in the image of $f$. Then, $f(c)=f(-c)=-c$ so $-c\ne 0$ is also in the image of $f$. However, this means that $f(-c)=f(c)=c$. Since $c\ne -c$, this is a contradiction, as desired.