I claim $f(x)=x-1$ for all $x$, or $f(x)=-x-1$ for all $x$. I sketch the steps as follows. As usual, denote by $P(x,y)$ the given assertion. 1) $P(x,1)$ implies $f$ is injective: if $f(x_1)=f(x_2)$ for some $x_1,x_2$ then $x_1=x_2$. 2) $P(1,1)$ gives $f(0)=-1$. 3) $P(0,x)$ gives $x-1=f(f(x) +1)$. Likewise, $P(-x,0)$ gives $x-1=f(-1-f(-x))$. 4) Steps 2 and 3 together with the injectivity implies $f(x)+f(-x)=-2$ for all $x$. 5) Now inspecting $P(x,1)$ we get $f(f(1)-f(x)) = -x$. Likewise, $P(1,x)$ gives $f(x) = xf(1)+1 +f(f(x)-f(1))$. This, together with $f(f(1)-f(x))+f(f(x)-f(1))=-2$ brings: for some $k\in\mathbb{R}$, it holds that $f(x)=kx-1$ for every $x$. 6) Insert $f(x)=kx-1$, and study to obtain $k\in\{\pm 1\}$. The end.

grupyorum wrote: I claim $f(x)=x-1$ for all $x$, or $f(x)=-x-1$ for all $x$. I sketch the steps as follows. As usual, denote by $P(x,y)$ the given assertion. 1) $P(x,1)$ implies $f$ is injective: if $f(x_1)=f(x_2)$ for some $x_1,x_2$ then $x_1=x_2$. 2) $P(1,1)$ gives $f(0)=-1$. 3) $P(0,x)$ gives $x-1=f(f(x) +1)$. Likewise, $P(-x,0)$ gives $x-1=f(-1-f(-x))$. 4) Steps 2 and 3 together with the injectivity implies $f(x)+f(-x)=-2$ for all $x$. 5) Now inspecting $P(x,1)$ we get $f(f(1)-f(x)) = -x$. Likewise, $P(1,x)$ gives $f(x) = xf(1)+1 +f(f(x)-f(1))$. This, together with $f(f(1)-f(x))+f(f(x)-f(1))=-2$ brings: for some $k\in\mathbb{R}$, it holds that $f(x)=kx-1$ for every $x$. 6) Insert $f(x)=kx-1$, and study to obtain $k\in\{\pm 1\}$. The end. Nice Solu

Arabian_Math wrote: @2above while proving the injectivity you need to handle the case when $f(1)=0$ separately because you get something like $af(1)=bf(1)$ which does not necessarily imply $a=b$ unless $f(1)\ne 0$. $P(x,1)$ $f(f(1)-f(x))=-x$ Now what? I mean from this $f$ is injective without having $f(1)=0$ Is it right?

GorgonMathDota wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[ f(xy) = yf(x) + x + f(f(y) - f(x)) \]for all $x,y \in \mathbb{R}$. Nice and easy problem! My Solution $P(1,1)\implies f(0)=-1$ $P(0,1)\implies f(f(1)+1)=0$ $P(0,x)\implies f(f(x)+1)=x-1\implies f-$bijective $P(f(1)+1,f(1)+1)\implies f((f(1)+1)^2)=f(1)\implies (f(1)+1)^2=1\implies f(1)=0,f(1)=-2$ Case 1.$f(1)=0$ $P(1,x)\implies f(f(x))=1-f(x),f(x)_{surjective}=t\in R\implies f(t)=1-t$ Case 2.$f(1)=-2$ $P(x,x)\implies f(x^2)=x(f(x)+1)-1$ $f(x^2)=f((-x)^2)\implies x(f(x)+1)-1=(-x)\cdot (f(-x)+1)-1\implies f(-x)=-2-f(x)$ $P(-1,x)\implies f(f(x))=1+f(-x)=1-2-f(x)=-1-f(x),f(x)_{surjective}=t\in R\implies f(t)=-1-t$

Answer

A nice FE! Ans: $f(x)=x-1$, $f(x)=-x-1$. It is easy to see that these satisfy the above FE. Let $P(x,y)$ denote the given assertion, using $P(x,1)\implies f(f(1)-f(x))=-x$, we observe that $f$ is bijective. Also, $P(1,1)\implies f(0)=-1$. Using $P(0,x)$ and $P(-x,0)$, we get that \[f(f(x)+1)=-1+x, \ f(-1-f(-x))=-1+x\]and by injectivity, we have $f(x)+f(-x)=-2$. Let $\lambda \in \mathbb{R}$ such that $f(\lambda )=0$, we have \[P(\lambda ,\lambda )\implies f(\lambda ^2)=\lambda -1\]\[P(\lambda ,0)\implies f(f(0))=f(-1)=-\lambda -1\implies f(1)=-2-f(-1)=\lambda -1=f(\lambda ^2)\]and $\lambda =\pm 1$. Case 1: $\lambda =-1$. We have $f(0)=-1, f(1)=-2, f(-1)=0$, using $P(-1,x)$, we have $-2-f(x)=f(-x)=-1+f(f(x)) \implies f(f(x))+f(x)=-1$. Also, $P(x,-1)\implies -2-f(x)=f(-x)=-f(x)+x+f(-f(x)) \implies f(-f(x))=-2-x$. Thus, \[-1=f(f(-f(x)))+f(-f(x))=f(-2-x)-2-x \implies f(-2-x)=x+1 \implies f(x)=-x-1.\]Case 2: $\lambda =1$. We have $f(0)=-1, f(1)=0, f(-1)=-2$, using $P(1,x)$, we have $f(f(x))=f(x)-1$. Also, $P(x,1)\implies f(-f(x))=-x$. Thus, \[f(-x)=f(f(-f(x)))=f(-f(x))-1=-x-1 \implies f(x)=x-1.\]And we are done. $\blacksquare$

i just have a question... Why do i see $\blacksquare$ after some solutions? what does it mean?

angrybird_r wrote: i just have a question... Why do i see $\blacksquare$ after some solutions? what does it mean? It`s often useful to break the problem up into parts called Claims, you claims something, and then prove it, it makes solution easuer to understand. After the proof of certain claim is completed, you write $\square$, which means "This proof is complete, and i am ready to get back to the solution". And when the whole solution is done you may add $\blacksquare$. It`s useful when author wants to add some comments/tips/motivation about the problems under the solution, but these are not realted to solution process itself.

Nice problem, but definitely too easy for A2. Let $P(x,y)$ denotes assertion of given functional equation. Note that $P(1,1)$ gives us that $f(0) =-1$. Now observe that $P(x,0)$ and $P(0,x)$ yields to: $$ f(-1-f(x)) = - x -1 \quad \text{and} \quad x- 1 = f(f(x) +1)$$Adding these two relations together we obtain that $f(f(x) + 1) + f( -1 - f(x)) =-2 $. Since $f(f(x)+1) = x -1$ we see that $f(x)$ is surjective function, so $f(x)+1$ varies through all real numbers and consequently $f(x)+f(-x) =-2$. Now $P(x,y)+P(y,x)$ gives us that: $$ 2f(xy) = xf(y) + yf(x) +x+y -2 \quad (1) $$Setting $y=1$ in (1) reveals that: $$ f(x) =x(f(1) +1) - 1 $$This implies that $f(x) =ax -1 $. But since $f(f(x) + 1) = x-1$ we get that $a^2x = x \implies a =1 \quad \text{or} \quad a =-1$. Thus $f(x) =-x-1$ and $f(x) = x -1 $, which both clearly work.

Here is my solution. $P(1,1): f(0)=-1$ $P(0,x): f(f(x)+1)=x-1$ and so it is obvious that $f$ is bijective. $P(x,0): f(-1-f(x))=-x-1$ Combined with the previous line and using surjecitvity, $f(x)+f(-x)=-2$ Comparing $P(1,x),P(x,1)$ and using the previous result gives $f(x)=cx-1$ for some constant $c$ and we are done.

Solution. I claim that the only solution to this functional equation is $f(x)=\pm x-1, \quad\forall x \in \mathbb{R}$. $P(1, 1)\implies f(0)=-1, P(0,x)\implies f(f(x+1))=x-1 \implies f$ is bijective. $P(0, x)+P(x, 0) \implies f(f(x) + 1) + f( -(f(x)+1) =-2\implies f(x)+f(-x)=-2.$ $P(x, y)+P(y, x) \implies 2f(xy)=xf(y)+yf(x)+x+y-2.$ $y=1 \implies f(x)=kx-1, f(f(x) + 1) = x-1\implies k^2x=x \implies k=\pm 1, \implies \boxed{f(x)=\pm x-1, \forall x \in \mathbb{R}.}\quad\blacksquare$

BMO SL 2019 A2 wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[ f(xy) = yf(x) + x + f(f(y) - f(x)) \]for all $x,y \in \mathbb{R}$. Let $P(x,y)$ be the given FE. To start, $P(1,0) \Rightarrow f(0)=-1$, hence $P(x,x) \Rightarrow f(x^2)=xf(x)+x-1$. We proceed now with a Claim: Claim: $f(x)+f(-x)=2$ for all $x$. Proof: Take $x \rightarrow -x$ at $f(x^2)=xf(x)+x-1$ and obtain $-xf(-x)-x-1=f(x^2)=xf(x)+x-1$ hence $f(x)+f(-x)=-2$ for non-zero $x$. Since $f(0)=-1$ we can actually extend the previous relation to all reals $\blacksquare$. Back to the problem, $P(x,1)$ implies $$f(f(1)-f(x))=-x$$which using the Claim rewrites as $$f(f(x)-f(1))=x-2$$Hence, $$P(1,x) \Rightarrow f(x)=f(1)x+1+f(f(x)-f(1))=(f(1)+1)x-1,$$so $f(x)$ equals $cx-1$ for a constant $c$. Putting this back into $P(x,y)$ we easily obtain that $c= \pm 1$, therefore the two solutions are $f(x) \equiv x-1$ and $f(x) \equiv -x-1$.

Nice FE The solutions are $f(x)=x-1$ and $f(x)=-x-1$, $\forall x \in \mathbb{R}$ First we plug in $y=1$, to get that $f(f(1)-f(x))=-x$, this implies that $f$ is a bijection, also this means that $f(0)=-1$ Next we plug in $x=y$, to get that $f(x^2)=xf(x)+x-f(0)=xf(x)+x-1$. Since $x^2=(-x)^2$ we have that: $$xf(x)+x-1=-xf(-x)-x-1$$this implies that $f(-x)=-f(x)-2$ This implies that $f(f(x)-f(1))=x-2$ Plugging in $x=0$ into the above result we have that: $$f(-1-f(1))=-2$$this implies that: $$f(1+f(1))=0$$ We plug in $x=y=1+f(1)$, to get that: $$f((1+f(1))^2)=f(1)$$by injectivity this implies that $(1+f(1))^2=1$ Thus we have a first case when $f(1)=0$. Now we just plug in $x=1$ to get that $f(y)-1=f(f(y))$, by surjectivity we have that $f(t)=t-1$ The second case is when $f(1)=-2$, this implies that $f(-1)=0$, and in turn we have that $f(f(x)+2)=x-2$. Now we set $x=1$ to get that $f(y)=-y-1$ By checking the solutions we confirm that these are the only two valid solutions.

What a fool am I? I obviously made that too hard for that problem. Let $P(x,y)$ be the assertion of $f(xy) = yf(x) + x + f(f(y) - f(x))$. $P(x,0)\implies f(0)=x+f(f(0)-f(x))\implies f(f(0)-f(x))=f(0)-x\implies f\text{ is bijective}$. $P(x,1)\implies f(x)=f(x)+x+f(f(1)-f(x))\implies f(f(1)-f(x))=-x\implies f(0)=-1$. $f(f(0)-f(x))=f(0)-x$ takes the form $f(-f(x)-1)=-x-1$. $P(1,x)\implies f(x)=xf(1)+1+f(f(y)-f(1))$ Since $f\text{ is bijective}$, we have: let $f(1)=a$, $f(b)=a+1$. Set $x=b$ to the latter we have that, $a+1=ab+1+a\implies ab=0$, thus $f(0)=a+1$ or $f(1)=0$. $P(x,x)\text{ and } P(-x,-x)\implies -xf(-x)-x-1=f(x^2)=xf(x)+x-1\implies f(x)+f(-x)=-2\,\forall x\in \mathbb R-0$. Thus, $f(-1)=-2-f(1)$. $\diamondsuit$: $P(x,x+1)\implies f(x^2+x)=(x+1)f(x)+x+ f(f(x+1)-f(x))$ $P(x+1,x)\implies f(x^2+x)=xf(x+1)+x+1+ f(f(x)-f(x+1))$ $f(-f(x)-1)=-x-1=f(f(1)-f(x+1)$, thus by injectivity, $-f(x)-1=f(1)-f(x+1)\implies f(x+1)-f(x)=f(1)+1$ Case 1. $f(0)=a+1$. Since $f(0)=-1$, then $a=-2$ and we have that $f(1)=-2$. Thus, $f(x+1)-f(x)=f(1)+1=-2+1=-1$. We also have that $f(-1)=-2-f(1)=0$. Thus by $\diamondsuit$ and last two, we have $xf(x+1)+x+1+ f(1)=(x+1)f(x)+x+ f(-1)$; $xf(x+1)+x+1-2=(x+1)f(x)+x\implies f(x)+x=x-1-x\implies f(x)=-(x+1)\,\forall x \in\mathbb R-0$ (notice that $f(0)$ also satisfies this). Case 2. $f(1)=0$. Thus, $f(x+1)-f(x)=f(1)+1=0+1=1$. We also have that $f(-1)=-2-f(1)=-2$. Thus by $\diamondsuit$ and last two, we have $xf(x+1)+x+1+ f(-1)=(x+1)f(x)+x+ f(1)$; $xf(x+1)+x+1-2=(x+1)f(x)+x\implies f(x)+x=x-1+x\implies f(x)=x-1\,\forall x \in\mathbb R-0$ (notice that $f(0)$ also satisfies this). Those two are indeed solutions by checking. Answer. $\boxed{f(x)=x-1\,\forall x \in\mathbb R\text{ and } f(x)=-(x+1)\,\forall x \in\mathbb R}$.

Let $P(x,y)$ be the assertion. $P(x,1) \implies f(f(1) - f(x)) = -x \implies f$ is bijective, $f(0) = -1$. $P(x,x) \implies f(x^2) = -1 + xf(x) + x = -1 - xf(-x) - x \implies f(-x) = -f(x) - 2$. $P(0,x) \implies f(f(x) + 1) = x-1$ and so $f(f(1) + 1) = 0$. Plugging $x=f(1) + 1$ in the first we got, we have $f(f(1)) = -f(1) - 1$. $P(1,f(1) + 1) \implies 0 = f(1)^2 + f(1) + 1 + f(-f(1))$. If $f(1)\neq 0$, then $f(-f(1)) = -f(f(1)) - 2 = f(1) - 1$ and the above equation becomes $f(1)^2 + 2f(1) = 0 \implies f(1) = -2$. $P(x,1) \implies f(-2 -f(x)) = -x \implies f(f(x) + 2) = x-2$ for all $x\neq 1$ but since it is also true for $x=1$, we have this for all $x$. $P(1,x) \implies f(x) = -2x + 1 + f(f(x) +2) = -x - 1$ and so $\boxed{f(x) = -x - 1 \ \forall x \in \mathbb{R}}$ is a solution. If $f(1) = 0$, then $P(1,x) \implies f(x) = 1 + f(f(x))$ and since $f$ is surjective, we have $\boxed{f(x) = x - 1 \ \forall x\in \mathbb{R}}$ as our last solution.

Let $P(x,y)$ be the assertion $f(xy) = yf(x) + x + f(f(y) - f(x))$. $P(x,1)\Rightarrow 0=x+f(f(1)-f(x))\Rightarrow f$ is bijective. $P(1,1)\Rightarrow f(0)=-1$ Let $k=f(1)+1$ so that $f(k)=0$. $P(k,k)\Rightarrow f\left(k^2\right)=f(1)\Rightarrow k^2=1\Rightarrow f(1)=\in\{-2,0\}$ $\textbf{Case 1: }f(1)=-2$ So $k=-1$, hence $f(-1)=0$ $P(0,x)-P(-x,0)\Rightarrow f(-1-f(-x))=f(f(x)+1)\Rightarrow f(-x)=-2-f(x)$ $P(-1,x)\Rightarrow -1-f(x)=f(f(x))$, surjectivity gives $\boxed{f(x)=-x-1}$, which works. $\textbf{Case 2: }f(1)=0$ $P(1,x)\Rightarrow f(x)=1+f(f(x))$, surjectivity again implies $\boxed{f(x)=x-1}$, which also works.

GorgonMathDota wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[ f(xy) = yf(x) + x + f(f(y) - f(x)) \]for all $x,y \in \mathbb{R}$. $f(xy)=yf(x)+x+f(f(y)-f(x))$ Let $P(x,y)$ be the assertion $P(1,1) \implies f(0)=-1$ $\exists a,b: f(a)=f(b) , a \neq b$ $P(a,1)-P(b,1) \implies a=b \implies f injective$ $P(-x,0)-P(0,x) \implies f(f(x)+1)=f(-1-f(-x)) \implies f(x)+f(-x)=-2$ $P(1,x)-P(x,1) \implies f(x)=x(f(1)+1)-1 \implies f(x)=cx-1$ Sub it in $P(x,y)$ we get : $c=\pm x$ Therefore, $f(x)=\pm x -1$ $\forall x \in \mathbb{R}$

GorgonMathDota wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[ f(xy) = yf(x) + x + f(f(y) - f(x)) \]for all $x,y \in \mathbb{R}$. Let $P(x,y)$ the assertion of the given F.E. $P(1,1)$ $$f(0)=-1$$$P(x,1)$ $$f(f(1)-f(x))=-x \implies f \; \text{bijective}$$$P(0,x)$ $$f(f(x)+1)=x-1 \overset{x=1}{\implies} f(f(1)+1)=0$$$P(f(1)+1,f(1)+1)$ $$(f(1)+1)^2=1 \implies f(1)=0 \; \text{or} f(1)=-2$$Case 1: $f(1)=0$ $P(1,x)$ $$f(f(x))=f(x)-1 \implies f(t)=t-1$$Case 2: $f(1)=-2$ Then $f(-1)=0$ and by $P(-1,x)$ $$f(-x)+1=f(f(x))$$$P(-1,1)$ $$f(-2)=1$$$P(x,-1)$ $$f(x)+f(-x)-x=f(-f(x))$$$P(-f(x),1)$ $$f(-x)+f(x)=-2$$And replacing this on $P(-1,x)$ $$-f(x)-1=f(f(x)) \implies f(t)=-t-1$$Thus we are done

I claim the solutions are $\boxed{f(x)=x-1}$ and $\boxed{f(x)=-x-1}$, which both work. Let $P(x,y)$ denote the given assertion. Step 1: $P(x,1): -x=f(f(1)-f(x))$, which implies $f$ is injective. Step 2: $P(1,1): f(0)=-1$ Step 3: $P(0,x): x-1=f(f(x)+1)$ Step 4: $P(-x,0): x-1=f(-1-f(-x))$ Step 5: So $f(x)+1=-1-f(-x)$, so $f(x)+f(-x)=-2$. Step 6: $P(1,x): f(x)=xf(1)+1+f(f(x)-f(1))$. Step 7: Now we also have $f(f(1)-f(x))+f(f(x)-f(1))=-2$. Step 8: So adding the two equations $-x=f(f(1)-f(x))$ and $f(x)=xf(1)+1+f(f(x)-f(1))$ gives $f(x)-x=xf(1)-1$, so $f(x)=(f(1)+1)x-1$. Step 9: So the functions must be of the form $kx-1$. Step 10: This gives $kxy-1=kxy-y+x+k^2y-k^2x-1$, so $-1=x-y-k^2(x-y)-1=(1-k^2)(x-y)-1$. Step 11: So $(1-k^2)(x-y)=0$, which implies that $k=\pm1$.

We prove that $f(x)=x-1$ and $f(x)=-x-1$ are the only solutions. Let $P(m,n)$ be the assertion in $f(xy) = yf(x) + x + f(f(y) - f(x))$. Claim 1 : $f(0)=-1$. Proof : $P(1,1) : 0=1+f(0)$. $\square $ Claim 2 : $f$ injective. Proof : $P(x,1) : x=-f(f(1)-f(x))...\diamondsuit$. Hence, if $f(x_1)=f(x_2)\implies x_1=x_2$. $\square $ Claim 3 : $f(x)+f(-x)=-2$. Proof : $P(x,0)\hspace{0.3cm}: -1=x+f(-1-f(x)) \implies -1-x=f(-1-f(x))....(\clubsuit)$. $P(0,-x) : -1=x+f(f(-x)+1) \implies -1-x=f(f(-x)+1)....(\spadesuit)$. $(\clubsuit)-(\spadesuit) : f(-1-f(x))=f(f(-x)+1) \overset{\text{Claim2}}{\implies} -1-f(x)=f(-x)+1 \implies f(x)+f(-x)=-2$. $\square$ Claim 4 : $f(y)=y (f(1)+1)-1$. Proof : $P(1,y) : f(y)=yf(1)+1+f(f(y)-f(1))$. But, $f(f(y)-f(1))\overset{\text{Claim3}}{=} -2-f(f(1)-f(y)) \overset{\diamondsuit}{=} -2+y$. Substituting this in earlier equation gives : $f(y)=yf(1)+1+(-2+y)$. Hence, $$f(y)=y(f(1)+1)-1 ....[\bigstar]. \square$$ Claim 5 : $f(1)=0$ or $-2$. Proof : $P(0,1) : 0=f(f(1)+1)$. $y \rightarrow f(1)+1$ in $\bigstar$ : $0=\left(f(1)+1 \right)^{2}-1 \implies f(1)+1= \pm 1 \implies f(1)=0 \text{ or}-2 $. $\square$ From claim 4 and claim 5, we get that $\boxed{f(x)=x-1}$ and $\boxed{f(x)=-x-1}$. It can be easily verified that both satisfy given equation; and hence are our only solutions!

$P(x,1) : f(x) = f(x) + x + f(f(1)-f(x)) \implies -x = f(f(1)-f(x))$. Let $a,b$ be such that $f(a) = f(b)$ now $:$ $P(a,1) , P(b,1) : -a = -b \implies a = b \implies f$ is injective. $P(1,1) : f(0) = -1$. $P(0,x) : f(f(x)+1) = x-1 \implies f$ is bijective. Let $f(a) = 0$. $P(a,a) : f(a^2) = a - 1 = f(f(a) + 1) \implies 1 = a^2 \implies a = \pm 1$. Case $1 : a = 1$. $P(1,x) : f(y) = 1 + f(f(y)) \implies f(y) - 1 = f(f(y)) \implies f(x) = x - 1$. Case $2 : a = -1$. $P(0,x) , P(-x,0) : f(x) + 1 = -f(-x) - 1 \implies f(-x) = -f(x) - 2$. $P(-1,x) : f(-x) = -1 + f(f(x)) \implies f(f(x)) + f(x) = -1 \implies f(x) = -(x+1)$. Answers $: f(x) = x-1 , f(x) = -(x+1)$.

We claim that $f\equiv id -1$ and $f\equiv -id-1.$ Clearly both fit. Let $P(x,y)$ be the given functional equation. Clearly $f$ is bijective. $P(1,1)$ gives $f(0)=-1$ and $P(0,x)$ gives $f(f(x)+x)=x-1.$ Take $f(k)=0,$ then $P(k,k)$ forces $k\in \{1,-1\}.$ If $k=1$ then $P(1,x)$ implies $f(f(x))=1-x,$ as required. If $k=-1$ then combining $P(-x,0)$ with $P(0,x)$ yields $f(x)=-2-f(-x).$ Then $P(-1,x)$ gives $f(f(x))=-1-f(x)$ as required.

Let $P(x,y)$ be the assertion of given f.e. \begin{align*} &P(1,1):f(0)=1 \\ &P(0,x):x-1=f(f(x)+1) \ \forall \Rightarrow f \ \text{ is bijective} \\ &P(x,0)+P(0,x): -2=f(x)+f(-x) \ \triangle\\ &P(0,1): f(f(1)+1)=0 \\ &P(f(1)+1,f(1)+1): (f(1)+1)^2=1 \Rightarrow f(1)=0 \ or \ f(1)=-2 \\ \end{align*} Case 1. $f(1)=1$. $P(x,1)+P(1,x): f(x)-x=1+f(-f(x))+f(f(x))\stackrel{\triangle}{=}-1 \Rightarrow f(x)=x-1$ Case 2. $f(1)=-2$ From $\triangle$, $f(-1)=0$. $P(x,-1)+P(-1,x): 2f(-x)=-1+x-f(x)+f(f(x)+f(-f(x))) \stackrel{\triangle}{=} x-f(x)-1$ Therefore $2f(-x)=x-f(x)-3$ or $f(x)= x-3-2f(-x)$ Then by $\triangle$ we have $$ -2= (x-3-2f(-x))+f(-x)\Rightarrow 1=x-f(-x)$$ It simplifies to $f(-x)=x-1 \Rightarrow f(x)= -x-1$ Therefore there is 2 functions $f(x)=x-1,f(x)=-x-1$ satisfying the problem.One may check these functions indeed holds

GorgonMathDota wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[ f(xy) = yf(x) + x + f(f(y) - f(x)) \]for all $x,y \in \mathbb{R}$. $$f(xy) = yf(x) + x + f(f(y) - f(x))...(\alpha)$$In $(\alpha) y=1:$ $$\Rightarrow f(f(1)-f(x))=-x...(\beta)$$$$\Rightarrow f \text{ is surjective}...(I)$$If $\exists \text{ }a\neq b/ f(a)=f(b)=k:$ In $(\beta) x=a:$ $$\Rightarrow f(f(1)-k)=-a$$In $(\beta) x=b:$ $$\Rightarrow f(f(1)-k)=-b$$$$\Rightarrow a=b (\Rightarrow \Leftarrow)$$ $$\Rightarrow f\text{ is inyective}...(II)$$By $(I)$ and $(II):$ $$\Rightarrow f\text{ is biyective}...(III)$$In $(\beta) x=1:$ $$\Rightarrow f(0)=-1$$In $(\alpha) x=0:$ $$\Rightarrow -1=-y+f(f(y)+1)$$$$\Rightarrow f(f(y)+1)=y-1...(\omega)$$In $(\alpha) y=0:$ $$\Rightarrow -1=x+f(-1-f(x))$$$$\Rightarrow f(-f(x)-1)=-x-1...(\phi)$$In $(\phi) x=-y:$ $$\Rightarrow f(-f(-y)-1)=y-1$$By $(\omega)$ and $(III):$ $$\Rightarrow -f(-y)-1=f(y)+1$$$$\Rightarrow f(-y)=-f(y)-2...(\lambda)$$In $(\alpha) x=1:$ $$\Rightarrow f(y)=yf(1)+1+f(f(y)-f(1))$$By $(\lambda):$ $$\Rightarrow f(y)=yf(1)+1-f(f(1)-f(y))-2$$By $(\beta):$ $$\Rightarrow f(y)=yf(1)+1+y-2$$$$\Rightarrow f(y)=y(f(1)+1)-1$$$$\Rightarrow f(y)=py-1, p\in\mathbb{R}...(IV)$$$(IV)$ in $(\alpha):$ $$\Rightarrow pxy-1=pxy-y+x+p(py-px)-1$$$$\Rightarrow 0=x-y+p^2y-p^2x$$$$\Rightarrow (p^2-1)x=(p^2-1)y...(V)$$In $(V) x=0, y=1:$ $$\Rightarrow p^2-1=0$$$$\Rightarrow p=\pm 1$$ $$\Rightarrow f(x)=x-1 \text{ and }f(x)=-x-1 \text{ are the only solutions}_\blacksquare$$

$P(1,1)$ $f(0)=-1$ $P(0,x)$ $x-1=f(f(x)+1)$ $f$ is bijective . $P(0,1)$ $f(f(1)+1)=0$ $P(f(1)+1,f(1)+1)$ $f((f(1)+1)^2)=f(1)$ $f(1)+1=1(*)$ $f(1)+1=-1(**)$ From $(*)$ $f(1)=0$ $P(x,1)$ $f(-f(x))=-x$ $P(1,-f(x))$ $f(x)=x-1$ which is a solution. From $(**)$ $P(x,x)$ $f(x^2)=xf(x)+x-1$ $P(-x,-x)$ $f(x^2)=-xf(-x)-x-1$ Thus $f(x)+f(-x)=2$ from $P(-1,x)$ with surjectivity implies $f(x)=-x-1$ so we are done

Let $P(x, y)$ be the given assertion. $P(1, 1)$ gives $f(1) = f(1) + 1 + f(0) \implies f(0) = -1$. $P(x, x)$ gives $f(x^2) = xf(x) + x - 1$, $P(-x, -x)$ gives $f(x^2) = -xf(-x) - x - 1$, subtracting one from the other yields $f(x) + f(-x) = -2 \; (1)$ (being true for $f(0)$ as well). $P(x, 1)$ and $P(1, x)$ give: $$f(x) = xf(1) + 1 + f(f(x) - f(1))$$$$0 = x + f(f(1) - f(x))$$Adding these two, we have $f(x) = xf(1) + x + 1 + f(f(x) - f(1)) + f(f(1) - f(x))$. Since $f(f(x) - f(1)) + f(f(1) - f(x)) = -2$ from $(1)$, we have $f(x) = (f(1) + 1)x - 1 = cx - 1$. Substituting this back into the original equation, we have: $$cxy - 1 = cxy - y + x + c^2y - c^2x - 1 \implies c^2(y - x) = y - x \implies c = \pm 1$$So, $f(x) = x + 1$ or $f(x) = - x - 1$.

We claim that the only solutions are $f(x)=\boxed{x-1}$ and $f(x)=\boxed{-x-1}$, which obviously work. We now prove that they are the only ones. Let $P(x,y)$ denote the assertion. From $P(1,1)$, we have that $f(1)=f(1)+1+f(0)$, which means $f(0)=-1$. Then, from $P(x,x)$ and $P(-x,-x)$, we have that $f(x^2)=xf(x)+x-1$. and $f(x^2)= -xf(-x)-x-1$. Thus, $xf(x)+x-1 = -xf(-x)-x-1$, which means $f(x)+f(-x)=-2$ for all nonzero $x$. Comparing $P(x,y)$ and $P(y,x)$, we have \begin{align*} f(xy)&=yf(x)+x+f(f(y)-f(x))\\ f(xy)&=xf(y)+y+f(f(x)-f(y)). \end{align*}Adding the two equations, we have \[2f(xy)=xf(y)+yf(x)+x+y-2.\]Now, letting $y=1$, we have $2f(x) = xf(1)+f(x)+x+1-2$, or $f(x) = x(f(1)+1) - 1$ for all $x\ne 1$. Let $c = f(1)+1$. Then, we must have \begin{align*}cxy-1 &= y(cx-1) + x + f(cy-cx) \\ &= cxy - y + x + c^2y-c^2x -1.\end{align*}Thus, we must have $y-x = c^2(y-x)$. Thus, $c=\pm 1$, as desired. $\blacksquare$

Let $P(x,y)$ be the assertion $f(xy) = yf(x) + x + f(f(y) - f(x))$ $P(x,1) \implies f(x)=f(x)+x+f(f(1)-f(x)) \implies f(f(1)-f(x))=-x$ so, $f$ is bijective $P(1,1) \implies f(0)=-1$ $P(0,1)\implies f(f(1)+1)=0$ $P(x,0)+P(0,x)\implies f(-1-f(x))+f(1+f(x))=-2$ from surjectivity we have that $f(x)+f(-x)=-2$ $P(f(1)+1, f(1)+1) \implies f((f(1)+1)^2)=f(1)$, from injectivity $f(1)=0$ or $f(1)=-2$ If $f(1)=-2$, then $P(-1,-1) \implies f(-1)=0 $ $P(-1,x)\implies -2-f(x)=f(-x)=-1+f(f(x))$ and from surjectivity $\boxed {f(x)=-x-1}$ If $f(1)=0$, then $P(1,x)\implies f(x)=1+f(f(x))$ again from surjectivity $\boxed {f(x)=x-1}$