YanYau wrote: Find all functions $f:\mathbb R \rightarrow \mathbb R$ such that $$f(f(xy-x))+f(x+y)=yf(x)+f(y)$$ for all real numbers $x$ and $y$. Let $P(x,y)$ be the assertion $f(f(xy-x))+f(x+y)=yf(x)+f(y)$ $P(0,0)$ $\implies$ $f(f(0))=0$ $P(0,1)$ $\implies$ $f(0)=0$ $P(x-1,0)$ $\implies$ $f(f(1-x))+f(x-1)=0$ $P(-1,x)$ $\implies$ $f(f(1-x))+f(x-1)=xf(-1)+f(x)$ Subtracting, we get $f(x)=ax\quad\forall x$ where $a=-f(-1)$ Plugging this back in original equation, we get $a\in\{0,1\}$ and the two solutions : $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ and $\boxed{\text{S2 : }f(x)=x\quad\forall x}$

let $P(x,y)$ denote the assertion $f(f(xy-x))+f(x+y)=yf(x)+f(y)$ $P(0,0)$ implies $f(f(0))=0$ $P(0,1)$ implies $f(0)=0$ $P(x,2)$ implies $f(f(x))+f(x+2)=2f(x)+f(2)$ and $P(x,1)$ implies $f(x+1)=f(x)+f(1)$ so $f(f(x))=f(x)+f(2)-2f(1)$ taking $x=0$ we get $f(2)=2f(1)$ so $f(f(x))=f(x)$ thus $P(x,y)$ is equivilant to $f(xy-x)+f(x+y)=yf(x)+f(y)$ taking $x=1$ we get $f(y-1)+f(y+1)=yf(1)+f(y)$ or $f(x+1)=f(x)+f(1)$ and $f(x)=f(x-1)+f(1)$ so $f(y)=yf(1)$ substracting we get $f(x)=0$ or $f(x)=x$

Medjl wrote: ...we get $f(y-1)+f(y+1)=yf(1)+f(y)$ so $f(y)=yf(1)$ Hemmm, this seems a quite hazardous deduction ...

pco wrote: Medjl wrote: ...we get $f(y-1)+f(y+1)=yf(1)+f(y)$ so $f(y)=yf(1)$ Hemmm, this seems a quite hazardous deduction ... ok corrected because $f(x+1)=f(x)+f(1)$

My solution: Let $P(x,y)$ be the assertion of $f(f(xy-x))+f(x+y)=yf(x)+f(y).$ $P(0,0)\implies f(f(0))=0.$ $(1)$ $P(x,1)\implies f(x+1)=f(x)+f(1).$ $(2)$ $$P(1,y+1)\implies f(f(y))+f(y+2)=^{(2)} f(f(y))+f(y)+2f(1)=yf(1)+f(1)+f(y+1)=^{(2)} yf(1)+2f(1)+f(y),$$then we find $f(f(y))=yf(1).$ $P(0,x)\implies f(0)=0.$ Case 1: $f(1)=0,$ then in we have $f(f(y))=0.$ From $P(x,0)\implies f(f(-x))+f(x)=f(x)=f(0)=0.$ Then $f(x)=0, \forall x\in\mathbb{R}.$ This solution is work. Case 2: $f(1)\not = 0,$ then $f(f(y))=yf(1).$ $(3)$ $P(x,0)\implies f(f(-x))=^{(3)} -xf(1)=f(0)-f(x)\to f(x)=cx.$ We can find easily $f(x)=x,\forall x\in\mathbb{R}$ where this solution is work.

My solution: Let $P(x,y)$ denote the functional equation $P(0,y): f(f(0))=yf(0)$ ,$f(0)=0$ $P(y,1): f(y+1)=f(y)+f(1)$ $P(y,2): f(f(y))=2f(y)+f(2)-f(x+2)=f(y)$ (from $P(y,1)$) $P(1,y): f(f(y-1))+f(y+1)=yf(1)+f(y)$ Case $1$ $f(1)$ is not 0 From $P(1,y) $and$ P(y,1)$ implies that f is injective From $P(y,2)$ implies that $f(x)=x$ Case $2$ $f(1)=0$ Therefore, $f(x+1)=f(x)$ and $f(f(y-1))+f(y+1)=f(y)$ Then, $f(f(y-1))=f(y-1)=f(y)=0$ and we are done!

$P(x,1): f(f(0))+f(x+1)=f(1)+f(x) \hspace{5mm}(1).$ Plugging $x=0$ in $(1)$ gives $f(f(0))=f(0).$ Now, $P(0,y) : f(f(0))=yf(0) \hspace{5mm} \forall y.$ It follows that $f(0)=0.$ Going back to $(1):$ $$\boxed {f(x+1)=f(x)+f(1).}$$$P(x,2): f(f(x))+f(x+2)=2f(x)+f(2)=2f(x)+2f(1)=f(x)+f(x+1)+f(1)=f(x)+f(x+2).$ It follows that $f(f(x))=f(x).$ $$P(1,y): f(f(y-1))+f(y+1)=yf(1)+f(y)\Longrightarrow$$$$f(y)-f(1)+f(y)+f(1)=f(y)+yf(1).$$ So $f(x)=cx.$ Trying this in $P$ gives $c=0$ and $c=1$ as solutions.

Sketch of another idea: we can get from given equation that for all $x\in R$ equality $f(x^2)=xf(x)$ holds. So for every $x\in R_+$ we have $f(x^2)=xf(x)=...=x^{1+\frac{1}{2}+...+\frac{1}{2^n}}f(\sqrt[2^n]{x})$. Taking $n\rightarrow \infty$ we get $f(x^2)=x^2f(1)$ Because $f(0)=0\wedge f(x)=-f(-x)$ it's easy to get the answer.

WolfusA wrote: Sketch of another idea: we can get from given equation that for all $x\in R$ equality $f(x^2)=xf(x)$ holds. So for every $x\in R_+$ we have $f(x^2)=xf(x)=...=x^{1+\frac{1}{2}+...+\frac{1}{2^n}}f(\sqrt[2^n]{x})$. Taking $n\rightarrow \infty$ we get $f(x^2)=x^2f(1)$ Because $f(0)=0\wedge f(x)=-f(-x)$ it's easy to get the answer. You need to prove continuity of the function $f(x)$ at $x=1$ to get $f(x^2)=x^2f(1)$. This is because all you get is that $f(x^2) = x^2 \lim_{n \to \infty} f(\sqrt[2^n]{x})$, if the limit exists.

You're right. Thanks for checking.

f(f(xy – x)) + f(x+y) = yf(x) + xf(y) .................(a) put x = y= 0 in (a) to get f(f(0)) + f(0) =f(0) put y = 1 in (a) to get f(f(x-x)) + f(x+1) = f(x) + f(1) f(f(0)) + f(x+1) = f(x) + f(1) f(x+1) – f(x) = f(1) .................(b) put x =1 to get; f(2) = 2f(1) , put x=2 to get; f(3) =3f(1) [Inductively or otherwise, it could be shown that f(n) = nf(1) ∀ v∈R].................(*) Put y =2 in (a) to get; f(f(x)) + f(x+2) = 2f(x) + f(2) ...........................(c) from (b) it’s easy to get that : f(x+2) – f(x) = 2f(1) and that f(2) = 2f(1) f(f(x)) + f(x+2) – f(x) = f(x) + f(2) f(f(x)) + 2f(1) = f(x) + 2f(1) ; f(f(x)) = f(x) ...........(d) from (d) it is obvious that f(m) = c {c, m ∈R} or equivalently f(g) = f(h) ∀ g,h∈R. The above assertion implies that from (b) LHS equals 0 implying that f(1) = 0 f(n) = nf(1) {as in (*)} = n∙0 = 0 let n = f(x) in (d) { x, f(x), n∈R by implication} thus we have; f(n) =0 or n ∀ n∈R

YanYau wrote: Find all functions $f:\mathbb R \rightarrow \mathbb R$ such that $$f(f(xy-x))+f(x+y)=yf(x)+f(y)$$ for all real numbers $x$ and $y$. an exellent problem??

An other solution: write $f(x+y)=yf(x)+f(y)-f(f(xy-x))=xf(y)+f(x)-f(f(xy-y))$ or ie $(x-1)f(y)+(1-y)f(x)=f(f(xy-y))-f(f(xy-x))$ set $y=1$ so $f(f(x-1))=f(f(0))+(x-1)f(1)$ or $f(f(x))=f(f(0))+xf(1)$ now our functional equation is ie to $yf(x)+f(y)=f(x+y)+f(f(0))+(xy-x)f(1)$ set $y=0$ so $f(0)=f(x)+f(f(0))-xf(1)$ hence $f(x)=ax+b$ plug in the original function to find $a=1$ and $b=0$ or $a=b=0$

P(0,y): ff(0)=yf(0) then f(0)=0 p(x,1): f(x+1)=f(x)+f(1) p(x,0): ff(-x)=-f(x) then the main equation becomes : f(x+y)=yf(x)+f(y)+f(x-xy) . P(x,y+1): f(x+y+1)-f(-xy)-f(1)=(y+1)f(x)+f(y). Then (y+1)f(x)+f(y)=(x+1)f(y)+f(x) then : f(x)=cx substitute in the original equation we get that c=0 or c=1 then: f(x) =x Or f(x)=0.

$P(0,0)\Rightarrow f(f(0))=0.$ $P(0,1)\Rightarrow f(f(0))+f(1)=f(0)+f(1)\Rightarrow f(0)=0.$ $P(x,0)\Rightarrow f(f(-x))+f(x)=0\Rightarrow f(f(-x))=-f(x).$ The above tells us that $f(f(xy-x))=-f(x-xy),$ which we can plug into $P.$ Let $$Q(x,y):-f(x-xy)+f(x+y)=yf(x)+f(y).$$Immediately, $Q(x,1)\Rightarrow{f(x+1)=f(x)+f(1)}.$ $Q(1,-x):-f(1+x)+f(1-x)=-yf(1)+f(-x).$ With $Q(x,1),$ we have $-(f(1)+f(x))+f(1)+f(-x)=-xf(1)+f(-x).$ Simplifying gives:$$f(x)=xf(1).$$Let $f(1)=c$ so we have $f(x)=xc.$ Plugging this in to $P,$ we $(xy-x)c(c-1)=0,$ so $c=0,1.$ This tells us that $c=0,1,$ so $\boxed{f(x)=0\text{ or }f(x)=x}.$

Solved this problem without writing.

The only answers are $f(x)=0$ and $f(x)=x$. Let $P(x,y)$ be the given assertion. From $P(0,y)$ we get that $f(f(0))=yf(0)=0$, which clearly implies that $f(0)=0$. From $P(x,0)$ we get that $f(f(-x))=-f(x)$. Now our assertion is transformed into the following: $$f(x+y)=yf(x)+f(y)+f(x-xy)$$Setting $y=1$ into the above equation we get that $f(x+1)=f(x)+f(1)$. Now set $x=1$ into our equation, we get that: $$f(1)=yf(1)+f(1-y)$$$$f(1-y)=f(1)(1-y)$$thus giving us $f(x)=xc$, where $c$ is a real number. Plugging this back into our original equation we get $c(c-1)(xy-x)=0$. Which clearly implies that $c \in \{0,1\}$. Which gives us our solutions.

Let $P(x,y)$ denote the assertion $f(f(xy-x))+f(x+y)=yf(x)+f(y)$. $P(0,0)\Rightarrow f(f(0))=0$ $P(0,1)\Rightarrow f(1)=f(0)+f(1)$, so $f(0)=0$ $P(x,1)\Rightarrow f(x+1)=f(x)+f(1)$ From this, we have $f(x+2)=f(x+1)+f(1)=f(x)+2f(1)$. $P(x,2)\Rightarrow f(f(x))=f(x)$ $P(x,0)\Rightarrow f(f(-x))+f(x)=0$ so $f$ is odd (using the above claim) $P(-1,x)\Rightarrow f(f(1-x))+f(x-1)=xf(-1)+f(x)\Rightarrow 0=f(1-x)+f(x-1)=f(x)+xf(-1)$. Thus, $f(x)=cx$ for some constant $c\in\mathbb R$. We see that the only solutions are $\boxed{f(x)=x}$ and $\boxed{f(x)=0}$.

Denote P(x,y) is the given assertion P(0,y) => f(f(0)) = y.f(0) P(x,1) => f(f(0))+f(x+1) = f(x)+f(1) => y.f(0)+f(x+1) = f(x)+f(1) => f(1) = f(0) + f(1) (this is because we replace (x,y) by (0,0)) Thus we get f(0)=0 and f(x) = f(x+1) - f(1). This means that f(y) = f(y+1)-f(1) (*) P(x,0) => f(f(-x))=-f(x) or f(f(x))=-f(-x) (**) From (*) => f(f(xy-x))+f(x+y)=y.f(x)+f(y+1)-f(1) => f(f(y-1))=y.f(1)-f(1) => f(1-y)=f(1)(1-y) (because of (**)) =>f(y)=f(1).y or f(y)=ay Test it back to the original equation, we get f(x)=0 and f(x)=x

We claim that the only possible solutions are $\fbox{f(x)=0}$ and $\fbox{f(x)=x}$. Claim 1:- $f(0)=0$ Let $P(x,y)$ be our assertion.Then, $P(0,0) \rightarrow f(f(0))=0$ $P(0,x) \rightarrow f(x)+f(f(0))=xf(0)+f(x) \Rightarrow f(0)=0$ $P(x-1,0) \rightarrow f(f(1-x))+f(x-1)=0$ $P(-1,x) \rightarrow f(f(1-x))+f(x-1)=xf(-1)+f(x)$ so $xf(-1)+f(x)=f(0)=0 \Rightarrow f(x)=-xf(-1)=cx$ Plugging in the original equation we get, $c=0,1$ So,$\fbox{f(x)=0}$ and $\fbox{f(x)=x}$. our only solutions.

YanYau wrote: Find all functions $f:\mathbb R \rightarrow \mathbb R$ such that $$f(f(xy-x))+f(x+y)=yf(x)+f(y)$$ for all real numbers $x$ and $y$. Let $P(x,y)$ the assertion of this F.E. $P(0,x)$ $$f(f(0))=xf(0) \implies f(0)=0$$$P(x,1)$ $$f(x+1)=f(x)+f(1)$$Case 1: $f(1)=0$ $P(1,x+1)$ $$f(f(x))=0$$$P(x,0)$ $$f(x)=0$$Case 2: $f(1) \ne 0$ $P(1,x+1)$ $$f(f(x))=xf(1)$$From here u have $f(f(1))=f(1)$ so $f(1)=f(f(1))=f(f(f(1)))=f(1)^2$ so $f(1)=1$ so $$f(f(x))=x \implies f \; \text{involution}$$And now we re-write the F.E. as $$xy+f(x+y)=yf(x)+f(y)+x$$And we also have $f(x+1)=f(x)+1$ so now by symetry $$yf(x)+f(y)+x=xy+f(x+y)=xf(y)+f(x)+y$$The trick is to add $-xy$ to both sides to get $$(y-1)(f(x)-x)=(x-1)(f(y)-y)$$Which means that the function $g(x)=\frac{f(x)-x}{x-1}$ is constant for all $x \ne 1$ so for all $x \ne 1$ we have $$f(x)=(c+1)x-c$$And using that $f$ is an involution we get $$(c+1)((c+1)x-c)-c=x \implies (c+1)^2x-c^2-2c=x \implies (c^2+2c)x=(c^2+2c) \implies c=-2 \; \text{or} \; c=0$$If $c=-2$ then $f(x)=2-x$ which is not possible as this means $0=f(0)=2$ so now we got $c=0$ which means that $f(x)=x$ which is indeed another answer. Hence $f(x)=x$ and $f(x)=0$ work, thus we are done

Let $P(x,y)$ denote the given assertion, namely: $f(f(xy-x))+f(x+y)=yf(x)+f(y).$ The only solutions are the zero function and the identity. To see this: $P(0,0)$ and $P(0,x)$ implies $f(0)=0.$ Then $P(x,0)$ implies $f(f(-x))=-f(x)$ and $P(1-x,1)$ implies $f(2-x)=f(1-x)+f(1).$ Finally, $P(1,1-x)$ implies $f(x)=xf(1).$

ezz $ff(x)=f(x)=f(1)x$