Let $ P(x,y) $ be the assertion of $f(xf(y)-yf(x))=f(xy)-xy$ $ P(x,0) \Rightarrow f(xf(0))=f(0), \forall x \in \mathbb{R} $ If $ f(0) \neq 0 $, then $ f $ is a constant, which isn't a solution. If $ f(0) = 0 $, then $ P(x,x) \Rightarrow f(x^2)=x^2, \forall x \in \mathbb{R} \rightarrow f(x)=x, \forall x \in \mathbb{R^{+}} $ Now, suppose there is a negative real $ YlOwOlY $ such that $ f(YlOwOlY)<YlOwOlY $ $ P(YlOwOlY,1) \Rightarrow f(YlOwOlY-f(YlOwOlY))=f(YlOwOlY)-YlOwOlY \rightarrow f(YlOwOlY)=YlOwOlY $, which is impossible. Thus, we have $ f(x) \ge x, \forall x \in \mathbb{R} $ Then, for any negative real $ a $ $ P(a,-1) \Rightarrow f(-a)+a=f(af(-1)+a)=af(-1)+a \rightarrow f(-a)=af(-1) $ Finally, we have $ \left\{\begin{matrix} f(x)=x, \forall x \in \mathbb{R^{*}} & & \\ f(x)=ax, \forall x \in \mathbb{R^{-}} & & \end{matrix}\right.$, where $ a \ge -1 $ is a constant.

Can you tell me what is $YIOwOIY$ please?

mathtastic wrote: Can you tell me what is $YIOwOIY$ please? I guess it's just a real number. BTW it's cute \(OuO)/

$YIOwOIY$ is similar to $Y(O\_O)Y$

mathtastic wrote: Can you tell me what is $YIOwOIY$ please? YlOwOlY wrote: Now, suppose there is a negative real $ YlOwOlY $ such that $ f(YlOwOlY)<YlOwOlY $ Rather it had been me to post the solution, I would have used $YESMAths$ instead of $YlOwOlY$.

My solution Let $A(x,y)$ be assertion of $f(xf(y)-yf(x)=f(xy)-xy$. $A(x,0)$ implies $f(xf(0))=f(0)$ so if $f(0)\neq 0$ we would have $f(x)=f(0)\forall x\in \mathbb R$ which is contradiction. $\implies \boxed{f(0)=0}$. $A(x,x)$ implies $f(x^2)=x^2$ $\implies \boxed{f(x)=x\forall x >0}$. Let $a$ be such that $f(a)=0$. $A(a,y)$ implies $f(af(y))=f(ay)-ay$ now choose $y$ such that $ay\ge 0$. We have $f(af(y))=0$. Clearly $a\le 0$ so we have that $y<0\implies f(y)>0$(except for $a=0$, it can't be equal zero as proceed above) but now $af(y)<0\implies f(af(y))>0$-contradiction. So $f(a)=0\implies a=0$. $A(x,y),x<0,y<0$ implies that $xf(y)-yf(x)=0$ and for $x=-1$ it implies $f(y)=-yf(-1),y<0$. $A(1,-1)$ implies that either $f(-1)=1$ or $f(-1)=-1$ so the solutions are $\boxed{f(x)=x}$ $\boxed {f(x)=|x|}$. Checking we see that they work.

YlOwOlY wrote: Let $ P(x,y) $ be the assertion of $f(xf(y)-yf(x))=f(xy)-xy$ $ P(x,0) \Rightarrow f(xf(0))=f(0), \forall x \in \mathbb{R} $ If $ f(0) \neq 0 $, then $ f $ is a constant, which isn't a solution. If $ f(0) = 0 $, then $ P(x,x) \Rightarrow f(x^2)=x^2, \forall x \in \mathbb{R} \rightarrow f(x)=x, \forall x \in \mathbb{R^{+}} $ Now, suppose there is a negative real $ YlOwOlY $ such that $ f(YlOwOlY)<YlOwOlY $ $ P(YlOwOlY,1) \Rightarrow f(YlOwOlY-f(YlOwOlY))=f(YlOwOlY)-YlOwOlY \rightarrow f(YlOwOlY)=YlOwOlY $, which is impossible. Thus, we have $ f(x) \ge x, \forall x \in \mathbb{R} $ Then, for any negative real $ a $ $ P(a,-1) \Rightarrow f(-a)+a=f(af(-1)+a)=af(-1)+a \rightarrow f(-a)=af(-1) $ Finally, we have $ \left\{\begin{matrix} f(x)=x, \forall x \in \mathbb{R^{*}} & & \\ f(x)=ax, \forall x \in \mathbb{R^{-}} & & \end{matrix}\right.$, where $ a \ge -1 $ is a constant. As wool wizard said ,a is 1 or -1 not real number which is larger than -1 / equal-1. When we get $f(x)=x $for all positive real number and 0 ,and $f(x)=-f(-1)x $for all negative real number If for all real number we have the equation $x f(y)=yf(x) $,we can get $f(x)=x $for all real number ,if there exist real number x and y such that $xf(y)-yf(x)$ is not zero ,suppose it is $c.$ we can get$f(c)=f(-c).$ So for positive real number c ( if c is a negative real number we will change c to -c) we have$ c=f(-c)=cf(-1). $So $f(-1)=1 $we have the solution $f(x)=|x| $:blush:

Wolowizard wrote: $A(1,-1)$ implies that either $f(-1)=1$ or $f(-1)=-1$ Check if it is wrong or too abbreviated.

Can someone tell me where I went wrong here? I found the solution $f(x)=x,$ but not $f(x)=|x|.$ I somehow managed to prove that $f$ was odd, but this clearly is not true because of the absolute value solution. Can someone point out the hole in my proof below? Claim: $f$ is odd As per usual, denote the assertion by $P(x,y)$. Then, the given equation rearranges to $f(xf(y)-yf(x))+xy=f(xy),$ and $P(y,x)$ gives that $f(yf(x)-xf(y))+xy=f(xy)$. Subtracting the two and substituting $x=1$ gives that $f(f(y)-yf(1))=f(yf(1)-f(y))$. It is easy to see from the given equation that $f(x^2)=x^2+f(0),$ so substituting we get that $f(y(1-f(1))+f(0))=f(y(f(1)-1)-f(0))$. Thus, $\forall y\geq 0,$ we have that $f(yc+g)=f(-yc-g)$(for some constant $f,g$). This is enough to imply that $f$ is odd. OR APPARENTLY NOT?

@above, it doesn't work if $c=f(1)-1=0$, which, in fact, has to hold. Let $P(x,y)$ be the assertion $f(xf(y)-yf(x))=f(xy)-xy$. $P(x,0)\Rightarrow f(xf(0))=f(0)$, so either $f$ is constant (no such solutions) or $f(0)=0$. $P(x,x)\Rightarrow f\left(x^2\right)=x^2$, so $f$ fixes the positive reals. From now on, let $x,y\in\mathbb R^+$. $P(-x,-y)\Rightarrow f(yf(-x)-xf(-y))=f(xy)-xy=0$ $P(yf(-x)-xf(-y),1)\Rightarrow yf(-x)-xf(-y)=0\Rightarrow\frac y{f(-y)}=\frac x{f(-x)}=c$ for some constant $c$. Thus $f(-x)=cx$ for some constant $c$. Testing this in the original equation, we find $c\in\{-1,1\}$, hence the solutions $\boxed{f(x)=x}$ and $\boxed{f(x)=|x|}$ which both work.

Nice problem! Let $P(x,y)$ denote the assertion in the given problem. Clearly no constant function works. $P(x,0) \implies f(xf(0)) = f(0)$. So if $f(0) \neq 0$, then shift $x$ to $\frac{x}{f(0)}$ so the equation becomes $f(x) = f(0)$ and so $f$ is constant which is a contradiction. And so $f(0) = 0$. $P(\sqrt{x},\sqrt{x}) \implies f(x) = x \forall x \in \mathbb{R}_{\geq 0}$ Suppose there exists a negative real $t$ such that $f(t) \neq t$. Let $x$ be any non-negative real. Consider the two assertions: $P(x,t) \implies f(x(f(t)-t)) = f(xt) - tx$ $P(t,x) \implies f(x(t-f(t)) = f(xt) -tx$. Comparing them yeilds that $f(x(f(t)-t)) = f(x(t-f(t))$. Now if $f(t) > t$ then shifting $x$ to $\frac{x}{f(t)-t}$ makes the equation $f(x) = f(-x)$. And so $f(x) = \mid x \mid$ And if $f(t) < t$ then shifting $x$ to $\frac{x}{t-f(t)}$ makes the equation $f(x) = f(-x)$. And so again $f(x) = \mid x \mid$. So the existence of such a $t$ implies that $\boxed{f(x) = \mid x \mid \forall x \in \mathbb{R}}$. If there doesn't exist such a $t$ then $f(t) = t \forall t \in \mathbb{R}_{<0}$. So $\boxed{f(x) =x \forall x \in \mathbb{R}}$.

mihajlon wrote: Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x$, $y \in \mathbb{R}$ hold: $$f(xf(y)-yf(x))=f(xy)-xy$$ Proposed by Dusan Djukic $$f(xf(y)-yf(x))=f(xy)-xy...(\alpha)$$In $(\alpha) y=0:$ $$\Rightarrow f(xf(0))=f(0)$$If $f(0)\neq 0:$ $$\Rightarrow f(x)=f(0), \forall x \in \mathbb{R}$$Replacing in $(\alpha):$ $$\Rightarrow 0=-xy, \forall x,y \in \mathbb{R} (\Rightarrow \Leftarrow)$$ $$\Rightarrow f(0)=0$$In $(\alpha) y=x:$ $$\Rightarrow f(x^2)=x^2 \Rightarrow f(x)=x, \forall x \in \mathbb{R}^{+}$$In $(\alpha) x<0, y<0:$ $$\Rightarrow f(xf(y)-yf(x))=0$$$$\Rightarrow \exists \text{ } a / f(a)=0$$In $(\alpha) x=a, y=1:$ $$f(a)=-a \Rightarrow a=0$$$\Rightarrow f$ is injective at $0$ $$\Rightarrow xf(y)=yf(x), \forall x,y \in \mathbb{R}^{-}$$As $f(x)\neq 0$ and $f(y)\neq 0$ $$\Rightarrow f(x)=cx, \forall x\in \mathbb{R}^{-}$$In $(\alpha) x<0, y>0:$ $$\Rightarrow f(xy-cxy)=cxy-xy$$$-x>0:$ $$\Rightarrow f((c-1)(-xy))=(1-c)(-xy)$$$$f((c-1)x)=(1-c)x, \forall x \in \mathbb{R}^{+}$$If $c-1\ge 0:$ $$\Rightarrow c-1=1-c \Rightarrow c=1 \Rightarrow f(x)=x, \forall x \in \mathbb{R}$$If $c-1<0:$ $$\Rightarrow c(c-1)=1-c \Rightarrow c=-1 \Rightarrow f(x)=|x|, \forall x \in \mathbb{R}$$ $$\Rightarrow f(x)=x, f(x)=|x| \text{ are the only solutions} _\blacksquare$$