Raja Oktovin wrote: Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $$f(f(x + y)) = f(x + y) + f(x)f(y) - xy$$ for all real numbers $x$ dan $y$. Let $P(x,y)$ be the assertion $f(f(x+y))=f(x+y)+f(x)f(y)-xy$ Let $f(0)=a$ It's elementary first (just plug in the original equation) to see that the only solution of the type $f(x)=bx+a$ is $f(x)=x$ Subtracting $P(x,y)$ from $P(x+y,0)$, we get the new assertion $Q(x,y)$ : $af(x+y)=f(x)f(y)-xy$ Then $Q(a,-a)$ $\implies$ $f(a)f(-a)=0$ and so $\exists u\in\{f(-a),f(a)\}$ such that $f(u)=0$ Then $Q(x-u,u)$ $\implies$ $af(x)=-u(x-u)$ If $a=0$, $Q(x,y)$ implies $f(x)f(y)=xy$ and so either $f(x)=x$ $\forall x$, which indeed is a solution, either $f(x)=-x$ $\forall x$, which is not a solution. If $a\ne 0$, this implies $f(x)=-\frac uax +\frac{u^2}a=bx+a$ for some $b$ and so (see beginning lines), $f(x)=x$, which is impossible since $f(0)\ne 0$ in this subcase. Hence the unique solution $\boxed{f(x)=x}$ $\forall x$

I have another solution with much more calculation the the one of Mr.pco. Let $P(x,y)$ be the assertion $f(f(x+y)) = f(x+y) + f(x)f(y) - xy$ and let $a: =f(0)$. We have $P(x,0) \Rightarrow f(f(x)) = (a+1)f(x),\forall x\in\mathbb{R}$. Thus, let $x=0$ we have $f(a) = a(a+1)$. Moreover, the assertion $P(x,y)$ becomes: $af(x+y) = f(x)f(y) - xy,\forall x,y,\in\mathbb{R}$. Suppose that $a\neq 0$. We have: $P(a,a) \Rightarrow af(2a) = a^2(a+1)^2 - a^2$, yielding $f(2a) = a^2(a+2)$. $P(2a,a) \Rightarrow af(3a) = a(a+1)a^2(a+2) - 2a^2$, yielding $f(3a) = a^2(a+1)(a+2) - 2a = a(a^3 + 3a^2 + 2a -2)$. $P(3a,a) \Rightarrow af(4a) = a(a+1)a(a^3 + 3a^2 + 2a -2) - 3a^2$ $P(2a,2a) \Rightarrow af(4a) = a^4(a+2)^2 - 4a^2$. This means that $a(a+1)a(a^3 + 3a^2 + 2a -2) - 3a^2 = a^4(a+2)^2 - 4a^2\Leftrightarrow a^2(a-3)(a+3) = 0$ But we have supposed that $a\neq 0$ then $a=3$ or $a=-3$. 1. If $a=3$ then $P(x,y)$ becomes $3f(x+y) = f(x)f(y) - xy$ and hence $f(3) = 3\cdot (3+1) = 12$. $P(x,1) \Rightarrow 3f(x+1) = f(1)f(x) - x$. $P(x+1,1) \Rightarrow 3f(x+2) = f(1)\frac{f(1)f(x) - x}{3} - (x+1)$ $P(x+2,1) \Rightarrow 3f(x+3) = f(1)\frac{f(1)\frac{f(1)f(x) - x}{3} - (x+1)}{3} - (x+2)$. $P(x,3) \Rightarrow 3f(x+3) = f(x)f(3) - 3x = 12f(x) - 3x$. Therefore, $12f(x) - 3x = f(1)\frac{f(1)\frac{f(1)f(x) - x}{3} - (x+1)}{3} - (x+2),\forall x\in\mathbb{R}$, contradiction. 2. If $a= -3$, the same argument yields another contradiction. Thus $a=0$ and we have $f(x)f(y) = xy,\forall x,y\in\mathbb{R}$. Put $y=1$ then $f(x) = kx,\forall x\in\mathbb{R}$, where $k=\frac{1}{f(1)}$. Now the assertion $P(x,y)$ yields that $k=1$. Therefore, the only solution is $f(x) = x$.

Hello!!!!!!

Roobiks [/hide]

RobRoobiks wrote: Hello!!!!!! Let $y = 0$ $f(f(x)) = f(x) + f(0)f(x)$ $f(f(x)) = (f(0) + 1)f(x)$ thus its easy to see $f(x) = (f(0) + 1)x$ Yes, easy to see, but wrong. You got $f(x) = (f(0) + 1)x$ $\forall x\in f(\mathbb R)$ and not $\forall x\in\mathbb R$ (you did not show that $f(x)$ is a surjection).

Mashimaru wrote: Therefore, $12f(x) - 3x = f(1)\frac {f(1)\frac {f(1)f(x) - x}{3} - (x + 1)}{3} - (x + 2),\forall x\in\mathbb{R}$, contradiction. How do you reach the contradiction? For example if $f(1) = 0$ then it is equivalent to $f(x) = \frac {x - 1}{6}$. If $f(1) \neq \sqrt [3]{108}$ then it is equivalent to $f(x) = ax + b$ for some $a,b \in \mathbb{R}$. If $f(1) = \sqrt [3]{108}$ then you can reach the contradiction at once. So you can only use it to see that $f$ is a linear function. (Sorry if this was obvious)

$f(f(x + y)) = f(x + y) + f(x)f(y) - xy$ (1) Denote $f(0) = c$. Take $y = 0$ to get $f(f(x)) = f(x)(1 + c)$. (2) Set $y = - x$ in (1): $f(f(0)) - f(0) = x^{2} - f(x)f( - x)$, but by (2), $f(f(0)) - f(0) = c^{2}$. Hence, $f( - x)f(x) = c^{2} - x^{2}$ (3). Take $x = c$ in (3) to conclude there is a real number $u$ satisfying $f(u) = 0$. Take $x = u$ in (2): $f(0) = f(f(u)) = f(u)(1 + c) = 0$. It means $f(f(x)) = f(x)$, as $c = 0$, but than from (1), $f(x)f(y) = xy$ for all $x,y$. Easy to verify $f(x) = 0$, for all $x$ is not a solution. Take $t$ such that $f(t) = w$ and $w$ is nonzero. $wf(x) = xt$, which can be rewritten as $f(x) = kx$, where $k$ is a fixed real number. Plugging it in (1), $(x + y)(k^{2} - k) = xy(k^{2} - 1)$, for all $x,y$. If $x = - y$ and they are both nonzero, the last equation would give $k^{2} - 1 = 0$. Assume $k = - 1$; Than, for all pairs $(x,y)$, $2(x + y) = 0$, which is clearly false. It means that $k = 1$ and $f(x) = x$, which obviously satisfies the given conditions.

Let $P(x,y):=f(f(x+y))=f(x+y)+f(x)f(y)-xy$ $P(0,0)$ yields $f(f(0))=f(0)+f(0)^2$ $P(-f(0),f(0))$ yields $f(f(0))=f(0)+f(-f(0))f(f(0))+f(0)^2\overset{\text{ from }P(0,0)}{\Longrightarrow}f(0)+f(0)^2=f(0)+f(0)^2+f(-f(0))f(f(0))\Longrightarrow f(f(0))f(-f(0))=0$ Thus $f(f(0))=0\text{ or }f(-f(0))\text{ or both of them are equal to }0$ Case 1: $f(-f(0))=0$ $P(0,-f(0))$ yields $f(f(-f(0))=f(-f(0))+f(0)f(-f(0))\Longrightarrow f(0)=0$ Case 2: $f(f(0))=0$ Plugging this in $P(0,0)$ we obtain $f(0)+f(0)^2=0\Longrightarrow f(0)(f(0)+1)=0$ thus either $f(0)=0\text{ or }f(0)=-1$ Subcase: $f(0)=-1$ This implies $f(-1)=0$ $P(x,0)$ yields $f(f(x))=f(x)+f(x)f(0)\Longrightarrow f(f(x))=0$ plugging this in $P(x,y)$ yields $f(x+y)+f(x)f(y)=xy:=Q(x,y)$ $Q(x+1,-1)$ yields $f(x)+f(x+1)f(-1)=-(x+1)\Longrightarrow f(x)=-(x+1)$ however plugging this into $P(x,y)$ yields a contradiction. Therefore since $f(0)=0$ in both of these cases we can conclude that $f(0)=0$ $P(x,0)$ yields $f(f(x))=f(x)$ plugging this in $P(x,y)$ yields $f(x+y)=f(x+y)+f(x)f(y)-xy\Longrightarrow f(x)f(y)=xy:=T(x,y)$ $T(x,1)$ yields $f(x)f(1)=x\Longrightarrow f(x)=\frac{x}{c}\text{ where }c=f(1)$ Pugging this in $P(x,y)$ yields $\frac{x+y}{c^2}=\frac{x+y}{c}+\frac{xy}{c^2}-xy\Longrightarrow(c-1)(x+y)=(c-1)xy\Longrightarrow(c-1)(x+y-xy)=0$ thus either $c=1\text{ or }x+y=xy$ However the latter clearly yields a contradiction. Therefore $c=1$ which implies that $f(x)=x$ In conclusion $\boxed{f(x)=x, \forall x\in\mathbb{R}}$ $\blacksquare$.