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 $\boxed{f(x)=x, \forall x\in\mathbb{R}}$ $\blacksquare$.