WakeUp wrote: Let $\mathbb{R}$ be the set of all real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfying for all $x,y\in\mathbb{R}$ the equation $f(x)+f(y)=f(f(x)f(y))$. Let $P(x,y)$ be the assertion $f(x)+f(y)=f(f(x)f(y))$ Let $A=f(\mathbb R)$ Let $u\in A$ Functional equation implies that $a,b\in A\implies a+b\in A$ and so $nu\in A$ $\forall n\in\mathbb N$ Let $x_1$ such that $f(x_1)=u$ Let $x_2$ such that $f(x_2)=2u$ Let $x_4$ such that $f(x_4)=4u$ $P(x_1,x_4)$ $\implies$ $5u=f(4u^2)$ $P(x_2,x_2)$ $\implies$ $4u=f(4u^2)$ So $4u=5u$ and $u=0$ So $A=\{0\}$ and so $\boxed{f(x)=0}\forall x$ which indeed is a solution.

* Assume that there is not exist a real number $a$ such that $f(a)=0$. let $y=b$ ($f(b) \ne 0$), we get $f(x)+q=f(qf(x))$. So, we can prove that $f(x)$ is a one-to-one function. Hence $\exists t, f\left( t \right) = 0$, which is not true, because of our assumption. * Otherwise, $\exists t, f\left( t \right) = 0$. Let $y=t$, we get $f(x)=f(0)=k:const$ From the condition, we infer $k=0$ so $\boxed{f\left( x \right) = 0}$

nguyenhung wrote: * Assume that there is not exist a real number $a$ such that $f(a)=0$. let $y=b$ ($f(b) \ne 0$), we get $f(x)+q=f(qf(x))$. So, we can prove that $f(x)$ is a one-to-one function. Hence $\exists t, f\left( t \right) = 0$, which is not true, because of our assumption. I'm sorry, but I understood nothing to this step. 1) you seem to conclude from "one to one" property that $\exists t, f\left( t \right) = 0$. I dont see why. "one to one" means injective. And I dont see how injectivty gives you the claimed result 2) I dont see how you can conclude anything from $f(x)+q=f(qf(x))$ where $x\in\mathbb R$ and $q\in f(\mathbb R)$ : neither injection, neither surjection.

Quote: I'm sorry, but I understood nothing to this step. 1) you seem to conclude from "one to one" property that $\exists t, f\left( t \right) = 0$. I dont see why. "one to one" means injective. And I dont see how injectivty gives you the claimed result 2) I dont see how you can conclude anything from $f(x)+q=f(qf(x))$ where $x\in\mathbb R$ and $q\in f(\mathbb R)$ : neither injection, neither surjection. Oh.. yes. You're right. My idea has a lot of troubles.

Let $P(x,y)$ denote the given assertion. $P(x,x)\implies f(f(x)^2)=2f(x)$ $P(f(x)^2,f(x)^2)\implies f(4f(x)^2)=4f(x)$ $P(x,4f(x)^2)\implies 5f(x)=4f(x)\implies f\equiv 0$; works.