j___d wrote: Find all real numbers $c$ for which there exists a function $f\colon\mathbb R\rightarrow \mathbb R$ such that for each $x, y\in\mathbb R$ it's true that $$f(f(x)+f(y))+cxy=f(x+y).$$ Let $P(x,y)$ be the assertion $f(f(x)+f(y))+cxy=f(x+y)$ Let $A=f(\mathbb R)$ Let $a=f(0)$ 1) If $c=0$ then such function exists Just choose for example $f(x)=x$ $\forall x$ Q.E.D. 2) If $c>0$, no such function $P(x,-x)$ $\implies$ $f(f(x)+f(-x))=cx^2+a$ and so $[a,+\infty)\subseteq A$ $P(x,0)$ $\implies$ $f(f(x)+a)=f(x)$ And so $f(x)=x-a$ $\forall x\ge 2a$ Let then $x,y$ such that : $x,y\ge 2a$ and $x+y\ge\max(2a,4a)$ $P(x,y)$ becomes $x+y-3a+cxy=x+y-a$ which obviously is impossible Q.E.D. 3) If $c<0$, no such function : $P(x,-x)$ $\implies$ $f(f(x)+f(-x))=cx^2+a$ and so $(-\infty,a]\subseteq A$ $P(x,0)$ $\implies$ $f(f(x)+a)=f(x)$ And so $f(x)=x-a$ $\forall x\le 2a$ Let then $x,y$ such that : $x,y\le 2a$ and $x+y\le\min(2a,4a)$ $P(x,y)$ becomes $x+y-3a+cxy=x+y-a$ which obviously is impossible Q.E.D. Hence the answer : $\boxed{c=0}$

Here is another approach for the case $ c \neq 0 $: Suppose that there are $ a \neq b \in \mathbb{R} $ such that $ f\left(a\right) = f\left(b\right) $. Compare $ P\left(x,a\right), P\left(x,b\right) $: $$ f\left(x+a\right)-cax = f\left(x+b\right)-cbx , \forall x \in \mathbb{R} $$Rewrite the identity as $$ f\left(x+T\right) = f\left(x\right)+cT\left(x-b\right), \forall x \in \mathbb{R} $$where $ T = a-b $. Now, compare $ f\left(x,2b-x\right), f\left(x+T,2b-x+T\right) $, $$ 2cTb+cT^2 = cTb \rightarrow T = -b \rightarrow a =0 $$Notice we have $ f\left(f\left(x\right)+f\left(0\right)\right) = f\left(x\right) $ ...

pco wrote: j___d wrote: Find all real numbers $c$ for which there exists a function $f\colon\mathbb R\rightarrow \mathbb R$ such that for each $x, y\in\mathbb R$ it's true that $$f(f(x)+f(y))+cxy=f(x+y).$$ Let $P(x,y)$ be the assertion $f(f(x)+f(y))+cxy=f(x+y)$ Let $A=f(\mathbb R)$ Let $a=f(0)$ 1) If $c=0$ then such function exists Just choose for example $f(x)=x$ $\forall x$ Q.E.D. 2) If $c>0$, no such function $P(x,-x)$ $\implies$ $f(f(x)+f(-x))=cx^2+a$ and so $[a,+\infty)\subseteq A$ $P(x,0)$ $\implies$ $f(f(x)+a)=f(x)$ And so $f(x)=x-a$ $\forall x\ge 2a$ Let then $x,y$ such that : $x,y\ge 2a$ and $x+y\ge\max(2a,4a)$ $P(x,y)$ becomes $x+y-3a+cxy=x+y-a$ which obviously is impossible Q.E.D. 3) If $c<0$, no such function : $P(x,-x)$ $\implies$ $f(f(x)+f(-x))=cx^2+a$ and so $(-\infty,a]\subseteq A$ $P(x,0)$ $\implies$ $f(f(x)+a)=f(x)$ And so $f(x)=x-a$ $\forall x\le 2a$ Let then $x,y$ such that : $x,y\le 2a$ and $x+y\le\min(2a,4a)$ $P(x,y)$ becomes $x+y-3a+cxy=x+y-a$ which obviously is impossible Q.E.D. Hence the answer : $\boxed{c=0}$ My solution is similar to pco's solution.

Another solution: If $c=0$ then such function exists (e.g. $f(x)=x \ \forall{x}$). Suppose that $c \neq 0$. Let $t=f(0)$. $$P(0,0) \implies f(2t)=t$$$$P(x,0) \implies f(f(x)+t)=f(x)$$$$P(x,2t) \implies f(x+2t)=f(x)+2ctx$$$$P(f(x)+t,f(x)+t) \implies f(2f(x))+c(f(x)+t)^2=f(2f(x))+4ctf(x) \implies c(f(x)-t)^2=0$$Thus, $f(x)=t \ \forall{x}$. However, plugging that into our equation gives us $cxy=0 \ \forall{x}$, contradiction. Hence, $c=0$ is the only answer.

only possible solution is $c=0$ : If $c=0$: $f(x)=0$ is a function that work in problem so $c \neq 0$ : let $P(x,y)$ be the assertion $f(f(x)+f(y))+cxy=f(x+y)$ $P(x,-x) \Rightarrow f(f(x)+f(-x))=f(0)+cx^2$ it means $\forall x\ge f(0) : \exists y_x : f(y_x)=x$ $(1)$ $P(x,0) \Rightarrow f(f(x)+f(0))=f(x) \Rightarrow \forall x\ge 2f(0): f(x)=x-f(0)$ now for very large $x,y$ we have : $x+y-3f(0)+cxy=x+y-f(0)$ so $c=0$