FelixD wrote: Find all functions $ f: \mathbb{R} \to \mathbb{R}$, such that \[ f(xf(y)) + f(f(x) + f(y)) = yf(x) + f(x + f(y))\] holds for all $ x$, $ y \in \mathbb{R}$, where $ \mathbb{R}$ denotes the set of real numbers. Let $ P(x,y)$ be the assertion $ f(xf(y))+f(f(x)+f(y))=yf(x)+f(x+f(y))$ Let $ f(0)=a$ $ f(x)=0$ $ \forall x$ is a solution. Let us from now consider $ \exists c$ such that $ f(c)\neq 0$ If $ f(y_1)=f(y_2)$, subtracting $ P(c,y_1)$ from $ P(c,y_2)$ implies $ 0=f(c)(y_2-y_1)$ and so $ y_1=y_2$ and $ f(x)$ is injective. $ P(0,1)$ $ \implies$ $ a+f(a+f(1))=a+f(f(1))$, so $ f(a+f(1))=f(f(1))$ and so, since $ f(x)$ is injective, $ a=0$ Then $ P(x,0)$ $ \implies$ $ f(f(x))=f(x)$ and, since $ f(x)$ is injective, $ f(x)=x$ Hence two solutions : $ f(x)=0$ $ \forall x$ $ f(x)=x$ $ \forall x$

let us suppose that f is non-constant($f(x)\equiv 0 $is a trivial solution)let $x=0$,$f(0)+f(f(y)+f(0))=yf(0)+f(f(y))$ let $y=0,f(xf(0))+f(f(x)+f(0))=f(x+f(0))$ it's trivial that f is injective. let $y=1$,then $f(f(1)+f(0))=f(f(1))$,hence $f(0)=0$ so $f(f(x))=f(x)$ hence $f(x)=x$.

littletush wrote: let us suppose that f is non-constant($f(x)\equiv 0 $is a trivial solution)let $x=0$,$f(0)+f(f(y)+f(0))=yf(0)+f(f(y))$ let $y=0,f(xf(0))+f(f(x)+f(0))=f(x+f(0))$ it's trivial that f is injective. let $y=1$,then $f(f(1)+f(0))=f(f(1))$,hence $f(0)=0$ so $f(f(x))=f(x)$ hence $f(x)=x$. Very interesting brand new solution ! Congrats !

is this solution ok? Let $P(x,y)$ be assertion of given equation.We can easily prove that the function is an injection(like in the post before). $P(x,x)$ ->$f(xf(x))+f(2f(x))=xf(x)+f(x+f(x))$ $P(f(x),x)$ ->$f(f(x)x)+f(2x)=f(x)x+f(f(x)+x)=f(xf(x))+f(2f(x))$ which implies $f(2x)=f(2f(x))$ so since the function is an injection we get $f(x)=x$ which is solution. Hence the only solutions are $f(x)=0,f(x)=x$

littletush wrote: let us suppose that f is non-constant($f(x)\equiv 0 $is a trivial solution)let $x=0$,$f(0)+f(f(y)+f(0))=yf(0)+f(f(y))$ let $y=0,f(xf(0))+f(f(x)+f(0))=f(x+f(0))$ it's trivial that f is injective. let $y=1$,then $f(f(1)+f(0))=f(f(1))$,hence $f(0)=0$ so $f(f(x))=f(x)$ hence $f(x)=x$. How did u get that $f$ is injective?

Let $P(x,y)$ be the assertion of the given equation. Suppose that $f(a)=f(b)$. Then, $P(x,a)$ and $P(x,b)$ gives $af(x)=f(xf(a))+f(f(x)+f(a))-f(x+f(a))=f(xf(b))+f(f(x)+f(b))-f(x+f(b))=bf(x)$. We have $a=b$ if there is some value of $x$ such that $f(x)\neq 0$. If an value of $x$ doesn't exist, then we have $\boxed{f(x)=0}$ as a solution, which indeed works. If an value of $x$ does exist, now we have $a=b$ or $f$ is injective. Let $P(0,1)$ so that $$f(0)+f(f(0)+f(1))=f(0)+f(f(1))\implies f(f(0)+f(1))=f(f(1)).$$Since $f$ is injective, we have $f(0)+f(1)=f(1)\implies f(0)=0.$ Now, let $P(x,0)$ so $$f(f(x))=f(x)$$and because $f$ is injective, we have $\boxed{f(x)=x}$, which is another solution. $\blacksquare$

pco wrote: FelixD wrote: Find all functions $ f: \mathbb{R} \to \mathbb{R}$, such that \[ f(xf(y)) + f(f(x) + f(y)) = yf(x) + f(x + f(y))\]holds for all $ x$, $ y \in \mathbb{R}$, where $ \mathbb{R}$ denotes the set of real numbers. Let $ P(x,y)$ be the assertion $ f(xf(y))+f(f(x)+f(y))=yf(x)+f(x+f(y))$ Let $ f(0)=a$ $ f(x)=0$ $ \forall x$ is a solution. Let us from now consider $ \exists c$ such that $ f(c)\neq 0$ If $ f(y_1)=f(y_2)$, subtracting $ P(c,y_1)$ from $ P(c,y_2)$ implies $ 0=f(c)(y_2-y_1)$ and so $ y_1=y_2$ and $ f(x)$ is injective. $ P(0,1)$ $ \implies$ $ a+f(a+f(1))=a+f(f(1))$, so $ f(a+f(1))=f(f(1))$ and so, since $ f(x)$ is injective, $ a=0$ Then $ P(x,0)$ $ \implies$ $ f(f(x))=f(x)$ and, since $ f(x)$ is injective, $ f(x)=x$ Hence two solutions : $ f(x)=0$ $ \forall x$ $ f(x)=x$ $ \forall x$ Could you explain some motivation,for such problems. this part (Let us from now consider $ \exists c$ such that $ f(c)\neq 0$)

itslumi wrote: Could you explain some motivation,for such problems. this part (Let us from now consider $ \exists c$ such that $ f(c)\neq 0$) I claimed that $f(x)\equiv 0$ is a solution So from then, I search only non allzero solutions. So it's normal to consider that $\exists c$ such that $f(c)\ne 0$

pco wrote: itslumi wrote: Could you explain some motivation,for such problems. this part (Let us from now consider $ \exists c$ such that $ f(c)\neq 0$) I claimed that $f(x)\equiv 0$ is a solution So from then, I search only non allzero solutions. So it's normal to consider that $\exists c$ such that $f(c)\ne 0$ Yes i undertand but how to now intuitivly as a costant to try this method at a right time and at a right problem

I dont know what to explain exactly ... I wanted to test / prove injectivity and comparing $P(x,y_1)$ with $P(x,y_2)$, it was obvious that finding an $x$ such that $f(x)\ne 0$ was enough to conclude injectivity. Hence the process : If $f(x)=0$ $\forall x$, we have a solution (non injective ). If $f\not\equiv 0$, then using an $x$ such that $f(x)\ne 0$ allows to prove injectivity. Hence my choice about $c$.

If $f\not\equiv 0$, it's clear that $f$ is injective. If we set $x=y=0$, we find that $f(2f(0))=0$. Then setting $x=y=2f(0)$ gives $f(0)=0$. Now by putting $y=0$, we find that $f(f(x))=f(x)$, from which we conclude $f(x)=x$ by injectivity. So the only functions are $f(x)=0$ and $f(x)=x$, which clearly work.