Notice that any constant function satisfies the condition. Let's look for non-constant solutions. Changing $x$ with $y$ we get that $f(f(y)+x)+f(f(x)+y)=2f(yf(x))$ and by the condition we get that $2f(xf(y))=2f(yf(x))\Leftrightarrow f(xf(y))=f(yf(x))$ (*), for every real $x$ and $y$. Setting $x=0$ to (*) we get that $f(yf(0))=f(0)$, implying that if $f(0)\ne 0$, then $f(x)=f(0)=c$, for every real $x$, that is to say $f$ is constant. So we focus on $f(0)=0$. Setting to the original $x=0$ and we obtain that $f(y)+f(f(y))=0$ (**), for every real $y$. Working on the above we get that $f(f(y))=-f(y)\Rightarrow f(f(f(y)))=f(-f(y))$ and using the above again $f(y)=-f(f(y))=f(f(f(y)))=f(-f(y))\Rightarrow f(-f(y))=f(y)$ (***), for every real $y$. Now set to (*) $y=1$ to get that $f(xf(1))=f(f(x))=-f(x)$ (1), by (**), for every real $x$. Note that (1) for $x=-1$ is $f(-f(1))=-f(-1)$ and using (***) $f(1)=-f(-1)$ Setting to (*) $y=-1$ and using the above and (***) we obtain that $f(xf(-1))=f(-f(x))=f(x)$ or $f(-xf(1))=f(x)\Leftrightarrow f(xf(1))=f(-x)$, for every real $x$ (2). Using (1) and (2) we get that $f(-x)=-f(x)$ for every real $x$, that is to say $f$ is odd. Setting to the original $y=x$ we get that $2f(x+f(x))=2f(xf(x))$ or $f(x+f(x))=f(xf(x))$. Now replacing $x$ with $-x$ we get that $f(-x+f(-x))=f(-xf(-x))$ and using that $f$ is odd we eventually get that $-f(x+f(x))=f(-x-f(x))=f(-x+f(-x))=f(-xf(-x))=f(xf(x))=f(x+f(x))$. Consequently $f(x+f(x))=f(xf(x))=0$, for every real $x$. Setting $x=1$ we get that $f(f(1))=0$ or by (**) $-f(1)=0\Rightarrow f(1)=0$. Now (1) implies that $f(x)=0$, for every real $x$. To sum up the only solution is $f(x)=c$, for every real $x$.

Blastzit wrote: Find all real-valued functions $f$ defined on the set of real numbers such that $$f(f(x)+y)+f(x+f(y))=2f(xf(y))$$for any real numbers $x$ and $y$. The same proof than above can be shortened (skipping proof that $f(x)$ is odd, for example) a bit :: Let $P(x,y)$ be the assertion $f(f(x)+y)+f(x+f(y))=2f(xf(y))$ Let $a=f(0)$ and $c=f(1)$ Subtracting $P(x,y)$ from $P(y,x)$, we get new assertion $Q(x,y)$ : $f(xf(y)=f(yf(x))$ 1) If $a\ne 0$ : $Q(\frac xa,0)$ $\implies$ $\boxed{\text{S1 : }f(x)=a\quad\forall x}$ which indeed is a solution, whatever is $a\in\mathbb R$ 2) If $a=0$ $P(x,0)$ $\implies$ $f(f(x)=-f(x)$ and so : a) $f(c)=f(f(1))=-f(1)=-c$ b) $f(f(c))=-f(c)=c$ c) $f(-c)=f(f(c))=c$ Subtracting $P(c,c)$ from $P(-c,1)$, we get $f(c+1)=0$ Then $P(1,1)$ $\implies$ $c=0$ And $Q(x,1)$ $\implies$ $f(f(x))=0$ and so $f(x)=-f(f(x))=0\quad\forall x$, already included in solution found in 1) above

is this right f(x+f(x))=f(xf(x)) x=f(y) , f(f(y)+ff(y))=0=f(-f(y)^2) Also we know if f(u)=0 then u can only be 0 So f(y)=0 for all reals( This is not the complete solution I what to know whether it's correct pls help)

NOOB1729 wrote: Also we know if f(u)=0 then u can only be 0 How ?

Oh ok I got my mistake

Blastzit wrote: Find all real-valued functions $f$ defined on the set of real numbers such that $$f(f(x)+y)+f(x+f(y))=2f(xf(y))$$for any real numbers $x$ and $y$. Clearly any $f\equiv \text{const} $ satisfies the equation.So now assume $f$ is non-constant. \[P(x,y)\implies f(f(x)+y)+f(x+f(y))=2f(xf(y))\]\[P(y,x)\implies f(f(x)+y)+f(x+f(y))=2f(yf(x))\]Thus $f(xf(y))=f(yf(x))$.Plugging $x=0\implies f(yf(0))=f(0)\implies f(0)=0$ since $f$ is non-constant. $P(0,y)\implies f(f(y))=-f(y)$.Now $P(0,f(y))\implies f(f(f(y)))=-f(f(y))$,but $f(f(f(y)))=f(-f(y))$.Thus $f(-f(y))=-f(f(y))$.But $-f(y)=f(f(y))=f(yf(1))$.Plug $y=-f(1)$ to get $-f(1)=f(f(1))=-f(-f(1))=f(-f(1)^2)=0\implies f(1)=0$.Now $f(f(y))=f(yf(1))=f(0)=0\implies -f(y)=0\implies f(y)=0$ as desired.$\square$ Whence we conclude that the only solution is $f\equiv \text{const}$

Pluto1708 wrote: Clearly if $f\equiv \text{const} \implies f\equiv 0$. Wrong. Any constant function fits.

Redacted

$ f(xf(y))=f(yf(x)) $ Case 1: $ f(0)\not=0,x=\frac{t}{f(0)},y=0 $ $ f(\frac{t}{f(0)}*f(0))=f(0)=f(t)=c,c\not=0 $ Case 2: $ f(0)=0 $ $ P(0,x)\implies f(f(x))+f(x)=0 $ $ P(f(x),y)\implies f(f(f(x))+y)+f(f(x)+f(y))=2f(f(x)f(y)) $ $ P(f(y),x)\implies f(f(f(y))+x)+f(f(y)+f(x))=2f(f(y)f(x)) $ $ Then $ $ f(y+f(f(x)))=f(x+f(f(y))) $ $ f(f(x))=-f(x) $ $ f(y-f(x))=f(x-f(y)) $ $ y=f(x) $ $ 0=f(f(x)-f(x))=f(x-f(f(x)))=f(x+f(x)) $ $ Then $ $ f(x+f(x))=0 $ $ P(x,x)\implies f(x+f(x))+f(x+f(x))=2f(xf(x))\implies f(xf(x))=0 $ $ f(xf(x))=0,x=1\implies f(f(1))=0=-f(1) $ $ f(1)=0 $ $ f(xf(y))=f(yf(x)),y=1\implies 0=f(0)=f(f(x)) $ $ Then $ $ 0=f(f(x))=-f(x)\implies f(x)=0 $

Solution

$\clubsuit \color{red}{\textit{\textbf{ANS:}}}$ $f(x)\equiv \textrm{const.} \quad \forall x\in \mathbb{R}$. $\spadesuit \color{blue}{\textit{\textbf{Proof:}}}$ It is easy to see that the above constant function is indeed a solution to the given FE. Let $P(x,y)$ be the given assertion, if $f(0)\ne 0$, then we have \[P\left(\frac{x}{f(0)},0\right) \quad \textrm{and} \quad P\left(0, \frac{x}{f(0)}\right): \boxed{f(x)=f(0)=c \quad \forall x\in \mathbb{R}, \textrm{where} \ c\in \mathbb{R}}.\]If $f(0)=0$, then \[P(x,0): f(f(x))=-f(x),\]\[P(f(x),0): f(-f(x))=f(x)\]and \[P(x,x): f(x+f(x))=f(xf(x)).\]Therefore, \[0=f(0)=f(f(x)-f(x))=f(f(x)+f(f(x)))=f(f(x)f(f(x)))=f(-f(x)^2).\]This leads us to \[P(-f(x),x): f(x+f(x))=f(f(-f(x))+x)=2f(-f(x)^2)=0.\]This means $f(1+f(1))=0$ and so \[P(1,1): f(f(1))=0\]together with $0=f(f(1))=-f(1) \implies f(1)=0$. Thus, \[P(1,x) \quad \textrm{and} \quad P(x,1): f(f(x))=0\]which implies $0=f(f(x))=-f(x) \implies \boxed{f(x)=0}$. $\quad \blacksquare$

The only answer is $f \equiv c$, where $c$ is an arbitrary real number. Let $P(x,y)$ denote the given assertion. From $P(x,x)$, we have that $f(x+f(x))=f(xf(x))$. Thus giving us $f(f(0))=f(0)$. Now notice that $P(0,x)=P(x,0)$, thus giving us $f(xf(0))=f(0)$. Now a case bash ensues. The first case is when $f(0) \neq 0$, this implies that we have that $xf(0)$ is surjective in the real numbers, thus giving us $f(t)=c$, for every real $t$. The second case is when $f(0)=0$. From $P(0,x)$, we get that $f(f(x))=-f(x)$. Thus the logical thing to do is to look at $P(x,f(t))$ and $P(f(t),x)$. Since they are both equal we have that $f(f(x)f(t))=f(-xf(t))=f(-tf(x))$. Thus from $f(-xf(t))=f(-tf(x))$ we must have that $f(t-f(x))=f(x-f(t))$. for every $x$ and $t$. This implies that $f(t+f(t))=0$, implying that $f(tf(t))=0$. This implies that $f(f(1))=0$, which implies that $f(1)=0$. Since we hav ethat $P(x,1)=P(1,x)$, then we must have that $f(f(x))=0$, which impluies that $-f(x)=0$, which gives us that $f(x)=0$.

Let $P(x,y)$ be the assertion $f(x+f(y))+f(y+f(x))=2f(xf(y))$. All solutions of the form $\boxed{f(x)=c},c\in\mathbb R$ work. Assume now that $f$ is nonconstant. $\textbf{1. }Q(x,y):f(xf(y))=f(yf(x))$ $P(x,y)-P(y,x)$. $\blacksquare$ $\textbf{2. }f(0)=0$ If $f(0)\ne0$ then $Q\left(\frac x{f(0)},0\right)\Rightarrow f(x)=f(0)$, contradiction $\textbf{3. }f(\pm f(x))=\mp f(x)$ $P(x,0)\Rightarrow f(f(x))=-f(x)$ In addition, $f(-f(x))=f(f(f(x)))=-f(f(x))=f(x)$. $\textbf{4. }f(1)=0$ $P(x,-f(x))\Rightarrow f(x+f(x))=2f(xf(x))$ $P(x,x)\Rightarrow 2f(x+f(x))=2f(xf(x))$ So $f(x+f(x))=0$, thus $f(xf(x))=0$. In particular, $x=1$ yields $0=f(f(1))=-f(1)$, so $f(1)=0$. $Q(x,1)\Rightarrow f(f(x))=0$ So $-f(x)=f(f(x))=0$, thus $f(x)=0$, which is constant.

I found another, somewhat miraculous solution. Denote by $P(x,y)$ the assertion $f(f(x)+y)+f(x+f(y))=2f(xf(y))$, and by $Q(x,y)$ the assertion $f(xf(y))=f(yf(x))$, which follows from comparing $P(x,y)$ and $P(y,x)$. As in the other solutions, $Q(x,0)$ shows that either $f$ is constant (indeed a solution), or $f(0)=0$. So WLOG $f(0)=0$. Then $$ P(x,0): f(f(x))+f(x)=0\implies f(f(x))=-f(x) $$and thus using $Q(x,1)$ also $f(f(1)x)=f(f(x))=-f(x)$ for all $x$. Therefore $$ P(f(x)f(1),1): f(\underbrace{f(f(x)f(1))}_{=-f(f(x))=f(x)}+1)+\underbrace{f(f(x)f(1)+f(1))}_{=-f(f(x)+1)}=\underbrace{2f(f(1)^2f(x))}_{=-2f(f(1)f(x))=2f(f(x))=-2f(x)} $$$$ \implies 0=f(f(x)+1)-f(f(x)+1)=-2f(x). $$So we obtain $f\equiv 0$, which is indeed a solution.

$P(x,y): f(f(x)+y)+f(x+f(y))=2f(xf(y))$. All constant solutions work, we assume $f$ is non-constant. Exchanging $x$ and $y$, we get $f(xf(y))=f(yf(x))$. Thus, $f(f(x))=f(xf(1))$ and $f(-f(x))=f(xf(-1))$. Furthermore, $f(xf(0))=f(0)$. Since non-constant, $f(0)=0$. $P(x,0): f(f(x))+f(x)=0$. Hence $-f(x)=f(f(x))=f(xf(1))$. $P(f(1),f(1)): f(f(1)f(f(1))=0$. But $f(f(1)f(f(1))=-f(f(f(1))=-f(-f(1))=f(-1)$, hence $f(-1)=0$. Since $f(-f(x))=f(xf(-1)$, $f(-f(x))=0$. $P(x,-f(x)): f(x)=0$.

I claim that all constant functions work. Let $P(x,y)$ be our assertion $P(0,0)\rightarrow f(f(0)=f(0)$ $P(x,0)\rightarrow f(f(x))+f(0)=2f(xf(0))$ $P(0,x)\rightarrow f(f(0)+x)+f(f(x))=2f(0)$ $P(y,x)-P(x,y) \implies f(xf(y))=f(yf(x))$ Now plug $(\frac{x}{f(0)},f(0))$ in the last equation and assume $f(0) \neq 0$ Then we have $f(x)=f(0)=c$ If $f(0)=0$ Then plug this value in the 2nd equation we get $f(f(x))=-f(x)$ and from the first equation we have $f(x)=0 \ \forall x \in \mathbb{R}$ and thus we conclude our proof.

Let $P(x,y)$ denote the assertion $f(f(x)+y)+f(x+f(y))=2f(xf(y)).$ Comparing $P(x,0)$ and $P(0,x)$ implies $f(xf(0))=f(0).$ So we have solution as constant function, which works. Or we have $f(0)=0.$ $P(x,0)$ implies $-f(x)=f(f(x))=f(xf(1))$ and also $f(-f(x))=f(x).$ $P(f(1),f(1))$ gives $f(-f(1)^2)=0.$ Now, $-f(-f(1))=-f(1)=f(-f(1)\cdot f(1))=0,$ so $f(1)=0.$ Comparing $P(x,1)$ and $P(1,x)$ implies $0=f(f(x))=-f(x).$ So $f$ is zero function, which works.

Fakesolver19 wrote: $P(x,0)\rightarrow f(f(x))+f(0)=2f(xf(0))$ No, it should be $f(f(x))+f(x+f(0))=2f(xf(0)).$