Very nice! Let $P(x,y)$ be the given assertion,. Clearly $f\equiv c$ works, so let's now assume that $f$ is non-constant. $P(0,0): f(0)=f(2f(0))$ $P(x,0): f(x)=f(f(x)+f(0))$ $P(x,2f(0)): f(x)+f(2f(0))=f(2f(0)x)+f(f(x)+f(2f(0))\Leftrightarrow f(2f(0)x)=f(0)$ so $f(0)=0$ otherwise $f$ must be constant, $P(x,0)$ becomes $f(f(x))=f(x)$. Observe that $f(a)=f(0)\Leftrightarrow a=0$ since $P(x,a)$ gives $f(ax)=f(0)=0$ and $f$ is not constant. Now lets define the set $A=\left \{k| f(x)=f(kx)\forall x \in \mathbb{R} \right \}$ If $f(a)=f(b)$ for some $a\neq b$ then from $P(x,a),P(x,b)$ we get that $f(xa)=f(xb)$ and thus $a/b=k\in A$ Also $f(f(1))=f(1)$ so $f(1)\in A$ ( note that $f(1)\neq 0$ since otherwise $P(x,1): f(1)=f(f(x)+f(1))\Rightarrow f(f(x))=0$ but $f(f(x))=f(x)$ and hence $f\equiv 0$ a contradiction.) Now $P(1,1): f(1)=f(2f(1))$ so $2f(1)\in A$ as well and hence for every $x$ we have that $f(2f(1)x)=f(x)=f(xf(1))\Rightarrow f(x)=f(2x)$ for all $x$, i.e $2\in A$. $P(x,x): 2f(x)=f(x^2)+f(2f(x))=f(x^2)+f(f(x))=f(x^2)+f(x)\Leftrightarrow f(x)=f(x^2)\Rightarrow x\in A,\forall x\neq 0$. Now pick any $x,y\neq 0$, then $x/y\in A$ and thus $f(y)=f(y\cdot x/y)=f(x)$. So all the solution are: $f(x)\equiv c,\forall x$ $f(0)=0,f(x)=c\neq 0,\forall x\neq 0$ and they clearly work.

Prod55 wrote: Very nice! Let $P(x,y)$ be the given assertion,. Clearly $f\equiv c$ works, so let's now assume that $f$ is non-constant. $P(0,0): f(0)=f(2f(0))$ $P(x,0): f(x)=f(f(x)+f(0))$ $P(x,2f(0)): f(x)+f(2f(0))=f(2f(0)x)+f(f(x)+f(2f(0))\Leftrightarrow f(2f(0)x)=f(0)$ so $f(0)=0$ otherwise $f$ must be constant, $P(x,0)$ becomes $f(f(x))=f(x)$. Observe that $f(a)=f(0)\Leftrightarrow a=0$ since $P(x,a)$ gives $f(ax)=f(0)=0$ and $f$ is not constant. Now lets define the set $A=\left \{k| f(x)=f(kx)\forall x \in \mathbb{R} \right \}$ If $f(a)=f(b)$ for some $a\neq b$ then from $P(x,a),P(x,b)$ we get that $f(xa)=f(xb)$ and thus $a/b=k\in A$ Also $f(f(1))=f(1)$ so $f(1)\in A$ ( note that $f(1)\neq 0$ since otherwise $P(x,1): f(1)=f(f(x)+f(1))\Rightarrow f(f(x))=0$ but $f(f(x))=f(x)$ and hence $f\equiv 0$ a contradiction.) Now $P(1,1): f(1)=f(2f(1))$ so $2f(1)\in A$ as well and hence for every $x$ we have that $f(2f(1)x)=f(x)=f(xf(1))\Rightarrow f(x)=f(2x)$ for all $x$, i.e $2\in A$. $P(x,x): 2f(x)=f(x^2)+f(2f(x))=f(x^2)+f(f(x))=f(x^2)+f(x)\Leftrightarrow f(x)=f(x^2)\Rightarrow x\in A,\forall x\neq 0$. Now pick any $x,y\neq 0$, then $x/y\in A$ and thus $f(y)=f(y\cdot x/y)=f(x)$. So all the solution are: $f(x)\equiv c,\forall x$ $f(0)=0,f(x)=c\neq 0,\forall x\neq 0$ and they clearly work. l hope that my solution is true Let $P(x,y)$ be the given assertion 1)$f(0)$$\neq$$0$ $P(x,0)$ $\rightarrow$ $f(x)=f(f(x)+f(0))$ $(1)$ $P(x,1)$ $\rightarrow$ $f(f(x)+f(1))=f(1)$ $(2)$ From $(1)$ we get $f(0)=f(2f(0))$ $P(\frac{1}{2f(0)},2f(0))$ $\rightarrow$ $f(0)=f(1)$ and from $(1)$ and $(2)$ we get f(x)=f(1)=c for all $x$$\in$$R$ $c$$\neq$$0$ 2)$f(0)=0$ $P(x,0)$ $\rightarrow$ $f(x)=f(f(x))$ $(1)$ $P(x,1)$ $\rightarrow$ $f(f(x)+f(1))=f(1)$ $(2)$ From $(2)$ we get $f(2f(1)$$=$$f(1)$ then $P(2,f(1))$ $\rightarrow$ $f(2)=f(1)$ $(3)$ From $P(f(x),y)$$=G(x,y)$ and $(1)$ we get $f(f(x)y)=f(xy)$ from $G(1,x)$,$G(2,x)$ and $(3)$ we get $f(x)=f(2x)$ from here we get $f(x)=f(\frac{x}{2^{\infty}})=f(0)=0$ Then for all $x$$\in$$R$ $f(x)=0$

Neat F.E., let $P(x,y)$ the assertion as usual, assume $f$ non-constant as the other case works. Main Weapon: If $f(a)=f(b)$, from $P(x,a)-P(x,b)$ we get $f(ax)=f(bx)$ for all $x \in \mathbb R$ call this $\star$ Now $P(0,0)$ gives $f(0)=f(2f(0))$ so from $\star$ we have $f(2f(0)x)=f(0)$ so if $f(0) \ne 0$ then $f$ is constant so we get that $f(0)=0$, now clearly $f$ is injective at $0$ or else $f(cx)=0$ for $c \ne 0$ such that $f(c)=0$ which again gives $f$ constant so nope. Now $P(1,1)$ gives $f(1)=f(2f(1))$ and $P(x,0)$ gives $f(x)=f(f(x))$ so by $\star$ we get $f(2f(1)x)=f(x)=f(f(1)x)$ so $f(x)=f(2x)$ and now by $P(x,x)$ we get that $f(x)=f(x^2)$ so by $\star$ we get that $f(xy)=f(xy^2)$, now pick $x,y \ne 0$ and let $x=\frac{a}{y}$ and $y=\frac{b}{a}$ for any reals $a,b \ne 0$ and we get $f(a)=f(b)$ so $f$ constant for all values but $0$. Hence the set of solutions is $f(x)=c$ for all reals $x$ and $f(x)=c \ne 0$ for all $x \ne 0$ and $f(0)=0$, thus we are done .

Let P(x,y) be the assertion: f(x)+f(y)=f(xy)+f(f(x)+f(y)) assume f(a)=f(b). If we substitute y=a or b, the only term that can change is f(ax). Therfore f(ax)=f(bx) for all x. P(0,0): f(2f(0))=f(0) => f(2f(0)x)=f(0). So either f is constant (which works) or f(0)=0. P(x,0): f(x)=f(f(x)) => f(2x)=f(2f(x)) (for x,y≠0) f(P(x,y)): f(f(x)+f(y))=f(f(xy)+f(f(x)+f(y)))= =f(f(x)+f(y))+f(xy)-f(xy(f(x)+f(y))) => f(f(x)+f(y))=f(1). For x=y≠0 we have f(1)=f(2f(x))=f(2x). So f is constant except 0 which it is 0 at (it works). But what if f was from R+ to R+, you might ask. Well: the equation f(f(x)+f(y))=f(1) still holds, so if we can rewrite the equation as f(x)+f(y)=f(xy)+f(1). Denote m=Inf(Im(f)). Suppose f(x)<m+ε for arbitrarily small ε>0. P(x,x): f(1)=2f(x)-f(x^2)<m+2ε. so f(1)=m. P(x,1/x): f(x)+f(1/x)=2m => f(x)=m for all x. So the only solution is a positive constant function.

Mathlover_1 wrote: Prod55 wrote: Very nice! Let $P(x,y)$ be the given assertion,. Clearly $f\equiv c$ works, so let's now assume that $f$ is non-constant. $P(0,0): f(0)=f(2f(0))$ $P(x,0): f(x)=f(f(x)+f(0))$ $P(x,2f(0)): f(x)+f(2f(0))=f(2f(0)x)+f(f(x)+f(2f(0))\Leftrightarrow f(2f(0)x)=f(0)$ so $f(0)=0$ otherwise $f$ must be constant, $P(x,0)$ becomes $f(f(x))=f(x)$. Observe that $f(a)=f(0)\Leftrightarrow a=0$ since $P(x,a)$ gives $f(ax)=f(0)=0$ and $f$ is not constant. Now lets define the set $A=\left \{k| f(x)=f(kx)\forall x \in \mathbb{R} \right \}$ If $f(a)=f(b)$ for some $a\neq b$ then from $P(x,a),P(x,b)$ we get that $f(xa)=f(xb)$ and thus $a/b=k\in A$ Also $f(f(1))=f(1)$ so $f(1)\in A$ ( note that $f(1)\neq 0$ since otherwise $P(x,1): f(1)=f(f(x)+f(1))\Rightarrow f(f(x))=0$ but $f(f(x))=f(x)$ and hence $f\equiv 0$ a contradiction.) Now $P(1,1): f(1)=f(2f(1))$ so $2f(1)\in A$ as well and hence for every $x$ we have that $f(2f(1)x)=f(x)=f(xf(1))\Rightarrow f(x)=f(2x)$ for all $x$, i.e $2\in A$. $P(x,x): 2f(x)=f(x^2)+f(2f(x))=f(x^2)+f(f(x))=f(x^2)+f(x)\Leftrightarrow f(x)=f(x^2)\Rightarrow x\in A,\forall x\neq 0$. Now pick any $x,y\neq 0$, then $x/y\in A$ and thus $f(y)=f(y\cdot x/y)=f(x)$. So all the solution are: $f(x)\equiv c,\forall x$ $f(0)=0,f(x)=c\neq 0,\forall x\neq 0$ and they clearly work. l hope that my solution is true Let $P(x,y)$ be the given assertion 1)$f(0)$$\neq$$0$ $P(x,0)$ $\rightarrow$ $f(x)=f(f(x)+f(0))$ $(1)$ $P(x,1)$ $\rightarrow$ $f(f(x)+f(1))=f(1)$ $(2)$ From $(1)$ we get $f(0)=f(2f(0))$ $P(\frac{1}{2f(0)},2f(0))$ $\rightarrow$ $f(0)=f(1)$ and from $(1)$ and $(2)$ we get f(x)=f(1)=c for all $x$$\in$$R$ $c$$\neq$$0$ 2)$f(0)=0$ $P(x,0)$ $\rightarrow$ $f(x)=f(f(x))$ $(1)$ $P(x,1)$ $\rightarrow$ $f(f(x)+f(1))=f(1)$ $(2)$ From $(2)$ we get $f(2f(1)$$=$$f(1)$ then $P(2,f(1))$ $\rightarrow$ $f(2)=f(1)$ $(3)$ From $P(f(x),y)$$=G(x,y)$ and $(1)$ we get $f(f(x)y)=f(xy)$ from $G(1,x)$,$G(2,x)$ and $(3)$ we get $f(x)=f(2x)$ from here we get $f(x)=f(\frac{x}{2^{\infty}})=f(0)=0$ Then for all $x$$\in$$R$ $f(x)=0$ the last step is wrong, $f(\frac{x}{2^{\infty}})=f(0)$ works if $f$ is continuous at 0 but it's not always true.

Solved with hints. Really nice and instructive problem. I think I learnt a lot from this one. The answers are $f(x) = c$ for all $x \in \mathbb{R}$ for a fixed constant $c \in \mathbb{R}$ and \[f(x) = \begin{cases} 0 & x=0\\ c & x\neq 0 \end{cases}\]for a fixed constant $c \in \mathbb{R}$. It’s easy to see that these functions satisfy the given conditions. We now show these are the only solutions. Let $P(x,y)$ denote the assertion that $f(x)+f(y)=f(xy)+f(f(x)+f(y))$ for real numbers $x$ and $y$. We first start off with fixing certain values of $f$. Claim : $f(0)=0$ or $f$ is constant. Proof : Say there exists some $\alpha \neq 0 \in \mathbb{R}$ such that $f(\alpha)=f(0)$. Then, comparing $P(x,0)$ and $P(x,\alpha)$ yields, \[f(\alpha x)=f(x)+f(\alpha)+f(f(x)+f(\alpha))=f(x)+f(0)-f(f(x)+f(0))=f(0)\]Thus, letting $x= \frac{t}{\alpha}$ implies that $f(t)=f(0)$ for all real numbers $t$. Thus, if $f$ is non-constant there does not exist such $\alpha \neq 0$ such that $f(x)=f(\alpha)$. Now, $P(0,0)$ implies $f(0)=f(2f(0))$ which according to our previous observation requires, $2f(0)=0$ so we have $f(0)=0$ indeed as claimed. In what proceed, we assume that $f$ is not constant. First of all, note that $P(x,0)$ implies, \[f(x)=f(x)+f(0)=f(0)+f(f(x)+f(0))=f(f(x))\]and $P(x,1)$ implies \[f(1)=f(f(x)+f(1))\]for all real numbers $x$. This will come in useful later. Note that $P(f(x),2)$ gives, \[f(2f(x)) = f(f(x))+f(2)-f(f(f(x))+f(2))=f(x)+f(2)-f(f(x)+f(2))\]and $P(x,x)$ also gives \[f(2f(x))=2f(x)-f(x^2)\]Thus, setting these two expressions for $f(2f(x))$ equal we have \[f(x^2) + f(2) = f(x)+ f(f(x)+f(2))\]and further, when $x=1$ in the above result, \[f(2)=f(f(1)+f(2))=f(1)\]Thus, $f(2)=f(1)$. Now, returning to $P(x,2)$ we have, \[f(x)+f(1)=f(x)+f(2)=f(2x)+f(f(x)+f(2)) = f(2x)+f(f(x)+f(1))=f(2x)+f(1)\]which implies that $f(2x)=f(x)$ for all real numbers $x$. Then, applying this result to $P(x,x)$ gives \[2f(x)=f(x^2)+f(2f(x))=f(x^2)+f(f(x))=f(x^2)+f(x)\]so, $f(x^2)=f(x)$ for all reals $x$. Now, from this based on our previous observation, it follows that $f(xy)=f(xy^2)$ for all real numbers $x$ and $y$. Considering $x= \frac{b}{a}$ and $y= \frac{a^2}{b}$ for any two non-zero $a\neq b $ implies that $f$ is indeed constant over the non-zero reals. Thus, all solutions are indeed as claimed.