proof :Check $P(1,y) $ we get $f(1+f(y))=yf(1)+f(f(1)) (*)$ so if exist $a,b$ such $f(a)=f(b)$ and if we put $y=a , y=b$ in $(*)$ we get $a=b$ or $f(1)=0$ case 1:$f$ be injective. by checking $P(0,0) \implies f(f(0))=f(0) \implies f(0)=0$ $P(x,0) \implies f(f(x))=f(x) \implies f(x)=x $ case 2: $f(1)=0$ also for case $f(1)=0$ solution is simple you can check $P(1,y) \implies f(f(x))=0 (**) $ we are going to proof $f $ bejective : proof :according $(*)$ $(*)+(**) \implies f(x)=0 $ $DONE$

arzhang2001 wrote: $f$ is Injective. and easy to prove $f(0)=0$ proof :proof :Check $P(1,y) $ we get $f(1+f(y))=yf(1)+f(f(1)) *$ so if exit $a,b$ such $f(a)=f(b)$ if we put $y=a , y=b$ in $*$ we get $a=b$ $P(x,0) \implies f(f(x))=f(x) \implies f(x)=x $ also easily we can show that $f$ is bejective but not help us for solving this problem $f$ is not necessarily injective, $f(1) = 0$ could work.

arzhang2001 wrote: $f$ is Injective. and easy to prove $f(0)=0$ proof :proof :Check $P(1,y) $ we get $f(1+f(y))=yf(1)+f(f(1)) *$ so if exit $a,b$ such $f(a)=f(b)$ if we put $y=a , y=b$ in $*$ we get $a=b$ $P(x,0) \implies f(f(x))=f(x) \implies f(x)=x $ also easily we can show that $f$ is bejective but not help us for solving this problem $f(x)=0$ for all real $x$ is also a solution

ak_47 wrote: arzhang2001 wrote: $f$ is Injective. and easy to prove $f(0)=0$ proof :proof :Check $P(1,y) $ we get $f(1+f(y))=yf(1)+f(f(1)) *$ so if exit $a,b$ such $f(a)=f(b)$ if we put $y=a , y=b$ in $*$ we get $a=b$ $P(x,0) \implies f(f(x))=f(x) \implies f(x)=x $ also easily we can show that $f$ is bejective but not help us for solving this problem $f(x)=0$ for all real $x$ is also a solution OK actually from $af(1)+f(f(1))=bf(1)+f(f(1))$ we can get two result: case one: f injective . two $f(1)=0$ which i forget to say. also for case $f(1)=0$ solution is simple you can check $P(1,y) \implies f(f(x))=0 \implies f(x)=0$

^$P(x,1)$ gives $f(x+f(1))+f(x)=f(x)+f(1)+f(f(x))\implies f(x)=f(f(x)).$

This is simple-looking but not simple at all . The answers are $x\mapsto 0$ and $x\mapsto x$. These clearly work so we prove that these are the only ones. Denote the functional equation by $P(x,y)$. Case 1: $f(1)\ne 0$ This case is easy. From $P(1,y)$, we get $f(f(y)+1)=yf(1)+f(f(1))$ or $f$ is injective. Hence from $P(x,0)$, we get $f(x+f(0))=f(f(x))$ or $f(x)=x+f(0)$ from injectivity. It's easy to check that $f(0)=0$. Case 2: $f(1)=0$ This is going to be tough so we step back and gain some more observations. From $P(x,1)$, we get $f(x)=f(f(x))$. Hence plugging in $x=1$ gives $f(0)=0$. From $P(1,y)$, we get $f(f(y)+1)=0$. Keeping the above observations in mind, we find that $P(x,f(k)+1)$ implies $f(x(f(k)+1)) = (f(k)+1)f(x)$. Now we are ready to land a major attack: comparing $P(x,y)$ and $P((f(k)+1)x, (f(k)+1)y)$, we get \begin{align*} f(x+f(y)) - f(y) - f(x) &= yf(x)-f(xy) \\ (f(k)+1)[f(x+f(y)) - f(y) - f(x)] &= (f(k)+1)^2[yf(x)-f(xy)] \end{align*}Thus we deduce that $f(k)(f(k)+1)(yf(x)-f(xy))=0$. If $f(xy)=yf(x)$ for any $x,y$, then it's easy to see that $f$ must be the zero function. Otherwise, we deduce that $f(k)\in\{0,-1\}$ for any $k$. Finally, we have to deal with a silly pointwise trap. Just consider $P(x,2020)$ and see that $f(x)$ must be $0$, otherwise the term $yf(x)$ will be too negative.

arzhang2001 wrote: proof :Check $P(1,y) $ we get $f(1+f(y))=yf(1)+f(f(1)) (*)$ so if exist $a,b$ such $f(a)=f(b)$ and if we put $y=a , y=b$ in $(*)$ we get $a=b$ or $f(1)=1$ case 1:$f$ be injective. by checking $P(0,0) \implies f(f(0))=f(0) \implies f(0)=0$ $P(x,0) \implies f(f(x))=f(x) \implies f(x)=x $ case 2: $f(1)=0$ also for case $f(1)=0$ solution is simple you can check $P(1,y) \implies f(f(x))=0 (**) $ we are going to proof $f $ bejective : proof :according $(*)$ $(*)+(**) \implies f(x)=0 $ $DONE$ Is this a fakesolve

The solutions are $f(x)\equiv 0,x$, both of which can be seen to work. Furthermore, it is easy to verify that these are the only linear solutions. So we'll just show that any solution to the FE must be linear. Let $P(x,y)$ denote the given FE. Note that \[P(x,0)\implies f(x+f(0))=f(f(x)),\]so $f$ injective implies that $f(x)=x+a$ for some constant $a$, or that $f$ is linear, so we're done. Note that \[P(1,y)\implies f(1+f(y))=yf(1)+f(f(1)),\]so if $f(1)\ne 0$, then $f(a)=f(b)\implies a=b$ by the above equation compared for $y=a$ and $y=b$. Thus, we may assume WLOG that $f(1)=0$, so in particular, \[f(1+f(y))=f(0).\]We see that \[P(x,1)\implies f(x)=f(f(x)),\]so in particular, $f(0)=0$, so $f(1+f(y))=0$ for all $y$. Let $T$ be the set of roots of $f$. The above implies that $1+f(y)\in T$ for any $S$. Furthermore, for any $t\in T$, we have \[P(x,t)\implies f(x)+f(xt)=tf(x)+f(t)+f(f(x))\implies f(tx)=tf(x).\]This allows us to solve the problem. In particular, for any $t\in T$, comparing $P(x,y)$ and $P(tx,ty)$ tells us that \[(t^2-t)f(xy)=(t^2-t)yf(x),\]so either $t\in\{0,1\}$, or $f(xy)=yf(x)$ for all $x,y\in \mathbb{R}$. If $T\ne\{0,1\}$, then we have $f(xy)=yf(x)$ for all $x,y$, so in particular, \[yf(x)=xf(y).\]Thus, $f(x)/x=C$ for all nonzero $x$ and some constant $x$, so $f$ is again linear, so we're done. Otherwise, we have $T=\{0,1\}$, so $1+f(y)\in\{0,1\}$ for all $y$. Now we have $f(x)\in\{-1,0\}$ for each real $x$. If $f(a)=-1$, then \[P(a,y)\implies f(a+f(y))+f(ay)=-y+f(y)-1,\]so taking large $y$ gives a contradiction. Thus, we have $f\equiv 0$, so again linear, so we're done. Thus, in all cases, we've shown that $f$ is linear.

Suppose $f(1)\ne 0.$ Plugging $x=1$ $f(1+f(y))=f(1)y+f(f(1))\implies y=\frac{f(1+f(y))-f(f(1))}{f(1)} $ $f$ is injective. Plugging $y=0$ we obtain $f(x+f(0))=f(f(x)), $ from injectivity $f(x)=x+f(0). $ Substituting that in the original equation $yf(0)=0\,\forall y\implies f(0)=0,f(x)=x\,\forall\, x\in\mathbb R$ If $f(1)=0$ then $f $ is identically $0. $ (It is yet to be proved) Plugging $y=1\implies f(x)=f(f(x)). $ $f $ is identity on its image. Hence, $f(0)=0. $

The only solutions are $f \equiv 0$ and $f=\text{id}$, both of which clearly work. Now we show that these are the only solutions. Let $P(x,y)$ denote the given assertion. If $f(1) \neq 0$, then $$P(1,y) \Rightarrow f(f(y)+1)=yf(1)+f(f(1))$$easily gives that $f$ must be injective. In that case, $$P(x,0) \Rightarrow f(x+f(0))=f(f(x)) \Rightarrow x+f(0)=f(x)$$Plugging this back in $P(x,1)$, one can easily check that $f(0)=0$, giving the solution $f=\text{id}$. Now, suppose $f(1)=0$, so that $P(1,y)$ gives $f(f(y)+1)=0$ and $P(x,1)$ gives $f(f(x))=f(x)$. FTSOC assume $f \not \equiv 0$, so that there exists some $x_1$ with $f(x_1)=r \neq 0$. Then the above results give $f(r)=r$ and $f(r+1)=0$. Then $$P \left(\frac{1}{r},r+1 \right) \Rightarrow f \left(\frac{r+1}{r} \right)=(r+1)f \left(\frac{1}{r} \right)=(r+1)z$$where $f \left(\frac{1}{r} \right)=z$. Now, using $f(f(x))=f(x)$, we compare the following two equations- $$P \left(\frac{1}{r},r \right) \Rightarrow f \left(r+\frac{1}{r} \right)=r+(r+1)z \Rightarrow f(r+(r+1)z)=r+(r+1)z$$$$\text{ and }$$$$P \left(r,\frac{r+1}{r} \right) \Rightarrow f(r+(r+1)z)=(r+1)z+(2r+1)$$Comparing gives $$r+(r+1)z=(r+1)z+(2r+1) \Rightarrow r=-1$$Thus, $\text{Im}(f) \in \{-1,0\}$ and $f(r)=r$ implies $f(-1)=-1$. Then $$P(x,-1) \Rightarrow f(x-1)+f(-x)=-1$$Putting $x=-1$, we have $f(-2)=-1$, and then $x=-2$ gives that either $f(2)$ or $f(-3)$ is zero. So suppose $y \in \{-3,2\}$ such that $f(y)=0$. Then $$P(x_1,y) \Rightarrow f(x_1y)=yf(x_1)=-y \Rightarrow f(x_1y) \in \{-2,3\}$$which is a contradiction to the fact that $f$ takes values in $\{-1,0\}$. Thus, we have our desired contradiction. $\blacksquare$

We claim that the only functions that work are $:=$ $$ \boxed {f(x)=x} \text{ } \forall x \in \mathbb R \quad \text {and} \quad \boxed {f(x)=0} \text{ } \forall x \in \mathbb R$$It is easy to see that these work. We now show that these are the only ones . Let $P(x,y)$ denote the assertion. We have the following assertions $:=$ $$P(0,x) \implies f(f(x)) = (x-1)f(0) + f(x) + f(f(0))$$$$P(0,x) \implies f(1+f(x)) = xf(1) + f(f(1))$$$$P(x,0) \implies f(f(x))=f(x+f(0))$$ Hence if one of $f(0)$ or $f(1)$ is not zero then $f$ is injective. First we assume that $f(0) \neq 0$ . Hence, from the third assertion, we get $f(x)=x+f(0)$ . Plugging back we get that $f(0)=0$ and this is a contradiction . Hence from now on we consider $f(0)=0$ . We have the following two cases $:=$ Case 1 : $f(1) \neq 0$ We claim that the only solution in this case is $f(x)=x$ for all reals $x$ . Proof : As noted before , we get that $f$ is injective and $f(x)=x+f(0)=x.$ Hence the claim holds. Case 2 : $f(1)=f(0)=0$ We claim that the only solution in this case is $f(x)=0$ for all reals $x$ . First we note that we have the following results after adapting the previous assertions $:=$ $f(f(x))=f(x)$ $f(f(x)+1)=0$ Note that $P(x,y)$ rewrites as $:=$ $$f(x+f(y))+ f(xy) =(y+1)f(x) +f(y)$$ FTSOC assume that there exists some reals $a,b$ with $a \neq 0,1$ and $b \neq 0$ so that we have $f(a)=b$ . Note that we also have $f(b)=b$ and $f(1+b)=0$ . Next we consider the case when $b \neq -1$ Write $c=b+1$ , then we have that $c\neq 0,1$ and $f(c)=0$ . Note that $$P(x,c) \implies f(cx)= cf(x)$$ Now we use a simple trick . Comparing $P(x,y)$ and $P(cx, cy)$ we get $f(xy)=yf(x)$ holds for all reals $x,y$ . This implies that $f(x) \equiv 0$ . Now we deal with the case $f(a)=-1 \leftrightarrow f(-1)=-1$ Note that we have $P(x,-1) \implies f(x-1)+f(-x) =-1.$ Plugging $x=-1$ , we get $f(-2)=-1$ Also $P(-1,y) \implies f(f(y)-1)+f(-y)=f(y)-y-1.$ Plugging $y=-2$ in the above equation yields $f(2)=1$ . Now we can just continue the above steps with $a=2$ and $b=1$ . Having exhausted all cases , we are done $\blacksquare$

Let $P(x,y)$ be the assertion $f(x+f(y))+f(xy)=yf(x)+f(y)+f(f(x))$. Case 1: $f$ is injective $P(x,0)\Rightarrow f(x+f(0))=f(f(x))$ So by injectivity, we have $f(x)=x+f(0)$. Testing, we see that only $\boxed{f(x)=x}$ fits. Case 2: $f$ is not injective $P(0,x)\Rightarrow f(f(x))+f(0)=xf(0)+f(x)+f(f(0))$ and since $f$ isn't injective, $f(0)=0$ Similarly: $P(1,x)\Rightarrow f(1+f(x))+f(x)=xf(1)+f(x)+f(f(1))\Rightarrow f(1)=0$ by non-injectivity $P(1,x)\Rightarrow f(1+f(x))=0$ Now let $k=1+f(a)$ for some $a$, so $f(k)=0$. $P(x,k)\Rightarrow f(kx)=kf(x)$ $P(kx,k)\Rightarrow f\left(k^2x\right)=kf(kx)=k^2f(x)$ $P(kx,ky)\Rightarrow kf(x+f(y))+k^2f(xy)=k^2yf(x)+kf(y)+kf(f(x))$ $kP(x,y)\Rightarrow kf(x+f(y))+kf(xy)=kyf(x)+kf(y)+kf(f(x))$ Comparing these last two, we obtain $k^2f(xy)+kyf(x)=kf(xy)+k^2yf(x)$. Setting $x=1$, it reduces to $k^2f(y)=kf(y)$. $\boxed{f(x)=0}$ works. Now assume that $f$ is not identically zero. By this, $\exists j:f(j)\ne0$. Setting here $y=j$, we get $k^2=k$, so $k\in\{0,1\}$. Substituting our definition of $k$, we have $f(x)\in\{-1,0\}\forall x$. Thus, we can express $f$ as: $$f(x)=\begin{cases}0&\text{if }x\in S\\-1&\text{if }x\notin S\end{cases}$$Where $\{0,1\}\subseteq S\subset\mathbb R$. $\exists p:f(p)=-1$ $P(p,y\in S)\Rightarrow f(py)=-y$ So since $\operatorname{Im}(f)=\{-1,0\}$, we must have that $-y\in\{-1,0\}\forall y\in S$. Thus, $S\subseteq\{0,1\}$, but we already know that $\{0,1\}\subseteq S$, thus we must have $S=\{0,1\}$. We see that $f(x)=\begin{cases}0&\text{if }x\in\{0,1\}\\-1&\text{if }x\in\mathbb R\setminus\{0,1\}\end{cases}$ doesn't work. $P(2,2)\Rightarrow f(4)=-4$, but since $4\notin\{0,1\}$ we should have $f(4)=-1$, contradiction. The proof is done. $\square$

let $P(x,y)$ denote the assertion in the problem , by $P(x,0)$ we know that $f(x+f(0)) = f(f(x))$ (1) so by this we get the motivation to show that $f$ is injective .Consider the following two cases : Case 1 : $f(1)$ is not equal to 0. Now we proceed showing that $f$ is injective , Let $f(a) = f(b)$ , then using $P(1,a)$ and $P(1,b)$ we get $af(1) = bf(1)$ and so $a = b$ and so $f$ is injective and by (1) we'll have $f(x) = x + C$ for some constant $C$ and it's easy showing that $C=0$ and so $f(x)=x$ for all $x$ and its clearly a solution to the equation. Case 2 : $f(1) = 0$ by $P(1,1)$ we'll get $f(0) = 0$ Now by $P(1,y)$ we have $f(1+f(y)) = f(0)$ so $f(1+f(y)) = 0$ this motivates us to define $g(x) = f(x) + 1$ so by $f(1+f(y)) = 0$ , we have $g(g(x)) = 1$ now rewriting the problem in terms of $g$ we'll have : $g(x+g(y)-1) = yg(x) - y + g(y) + g(g(y) - 1) - 1$ Let this assertion be $Q(x,y)$ , now using $Q(g(x),y)$ we have $g(g(x) + g(y) - 1) = g(y) + g(0) - 1$ and clearly the left hand side is symmetric with respect to $x,y$ so we have $g(y) + g(0) - 1 = g(x) + g(0) - 1$ therefore we must have $g(x) = g(y)$ and since $x,y$ where arbitrary real numbers we'll find out that $g$ is a constant function , that is : $g(x) = T$ for all real numbers $x$ and because $g(g(x)) = 1$ , $T = 1$ so $g(x) =1$ for all real numbers $x$ and we knew that $g(x) = f(x) + 1$ so $f(x) = 0$ for all real numbers $x$. So the solutions are $f(x) =x$ and $f(x) = 0$.

If $f(1)$ is not zero then $P(1,a)$ and $P(1,b)$ shows it is injective. $P(x,0)$ gives $f(x)=x+f(0)$, plugging in the equation gives $c=0$. If $f(1)$ is zero then $P(1,1)$ shows $f(0)=0$ which implies $f(x)=f(f(x))$ from $P(x,0)$. $P(x,f(y))$ and swapping $x$ and $y$ gives $f(xf(y))=f(yf(x))$. Putting $y=1$ in the last equation gives $f(x)=0$ which is a solution.

Why does $f(xf(1))=f(x)$ imply $f(x)=0$?

jasperE3 wrote: Why does $f(xf(1))=f(x)$ imply $f(x)=0$? f(1)=0 and f(0)=0.

Ok.. here's how my short solution goes Let $f(0) = c$ , then from $P(x,0)$ we will get $$f(x+c) = f(f(x))$$, suppose there exist a $z$ not equals $0$ such that $f(z) = c$ , then $P(0 , z)$ gives us $$c = zc + c$$this means $z=0$ ,a contradiction. Hence only $x$ satisfying $f(x) = c$ is $x=0$. Now as $f$ is surjective choose a $k$ not equals $0$ such that $f(k) = 0$ , then from $P(k,k)$ , we will get $f(0) = f(k^2) = c$ , a contradiction to above argument unless $k=0$. So only $x$ satisfying $f(x) =0$ is $x=0$. Claim - $f$ is injective . From $P(1,x)$ , we will got $$ f(1+f(x))=xf(1)+f(f(1)) $$, suppose $f(a) = f(b)$ , then substituting we will gt $a=b$ as $f(1)$ not equals $0$. Now $P(x,0)$ will give $$f(f(x)) = f(x)$$, and as $f$ is injective , we will get $f(x) = x$ and $f(x) = 0 $ are the only solutions. QED .

USJL wrote: Let $\mathbb{R}$ be the set of all real numbers. Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that for any $x,y\in \mathbb{R}$, there holds \[f(x+f(y))+f(xy)=yf(x)+f(y)+f(f(x)).\] Let $P(x, y)$ be the assertion. $f \equiv 0$ is the only constant solution, let us say $f$ is non-constant. $P(1, x) \implies f(1 + f(y)) = yf(1) + f(f(1))$ which means that $f$ is injective or $f(1) = 0$ (or possibly both), if $f$ is injective, inserting $y = 0$ gives that $f(0) = 0$ and therefore $P(0, k)$ in the original equation will give that $f(f(k)) = f(k) \implies f(k) = k$ for all reals $k$. If $f(1) = 0$, then $f(0) = 0$ follows identically. We get that $f(f(x)) = f(x)$. $P(x, f(y))$ and $P(y, f(x))$ gives that $f(xf(y)) = f(yf(x))$, so inserting $y = 1$ here gives $f(f(x)) = 0$ for all reals $x$ so $f(x) = f(f(x)) \equiv 0$. is a solution which was already discussed before. @above Your solution is incorrect, we cannot claim $f$ is injective unless $f(1) \neq 0$, if $f(1) = 0$ then you cannot claim anything about injectivity.

DapperPeppermint wrote: @above Your solution is incorrect, we cannot claim $f$ is injective unless $f(1) \neq 0$, if $f(1) = 0$ then you cannot claim anything about injectivity. I have edited

USJL wrote: Let $\mathbb{R}$ be the set of all real numbers. Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that for any $x,y\in \mathbb{R}$, there holds \[f(x+f(y))+f(xy)=yf(x)+f(y)+f(f(x)).\] $P(x,0)$ gives: $f(x+f(0))=f(f(x))$ (1) $P(1,y)$ gives:$f(1+f(y))+yf(1)+f(f(1))$ (2) now for $y=f(y),y+f(0)$ at (2) and using (1) we get:$f(1)[f(y)-y-f(0)]=0$ so $f(1)=0$ or $f(x)=x+f(0)$. if $f(1)=0$ $P(x,1)$ gives: $f(f(x))=f(x)$ (3) Using (3) and $P(x,f(y))$ and change at this $x,y$ we get $f(xf(y))=f(yf(x))$ for $y=0$ we have $f(0)=f(f(x))=f(x)$ so $f$ is constant. Te only constant sulotion is $f(x)=0$ if $f(x)=x+f(0)$ from the first $f(x)=x$

The title claims that this FE is simple but my solution says otherwise lol But maybe that's just because I'm too bad at FEs We claim that the only solutions are $f(x) = 0$ and $f(x) = x$. It's easy to see that both of them are solutions to the original problem, hence we are left to prove that they are the only ones. $P(0, y)\Rightarrow f(f(y)) + f(0) = yf(0) + f(y) + f(f(x))$ $P(x, 0)\Rightarrow f(x + f(0)) = f(f(x))$ Suppose injectivity holds, then we have \[x + f(0) = f(x)\]Hence $f(x) = x\forall x\in \mathbb{R}$. Suppose there exist $a$ and $b$ such that $f(a) = f(b)$. Then, we have $f(0) = 0$, or else $f$ is injective. Thus, we have $f(f(x)) = f(x)$. $P(1, y)\Rightarrow f(1 + f(y)) = yf(1) + f(f(1))$ Using the same reasoning above, we also know that $f(1) = 0$. Hence, \[f(1 + f(x)) = 0\forall x \in \mathbb{R}\]$P(1 + f(x), y)\Rightarrow f(1 + f(x) + f(y)) + f(y(1 + f(x))) = f(y)$ $P(x, 1 + f(y))\Rightarrow f(x(1 + f(y))) = f(x)f(y) + f(x)$ Substituting $f(x(1 + f(y)))$ using the equation above, we get \[f(1 + f(x) + f(y)) = -f(x)f(y)\]Plugging in $y = x$ yields \[f(1 + 2f(x)) = -f(x)^2\]Now, changing $y\mapsto f(y)$ in the original equation and comparing them with each other, we get that \[f(xy) - yf(x) = f(xf(y)) - f(x)f(y)\]Hence, \[yf(x) + f(xf(y)) = f(xy) + f(x)f(y) = xf(y) + f(yf(x))\text{ (*)}\]Since the equation is symmetric w.r.t. $x$ and $y$. $P(f(x), f(x))\Rightarrow f(2f(x)) + f(f(x)^2) = f(x)^2 + 2f(x)\Rightarrow f(2f(x)) + f(-f(1 + 2f(x))) = 2f(x) -f(1 + 2f(x))$ $P(-f(x), f(x))\Rightarrow f(-f(x)^2) = f(x)f(-f(x)) + f(x) + f(-f(x))\Rightarrow f(1 + 2f(x)) = f(x)f(-f(x)) + f(x) + f(-f(x))$ $\Rightarrow f(x)^2 + f(x)f(-f(x)) + f(x) + f(-f(x)) = 0$ Hence, \[(f(x) + 1)(f(x) + f(-f(x))) = 0\]So either $f(x) = -1$ or $f(x) = -f(-f(x))$. Suppose $f(1 + 2f(x)) = -1$, then we know \[f(2f(x)) = 2f(x) + 1\]and either $f(x) = 1$ or $f(x) = -1$. If $f(x) = 1$, then $f(3) = -1$ and $f(2) = 3$. However, this means that $f(3) = f(f(2)) = 3$, a contradiction. If $f(x) = -1$, then $f(-1) = -1$ and $f(-2) = -1$. $P(-1, -2)$ (in $(*)$)$\Rightarrow f(2) = 1$, meaning $f(1) = f(f(2)) = 1$, a contradiction. Hence, we know that $f(1 + 2f(x)) = -f(-f(1 + 2f(x)))\Rightarrow f(x)^2 = f(f(x)^2)$. Plugging $P(f(x), x)$ in $(*)\Rightarrow f(x)^2 + xf(x) = f(xf(x)) + f(x)^2\Rightarrow f(xf(x)) = xf(x)$ $P(x, x)\Rightarrow f(x + f(x)) + f(x^2) = xf(x) + 2f(x)$ $P(x, f(x))\Rightarrow f(x + f(x)) + xf(x) = f(x)^2 + 2f(x)$ Taking the difference of these two equations we have \[f(x)^2 + f(x^2) = 2xf(x)\]Hence, if $f(x^2) = x^2$, then $(f(x) - x)^2 = 0$ so $f(x) = x$. Suppose for some $\varepsilon$, $f(\varepsilon) = \varepsilon$. Then, we know that \[f(\varepsilon^2) = \varepsilon^2\]from then we have $f((-\varepsilon)^2) = (-\varepsilon)^2\Rightarrow f(-\varepsilon) = -\varepsilon$. Plugging $P(\varepsilon, -\varepsilon)$ into $f(1 + f(x) + f(y)) = -f(x)f(y)$, we know that \[f(\varepsilon)^2 = 0\Rightarrow \varepsilon = 0\]Since $f(f(x)^2) = f(x)^2$, so by the result above we know $f(x) = 0$ $\forall x\in \mathbb{R}$, hence we are done.

@above thanks for bumping and nice sol $P(0,x)$ $ff(x)+f(0)=xf(0)+f(x)+ff(0)$ Case1:$f(0) \ne 0$ Then $f$ is injective function $P(-f(0),0)$ $f(0)=ff(-f(0))$ $0=f(-f(0))$ $P(x,0)$ $f(x+f(0))=ff(x)$ $f(x)=x+f(0)$ $P(f(0),0)$ $ff(0)=2f(0)$ $P(0,-f(0))$ $2f(0)=-f(0)^2+ff(0)=-f(0)^2+2f(0)$ so $f(0)=0$ and contradicition. $P(0,x)$ $ff(x)=f(x)$ $P(1,x)$ $f$ injective if $f(1) \ne 0$ $P(f(x),1)$ $f(x)+f(1)=x+f(1)$ $f(x)=x$ $f(1)=0$ $P(x,y)$ , $P(x,f(y))$ ,and $P(f(x),y)$ Plug $y=1$ $f(x)=0$

I think it is simple enough: Let $P(x,y)$ be the assertion that $f(x+f(y))+f(xy)=yf(x)+f(y)+f(f(x))$