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Write assertion $P(x, y):f(x^2) + xf(y) = f(x) f(x + f(y)) $. $P(0, x):$ $f(0)=f(0)f(f(x))$. So $f(0)=0$ or $f(f(x))=1$ 1-case: $f(0)=0$ $P(x, 0):$ $f(x^2)=f(x)^2$.$ \Longrightarrow $ $P(x, y):f(x)^2 + xf(y) = f(x) f(x + f(y)) $$ \Longrightarrow $$f-$injective. $P(-f(x), x):f(x^2) = f(f(x)^2) $ Since injective $\boxed{f(x)=x}$ 2-case: $f(f(x))=1$ Let $f(a)=1$ for some $a$. So $f(f(a))=1$ $ \Longrightarrow $ $f(1)=1$. $P(1, 1):$ $f(2)=2$. But $f(f(2))=1$ and $f(2)=1$. Contradiction. $ \Longrightarrow $ Such $f(x)$ doesn't exist. Thus, the only solution is $f(x)=x$

For the assuption Quote: Let $f(a)=1$ for some $a$. You might have to prove the function is surjective first. Because $f(x)=0$ for all $x$ works as well. You also forgot the case $f(f(x))=0$

sualehasif996 wrote: For the assuption Quote: Let $f(a)=1$ for some $a$. You might have to prove the function is surjective first. Because $f(x)=0$ for all $x$ works as well. You also forgot the case $f(f(x))=0$ I don't have to prove that $f(x)$ is is surjective, because $f(f(x))=1$ and this yields that there is no such $a$ P.S: $f(f(x))=1$. Put $x \Rightarrow f(x):$ $f(f(f(x)))=1$ and $f(1)=1$

chessmaster4 wrote: $P(x, y):f(x)^2 + xf(y) = f(x) f(x + f(y)) $$ \Longrightarrow $$f-$injective. Why?

Do note that your solution still misses the constant solution $f(x)=0$ for all $x$.

Actually, if f(x) = 0 works, doesn't that imply that f is NOT injective?

But again why is this? danepale wrote: chessmaster4 wrote: $P(x, y):f(x)^2 + xf(y) = f(x) f(x + f(y)) $$ \Longrightarrow $$f-$injective.

We can easily find that $f(x)+1=f(f(x)+1)$. Maybe it helps.

We can choose $x=y$, and then we get that $f(x)+x=f(f(x)+x)$, and it's done.

abnunc wrote: $f(x)+x=f(f(x)+x)$, and it's done. No you need to prove that $g(x)=f(x)+x$ is surjective

abnunc wrote: We can choose $x=y$, and then we get that $f(x)+x=f(f(x)+x)$, and it's done. You just proved that there exist a fixed point, that doesn't mean that f(x) = x for all x I got that $ (1+f(x)) ; (x+f(x)) ; (n(x+f(x)) $ are all fixed points, i think i got more, but can't remember

But we have that $f(0)=0$ and $f(1)=1$

Yeah, you're right... We only get that $f(n)=n$, where $n$ is natural...

I will skip the part $f(n)=n$, and when $f(0)$ isn't zero. So let us define $g(x)=f(x)-x$. Before actualy using it, we will prove that either $f(x)=0$ for all $x$, or $f$ hasn't any other zero apart $f(0)=0$. Let's suppose $f(u)=0$ for some nonzero $u$. Than, by $P(u, y)$, we get $f(y)=0$, for all real $y$. Assume this is not the case, and so $g(x)$ is never $-x$, exept for $x=0$. Now, written in terms of $g(x)$, $P(x, y)$ becomes: $(x+g(x))g(x)=(x+g(x))g(x+y+g(y))+(y+g(y))g(x)$. So, by plugging in natural $y=n$, and assuming that for some $x$, $g(x)$ is nonzero we get$g(x+n)=g(x)(1+\frac{n}{x+g(x)})$. So, $g(x+1)=g(x)(1+\frac{1}{x+g(x)})$, and$g(x)(1+\frac{2}{x+g(x)})=g(x+2)=g(x+1+1)=g(x+1)(1+\frac{1}{x+1+g(x+1)})=g(x)(1+\frac{1}{x+g(x)})(1+\frac{1}{x+1+g(x+1)})$. By plugging in what $g(x+1)$ is and after lenghtly computations(if I am not mistaken), we get $\frac{1+x+g(x)}{x+g(x)}(1+\frac{x+g(x)}{x+x^2+2g(x)+2xg(x)+g(x)^2})=\frac{2+x+g(x)}{g(x)+x}$, and by simplifying, we have $g(x)=0$, contradiction. So $g(x)=0$ for all real $x$, and so $f(x)=x$

aleksam wrote: $P(x, y)$ becomes: $(x+g(x))g(x)=(x+g(x))g(x+y+g(y))+(y+g(y))g(x)$. So, by plugging in natural $y=n$, and assuming that for some $x$, $g(x)$ is nonzero we get$g(x+n)=g(x)(1+\frac{n}{x+g(x)})$ Hmmm, Isn't it should be $g(x+n)=g(x)(1-\frac{n}{x+g(x)})$ Anyway, I think same method work, but it left little case $g(x+1)=-2x$ which is not a solution

Another solution: We have $f(n)=n, $ for all integers $n.$ $P(x,n)$ gives $f^2(x)+nx=f(x)f(x+n)$ and $P(x+1,n-1)$ gives $f^2(x+1)+(n-1)(x+1)=f(x+1)f(x+n)$ So, for $x>0,$ we have $\frac{f^2(x)+nx}{f^2(x+1)+(n-1)(x+1)}=\frac{f(x)}{f(x+1)}.$ Letting $n\to \infty$ we get $\frac{f(x)}{f(x+1)}=\frac{x}{x+1}.$ Combining the last equality with $f^2(x)+x=f(x)f(x+1)$ we obtain $f(x)=x$ for all $x>0.$ Finally, taking $n\in \Bbb{N}$ such that $x+n>0,$ from $f^2(x)+nx=f(x)f(x+n)$ we have $f(x)=x$ for all $x.$

$ y=0 $ gives that $ f(0)=f(0)f(f(y)) $. The case $ f(f(y))=1 $ can be done as above. Let $ f(0)=0 $. If there is non-zero $ t $ such that $ f(t)=0 $ then $ f(x)=0 $ Now, let $ f(x) \neq 0 \text{ } \forall x \neq 0. f(x^2)=f(x)^2, f(1)=1 $ and $ x=1 $ gives that $ f(f(y)+1)=f(y)+1 $. Setting $ f(y)+1 $ for $ y $ we get that $ f(x)f(x+f(y)+1)=xf(y)+x+f(x)^2 $ and setting $ x+1 $ for $ x $ we get $ f(x+1)(f(x+f(y)+1)=(x+1)f(y)+f(x+1)^2 $ From the last two equations, we can obtain that $ \frac{f(x+1)}{f(x)}=\frac{(x+1)f(y)+f(x+1)^2)}{xf(y)+x+f(x)^2} $. Now, y=1 gives that $ 2xf(x+1)-(x+1)f(x)=f(x)f(x+1)(f(x+1)-f(x)) $ or $ 2x(f(x+1)-f(x))+f(x)(x-1)=f(x)f(x+1)(f(x+1)-f(x)) $, but from the given equation we have that $ f(x+1)-f(x)=\frac{x}{f(x)} $ by setting $ y=1 $, so substituting this in the last equation, together with $ f(x+1)=\frac{f(x)^2+x}{f(x)} $ we get that $ f(x)^2=x^2 $. $ f(x)=-x $ is not a solution and if there is $ y $ such that $ f(y)=-y $, we take $ x $ such that $ f(x)=x $ in the given equation, and we get a contradiction, so finally $ f(x)=x \text{ } \forall x $

I was sent this a few days ago and have a pretty bashy and probably overcomplicated solution. Still, it's mine.

chessmaster4 wrote: Write assertion $P(x, y):f(x^2) + xf(y) = f(x) f(x + f(y)) $. $P(0, x):$ $f(0)=f(0)f(f(x))$. So $f(0)=0$ or $f(f(x))=1$ 1-case: $f(0)=0$ $P(x, 0):$ $f(x^2)=f(x)^2$.$ \Longrightarrow $ $P(x, y):f(x)^2 + xf(y) = f(x) f(x + f(y)) $$ \Longrightarrow $$f-$injective. $P(-f(x), x):f(x^2) = f(f(x)^2) $ Since injective $\boxed{f(x)=x}$ 2-case: $f(f(x))=1$ Let $f(a)=1$ for some $a$. So $f(f(a))=1$ $ \Longrightarrow $ $f(1)=1$. $P(1, 1):$ $f(2)=2$. But $f(f(2))=1$ and $f(2)=1$. Contradiction. $ \Longrightarrow $ Such $f(x)$ doesn't exist. Thus, the only solution is $f(x)=x$ Your solution is wrong because you just have $f(x)=x$ for all $x>0$.

My solution: Not difficult to see that $f(0)=0$ is a solution. Suppose that $f(x)$ is not equail $0$. Pulg in $x=0$ $ \Longrightarrow $ $f(0)=f(0).f(f(y))$ $ \Longrightarrow $ $f(0)=0$. Pulg in $x=0$ $ \Longrightarrow $ $f(x^2)=(f(x))^2$. Pulg in $x=-f(y)$ $ \Longrightarrow $ $f(f^2(y)) =f^2(y)$. Pulg in $x=y$ and $f(f^2(y)) =f^2(y)$ $ \Longrightarrow $ $f^2(x) + xf(x)=f(x).f(x+f(x))$$ \Longrightarrow $ $f(x)+x=f(f(x)+x)$. Since $f(x)+x$ is full light $ \Longrightarrow $ $f(x)=x$ for all $x$.

First, taking $(x,y)=(0,y)$ yields $f(0)=f(0) \cdot f(f(y)) \implies f(0)=0$ or $f(f(y))=1.$ We first tackle the case that $f(f(x))=1$ for all real $x.$ We claim that $f(1)=1.$ If there exits $a\in \mathbb{R}-\{0\}$ for which $f(a)=0,$ then $1=f(f(a))=f(0),$ consequently $f(1)=f(f(0))=1.$ So we assume $f(x)\ne 0$ for all real $x.$ One one hand, taking $(x,y)=(f(x),y)$ yields that \[f(f(x)^2)+f(x)f(y)=f(f(x)+f(y)),\]set $(x,y)=(f(x),y)$ again and we get $f(1)+f(y)=f(1+f(y)).$ On the other hand, taking $(x,y)=(1,y)$ yields that $f(1)+f(y)=f(1)f(1+f(y)).$ Since $f(1+f(y))\ne 0$ so we conclude that $f(1)=1.$ Therefore $2=1+1=f(1+1)=f(2),$ but $1=f(f(2))=f(2)=2,$ contradiction. Then we tackle the case that $f(0)=0.$ If there exits $a\in \mathbb{R}-\{0\}$ for which $f(a)=0,$ then taking $(x,y)=(a,y)$ yields that $af(y)=0 \implies f(y)\equiv 0$ for all real $y.$ So we assume $f(x)\ne 0$ for all $x\ne 0.$ Taking $(x,y)=(x,0)$ we get \[f(x^2)=f(x)^2.\]Taking $(-f(y),y)$ we get \[f(f(y)^2)=f(y)^2.\qquad (\heartsuit)\]Taking $(x,y)=(x,x)$ we get \[f(x)^2+xf(x)=f(x)f(x+f(x))\implies x+f(x)=f(x+f(x)).\qquad (\spadesuit)\]In particular, set $(x,y)=(1,1)$ we get $2=f(1+f(1))=1+f(1) \implies f(1)=1.$ Moreover, taking $(x,y)=(1,y)$ yields that $1+f(y)=f(1+f(y)).$ Taking $(x,y)=(-f(x),1+f(x))$ we get \[f(f(x)^2)-f(x)(1+f(x))=f(f(-x)) \implies f(-f(x))=-f(x).\]here $f(f(x)^2)-f(x)^2=0$ follows by $(\heartsuit).$ Finally, taking $(x,y)=(-f(x),x+f(x))$ we get \[-xf(x)=f(x)^2-f(x)(x+f(x))=f(-f(x))\cdot f(x)=-f(x)\cdot f(x)\]Since $-f(x)\ne 0$ by our assumption, hence we conclude that $f(x)=x.$ In conclusion, we have two solution: $\boxed{f(x)\equiv 0}$ or $\boxed{f(x)\equiv x.}$ $\square$

Let $P(x,y)$ denote the problem statement. Clearly $\boxed{f\equiv 0}$ is a solution, so now we look for solutions that are not uniformly zero. $P(0,y)\implies f(0)=f(0)f(f(y))\implies f(0)=0$ or $f(f(y))=1.$ Case 1: $f(f(y))=1.$ Let $u$ satisfy $f(u)=1,$ clearly such $u$ exists since $f(f(y))=1$ and the image of $f$ is nonempty. Then $1=f(f(u))=f(1)\implies f(1)=1.$ $P(1,1)\implies 2f(1)=f(1)f(f(1)+1)\implies f(2)=2.$ But $f(f(2))=f(2)=1,$ contradiction. (motivation for this contradiction is by noting that the only fixed point for $f$ in this case is $1.$) Case 2: $f(f(0))=0.$ Fixing $x$ and comparing $P(x,u_1)$ and $P(x,u_2)$ with $f(u_1)=f(u_2)$ yields $f$ injective. $P(x,0)\implies f(x^2)=[f(x)]^2,$ so rewrite $P(x,y)$ as $[f(x)]^2+xf(y)=f(x)f(x+f(y)).$ $P(f(x),y)\implies [f(f(x))]^2+f(x)f(y)=f(f(x))f(f(x)+f(y)).$ $P(f(y),x)\implies [f(f(y))]^2+f(y)f(x)=f(f(y))f(f(x)+f(y)).$ Subtracting, $\left(f(f(x))+f(f(y))\right)\left(f(f(y))-f(f(x))\right)=f(f(x)+f(y))\left(f(f(y))-f(f(x))\right).$ Since $f$ is injective, for $y\neq x,$ $f(f(y))-f(f(x))\neq 0$ so dividing yields $f(f(x)+f(y))=f(f(x))+f(f(y)).$ Multiplying by $f(f(x))$ and comparing with $P(f(x),y)$ gives $f(x)f(y)=f(f(x))f(f(y))\implies f(f(x))=cf(x).$ Now $x\to f(x)$ in $P(f(x),y)$ gives $c^2[f(x)]^2+f(x)f(y)=cf(x)f(f(x)+f(y)).$ For $x$ such that $f(x)\neq 0,$ we have $c^2f(x)+f(y)=cf(f(x)+f(y))=c^2f(x)+c^2f(y)\implies c=\pm 1.$ If $c=1,$ from injectivity we get $\boxed{f\equiv \text{Id}}.$ We now consider the case $c=-1.$ From $P(-f(x),x)$ we get $f([f(y)]^2)=[f(y)]^2,$ or that $f(\text{nonnegative})=\text{nonnegative},$ contradiction unless $f$ is identically $0.$ We may now conclude.

abnunc wrote: But again why is this? danepale wrote: chessmaster4 wrote: $P(x, y):f(x)^2 + xf(y) = f(x) f(x + f(y)) $$ \Longrightarrow $$f-$injective. Because, We write X=0 and we get f(0)=f(0)×f(f(y)) and 1=f(f(y))

Bruuuuuuuuuuuute force F.E. Case 1: $f$ is constant (Here its obvious that $f(x)=0$ so well skip it) Case 2: $f$ is non-constant (so there exists $d$ such that $f(d) \ne 0$) Let $P(x,y)$ the assertion of the given F.E. $P(0,x)$ $$f(0)=f(0)f(f(x)) \implies f(f(x))=1 \; \text{or} \; f(0)=0$$Case 2.1: $f(f(x))=1$ Here we get that $f(1)=f(f(f(x)))=1$ thus by $P(1,x)$ $$f(x)+1=f(f(x)+1) \implies f(x)+1=f(f(x)+1)=f(f(f(x)+1))=1 \implies f(x)=0 \; \text{contradiction!!}$$Case 2.2: $f(0)=0$ First we will prove that $f$ is injective at $0$ so assume that exists $c$ such that $f(c)=0$ then by $P(c,d)$ $$cf(d)=0 \implies c=0 \implies f \; \text{injective at} \; 0$$$P(x,0)$ $$f(x^2)=f(x)^2 \implies f(1)=1$$$P(1,x)$ $$f(x)+1=f(f(x)+1) \implies f(n)=n \; \text{and} \; f(f(x)+n)=f(x)+n \; \; \forall n \in \mathbb Z^+$$$P(-f(x),x)$ $$f(f(x))^2=f(f(x)^2)=f(x)^2=f(x^2)$$$P(f(x),n)$ for any $x \ne 0$ and for any $n \in \mathbb Z^+$ $$f(f(x))=f(x)$$$P(f(x),y)$ for any $x \ne 0$ $$f(x)+f(y)=f(f(x)+f(y)) \implies f(2f(x))=2f(x)$$$P(-f(x),f(x)+n)$ for any $x \ne 0$ and any $n \in \mathbb Z^+$ $$f(x)^2-f(x)(f(x)+n)=nf(-f(x)) \implies f(-f(x))=-f(x)$$$P(x,-f(x))$ $$f(x-f(x))=f(x)-x$$Assume that there exists $a$ such that $f(a)-a \ne 0$ hence by $P(a-f(a),a)$ $$(f(a)-a)^2-(f(a)-a)f(a)=(f(a)-a)f(a) \implies f(a)-a=2f(a) \implies f(a)=-a$$But remember that $f(-f(a))=-f(a)$ or in other words $f(a)=a$ which means $a=0$ which is not possible since $f(0)-0=0$ Thus $f(x)=x$ is the only non-constant solution and we are done

danepale wrote: Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that for all real $x,y$: $$ f(x^2) + xf(y) = f(x) f(x + f(y)) \, . $$ $P(0,y)$ gives $f(0)=0$ or$f(f(y))=1$ if $f(f(y))=1$ then$P(0,f(y))$ and$P(1,1)$ gives $f(2)=2$ contradiction. if $f(0)=0$ then $P(x,0)$ gives $f(x^2)=f(x)^2$ (4) if $f(t)=0$ then $P(t,x)$ gives$t=0$.(1) $P(-f(y),y)$ gives $f(f(y)^2)-f(y)^2=0$. (2) $P(x,x)$ and (1) gives $f(x)+x+f(f(x)+x)$. (3) now $P(-f(y),y+f(y))$ with (1),(2),(3) gives $f(-f(y))=-y$ so $f$ is bijective and from (2) we have $f(x)=+-x$ using (4) we have that $f(x)=x$ for every oppositive number $x$. suppose that there is number $k$ such that $f(k)=-k$ then $P(k,y>0)$ gives $k+y=-f(k+y)$ wich is not true for $y>-k$. so the only sollution is $f(x)=x$.

Let $P(x,y)$ denote the given assertion as usual. $P(-1,1)\Rightarrow f(-1)f(f(1)-1)=0$ Case 1: $f(-1)=0$ $P(-1,-1)\Rightarrow f(1)=0$ $P(1,x)\Rightarrow\boxed{f(x)=0}$ Case 2: $f(-1)\ne0,f(f(1)-1)=0$ $P(1,f(1)-1)\Rightarrow f(1)\in\{0,1\}$ If $f(1)=0$ then $f(-1)=0$ which is a contradiction. So $f(1)=1$ and $f(0)=0$: $P(x,f(1)-1)\Rightarrow f(x^2)=f(x)^2$ (so $P(x,y)$ becomes $f(x)^2+xf(y)=f(x)f(x+f(y))$) $P(1,x)\Rightarrow 1+f(x)=f(1+f(x))$ $P(f(x),1)\Rightarrow f(f(x))^2+f(x)=f(f(x))+f(x)f(f(x))\Rightarrow(f(f(x))-f(x))(f(f(x))-1)\Rightarrow f(f(x))\in\{1,f(x)\}$ $P(-f(x),x)\Rightarrow f(f(x))^2=xf(x)$ If $f(f(x))=1$ then $f(x)=\frac1x$ ($f(f(0))\ne1$) whereas if $f(f(x))=f(x)$ then $f(x)\in\{0,x\}$. This means that $f(x)\in\left\{0,x,\frac1x\right\}$ for all $x$. If $f(k)=0$ for some $k$: $P(k,1)\Rightarrow k=0$ If $f(p)=\frac1p$ for some $p\notin\{-1,0,1\}$, then $f(f(p))^2=pf(p)=1$. If $f(f(p))=1$, then: $$1=f(1)^2=f(f(f(p)))^2=f(p)f(f(p))=\frac1p$$hence $p=1$, contradiction. If $f(f(p))=-1$, then $f(f(p))=f(p)$, hence $f(p)=-1$ and $p=-1$, contradiction. Finally, $\boxed{f(x)=x}$ for all $x$.

Here goes my solution: $(0,y)$ gives: \[f(0)=f(0)f(f(y))\], so we have 2 cases: case 1: $f(f(y))=1, \forall y$ It is easy to see that $1=f^3(x)=f(1)$ Plugging $(1,y)$ we get $1+f(y)=f(1+f(y))$. That also means that: \[1=f(f(1+f(y)))=f(1+f(y))=1+f(y)\]so $f$ is zero everywhere, contradiction to $f(f(y))=1$ for all $y$ Case 2: $f(0)=0$ Putting $(x,0)$, we get: $f(x^2)=f(x)^2$ Also, if $f(t)=0$ for $t \neq 0$, $(t,y)$ gives $f(y)=0 \forall y$ From now on, we assume that the only zero of $f$ is $0$. We have $f(1)=f(1)^2$, so $f(1)=1$. Plugging $(1,y)$ we have that $1+f(y)$ is a fixed point of $f$ for all $y$. $(x,1)$ gives $f(x)^2+x=f(x)f(x+1)$, which rewrites to $f(x)(f(x+1)-f(x))=x$ $(x,1+f(x))$ now gives $f(x)(f(x+f(x)+1)-x-f(x))=x$. From above 2, if x is not 0 we get $f(x+f(x)+1)=x+f(x+1)$, (*) which holds also for $x=0$. Finally, from $(x+1,x)$ we get: \[f(x+1)^2+(x+1)f(x)=f(x+1)f(x+1+f(x))\]\[f(x+1)^2+(x+1)f(x)=f(x+1)(x+f(x+1))\]\[(x+1)f(x)=xf(x+1)\]Plugging this back into $(x,1)$ and simplifying, we get $f(x)^2=x^2$. If there exists $t \neq 0$ such that $f(t)=-t$, putting such $t$ in (*) gives $f(t+1)=1-t$. Possibilities are $t+1=1-t$ which gives $t=0$ or $t+1=t-1$ which is impossible, contradiction. So only functions that satisfy are $f(x)=0$ and $f(x)=x$. One can easily verify that these functions indeed work

Since $f\equiv 0$ works, consider $f\not\equiv 0$. Let $P(x,y)$ denote the assertion $f (x^2) + xf(y) = f(x) f(x + f(y))$. If $f(f(x))=1$ for all $x$, then $f(1)=1$ and $P(1,1)$ gives $f(2)=2$, impossible. So $P(0,x)$ is $f(0)=0$. $P(x,0)$ gives $f(x)^2=f(x^2)$. Let $f(x_0)=0$, then $P(x_0,x)$ gives $x_0=0$. $P(x,x)$ implies $f(x+f(x))=x+f(x)$ and $P(-f(x),x)$ implies $f(f(x)^2)=f(x)^2$. So now $P(-f(x),x+f(x))$ shows surjectivity. So $f$ is identity for positive reals. $P(x,y)$ for $x>0$ is $f(x+z)=x+z$ for all $z$, this clearly works.

F10tothepowerof34 wrote: $P(x,0)\Longrightarrow f(x^2)=f(x)f(x)$ which implies that the function is multiplicative. Unfortunately, $f(x^2)=f(x)f(x)$ does not imply $f$ is multiplicative

Denote by $P(x,y)$ the assertion $ f(x^2) + xf(y) = f(x) f(x + f(y)) \, . $ $f(x)\equiv0$ obviously works, so suppose, we don't have a constant function. By $P(0,x)$ we get $f(0) = f(0)f(f(x))$, so either Case 1: $f(0)\neq0$ and $f(f(x))=1$ or Case 2: $f(0)=0 \, .$ For case 1 note that $P(0,f(x))$ and $P(1,1)$ yields a contradiction. So we have $f(0)=0$. Therefore, by $P(x,0)$ we have $f(x^2)=f(x)^2$. Now plugging $P(-f(x), x+f(x))$ and using $P(x,x): f(x^2)+xf(x)=f(x)f(x+f(x))$ gives $-x=f(-f(x))$, which also shows surjectivity, so there exists $a\in\mathbb{R}$ s.t. $f(a)=y, \forall y\in\mathbb{R}$. $P(-f(x),x)$ moreover shows $f(f(x)^2)=f(x)^2$. Using surjectivity, we write $f(x)=y, \forall y\in\mathbb{R}$. Therefore $f(y^2)=y^2 \Rightarrow f(x)=x, \forall x\geq0$. However, we know that $-x=f(-f(x))$. Therefore, for $x>0 \Rightarrow f(-x)=f(-f(x))=-x$, so $f(x)=x, \forall x<0$. We conclude $\boxed{f(x)=x, x\in\mathbb{R}}$ or $\boxed{f(x)\equiv 0}$ $\square$.