I'll give a few main ideas, but won't do it in entirety since many parts are bashy/pluggy-chuggy (they are pretty simple with enough effort) Plugging in $y = -f(x)$ shows that $f(c) = 0$ for some $c$. $y = c$ gives $f(f(x)^2+c^2) = (c+f(x)) \cdot x$ (*) Say $f(a) = f(b) \not= -c$. Then by the previous equation, $a = b$. So $f$ is ALMOST injective, unless $f(a) = -c$. Now we split into cases: $c = 0$. In this case $f(0) = 0$, and with the injectivity things above, (and plugging in nice values) gives that $f(x) = x, 0,$ or $-x$ for all $x$. Now it suffices to check that $f(x) = x$ and $f(y) = -y$ is impossible. This case gives the solutions $f(x) \equiv x$ and $f(x) \equiv -x$. Now let's look at $c \not= 0$. Let $f(0) = d$. Then $x = 0, y = -d$ gives $f(2d^2) = 0$. Then $x = 2d^2, y = 0$ gives $f(4d^3) = 0$. Since $c \not= 0$, by the almost injectivity above, $2d^2 = 4d^3 \implies d = 1/2$ ($d = 0$ is impossible). So $f(0) = \frac{1}{2}$. Also, $f(2 \cdot (1/2)^2) = 0$, so $f(1/2) = 0$. After these, some more calculation similar to the $f(0) = 0$ case above gives $f(x) \equiv \frac{1}{2}-x$ is another solution. So we have 3 solutions. (Sorry for not wanting to type out all computations explicitly, I'm sure that you guys can do it yourselves!)

Call $S$ the set of roots of $f$. First use $y=-f(x)$ and we see that $2f(x)^2+2xf(-f(x)) \in S$. Now we see that $x \in S \Rightarrow 2xf(0) \in S$ using the above result. And also $x, y \in S \Rightarrow f(x^2)=f(y^2)=xy=x^2=y^2$, using the original equation four times. So we see $|S|=1$, since we know because of our result above that $S$ contains elements. Say $S = \{c\}$. So $2f(x)^2+2xf(-f(x))=c$ for all $x$. With $x=c$ we get $c=0$ or $f(0)=\frac{1}{2}$. In the first case we get $f(x)^2 = -xf(-f(x))$. In the original equation if $x=-f(y)$ we get that $x^2-2f(x)^2+f(-f(x))^2=0$. Using quadratic law we get that $f(x) = x$ or $-x$ for all $x$. Using $f(x)^2=-xf(-f(x))$ we get that $f$ is injective (also remembering $S$) so therefore $f=-x$ or $f=x$, which can be verified. Otherwise we get $f(0)=\frac{1}{2}$. Using the result we evaluate $c=\frac{1}{2}$. We also have $f(x)^2+xf(-f(x))=1/4$ and $x^2-2f(x)^2+f(-f(x))^2 = 1/2$. Using the quadratic law again we get that $f(x)^2 = (1/2-x)^2$. We use injectivity again and finish, finding the other solution $f(x)=1/2-x$.

Sketch of my solution: First prove $f(2f(0)^2)=0$. Assume $f$ is not injective, $f(x_1)=f(x_2)$ with $x_1\neq x_2$. Deduce that $f(x_1)=-\frac 1 2$, and $\frac 1 2$ is the only root. Then deduce that $\{x_1,x_2\}=\{1,-1\}$ and find a contradiction. Thus $f$ is injective and $x^2+2yf(x)+f(y)^2=y^2+2xf(y)+f(x)^2$. From there we can have $f(0)=0$ or $\frac 1 2$. Split into cases and we have the three solutions.

I did same as QuadraticReciprocity did. But once I get to $x^2+2yf(x)+f(y)^2=y^2+2xf(y)+f(x)^2$, I did something different. Plugging $y=0$ in that equation gives $f(x)^2=x^2-2xf(0)+f(0)^2$ so we have, $x^2+2yf(x)+y^2-2yf(0)+f(0)^2=y^2+2xf(y)-2xf(0)+f(0)^2\implies y(f(x)-f(0))=x(f(y)-f(0))$. Which implies $f$ must be a linear function and just expand. imo this problem is quite easy with enough bravity and confidence.

This inhumanly case-bashy and boring problem wasted my otherwise fine Saturday morning, so I'll retaliate with a justly case-bashy solution. Let $P(x,y)$ mean $P$lug in $x,y$ into the given equation. Now comparing $P(0,f(0))$ and $P(0,-f(0))$ gives $f(0)f(f(0))=0$. Now we have a case-fork: Case 1: $f(0)=0$. Now $P(x,0)$ implies $f(f(x)^2)=xf(x)$. If $x,y\in\mathbb R$ satisfy $f(x)=f(y)\ne 0$, then the above equation gives $f(f(y)^2)=f(f(x)^2)\implies xf(x)=yf(y)\implies x=y$. Call this property of $f$ awesomeness. The statements $P(x,0)$ and $P(0,x)$ give $$f(f(x)^2)=f(x^2)=xf(x)$$If $f(x)\ne 0$, then we need $x\ne 0$, so by awesomeness, $f(x)^2=x^2\implies f(x)=\pm x$. So for any $x$, we have $f(x)=x$ or $f(x)=-x$ or $f(x)=0$. Now we claim that we can't have $f(x)=0$ except for $x=0$. Let's say $f(c)=0$ for some $c\ne 0$. Take $x$ so that $x\not\in \left\{0,c,-c,\tfrac32 c,-\tfrac32 c\right\}$, and consider $P(c,x)$, which is $$f(x^2+2cf(x))=x(c+f(x)).$$This takes us a whole plethora of cases: If $f(x)$ is $0$, then $f(x^2)=xc\implies xc\in\{0,x^2,-x^2\}\implies x=\pm c$. If $f(x^2+2cf(x))=0$, we have $f(x)=-c\implies c\in\{0,x,-x\}.$ If $f(x)=x$, then $\pm (x^2+2cx)=x(c+x)\implies c=0 \text{ or }x=\frac32 c.$ If $f(x)=-x,$ then $\pm (x^2-2cx)=x(c-x)\implies c=0 \text{ or }x=\frac32 c.$ Each of these is a contradiction, so our claim is correct. So $f(x)=\pm x$ for all $x$. If $f(x)=x$ for some $x\ne 0$ and $f(y)=-y$ for some other $y\ne0$, then $P(x,y)$ means $\pm (x^2-2xy+y^2)=(x+y)(x-y)$, which is easily seen to be impossible. So this gives us two solutions: $\boxed{f(x)=x\;\forall x\in\mathbb R}$ and $\boxed{f(x)=-x\;\forall x\in\mathbb R}$, and they both work. Case 2: $f(0)\ne 0$. Then $f(0)f(f(0))=0$ gives $f(f(0))=0$. Now comparing $P(x,f(0))$ and $P(0,f(x))$ gives $$f(f(x))(f(x)+f(0))=x(f(x)+f(0)).\quad (\star)$$So if $f(x)=f(y)\ne -f(0)$, we have $f(f(x))(f(x)+f(0))=f(f(y))(f(y)+f(0))\implies x(f(x)+f(0))=y(f(y)+f(0)).$ So $$f(x)=f(y)\ne-f(0)\implies x=y.$$Call this property awesomeness as well because there's no risk of confusion. Now $P(f(0),0)$ gives $f(2f(0)^2)=0=f(f(0))\ne -f(0)$, so by awesomeness $2f(0)^2=f(0)\implies f(0)=\tfrac12.$ Then $f(\tfrac12)=f(f(0))=0.$ Now from $P(x,0)$ and $P(0,x)$ lead us to $$f(f(x)^2+x)=f\left(x^2+\frac14\right)=f(x)\left(x+\frac12\right).$$Now for each $x$, we can have $f(x)(x+\tfrac12)=-f(0)=-\tfrac12$, or if not, we can get by awesomeness that $f(x)^2+x=x^2+\tfrac14\implies f(x)=\pm (x-\tfrac12)$. So for every $x$, $$f(x)\in\left\{\frac{-\frac12}{x+\tfrac12},x-\frac12,\frac12-x\right\}.$$Now if for some $x$, we really have $f(x)$ equal to that eyesore that is $\frac{-\tfrac12}{x+\tfrac12}$ (we clearly need $x\ne-\tfrac12$), we'd have $f(x)(x+\tfrac12)=-\tfrac12$, so $f(x^2+\tfrac14)=-\tfrac12$, so we could have a bunch of cases: $$f(x^2+\frac14)=x^2+\frac14-\frac12=-\frac12\implies x^2=-\frac14,$$which is impossible; $$f(x^2+\frac14)=\frac12-x^2-\frac14=-\frac12\implies x=0,$$so $f(0)=-1$, which is a lie since $f(0)=\tfrac12$. $$f(x^2+\frac14)=\frac{-\frac12}{x^2+\frac14+\frac12}=-\frac12\implies x=\pm \frac12.$$But $x$ can't be $-\tfrac12$ , so $x=\tfrac12\implies f(\tfrac12)=-\tfrac12$, which another lie because $f(\tfrac12)=0$. Now we have narrowed it down to $f(x)\in\{x-\tfrac12,\tfrac12-x\}$. If we have $x\ne\tfrac12$ so that $f(x)=x-\tfrac12$ then $x\ne 0$, so $f(x)+f(0)=x\ne 0,$ and $(\star)$ gives us $f(f(x))=x\implies f(x-\tfrac12)=x\implies \pm(x-\tfrac12-\tfrac12)=x$, which is impossible. Therefore $\boxed{f(x)=\tfrac12-x\;\forall x\in\mathbb R}$, which is another solution. $\blacksquare$ Let me know if there's a typo. Or don't. I'm not a huge fan of this monstrosity anyway.

Ankoganit wrote: Case 1: $f(0)=0$. Now $P(x,0)$ implies $f(f(x)^2)=xf(x)$. If $x,y\in\mathbb R$ satisfy $f(x)=f(y)\ne 0$, then the above equation gives $f(f(y)^2)=f(f(x)^2)\implies xf(x)=yf(y)\implies x=y$. Call this property of $f$ awesomeness. How does $ xf(x)=yf(y)\implies x=y$

In your quote it is assumed that f has equal values at x,y as we try to show injection

Let $P(x,y)$ denote the assertion \[f(y^2+2xf(y)+f(x)^2)=(y+f(x))(x+f(y)).\]Further let $f(0)=c.$ We claim that the only solutions are $f(x)\equiv x, -x,1/2-x$ which are all easily checked to work. $P(x,-f(x))\implies \text{ker}(f)\neq\emptyset.$ Let $x_0\in\text{ker}(f).$ If $f$ is injective, $P(x,y)-P(y,x)$ gives $y^2+2xf(y)+f(x)^2=x^2+2yf(x)+f(y)^2\implies (f(y)-x)^2=(f(x)-y)^2$ so either $f(x)+x=f(y)+y$ or $f(x)-x=-f(y)+y$ so $f\equiv -x+c,x.$ Now assume $f$ is not injective so that $f(a)=f(b),a\neq b.$ Then $P(a,x_0)-P(b,x_0)\implies (x_0+f(a))a=(x_0+f(b))b\implies x_0=-f(a).$ This immediately implies that $|\text{ker}(f)|=1.$ $P(x_0,0)\implies f(2x_0c)=0,$ $P(0,x_0)\implies f(c^2+x_0^2)=0$ These statements imply $2x_0c=c^2+x_0^2=x_0,$ which have the solutions $(x_0,c)=(0,0),(1/2,1/2).$ In the first case, $P(0,x)\implies f(x^2)=xf(x)$ $P(x,0)\implies f(x)f(f(x))=f(f(x)^2)=xf(x)$ so $f(x)=0$ or $f(f(x))=x$ for each $x.$ The former can only happen when $x=0$ when the latter holds true, so $f(f(x))=x$ for all $x.$ Now this shows that $f$ is injective and we conclude as before. In the second case, $P(x,-f(x))\implies 2f(x)^2+2xf(-f(x))=1/2,$ $P(-f(x),x)\implies x^2-2f(x)^2+f(-f(x))^2=1/2.$ Chaining these gives $(f(-f(x))-x)^2=(2f(x))^2\implies f(-f(x))=x\pm 2f(x).$ Substituting back gives $f(x)^2+x(x\pm 2f(x))=1/4\implies (f(x)\pm x)^2=(1/2)^2$ so for each $x$ we either have $f(x)=1/2+x$ or $f(x)=1/2-x.$ Assume a particular $x\neq 0$ satisfies $f(x)=x+1/2.$ $P(x,0)\implies f(x+(1/2+x)^2)=(1/2+x)^2\implies x=-1/2$ or $1/2-x-2(1/4+x+x^2)=0\implies-3x-2x^2=0\implies x = -3/2.$ $P(0,x)\implies f(x^2+1/4)=x^2+x\implies x=3/4$ or $2x^2+x-1/4=0$ which yields none of the solutions above. Thus we conclude that no $x$ can satisfy $f(x)=x+1/2$ except $x=0,$ and we conclude this case as well.

Let $P(x,y)$ be the given FE. We claim the solutions are $f(x)\equiv x,-x,1/2-x$. It is easy to verify that these work. Now, suppose that $f$ is a solution. Let $S=\{x\in\mathbb{R}:f(x)=0\}$. If $a,b\in S$, then \[P(a,b)\implies f(b^2)=ab.\]Thus, if $a\ne b\in S$, then $b^2=ab=f(b^2)$ and $a^2=ab=f(a^2)$, so $ab=a^2=b^2$, which is a contradiction to $a\ne b$. Therefore, $|S|\le 1$. However, $P(x,-f(x))$ implies that $S$ is nonempty, so we have $S=\{k\}$ for some $k$. Now, $P(x,-f(x))$ and $P(-f(y),y)$ imply that \[2f(x)^2+2xf(-f(x))=y^2-2f(y)^2+f(-f(y))^2=k\]for all $x,y\in\mathbb{R}$. Setting $x=y$, we can use these equations to get two different expressions for $4y^2f(-f(y))^2$, and setting them equal, we get \[4y^4-8y^2f(y)^2+[k-2f(y)^2]^2-4y^2k=0.\]This is a quadratic equation in $f(y)^2$, so solving gives \[\boxed{f(y)^2=\frac{k}{2}+y^2\pm y\sqrt{2k}}.\]Plugging in $y=k$, we see that \[0=k/2+k^2+k\sqrt{2k},\]so $k=0$ or $1/2+k+\sqrt{2k}=0$, or $(1/2+k)^2=2k$, or $k=1/2$. Thus, $k\in\{0,1/2\}$, so we have two cases. Case 1: $k=0$. Here, we see that $f(y)^2=y^2$ for all $y$. Suppose $f(x)=x$ and $f(y)=-y$ where $x,y,0$ are all distinct. We see that \[P(x,y)\implies f(y^2-2xy+x^2)=(y+x)(x-y)=x^2-y^2,\]so \[x^2-2xy+y^2=\pm(x^2-y^2).\]This can easily be checked to be a contradiction to the statement that $x,y,0$ are all distinct, so we must have either $f(x)\equiv x$ or $f(x)\equiv -x$. Case 2: $k=1/2$. Here, we see that $f(y)^2=1/4+y^2\pm y=(1/2\pm y)^2$ for each value of $y$. Note that \[P(x,x)\implies f((x+f(x))^2)=(x+f(x))^2.\]Let $u=x+f(x)$. Thus, we have \[u^2=\pm 1/2 \pm u^2.\]We can't have $u^2=u^2\pm 1/2$, so in fact $u^2=\pm 1/2-u^2$, so $u^2=\pm 1/4$. Thus, $u=\pm 1/2$, so for each $x$, $f(x)\in\{1/2-x,-1/2-x\}$. Suppose $f(y)=-1/2-y$. Then, \[P(1/2,y)\implies f(y^2-y-1/2)=y(-y)=-y^2,\]so \[y^2-y-1/2=\pm 1/2+y^2,\]so $y=0,-1$. We already know that $f(0)=1/2\ne -1/2-0$, so we have $f(x)=1/2-x$ for all $x\ne -1$, and $f(-1)\in\{3/2,1/2\}$. Suppose that $f(-1)=1/2$. Then, \[P(-1,1/2)\implies f(1/4+1/4)=(1/2+1/2)(-1+1/2)=-1/2,\]so $f(1/2)=-1/2$, which is a contradiction. Thus, $f(x)\equiv 1/2-x$ in this case, so we're done.

Let $P(x,y)$ be the assertion. Let the original FE be denoted by $(*)$. $P(y,x) \longrightarrow f(y^2+2xf(y)+f^2(x))= (x+f(y))(y+f(x)) = f(x^2 +2yf(x)+f^2(y)) \dots (1)$ $P(0,y) \longrightarrow$, $0 \in$ range of $f$ Claim 1: $0$ has a unique input. Suppose $\exists a \neq b$ such that $f(a)=f(b)=0$. $P(a,b)$ in $(1) \longrightarrow f(b^2) =ab=f(a^2)$ $P(a,a)$ in $(*) \longrightarrow f(a^2) = a^2$. Similarly, $f(b^2)=b^2$. So $b^2 =ab=a^2$. If $a \neq 0$, then $b=a$, contradiction. If $a=0$, then $b=0$. Contradiction. So now $\exists ! t: f(t)=0$. $P(t,0)$ in $(1) \longrightarrow f(2tf(0)) = 0 = f(t^2+f^2(0))$. So $2tf(0)=t$. Then $t=0$ or $f(0)=1/2$. But then $t^2+f^2(0)=t^2 +1/4 =t \longrightarrow t=1/2$. So $t \in \{0, \frac{1}{2} \}$. We will look at each of these cases seperately. Case 1: $f(1/2)=0$ and $f(0)=1/2$. $P(x,1/2)$ in $(1) \longrightarrow f(1/4 +f^2(x))= x(1/2+f(x))= f(x^2+f(x)) \dots (2)$ Claim 2: $f$ is injective. Say $f(a)=f(b)$. $P(a), P(b)$ in $(2) \longrightarrow f(1/4+f^2(a)) = a(1/2 +f(a)) = f(1/4+f^2(b)) = b(1/2+f(b)) \longrightarrow a=b$ if $f(a)+1/2 \neq 0$. Now consider the inputs of $-1/2$. I claim that there is just one. Then $f$ is injective everywhere. Say $f(a)=-1/2$. $P(a)$ in $(2) \longrightarrow f(a^2-1/2)=0 \longrightarrow a^2-1/2 = 1/2 \longrightarrow a= \pm 1.$ $P(x,0)$ in $(1) \longrightarrow f(2xf(0)+f^2(x)) = f(x)(x+f(0)) = f(x^2 +f^2(0)) \dots (3)$ $P(1), P(-1)$ in $(3) \longrightarrow f(1)(1+f(0)) = f(-1)(-1+f(0)) \longrightarrow 3f(1) =-f(-1)$. So $f(1) \neq f(-1)$. Hence $f$ is injective. Now from $(2), \ x^2+f(x) = 1/4 +f^2(x) $. From $(3), \ f^2(x) + x =x^2 + 1/4$. Subtracting both equations we get $\boxed{f(x) = -x+\frac{1}{2}}$. Case 2: $f(0)=0$. Now $(3)$ is $f(f^2(x)) = xf(x) = f(x^2)$. So $f$ is injective, since if $f(a)=f(b) \neq 0$ then $af(a)=bf(b) \longrightarrow a=b$. And $f$ is already injective at $0$ too. So $(3)$ gives $f^2(x) =x^2$. $P(1,y)$ in $(1) \longrightarrow f(y^2 + 2f(y) +1) = f(1+2yf(1)+y^2)\longrightarrow f(y) = yf(1)$. So $\boxed{f(x) = x}$ or $\boxed{f(x)=-x}$. So done.

Denote the assertion as $P(x,y)$. We first show that there exists exactly one $z$ s.t. $f(z)=0$. Suppose on the contrary that $f(z_1)=f(z_2)=0,z_1\neq z_2$. $P(z_1,z_1),P(z_1,z_2),P(z_2,z_1),P(z_2,z_2)$ quickly give a contradiction. Now, $P(x,-f(x))$ and $P(-f(y),y)$ yield that $$z=2f(x)^2+2xf(-f(x))=x^2-2f(x)^2+f(-f(x))^2$$. Substituting $x=z$ gives $z=f(0)=z^2+f(0)^2=2z^2$. So, $z\in\{0,1/2\}$. Note that $z=2f(x)^2+2xf(-f(x))$ nearly implies injectivity. Namely, if $f(a)=f(b)$, we must have $f(a)=f(b)=-z$. Of course, this gives $a=b$ if $z=0$, so the only case where we don't have injectivity is when $f(a)=f(b)=-1/2=-z$. In this case, we have $z=1/2=a^2-1/2$, so $a^2=b^2=1$. Therefore, the only case in which injectivity doesn't hold is if $(a,b,z)=(-1,1,1/2)$. However, $P(0,1)$, $P(-1,0)$ quickly give a contradiction. So, $f$ is injective. Combining injectivity with $P(x,y)$, $P(y,x)$, we get $$y^2+2xf(y)+f(x)^2=x^2+2yf(x)+f(y)^2$$Substituting $y=z$, $z^2+f(x)^2=x^2+2zf(x)\implies f(x)=z\pm x$. If $z=0$, $f(x)=\pm x$, so we get $xf(y)=yf(x)$. This means that we choose the same sign for all $f$. Both solutions generated work. If $z=1/2$, suppose that we have nonzero $x,y$ such that $f(x)=1/2-x$, $f(y)=1/2+y$. Our assertion gives $2xy=y$. So, we must have $x=1/2$. Of course, the function $f(x)=\begin{cases} 1/2+x&x\neq 1/2\\ 0&x=1/2\end{cases}$ doesn't work, so we must have either $f(x)=1/2-x$ or $f(x)=1/2+x$. Only $f(x)=1/2-x$ works. Thus, our complete set of solutions is $f(x)=1/2-x$, $f(x)=x$, $f(x)=-x$.

Solved with dantaxyz. Let $P(x,y)$ denote the FE. Lemma: There is exactly one $a$ such that $f(a)=0$. Proof: First, note that $P(42,-f(42))$ gives $f(c)=0$ for some $c$, so $f$ has at least one root. Now suppose $f(a)=0$ and $f(b)=0$. Then $P(a,a)$ gives $f(a^2)=a^2$, while $P(b,a)$ gives $f(a^2)=ab$; hence $a^2=ab$, i.e. $a=b$. $\blacksquare$ Now $P(x,-f(x))$ gives \[ f(2f(x)^2+2xf(-f(x)))=0 \implies 2f(x)^2+2xf(-f(x))=C \qquad (\heartsuit) \]for some constant $C$ such that $f(C)=0$, by the lemma. Now, suppose $f(a)=f(b)$ for some $a,b$. Then $af(-f(a))=bf(-f(b))$ by $(\heartsuit)$, so $f(-f(a)) (a-b)=0$. Case 1: $a\not = b$ for some $f(a)=f(b)$. Then $f(-f(a))=f(-f(b))=0$ by the above. Then $C=-f(a)$, since $f(C)=0$ and $f(-f(a))=0$. But also, plugging in $a$ into $(\heartsuit)$ gives $2f(a)^2=C$. Combining these two gives $2C^2=C$, so $C=0$ or $C=1/2$. $~$ Subcase 1.1: $C=0$. Then $0=-f(a)$ and $0=-f(b)$ similarly, so $f(a)=f(b)=0$. But by the lemma, this implies $a=b$, contradiction. $~$ Subcase 1.2: $C=1/2$. So $f(a)=f(b)=-1/2$. We know $f(1/2)=0$, and plugging in $1/2$ into $(\heartsuit)$ gives $f(0)=1/2$. For $f(a)=f(b)$, we saw $f(-f(a))(a-b)=0$, so if $f(-f(a))\not = 0$, i.e. $f(a)\not = -1/2$, then $a=b$. Now, comparing the LHS's of $P(x,1/2)$ and $P(1/2,x)$ gives \begin{align*} f(1/4+f(x)^2) = f(x^2+f(x)) \qquad (\diamondsuit). \end{align*}We know $f$ is injective except when the equal outputs are $-1/2$. We will show that there is at most one $d$ such that $f(d)=-1/2$; this would show $f$ is injective everywhere. Suppose $f(d)=-1/2$. Then plugging in $d$ into $(\diamondsuit)$ gives $f(1/2)=f(d^2-1/2)$. Since $f(1/2)=0 \not = -1/2$, we can use injective to conclude $1/2=d^2-1/2$, i.e. $d=\pm 1$. Now we just need to show $f(1)\not = f(-1)$; suppose $f(1)=f(-1)=-1/2$. But \[ P(0,x): f(x^2+f(0)^2) = [x+f(0)]f(x). \]Plugging in $1$ and $-1$ clearly has equal LHS, and since $f(1)=f(-1)$, actually $1+f(0)=-1+f(0)$, a clear contradiction. Therefore, $f$ is completely injective, contradicting the assumption for Case 1. $~$ Case 2: $a=b$ for all $f(a)=f(b)$. So $f$ is injective. Then, $P(x,y)$ with $P(y,x)$ have the same RHS, and using injective on the LHS gives \begin{align*} &y^2+2xf(y)+f(x)^2 = x^2+2yf(x)+f(y)^2 \\\implies& (f(x)-y)^2 = (f(y)-x)^2 \qquad (\clubsuit). \end{align*}Let $a$ be the unique root of $f$. Then plugging in $y=a$ above gives $(f(x)-a)^2=x^2$, so $f(x)=a\pm x$ for each $x$. By the same analysis as in the beginning of Case 1, we know $a=0$ or $a=1/2$. Subcase 2.1: $a=0$, so $f(x)=\pm x$ for each $x$. Suppose $f(x)=x$ and $f(y)=-y$ for some nonzero $x,y$. Then $(\clubsuit)$ gives $(x-y)^2 = (-y-x)^2$, so $xy=0$, contradiction. So $f(x)=x$ for all $x$ or $f(x)=-x$ for all $x$, both of which work. $~$ Subcase 2.2: $a=1/2$, so $f(x)=1/2 \pm x$ for each $x$. Suppose $f(x)=1/2+x$ and $f(y)=1/2-y$ for some nonzero $x,y$. Then $(\clubsuit)$ gives $(1/2+x-y)^2 = (1/2-x-y)^2$. So $x=0$ or $y=1/2$. Subcase 2.2.1: If $x=0$, then $f(0)=1/2$, and $f(y)=1/2-y$ for all other $y$. This is just $f(x)=1/2-x$ for all $x$, which works. Subcase 2.2.2: If $y=1/2$, then $f(1/2)=0$, and $f(x)=x+1/2$ for all other $x$. So we have to test the function $f(x)=\begin{cases} 1/2+x & x\not = 1/2 \\ 0 & x=1/2 \end{cases}$. This function does not work. In conclusion, the solutions are $f(x)=x$, $f(x)=-x$, $f(x)=1/2-x$, all of which can be checked to work.

Let $P(x, y)$ denote the assertion. Comparing $P(x, y)$ with $P(y, x)$ and exploiting symmetry yields\[f(y^2 + 2xf(y) + f(x)^2) = f(x^2 + 2y(f(x)) + f(y)^2)\]which we denote as a new assertion $Q(x, y)$. We will prove that $f$ is injective, which clearly will be very useful. First, we show that $f$ is injective at $0$. By $P(x, -f(x))$, there must clearly exist some $c$ for which $f(c) = 0$. Suppose there were two such $c$: some $c_1 \neq c_2$ for which $f(c_1) = f(c_2) = 0$. Note that $P(c_1, c_2)$ and $P(c_2, c_1)$ give $f(c_1^2) = c_1c_2 = f(c_2^2)$. Next $P(c_1, c_1)$ and $P(c_2, c_2)$ give $f(c_1^2) = c_1^2$ and $f(c_2^2) = c_2^2$ so\[c_1^2 = c_1c_2 = c_2^2 \implies c_1 = c_2\]as desired. Hence there is a unique $c$ for which $f(c) = 0$. Note that $P(c, 0)$ and $P(0, c)$ yields\[0 = f(2cf(0)) = f(c^2 + f(0)^2) \implies c = 2cf(0) = c^2 + f(0)^2\]which upon solving, either yields $(c, f(0)) = (0, 0)$ or $(c, f(0)) = (\tfrac12, \tfrac12)$. Note that $P(x, -f(x))$ along with injectivity at $0$ tells us that $2xf(-f(x)) + f(x)^2 = c$, which we call relation $R(x)$. Next, we will almost show that $f$ is injective, but fail at $-c$. Indeed, suppose $f(a) = f(b) \neq -c$. Note that $R(a)$ and $R(b)$ tell us $2af(-f(a)) = c - f(a)^2 = c - f(b)^2 = 2bf(-f(b))$, and since $f(a), f(b) \neq -c$, injectivity at $0$ tells us that $f(-f(a)), f(-f(b)) \neq 0$, so we may indeed conclude $a = b$. If $(c, f(0)) = (0, 0)$, then we have already proven universal injectivity because we know $f$ is injective at $0$. Hence $Q(x, y)$ upon unwrapping becomes\[y^2 + 2xf(y) + f(x)^2 = x^2 + 2yf(x) + f(y)^2 \implies (f(x) - y)^2 = (f(y) - x)^2\]and setting $y = 0$ yields $\boxed{f(x) = \pm x}$. If $(c, f(0)) = (\tfrac12, \tfrac12)$ then we are not quite done with injectivity yet. For contradiction, suppose there are $a \neq b$ for which $f(a) = f(b) = -\tfrac12$. Then $P(\tfrac12, a)$ and $P(\tfrac12, b)$ along with $f(\tfrac12) = 0$ yield\[f(a^2 - \tfrac12) = 0 = f(\tfrac12) \text{ and similarly } f(b^2 - \tfrac12) = 0 = f(\tfrac12)\]so injectivity at $0$ tells us that $a^2 = b^2 = 1 \implies (a, b) = (-1, 1)$ or $(1, -1)$ hence $f(-1) = f(1) = -\tfrac12$ and no other $x$ may satisfy $f(x) = -\tfrac12$. Note that $Q(-1, 1)$ yields $f(\tfrac94) = f(\tfrac14)$ and since $f$ is already shown to be injective at all points other than $-\tfrac12$, it must follow that $f(\tfrac94) = f(\tfrac14) = -\tfrac12$. However, also, $\tfrac94, \tfrac14 \not \in \{-1, 1\}$, which is our desired contradiction. Thus $f$ is universally injective. Next, note that unwrapping $Q(x, y)$ yields\[y^2 + 2xf(y) + f(x)^2 = x^2 + 2yf(x) + f(y)^2 \implies (f(x) - y)^2 = (f(y) - x)^2.\]Setting $y = 0$ yields $f(x) = \pm(x - \tfrac12)$ but $f(0) = \tfrac12$ so the only valid solution here is $f(x) =\boxed{\tfrac12 - x}$. Lastly, we may easily check that the aforementioned functions all work, so we are done. $\blacksquare$

Ankoganit wrote: This inhumanly case-bashy and boring problem wasted my otherwise fine Saturday morning, so I'll retaliate with a justly case-bashy solution. Let $P(x,y)$ mean $P$lug in $x,y$ into the given equation. Now comparing $P(0,f(0))$ and $P(0,-f(0))$ gives $f(0)f(f(0))=0$. Now we have a case-fork: Case 1: $f(0)=0$. Now $P(x,0)$ implies $f(f(x)^2)=xf(x)$. If $x,y\in\mathbb R$ satisfy $f(x)=f(y)\ne 0$, then the above equation gives $f(f(y)^2)=f(f(x)^2)\implies xf(x)=yf(y)\implies x=y$. Call this property of $f$ awesomeness. The statements $P(x,0)$ and $P(0,x)$ give $$f(f(x)^2)=f(x^2)=xf(x)$$If $f(x)\ne 0$, then we need $x\ne 0$, so by awesomeness, $f(x)^2=x^2\implies f(x)=\pm x$. So for any $x$, we have $f(x)=x$ or $f(x)=-x$ or $f(x)=0$. Now we claim that we can't have $f(x)=0$ except for $x=0$. Let's say $f(c)=0$ for some $c\ne 0$. Take $x$ so that $x\not\in \left\{0,c,-c,\tfrac32 c,-\tfrac32 c\right\}$, and consider $P(c,x)$, which is $$f(x^2+2cf(x))=x(c+f(x)).$$This takes us a whole plethora of cases: If $f(x)$ is $0$, then $f(x^2)=xc\implies xc\in\{0,x^2,-x^2\}\implies x=\pm c$. If $f(x^2+2cf(x))=0$, we have $f(x)=-c\implies c\in\{0,x,-x\}.$ If $f(x)=x$, then $\pm (x^2+2cx)=x(c+x)\implies c=0 \text{ or }x=\frac32 c.$ If $f(x)=-x,$ then $\pm (x^2-2cx)=x(c-x)\implies c=0 \text{ or }x=\frac32 c.$ Each of these is a contradiction, so our claim is correct. So $f(x)=\pm x$ for all $x$. If $f(x)=x$ for some $x\ne 0$ and $f(y)=-y$ for some other $y\ne0$, then $P(x,y)$ means $\pm (x^2-2xy+y^2)=(x+y)(x-y)$, which is easily seen to be impossible. So this gives us two solutions: $\boxed{f(x)=x\;\forall x\in\mathbb R}$ and $\boxed{f(x)=-x\;\forall x\in\mathbb R}$, and they both work. Case 2: $f(0)\ne 0$. Then $f(0)f(f(0))=0$ gives $f(f(0))=0$. Now comparing $P(x,f(0))$ and $P(0,f(x))$ gives $$f(f(x))(f(x)+f(0))=x(f(x)+f(0)).\quad (\star)$$So if $f(x)=f(y)\ne -f(0)$, we have $f(f(x))(f(x)+f(0))=f(f(y))(f(y)+f(0))\implies x(f(x)+f(0))=y(f(y)+f(0)).$ So $$f(x)=f(y)\ne-f(0)\implies x=y.$$Call this property awesomeness as well because there's no risk of confusion. Now $P(f(0),0)$ gives $f(2f(0)^2)=0=f(f(0))\ne -f(0)$, so by awesomeness $2f(0)^2=f(0)\implies f(0)=\tfrac12.$ Then $f(\tfrac12)=f(f(0))=0.$ Now from $P(x,0)$ and $P(0,x)$ lead us to $$f(f(x)^2+x)=f\left(x^2+\frac14\right)=f(x)\left(x+\frac12\right).$$Now for each $x$, we can have $f(x)(x+\tfrac12)=-f(0)=-\tfrac12$, or if not, we can get by awesomeness that $f(x)^2+x=x^2+\tfrac14\implies f(x)=\pm (x-\tfrac12)$. So for every $x$, $$f(x)\in\left\{\frac{-\frac12}{x+\tfrac12},x-\frac12,\frac12-x\right\}.$$Now if for some $x$, we really have $f(x)$ equal to that eyesore that is $\frac{-\tfrac12}{x+\tfrac12}$ (we clearly need $x\ne-\tfrac12$), we'd have $f(x)(x+\tfrac12)=-\tfrac12$, so $f(x^2+\tfrac14)=-\tfrac12$, so we could have a bunch of cases: $$f(x^2+\frac14)=x^2+\frac14-\frac12=-\frac12\implies x^2=-\frac14,$$which is impossible; $$f(x^2+\frac14)=\frac12-x^2-\frac14=-\frac12\implies x=0,$$so $f(0)=-1$, which is a lie since $f(0)=\tfrac12$. $$f(x^2+\frac14)=\frac{-\frac12}{x^2+\frac14+\frac12}=-\frac12\implies x=\pm \frac12.$$But $x$ can't be $-\tfrac12$ , so $x=\tfrac12\implies f(\tfrac12)=-\tfrac12$, which another lie because $f(\tfrac12)=0$. Now we have narrowed it down to $f(x)\in\{x-\tfrac12,\tfrac12-x\}$. If we have $x\ne\tfrac12$ so that $f(x)=x-\tfrac12$ then $x\ne 0$, so $f(x)+f(0)=x\ne 0,$ and $(\star)$ gives us $f(f(x))=x\implies f(x-\tfrac12)=x\implies \pm(x-\tfrac12-\tfrac12)=x$, which is impossible. Therefore $\boxed{f(x)=\tfrac12-x\;\forall x\in\mathbb R}$, which is another solution. $\blacksquare$ Let me know if there's a typo. Or don't. I'm not a huge fan of this monstrosity anyway. Nice one

A different solution: (we will do no case work on $f^{-1}(0)$, we will just directly show $f$ is a involution) The answer is $f(x) \equiv x$ ; $f(x) \equiv -x$ and $f(x) \equiv \frac{1}{2} - x$. It is not hard to verify that these work. Now we prove these are the only solutions. Let $P(x,y)$ be the given assertion. Claim: There is a unique $\alpha$ such that $f(\alpha) = 0$. Proof: FTSOC, let $f(a) = f(b) = 0$ with $a \ne b$. At least one of $a,b$ is non-zero, say $a \ne 0$. Comparing $P(a,a)$ and $P(a,b)$ gives $a^2 = ab$, which is a contradiction. Finally, $P(-f(y),y)$ gives $f$ is surjective at $0$, which proves our claim. $\square$ $$P(-f(y),y) ~:~ y^2 - 2f(y)^2 + f(-f(y))^2 = \alpha \qquad{(1)}$$ Claim: $f$ is a involution, and consequently a bijection. Proof: $(1)$ gives that if $f(a) = f(b)$, then $a^2 = b^2$. Now fix any $x \in \mathbb R$. Comparing $P(x,f(x))$ and $P(f(x),x)$ we obtain $$ f(2f(x)^2 + 2xf(f(x))) = f(x^2 + 2f(x)^2 + f(f(x))^2 ) $$ If the sum of the two things inside $f$ is zero, then that would force $f(x) = 0$ and $x + f(f(x)) = 0$, implying $x=\alpha$ and $f(0) = -\alpha$. Then taking $x = 0,\alpha$ in $(1)$ and adding the two things gives $\alpha = 0$, i.e. $f(0) = 0$. But in that case, $f(f(0)) = 0$ is of course true. If the difference of the two things inside $f$ is zero, then that gives $\Big(f(f(x)) - x \Big)^2 = 0$, implying $f(f(x)) = x$. This proves our claim. $\square$ Now comparing $P(x,y)$ , $P(y,x)$ and using injectivity of $f$ we obtain $$(f(x) - y)^2 = (f(y) - x)^2 \qquad{(2)}$$ $\alpha = 0$. Then $f(x) \in \{x,-x\}$. FTSOC, $f(a) = a$ and $f(b) = -b$ with $a,b \ne 0$. Then $x=a,y=b$ in $(2)$ gives a contradiction. Hence, $f(x) \equiv x$ or $f(x) \equiv -x$. $\alpha \ne 0$. Then $y=0$ gives $$f(x)^2 = (x - \alpha)^2 \qquad{(3)}$$Replacing $x$ by $f(x)$ in $(3)$ gives $$x^2 = (f(x) - \alpha)^2 \qquad{(4)}$$So adding $(3),(4)$ we obtain $$x + f(x) = \alpha ~ \implies ~ f(x) \equiv \alpha - x$$It is not hard to verify that $\alpha = \frac{1}{2}$ is the only non-zero $\alpha$ which works. This completes the proof of the problem. $\blacksquare$

Let $S=f^{-1}(\{0\})$ be the set of roots of $f$, and $P(x,y)$ denotes the equation. Case 1: $|S|=0$ $P(x,-f(x))\Rightarrow 2f(x)^2+2xf(-f(x))\in S$, contradiction. Case 2: $|S|\ge2$ Let $x,y\in S$ be distinct. $P(x,x)\Rightarrow f(x^2)=x^2$ $P(y,y)\Rightarrow f(y^2)=y^2$ $P(x,y)\Rightarrow y^2=xy$ $P(y,x)\Rightarrow x^2=xy$ If $y=0$, then $x=0$, contradiction. Then $y\ne0$ and $x=y$, contradiction. Case 3: $|S|=1$, $0\notin S$ Let $S=\{c\}$, where $c\ne0$. Thus $f(x)=0$ iff $x=c$. $P(c,0)\Rightarrow 2cf(0)=c\Rightarrow f(0)=\frac12$ $P(0,c)\Rightarrow c^2-c+\frac14=0\Rightarrow c=\frac12$ $P(x,-f(x))\Rightarrow f(x)^2+xf(-f(x))=\frac12$ Now $f(-f(x))=0$ would imply $-f(x)=\frac12$, so $f(x)=-\frac12$. But if $f(a)=f(b)\ne-\frac12$ then $a=b$. $P(-f(x),x)\Rightarrow x^2-2f(x)^2+f(-f(x))^2=\frac12$ Then $f(x)^2=\left(\frac12-x\right)^2$. Assume that there is $a,b$ with $f(a)=a-\frac12$ and $f(b)=\frac12-b$ and $a+b\ne1$. $P(a,b)\Rightarrow f\left((a-b)^2+\frac14\right)=\left(a+b-\frac12\right)\left(a-b+\frac12\right)$ $P(b,a)\Rightarrow f\left((a+b)^2-2b+\frac14\right)=\left(a+b-\frac12\right)\left(a-b+\frac12\right)$ Thus $f\left((a+b)^2-2b+\frac14\right)=f\left((a-b)^2+\frac14\right)$, and since $(a-b)^2+\frac14\ge\frac14>-\frac12$ we have $(a+b)^2-2b+\frac14=(a-b)^2+\frac14$, so $b=0$. But since $f(x)=\begin{cases}x-\frac12&\text{if }x\ne0\\\frac12&\text{if }x=0\end{cases}$ is not a solution, we have a contradiction. Since $f(x)=x-\frac12$ is not a solution, we must have $\boxed{f(x)=\frac12-x}$, which fits. Case 4: $|S|=\{0\}$ $P(0,x)\Rightarrow f(x^2)=xf(x)$ $P(x,0)\Rightarrow f(f(x)^2)=xf(x)\Rightarrow f(f(x))f(x)=xf(x)\Rightarrow f(f(x))=x\forall x\ne0$, but $f(f(0))=0$ obviously implies $f(f(x))=x$ for all $x$. Then $f$ is injective, so $f(x)^2=x^2$. Assume that there is $a,b\ne0$ with $f(a)=a$ and $f(b)=-b$. Note that $a\ne b$ since $a=b$ would imply $a=-a$ and $a=0$. $P(a,b)\Rightarrow f((a-b)^2)=a^2-b^2\Rightarrow (a-b)f(a-b)=a^2-b^2\Rightarrow f(a-b)=a+b$ But $f(a-b)$ is either $a-b$ or $b-a$. In either of these cases, $ab=0$, so contradiction. Thus either $\boxed{f(x)=x}$ or $\boxed{f(x)=-x}$, which both are solutions.

Let $P(x,y)$ denote the assertion that \[f(y^2+2xf(y)+f(x)^2)=(y+f(x))(x+f(y))\]for all reals $x$ and $y$. Let $S$ be the set of the roots of $f$. Claim: $|S|<2$ Proof: Suppose $x,y\in S$ and $x\ne y$. $P(x,x): f(x^2)=x^2$. $P(y,y): f(y^2)=y^2$. $P(x,y): f(y^2)=xy$. $P(y,x): f(x^2)=xy$. If $x\ne0$, then since $x^2=xy$, $x=y$. If $x=0$, then $f(y^2)=y^2$ and $f(y^2)=0$, so $y=0$. Case 1: $|S|=0$. $P(x,-f(x)): f(2f(x)^2+2xf(-f(x)))=0$, a contradiction. So $|S|=1$. Case 2: $S=\{0\}$. $P(0,x): f(x^2)=xf(x)$ $P(x,0): f(f(x)^2)=xf(x)$. But we also have $P(0,f(x)): f(f(x)^2)=f(x)f(f(x))$. This implies $f(x)f(f(x))=xf(x)\implies f(f(x))=x$ if $x\ne0$, but since $f(f(0))=0$, we have $f(f(x))$ for all $x\in\mathbb{R}$. Thus, $f$ is injective, so $f(x)^2=x^2\implies f(x)=\pm x$. Now suppose there exist $a$ and $b$ such that $f(a)=a$ and $f(b)=-b$ with $a,b\ne0$. $P(a,b): f(b^2-2ab+a^2)=f((a-b)^2)=(a+b)(a-b)$. Then we either have $(a-b)^2=(a+b)(a-b)$ or $(b-a)^2=(b+a)(b-a)$. Since $a+b\ne a-b$ and $b+a\ne b-a$, we must have $a-b=0$, a contradiction. So we must have $\boxed{f(x)=x\text{ for all }x\in\mathbb{R}}$ or $\boxed{f(x)=-x\text{ for all }x\in\mathbb{R}}$, which both work. Case 3: $|S|=1$ and $S\ne \{0\}$. Suppose $S=\{k\}$. $P(k,k): f(k^2)=k^2$. $P(k,0): f(2kf(0))=0\implies 2kf(0)=k \implies f(0)=\frac{1}{2}$. $P(0,k): f(k^2+\frac{1}{4})=0\implies k^2+\frac{1}{4}=k$. Now we have $k^2-k+\frac{1}{4}=0$, so $(k-\frac{1}{2})^2=0\implies k=\frac{1}{2}$ and $f(\frac{1}{4})=\frac{1}{4}$. Claim: $f$ is injective. Proof: Suppose $f(b)=f(c)$. $P(b,\frac{1}{2}): f(\frac{1}{4}+f(b)^2)=(\frac{1}{2}+f(b))b$ $P(c,\frac{1}{2}): f(\frac{1}{4}+f(b)^2)=(\frac{1}{2}+f(b))c$. Thus, $(\frac{1}{2}+f(b))b=(\frac{1}{2}+f(b))c$. So either $b=c$ or $f(b)=-\frac{1}{2}$. $P(x,-f(x)): f(2f(x)^2+2xf(-f(x)))=0$, so $2f(x)^2+2xf(-f(x))=\frac{1}{2}$. $P(-f(x),x): f(x^2-2f(x)^2+f(-f(x))^2)=0\implies x^2-2f(x)^2+f(-f(x))^2=\frac{1}{2}$. Thus, $-x^2+4f(x)^2+(2x-f(-f(x))(f(-f(x))=0$. Now we will take $x$ such that $f(x)=-\frac{1}{2}$. We have \[-x^2+1=0\implies x=\pm1.\] Suppose that $f(1)=f(-1)=-\frac{1}{2}$. $P(1,-1): f(1-1+1)=f(1)=(-1-\frac{1}{2})(1-\frac{1}{2})=-\frac{3}{4}$, a contradiction. Since there exist no distinct $b$ and $c$ so that $f(b)=f(c)=-\frac{1}{2}$, $f$ is injective. $P(0,x): f(x^2+\frac{1}{4})=f(x)(x+\frac{1}{2})$. $P(x,0): f(x+f(x)^2)=f(x)(x+\frac{1}{2})$. We have $x^2+\frac{1}{4}=x+f(x)^2$, so $(x-\frac{1}{2})^2=f(x)^2$, so $f(x)=x-\frac{1}{2}$ or $\frac{1}{2}-x$. Suppose $a$ satisfies $f(a)=a-\frac{1}{2}$. $P(a,\frac{1}{2}): f(a^2-a+\frac{1}{2})=a^2$. So either $a^2-a+\frac{1}{2}-\frac{1}{2}=a^2$ or $\frac{1}{2}-a^2+a-\frac{1}{2}=a^2$. The former implies $a=0$. The latter implies $2a^2=a$, so $a=\{0,\frac{1}{2}\}$. But we know that for both these value of $a$, $f(a)=\frac{1}{2}-a$, so we can conclude that the only solution for this case is $\boxed{f(x)=\frac{1}{2}-x\text{ for all x }\in\mathbb{R}}$.

Let $P(x,y)$ be the given assertion. Claim. 1: $f$ is injective at 0. Proof. Note that $f$ attains 0 by $P(x,-f(x)).$ So assume $\exists$ some $a,b$ such that $f(a)=f(b)=0.$ Compare $P(a,b), P(b,a), P(a,a), P(b,b)$ gives that $a=b.$ So we're done. Now let $f(k)=0,$ then by $P(x,-f(x)):$ $$k=2f(x)^2+xf(-f(x))~~~(*)\; \blacksquare$$ Assume $f(a)=f(b)\neq 0,$ for some $a,b.$ By $(*)$ it follows that either $a=b$ or $f(a)=f(b)=-k.$ Comparing $P(k,a)$ and $P(k,b)$ gives $a=b$ or $a=-b.$ Claim. 2: $x^2+2yf(x)+f(y)^2=y^2+2xf(y)+f(x)^2 \qquad (\dagger)$ Proof. Compare $P(x,y)$ and $P(y,x),$ to get the symmetry. Assume the contrary to the claim. Then it must be that $$x^2+2yf(x)+f(y)^2+y^2+2xf(y)+f(x)^2=0 \implies (f(x)+y)^2+(f(y)+x)^2=0,$$absurd. $\blacksquare$ In $(\dagger)$ set $y\mapsto 0$ to get $f(x)^2=x^2-2xf(0)+f(0)^2.$ Putting this into $(\dagger)$ we get $x(f(y)-f(0))=y(f(x)-f(0)),$ so $f$ must be linear. Hence $f(x)=ax+b.$ Checking yields that $f(x)=x$, $f(x)=-x$ or $f(x)=-x+.5,$ all of these work.

Denote the assertion with $P(x, y)$. Claim: $\ker f$ has cardinality $1$. Proof. If $a, b \in \ker f$ then by $P(a, b)$ it follows that \[ f(a^2) = f(b^2) = ab \]which implies that the kernel has at most one element. By $P(x, -f(x))$ it follows that \[ f(2f(x)^2 + 2xf(-f(x))) = 0 \]so $\ker f$ is nonempty. $\blacksquare$ Claim: $f(0) = 0$ or $\frac{1}{2}$, and $f(x) = \pm (f(0) - x)$ pointwise. Proof. Now, if we let $c$ be the one element in the kernel, then \[ 2f(x)^2 + 2xf(-f(x)) = c \]This implies that $f$ is injective. By comparing $P(x, y)$ and $P(y, x)$, it follows that \[ f(y^2 + 2xf(y) + f(x)^2) = f(x^2 + 2yf(x) + f(y)^2) \]As such, \[ y^2 + 2xf(y) + f(x)^2 = x^2 + 2yf(x) + f(y)^2 \]\[ (f(x) - y)^2 = (f(y) - x)^2 \]so either $f(x) + x = f(y) + y$ or $f(x) + f(y) = x + y$. Thus, $f(x) = \pm(f(0) - x)$ holds pointwise. As such, it follows that \[ 2(f(0) - x)^2 + 2x \pm(f(0) \pm(f(0) + x)) = c \]or \[ 2f(0)^2 - 4xf(0) + 2x^2 \pm 2xf(0) \pm 2x(f(0) + x) = c \]which means that $x$ satisfies one of four polynomials. Since $x$ takes on infinite values, at least one of the polynomials must be the zero polynomial, which means by degree that one of \[ 2f(0)^2 - 4xf(0) \pm 2xf(0) - 2xf(0) = c \]is the zero polynomial. It can be seen that this only holds if $f(0) = 0$ or $f(0) = \frac{1}{2}$. $\blacksquare$ Claim: If $f(0) = 0$, then $f(x) = x$ or $f(x) = -x$ Proof. Now, let $A$ be the set such $f(x) = - x$ and $B$ the set such $f(x) = x$. Then, for $x \in A$ and $y \in B$ it follows that \[ f((x+y)^2) = f((x-y)^2) \]which can only hold if one of $A$ or $B$ is $\{0\}$ or empty, giving the result. $\blacksquare$ Claim: If $f(0) = \frac{1}{2}$, then $f(x) = \frac{1}{2} - x$. Proof. Now, let $A$ be the set such $f(x) = \frac{1}{2} - x$ and $B$ the set such $f(x) = x - \frac{1}{2}$. Then, for $x \in A$ and $y \in B$, by $P(x, y)$ and $P(y, x)$ again \[ f\left(x^2 + y^2 + 2xy - 2x + \frac14\right) = f\left(x^2 + y^2 - 2xy + \frac14\right) \]which implies that either $4xy = 2x$ or that \[ 2x^2 + 2y^2 - 2x = 1/2 \]If $y \ne \frac{1}{2}$ this leads to a contradiction in the second equation by varying at least one of $x$ and $y$. Thus, $B$ must be $\left\{\frac{1}{2} \right\}$ and $f(x) = \frac{1}{2} - x$, which can be confirmed to work $\blacksquare$