Easy for P5

Let $P(x,y)$ be the assertion. $P(0,0)\to f(0)=0$ $P(0,x)\to f(f(x))=f(x^2)$ $\therefore P(x,y):\ f(x^2+f(y))=f(x^2)+f(y^2)+2f(xy)$ $P(1,x)-P(-1,x)\to f(x)=f(-x)$, so we just consider the case in $\mathbb{R}_{\ge 0}$ If $f(a)=f(b)$, $P(t,a),P(t,b)$ imply $f(ta)=f(tb),\ \forall t\in \mathbb{R}$. Suppose $f(a)=0$ for some $a \neq 0$, then $f(ta)=0\ \forall t\in \mathbb{R}$, i.e $f(x)\equiv 0$. Now suppose that $f(x)=0 \Leftrightarrow x=0$. $P(\sqrt{|x^2+f(z)|},y)\to f(x^2+f(y)+f(z))=f(x^2+f(z))+f(y^2)+2f(\sqrt{|x^2+f(z)|}y)=f(x^2)+f(y^2)+f(z^2)+2f(xz)+2f(\sqrt{|x^2+f(z)|}y)$. Change the variables $y,z$ we have $$f(xz)+f(\sqrt{|x^2+f(z)|}y)=f(xy)+f(\sqrt{|x^2+f(y)|}z) $$Supose there exist $y>z>0$ s.t. $f(z)y^2<f(y)z^2$. Let $x^2=\frac{z^2f(y)-y^2f(z)}{y^2-z^2}$, then $\sqrt{|x^2+f(z)|}y=\sqrt{|x^2+f(y)|}z$, implying $f(xz)=f(yz)\Rightarrow f(y)=f(z)$. However, since $f(z)y^2<f(y)z^2\Rightarrow y^2<z^2$, which is a contradiction. Hence $\forall y>z>0,\ \frac{y^2}{f(y)}\ge \frac{z^2}{f(z)}$. Now we prove that $f$ is injective. Suppose not, then $\exists t>1$ s.t. $f(x)=f(tx),\ \forall x\in\mathbb{R}_{\ge 0}$. Hence $\frac{t^2x^2}{f(tx)}=\frac{x^2}{f(x)}\Rightarrow f(x)>0,\ \forall x>0$. Choose $n\in\mathbb{N}$ sufficiently large such that there exists $c>0$ satisfying $c^2+f(y)=t^ny^2$. $P(c,y)\to f(x^2)+2f(xy)=0$, which is a contradiction. Hence $f$ is injective. Since $f(f(x))=f(x^2)\Rightarrow f(x)=x^2$, as desired.

My solution maybe a little complicated... My solution: Let $P(x,y)$ be the assertion. $P(0,0)\Rightarrow f(f(0))=f(f(0))+f(0)+2f(0)\Rightarrow f(0)=0 $ $P(0,y)\Rightarrow f(f(y))=f(y^2)......(1) $ $P(x,y)\Leftrightarrow f(x^2+f(y))=f(x^2)+f(y^2)+2f(xy)...(2) $ $P(y,x)\Leftrightarrow f(y^2+f(x))=f(x^2)+f(y^2)+2f(xy)...(2') $ By (2)(2'),we get$ f(x^2+f(y))=f(y^2+f(x))$ Consider$P(-x,y)$,we can get that $f(x)$is an even function. Now we have 2 case: case 1:$f $ exists a zero point $t\ne0$. $P(x,t)\Rightarrow f(x^2)=f(x^2+f(t))=f(f(x))+f(t^2)+2f(xt)\Rightarrow f(t^2)+2f(xt)=0 $ $P(0,t)\Rightarrow f(t^2)=0$ so we have $f(xt)=0$. So for all $x\ne0$,we have $f(x)=f(\frac{x}{t}\cdot t)=0$ The answer is$f(x)=0$in this case. case 2:$f $doesn't exist a zero point $t\ne0$. We prove 3 conclusions firstly. Conclusion 1:$f(xf(1))=f(x)......(*1)$ Let $y=1$in (1),we have$f(f(1))=f(1)......(3)$ Let $y=f(1)$in (1),we have$f(f(1)^2)=f(f(f(1)))=f(f(1))=f(1)......(4)$ $P(x,1)\Rightarrow f(x^2+f(1))=f(x^2)+f(1)+2f(x)......(5)$ $P(x,f(1))\Rightarrow P(x^2+f(f(1))=f(x^2)+f(f(1)^2)+2f(xf(1))$ By(3)(4),we have $f(x^2+f(1))=f(x^2)+f(1)+2f(xf(1))......(6)$ By(5)(6),we have$f(xf(1))=f(x)$,as desired. Conclusion 2: $f(x^2f(1)^2+f(1))=f(x^2f(1)+1)......(*2)$ $P(x,1)\Rightarrow f(x^2+f(1))=f(x^2)+f(1)+2f(x) $ $P(x,f(1))\Rightarrow f(x^2f(1)^2+f(1))=f(x^2f(1)^2)+f(1)+2f(xf(1))=f(x^2)+f(1)+2f(x)=f(x^2+f(1))$ We can get $f(x^2f(1)^2+f(1))=f(x^2f(1)+1)$ by conclusion 1,as desired. Conclusion 3:$f(x)>0,\forall x\ne0$ Firstly,we prove that $f(1)>0$. Proof by contradiction.Assume that $f(1)<0$,let $x=\sqrt{-f(1)}$in(*2),we get$$f(-f(1)^2+1)=f(-f(1)+f(1))=0$$Because 0 is the liquid zero point,so$$-f(1)^2+1=0\Rightarrow f(1)=-1$$$P(1,1)\Rightarrow f(1+f(1))=4f(1)......(7)$ Substituting $f(1)=-1$ in (7),we get that f(1)=0,a contradiction. So $f(1)>0$. Might as well assume that there is a $s$ satisfied that $s>0,f(s)<0$($f$is an even function) $P(f(1)\sqrt{-f(s)},s) and P(\sqrt{-f(s)},s)\Rightarrow -f(1)^2f(s)+f(s)=0\Rightarrow f(1)=1$ Now we prove that $f$is an injective function in $R_+$. Assume there are 2 numbers:u,v satisfied that $f(u)=f(v),u>v>0$. Then we have: $4f(u^2)=f(u^2+f(u))=f(u^2+f(v))=f(v^2+f(u))=f(v^2+f(v))=4f(v^2)\Rightarrow f(u^2)=f(v^2)$ $P(x,u)and P(x,v)\Rightarrow f(ux)=f(vx)$ Let $a=\frac{u}{v}>1$,then we can get:$f(x)=f(ax)$ Might as well assume that $a>\sqrt{2}$,or we can choose a sufficiently large integer k satisfied $a^k>\sqrt{2}$,and replace$ a$ with $a^k$ Let $x=1$,we have $f(a)=f(1)=1$ Let $x=a$,we have $f(a^2)=f(a)=1$ $P(x,1)and P(ax,1)\Rightarrow f(x^2+1)=f(a^2x^2+1)=f(x^2+\frac{1}{a^2})$ Let$x=\sqrt{a^2-2}$,we have$$f(a^2-1)=f(a^2-2+\frac{1}{a^2})=f(a^4-2a^2+1)=f((a^2-1)^2)......(8)$$$P(x,1)and P(\frac{x}{a},1)\Rightarrow f(x^2+1)=f(\frac{x^2}{a^2}+1)=f(x^2+a^2) $ Let $x=a^2-1$,we have$f((a^2-1)^2+1)=f((a^2-1)^2+a^2)=f(a^4-a^2+1)$. Let$x=\sqrt{a^4-a^2}$,we have$f(a^4-a^2+1)=f(a^4-a^2+a^2)=f(a^4)=f(a^3)=f(a^2)=f(a)=1$ So $f((a^2-1)^2+1)=1$ $P(a^2-1,1)\Rightarrow 1=f((a^2-1)^2+1)=f((a^2-1)^2)+f(1)+2f(a^2-1)$ $\Rightarrow f((a^2-1)^2)+2f(a^2-1)=0$ By(8),we have $f(a^2-1)=0\Rightarrow a^2-1=0$,but $a^2-1>0 $when$a>1$,a contradiction. So $f$is an injective function in $R_+$. $f(-f(s))=f(f(s))=f(s^2)$(by $f(f(x))=f(x^2)$) Because $-f(s)>0,s^2>0$,so we get that$f(s)=-s^2$(By $f$is an injective function in $R_+$ ) $P(s,s)\Rightarrow 0=f(s^2-s^2)=4f(s^2)\Rightarrow f(s^2)=0\Rightarrow s=0$,a contradiction. So the conclusion 3 is right. Back to the question. We prove $f$is an injective function in $R_+$ again. When$x\ne0$,$$f(x^2+f(y))=f(x^2)+f(y^2)+2f(xy)>f(y^2)=f(f(y))......(9)$$(by conclusion 3) Assume there are 2 numbers:u,v satisfied that $f(u)=f(v),u>v>0$. Similarly(the proof of conclusion 3),we have $f(xu)=f(xv)$. For any$s>0$,$f(s)>0$.Let$x=\frac{f(s)}{v}$,we get that$f(\frac{uf(s)}{v})=f(f(s))......(10)$ But when we let$x=\sqrt{(\frac{u}{v}-1)f(s)},y=s$,we can get$f(\frac{uf(s)}{v})>f(f(s))$,a contradiction with (10)! So $f$is an injective function in $R_+$. Because$ f(f(x))=f(x^2) $when$x\ne0$,so we have $f(x)=x^2$when $x\ne0$,because $f(0)=0$,so we have $f(x)=x^2$in this case. To sum up,all the answers is $f(x)=0$and$f(x)=x^2$.

The only answers are $f(x) \equiv 0$ and $f(x) \equiv x^2$. These are easily seen to work so we prove they are the only ones. If $P(x,y)$ denotes the assertion then $P(0,0) \implies f(0) = 0$ $P(x,0) \implies \boxed{f(x^2) = f(f(x))}$. $P(x,1)$ gives $f$ even. So we can rewrite $P(x,y)$ and $P(y,x)$ in the more verbose but symmetric form: \begin{align*} f(x^2+f(y)) = f(y^2+f(x)) &= f(x^2) + f(y^2) + 2f(xy) \\ &= f(f(x)) + f(f(y)) + 2f(xy). \end{align*} Claim: Whenever $f(a) = f(b)$, we have $f(ax) = f(bx)$ for any $x \in {\mathbb R}$. Proof. If $c = f(a) = f(b)$ then \begin{align*} P(x,a) \implies f(x^2+c) = f(c) + f(f(x)) + 2f(ax) \\ P(x,b) \implies f(x^2+c) = f(c) + f(f(x)) + 2f(bx) \end{align*}as desired. $\blacksquare$ In particular, if $f$ has a nontrivial root it is constant, so let's assume this is not the case and prove that we must have $f(x) \equiv x^2$. Claim: We have $f \ge 0$ everywhere. Proof. Suppose $f(s) = -t^2 \ge 0$ where $s,t \ge 0$, then \[ P(s,t) \implies f(s^2 + f(t)) = f(t^2 + f(s)) = f(0) = 0 \implies f(t) = -s^2. \]We then write \begin{align*} P(\sqrt{s^2+t^2}, s) &\implies f(s^2) = f(s^2+t^2) + f(s^2) + 2f(s\sqrt{s^2+t^2}) \\ P(\sqrt{s^2+t^2}, t) &\implies f(t^2) = f(s^2+t^2) + f(t^2) + 2f(t\sqrt{s^2+t^2}) \end{align*}and thus by the claim we have $f(sx) = f(tx)$ for all $s$ and $t$. In particular now $s = t$ and \[ P(s,s) \implies 0 = f(0) = f(s^2) + f(s^2) + 2f(s^2) \implies s = 0. \]$\blacksquare$ Claim: We have $f$ is injective. Proof. If $f(a) = f(b) \ne 0$ then let $r = a/b > 1$ with $f(r) = f(1)$ By repeated squaring we may even assume $r > f(1)$. Then if we let $x = \sqrt{r-f(1)}$ we get \[ P(x, 1) \implies f(1) = f(r) = f(f(x)) + f(1) + 2f(x) \]and hence $f(f(x)) + 2f(x) = 0$. However, this means we have $f(f(x)) = f(x) = 0$, or $x = 0$, and this means $r = f(1)$, contradiction. $\blacksquare$ Since $f$ is injective, from $f(x^2+f(y)) = f(y^2+f(x))$ together with $f(0) = 0$ we conclude $f(x) \equiv x^2$ as desired.

I tried to write the following solution during APMO, but I missed a subtle point: Let $P(x,y)$ denote the assertion. From $P(0,0)$ and $P(x,0)$ we get $f(0)=0$ and $f(x^2)=f(f(x))$. If $f(y)=0$ for some $y \neq 0$ then from $P(x,y)$ we get the solution $f=0$. Else $f$ is injective at $0$. If $f(x)=-x^2$ for some $x \neq 0$ then we get a contradiction from $P(x,x)$. Suppose $f(z) \neq z^2$ for some $z$. Then comparing $P(x,z^2)$ and $P(x,f(z))$ we get $$f(xz^2)=f(xf(z))$$Or $f(x)=f(kx)$ for some $k \neq -1,1$. WLOG $|k|>1$. From $P(x,y),P(y,x),P(x,ky),P(ky,x)$ we get $$f(k^2y^2+f(x))=f(x^2+f(ky))=f(x^2+f(y))=f(y^2+f(x))$$If $f(y) < 0$ for some $y$ then we can choose $x$ such that $x^2+f(y)=0$. Hence injectivity at 0 gives $$y^2+f(x)=k^2y^2+f(x)=0$$which is a contradiction since $k^2 \neq 1$. Hence $f(y) \geq 0$ for all $y$. If $f(t) < t^2$ for some $t$, then choose $x$ such that $x^2=t^2-f(t) > 0$. Then $P(x,t)$ gives $$f(t^2)=f(x^2)+f(t^2)+2f(xt) >f(t^2)$$Hence $f(t) \geq t^2$ for all $t$. But then, $f(x)=f(kx)= \dots f(k^nx) \geq k^{2n}x^2$ for any positive integer $n$, which is a contradiction for large $n$. Hence, we get the solution $f(z)=z^2$ for all $z$.

Supercali wrote: I tried to write the following solution during APMO, but I missed a subtle point: Let $P(x,y)$ denote the assertion. From $P(0,0)$ and $P(x,0)$ we get $f(0)=0$ and $f(x^2)=f(f(x))$. If $f(y)=0$ for some $y \neq 0$ then from $P(x,y)$ we get the solution $f=0$. Else $f$ is injective at $0$. If $f(x)=-x^2$ for some $x \neq 0$ then we get a contradiction from $P(x,x)$. Suppose $f(z) \neq z^2$ for some $z$. Then comparing $P(x,z^2)$ and $P(x,f(z))$ we get $$f(xz^2)=f(xf(z))$$. After comparing $P(x,z^2)$ and $P(x,f(z)$, how can we get that equation?

$P(x, z^2)$ implies $$f(x^2+f(z^2))=f(f(x))+f(z^4)+2f(xz^2)=f(f(x))+f(f(f(z))) +2f(xz^2)$$$P(x,f(z))$ implies $$f(f(x)) +f(f(f(z))) +2f(xf(z)) =f(x^2+f(f(z)))=f(x^2+f(z^2))$$Which implies the proposition

APMO 2019 P5 wrote: Determine all the functions $f : \mathbb{R} \to \mathbb{R}$ such that \[ f(x^2 + f(y)) = f(f(x)) + f(y^2) + 2f(xy) \]for all real numbers $x$ and $y$. Solution. The only solutions are $f(x)=x^2$ and $f(x)=0.$ It can be easily checked that these satisfy the functional equation. We prove that these are the only solutions. It is easy to see that the only constant solution is $f(x)=0.$ So assume that $f$ is non-constant. Let $P(x,y)$ denote the given functional equation. Call a function $g$ almost injective if it satisfies $g(a)=g(b)\implies |a|=|b|.$ Consider the following claim: Claim 1. If $f(c)=0$ then $c=0.$ Proof. Note that \[P(0,0)\implies f(f(0))=f(f(0))+f(0)+2f(0)\implies f(0)=0.\]Therefore, $P(x,0)\implies f(x^2)=f(f(x)).$ Now, \[P(x,c)\implies f(x^2)=f(f(x))+f(c^2)+2f(xc)\implies 0=f(x^2)+2f(xc)\]which means $f$ is constant unless $c=0.$ Hence the claim. $\square$ Claim 2. If $f(a)=f(b)$ then $f(x)=f(\tfrac{a}{b}\cdot x)=f(\tfrac{b}{a}\cdot x)$ holds for all $x\in\mathbb R.$ Proof. Using $P(a,0)$ and $P(b,0)$ we obtain $f(a^2)=f(b^2).$ Now, \begin{align*} P(x,a) &\implies f(x^2+f(a))=f(x^2)+f(a^2)+2f(ax)\\ P(x,b) &\implies f(x^2+f(b))=f(x^2)+f(b^2)+2f(bx) \end{align*}Therefore it follows that $f(ax)=f(bx).$ Hence the claim. $\square$ Suppose $f$ is almost injective. We already know $f(x^2)=f(f(x))$ hence $f(x)\in\{x^2,-x^2\}~\forall~x\in\mathbb R.$ Let $r\in\mathbb R$ such that $f(r)=-r^2$, then $P(r,r)\implies f(r^2+f(r))=4f(r^2)\implies f(r^2)=0$ now using Claim 1, we conclude that $r=0.$ Thus if $f$ is almost injective, then it follows that $f(x)=x^2$ for all $x\in\mathbb R,$ which indeed is a solution. So now let us assume that $f$ is not almost injective, which means by using Claim 2, there exists $\lambda\in\mathbb R$ such that $f(x)=f(\lambda x)$ where $\lambda\notin\{0,1,-1\}.$ Without loss of generality assume that $|\lambda|>1.$ Claim 3. $f(x)\ge 0$ for all $x\in\mathbb R.$ Proof. Assume the contrary that there exists a $t$ such that $f(t)<0.$ Note that \[P(\lambda x,y)\implies f(\lambda^2 x^2+f(y))=f(\lambda^2 x^2)+f(y^2)+2f(\lambda xy)=f(x^2)+f(y^2)+2f(xy)=f(x^2+f(y))\]So using Claim 1, \[P\left(\lambda\sqrt{-f(t)},t\right)\implies f(-\lambda^2f(t)+f(t))=f(-f(t)+f(t))=0\implies f(t)(1-\lambda^2)=0\implies f(t)=0\]which is a contradiction. Hence the claim. $\square$ Claim 4. $f(x)\ge x^2$ for all $x\in\mathbb R.$ Proof. Suppose there exist a $k$ such that $f(k)<k^2.$ Then consider the following, \begin{align*} P\left(\sqrt{k^2-f(k)},k\right)&\implies f(k^2)=f(k^2-f(k))+f(k^2)+2f\left(k\sqrt{k^2-f(k)}\right) \\ &\implies f(k^2-f(k))+2f\left(k\sqrt{k^2-f(k)}\right)=0 \end{align*}So using Claim 3 and Claim 1, we get that $f(k^2-f(k))=0\implies f(k)=k^2$. Contradiction ! Hence the claim. $\square$ Now, observe that \[f(x)=f(\lambda^nx)\ge \lambda^{2n}x^2.\]Since $|\lambda|>1$ so taking $n$ very large we get a contradiction. And we are done. $\blacksquare$

yay i finally solved this! uwu here's a solution:

personally this felt like tricky but really nice algebra. Harder than something like ISL 2016 A4, but not too hard (probably like imo 2 level) and probably easier than something like IMO 2017 2. congrats to proposer! also despite the fact that geo is by far my best subject APMO 2019 3 was a LOT harder than this

First we start with the obvious substitutions: $P(0,0): f(0)=0$ $P(x,0): f(x^2)=f(f(x))$. $P(x,y)-P(y,x): f(xy)=f(-xy)=>$ $f$ is even. It's pretty clear here that $f \equiv x^2$ will be the only non-trivial solution, and thus the whole problem boils down to proving injectivity for $x \geq 0$. Naturally, say for some $t \neq +-y$, take some $f(y)=f(t)$. Thus $f(x^2+f(y))=f(x^2+f(t))=>f(xt)=f(xy)$. As a corollary, this would imply $...=f((\frac ty)^2 \cdot y)=f(y)=f(t)=f(t \cdot \frac{y}t)=f(t \cdot (\frac{y}t)^2)=... =>f(y)=f(t)=f(t \cdot (\frac yt)^n)$, for integers $n$. Another important result is if $f(m)=0 =>P(x,m): f(mx)=0=>m=0$ or $f \equiv 0$. Thus from now on assume $f \not \equiv 0$. Thus we can keep $f(y)$ constant while making $y$ as large as we want, as if $t>y$, then we know $f(y)=f(y \cdot (\frac ty)^n)$, and we can make $n$ as large as we want, and if $t<y$, then we can make $n$ as small as want, hence we can, for some $x$, make $x^2-f(x)>0=>\sqrt{x^2-f(x)}$ is defined. Hence $P(\sqrt{y^2-f(y)},y): f(y^2)=f(y^2-f(y))+f(y^2)+2f(\sqrt{y^4-y^2f(y)}) =>f(y^2-f(y))+2f(\sqrt{y^4-y^2f(y)})=0$. Notice that, as we assumed $f(y)=f(t)$, $ t \neq +-y$, and that we can make $y^2-f(y)$ as large as we want, and $f(m)=0$ iff $m=0$, we can't have $f(y^2-f(y))$ or $f(\sqrt{y^4-y^2f(y)})=0$, thus we get through our assumption that there must exist some $y_0$ such that $f(y_0)<0$. Thus our problem has further boiled down to proving that $f$ can't have an output which is negative. Well, let's assume now that $f(y_0)=-k^2$. A natural exploitation of symmetry gives us, as $f(x^2+f(y))=f(y^2+f(x))$: $f(y_0^2+f(k))=f(k^2+f(y_0))=0 =>f(k)=-y_0^2$. Thus $f(y_0^2)=f(f(y_0))=f(-k^2)=f(k^2)$. It'd be a crime not to exploit the oodles of symmetry this provides us: $P(\sqrt{k^2+y_0^2},y_0)-P(\sqrt{k^2+y_0^2},k): f(k \sqrt{k^2+y_0})=f(y_0 \sqrt{k^2+y_0^2})=>f(k)=f(y_0)=>y_0=+-k$. Thus $f(y_0)=-y_0^2$- but $P(y_0,y_0)$ gives us $f(y_0^2+f(y_0))=4f(y_0^2)=>y_0=0=>$ Contradiction to $f(y_0)<0$. Hence we see that we can't have $f(y)=f(t)$ if $y \neq +-t$, hence $f(f(x))=f(x^2)=>f(x)=+-x^2$, for some $x$. Now say that $f(x_0)=-x_0^2$. $P(x_0,x_0)$ directly gives us $f(0)=0=4f(x_0)^2=>x_0=0$. Hence we see that $f(x)=x^2$ is the only possible non-trivial solution, and it of course works. Thus our solutions are: $f \equiv x^2$ and $f \equiv 0$. Fun problem to solve, for sure, but this deserves to be more of a P3, or maybe a P4... nonetheless, cool problem, and as @above pointed out, very comparable difficulty-wise to recent IMO 2/5 FEs.

Solution Let $P (0,0),P (a,0)\implies f (f (a))=f (a^2),f (0)=0$ If $f (x)=f (y)\implies f (xb)=f (yb)\implies f (a)=0\implies a=0$ or $f (a) $ is always $0$, which is a solution. If $\frac {x}{y}\neq\pm 1\implies f (at^n)=f (a)$ $\forall a>1, |t|>1$ $P (at^n,b)\implies f (aa t^{2n}+f (b))=f (aa+f (b)) $, set $f (y)=0$ If $f (b)<0$ take $a=-\sqrt {b} -><-\frac {aat^{2n}+f (b)}{aa+f (b)} $ can be any real number from $1$ to $t^{2n} $. Because $t^{2n} $ is unbounded. We get $f (a) $ is identical if $a \neq 0$. $P (1,1)\implies f (a)=0$ Hence, $f (a)=a^2$ or $-a^2$ if $f (a)=-a^2$, $P (a,a)\implies a=0$ $\implies f (a)=a^2$ $\forall a\in\mathbb R$.

We claim that the only solutions are $f(x) \equiv 0$ and $f(x) \equiv x^2$. It's easy to see both work by plugging them into the given equation. We now prove these are the only solution. Let $P(x, y)$ be the assertion. $P(0, 0) \implies f(0) = 0, P(x, 0) \implies f(f(x)) = f(x^2)$. Comparing $P(x, 1)$ and $P(-x, 1)$ gives that $f$ is even. Suppose $f(a) = f(b)$. Then comparing $P(x, a)$ and $P(x, b)$ gives $f(ax) = f(bx)$. Thus, if $f(a) = f(0) = 0$ for $a \neq 0$, then $f$ is identically zero. Suppose $f$ isn't identically zero for the rest of the proof. If $f(a) = f(b)$, and $|a| \neq |b|$, then comparing $P(b/a*x, y)$ and $P(x, y)$ gives, for $x^2 \neq 0$, $f((x^2+f(y))/((bx/a)^2+f(y))) = f(((bx/a)^2+f(y))/(x^2+f(y))) = f(1)$. Note that $b/a$ can be replaced by $(b/a)^n$ for any positive integer $n$. Choose $f(y) \neq 0$. Then for $c > 1$, set $x^2 = (c-1)f(y)/(b^2/a^2-c)$, where $|b| > |a|$ or $|b| < |a|$ according as $f(y) > 0$ or $f(y) < 0$ respectively, where $b/a$ can be replaced by $(b/a)^n$ for large enough positive integer $n$ to make the assignment valid. Then this along with evenness give $f(d) = f(1)$ for all $d \neq 0$. Choosing $xy \neq 0, x^2+f(y) > 0$ we get $f(1) = 0$ from $P(x, y)$, a contradiction. Therefore $f(a) = f(b) \iff |a| = |b|$. Thus $|f(x)| = x^2$. Then $P(1, 1)$ implies $f(1) = 1$. Then $P(1, y)$ implies that $f(y^2) = -y^4$ only if $y^2 = 1/2$, i.e. $f(x) = x^2$ for all $x$ with $|x| \neq 1/2$. $P(1/\sqrt2, \sqrt2) \implies f(1/2) = 1/4$, and we're done.

Solved with nprime06 We claim the only solutions are $f(x) = 0$ and $f(x) = x^2.$ Let $P(x,y)$ denote the assertion. Then, $P(0,0) \iff f(0) = 0$ and $P(x,0) \iff f(x^2) = f(f(x))$ and $P(x,y)-P(y,x) \iff f(xy) = f(-xy)$ Lemma 1: $f$ is injective at $0$ Proof. Assume $f\not\equiv 0$ and that there exists a $k\ne 0$ such that $f(k)=0$. Then by plugging $P(x,k)$ we see that $f(xk)=0$, and varying $x$ we have $f\equiv 0$, a contradiction. Hence $f(k)=0$ iff $k=0$. Let there be $f(a)=f(b)$. Then comparing $P(a,x)$ and $P(b,x)$ we have $f(ax)=f(bx)$ (1) for all real $x$. We present two more additional lemmas: Lemma 2: $f(x)\ge 0$ for all $x>0$ Proof. Suppose not; ie. there exists some $a$ such that $f(a)<0$. Let $b=\sqrt{-f(a)}$ ie. $b^2+f(a)=0$. Because $f(x^2+f(y))=f(y^2+f(x))$, we note that $f(a^2+f(b))=f(b^2+f(a))\iff f(a^2+f(b))=0\iff f(b)=-a^2$ from lemma 1. Comparing $P(\sqrt{a^2+b^2},a)$ and $P(\sqrt{a^2+b^2},b)$ we have that $0=f(a^2+b^2)+2f(a\sqrt{a^2+b^2})$ and $0=f(a^2+b^2)+2f(b\sqrt{a^2+b^2})$. Subtracting gives $f(a\sqrt{a^2+b^2})=f(b\sqrt{a^2+b^2})$; using (1) we have $f(a)=f(b)\iff -b^2=-a^2\iff a=b$, so $f(a)=-a^2$. Plugging $P(a,a)$ we have $0=f(a^2)+f(a^2)+2f(a^2)\implies a=0$, by lemma 1. A contradiction Lemma 3: $f(x)\ge x^2$ ie. stronger version of lemma 2. Proof. Suppose the contrary ie. there exists some $x>0$ such that $f(x)<x^2$. $P(\sqrt{x^2-f(x)},x)$ gives\[f(x^2)=f(x^2-f(x))+f(x^2)+2f(\sqrt{x^4-x^2f(x)}). \]Since $f(x)\ge 0$ by Lemma 2 we now see that\[f(x^2-f(x))+2f(\sqrt{x^4-x^2f(x)})=0\iff f(x^2-f(x))=f(\sqrt{x^4-x^2f(x)})\iff 0=x^2-f(x)=\sqrt{x^4-x^2f(x)}\iff f(x)=x^2\]by lemma 1. This is a contradiction, thus proved. Lemma 4: $f(x)$ is injective over positives (evenness extends this to the negatives). Proof. Suppose there are $a>b>0$ such that $f(a)=f(b)$, and let $\sigma=\frac ab>1$. Take some arbitrary real $c$ and some constant $d$; by lemma 3 we have that $f(a^c)=f(b^c)\iff f(d)=f(\sigma^cd)\ge \sigma^{2c}d^2$. then let $c\to \infty$; thus $f(d)$ becomes very large; varying $d$ we see that $f(r)$ for any real number $r$; contradiction. Thus $f$ is injective. To finish, note that $f(f(x)) = f(x^2)$ and $f$ is injective so $f(x) = x^2$ as desired.

The only solutions are $\boxed{f\equiv x^2}$ and $\boxed{f\equiv 0},$ which can be easily seen to work. Henceforth assume $f\not\equiv x^2,0.$ Let $P(x,y)$ denote the assertion. Remark that $P(0,0)$ gives $f(f(0)=f(f(0))+3f(0),$ so $f(0)=0.$ $\textbf{Claim: }$ For any real $c\ne 0,1,$ if $f(a)=f(ca)$ for some $a,$ then $f(x)=f(cx)$ for all $x.$ $\emph{Proof: }$ $P(a,0)$ and $P(ca,0)$ yield $$f(c^{2}a^{2})=f(f(ca))=f(f(a))=f(a^2).$$Additionally, from $P(x,a)$ and $P(y,a),$ we have \begin{align*} f(f(x))+f(a^2)+2f(ax) &=f(x^2+f(a))\\ &=f(x^2+f(ca))\\ &=f(f(x))+f(c^{2}a^{2})+2f(cax) \end{align*}Therefore, $f(ax)=f(cax)$ for all $x,$ giving the desired statement. $\blacksquare$ Let $S$ be the set of $c$ as above. $\textbf{Claim: }$ $S$ is nonempty. $\emph{Proof: }$ $P(x,0)$ gives $f(x^2)=f(f(x)).$ Therefore, if $f$ is injective, then $f\equiv x^2,$ contradiction. Thus, $f$ is not injective, so there exist $a,c$ with $c\ne 0,1$ and $f(a)=f(ca).$ $\blacksquare$ $\textbf{Claim: }$ $S\ne \{-1\}.$ $\emph{Proof: }$ If $S=\{-1\},$ then $P(x,0)$ implies that $f(x)=\pm x^2.$ But we assumed $f\not\equiv x^2$ and we can check that $f\equiv -x^2$ doesn't satisfy the FE, so we are done. $\blacksquare$ Now fix $c\in S.$ Since $c\in S\iff\frac{1}{c}\in S$ and $c\ne\pm 1,$ we may assume $|c|<1.$ The assertions $P(cx,y)$ and $P(x,y)$ yield \begin{align*} f(c^{2}x^{2}+f(y)) &= f(f(cx))+f(y^2)+2f(cxy)\\ &= f(f(x))+f(y^2)+2f(xy)\\ &= f(x^2+f(y)). \end{align*}Hence, for all $x,y,$ we have $M=\frac{c^{2}x^{2}+f(y)}{x^2+f(y)}\in S.$ Since $f\not\equiv 0,$ we can choose $y$ such that $f(y)\ne 0.$ If $f(y)<0,$ then $M$ takes all real values in the interval $(c^2,\infty)$ as $x$ varies. Since $c^2<1,$ this means $f(a)=f(b)$ for all $|b|>|a|.$ This implies that $f$ is constant on $\mathbb{R}^{+}$ and $\mathbb{R}^{-},$ but we can check that no such functions work. If $f(y)>0,$ then apply the same logic for $c'=\frac{1}{c}.$ Then $M'=\frac{c'^{2}x^{2}+f(y)}{x^2+f(y)}$ takes all real values in the interval $(1,c'^2)$ as $x$ varies. This means $f(a)=f(b)$ for all sufficiently small $b$ with $|b|>|a|,$ which again gives $f$ is constant on $\mathbb{R}^{+}$ and $\mathbb{R}^{-}.$ Having reached a contradiction, we conclude that $f\equiv 0$ and $f\equiv x^2$ are the only solutions.

We claim that $f \equiv 0$ and $f \equiv x^2$ are the solutions, which clearly work. Indeed, if $f(x)=c$ for all $x$, then $c=3c \implies c=0$, so assume that $f$ is non-constant. Let $P(x,y)$ the assertion $f(x^2 + f(y)) = f(f(x)) + f(y^2) + 2f(xy)$. $\implies P(0,0)$ gives us that $3f(0)=0 \implies f(0)=0 (*)$ Now, we prove that there exists an $M \in \mathbb{R}$ such that $f(M)>0$. Suppose that $f(x) \leq 0$ for all $x$. Let $g(x)=-f(x) \implies g(x^2-g(y))=g(-g(x))+g(y^2)+2g(xy) \implies$ picking $x=\sqrt{g(y)}$, since $g(x) \geq 0$, we must have that $f(y^2)=0$ for all $y$ $\implies f(x^2)=0 \forall x \in mathbb{R} \implies$ from $(*)$, $P(0,y)$ yields $f(f(y))=f(y^2)=0 (**)$. Subsituting at the original equation, $f(x^2+f(y))=2f(xy) \implies$ if $y <0$, choose $x$ large enough such that $x^2+f(y)>0 \implies 0=f(x^2+f(y))=2f(xy) \implies f(xy)=0$ and $xy<0$, so $f(x)=0 \forall x \in \mathbb{R}$, a contradiction. Hence, there exists an $M \in \mathbb{R}$ such that $f(M)>0$ $(\heartsuit)$. Now, we prove the following claim. Claim: $f(a)=f(b) \implies |a|=|b|$. Proof: Let $S=\{s \in \mathbb{R}|f(sx)=f(x) \forall x \in \mathbb{R}\}$. Observe that if $u,v \in S \implies uv \in S$. $(\spadesuit)$ Now, assume that there are $a,b \in \mathbb{R}$ with $|a| \neq |b|$ such that $f(a)=f(b)$. $\implies$ comparing $P(x,a)$ with $P(x,b)$, we have that $f(a^2)+2f(ax)=f(b^2)+2f(bx)$, but from $(**)$, $f(a^2)=f(b^2) \implies f(ax)=f(bx) \forall x \in \mathbb{R} \implies \frac{a}{b} \in S$. WLOG $s=\frac{a}{b}>1 \implies$ from $(\spadesuit)$, $s^n \in S$ for all $n \in \mathbb{N}$. Hence, we basically proved that if $f(a)=f(b) \implies \frac{a}{b} \in S$. Now, comparing $P(xs,y)$ with $P(x,y)$ we have that $$f(x^2s^2+f(y))=f(x^2+f(y)) \implies \frac{x^2s^2+f(y)}{x^2+f(y)} \in S \quad \forall x,y \in z\mathbb{R}$$$\implies$ from $(\heartsuit)$, choose $y$ to be $M \implies \frac{x^2s^2+f(M)}{x^2+f(M)} \in S \forall x \in \mathbb{R} (\heartsuit \heartsuit)$, and a positive real number $t$ is equals to $\frac{x^2s^2+f(y)}{x^2+f(y)} \iff x^2=\frac{f(M)(1-t)}{t-s^2} \implies$ if $t>1$, choose $n$ large enough such that $s^{2n}>t$ (this is possible since $s>1$). $\implies$ for all $t \in \mathbb{R}_{>0}$, it's possible to choose $x \in \mathbb{R}$ and $s \in S$ such that $t= \frac{x^2s^2+f(M)}{x^2+f(M)}$, so from $(\heartsuit \heartsuit)$, $S= \mathbb{R}_{>0}$. Thus, $f(x)= \lambda_1$ for all $x>0$ and $f(x)=\lambda_2$ for all $x<0$. From $(**)$, $$f(x^2+f(y))=f(x^2)+f(y^2)+2f(xy)=2(\lambda_1+f(xy))$$$\implies$ choosing $y$ arbitrarily negatibe and $x \in \mathbb{R}_{>0}$ such that $x^2+\lambda_2>0$, we have that $\lambda_1=-2\lambda_2$, and choosing $x,y$ to be positive, we have that $\lambda_1=\lambda_2=0 \implies f \equiv 0$, a contradiction by assumption. $\qquad \square$ Now, $P(x,0)$ yields $f(f(x))=f(x^2) \implies f(x)=x^2$ or $f(x)=-x^2$. If $f(a)=-a^2$ for some nonzero $a$, $P(a,a)$ gives us that $0=4f(a^2)$, a contradiction. Therefore, $f(x)=x^2 \quad \forall x \in \mathbb{R}$ is the only solution if $f$ is non-constant, as desired. $\blacksquare$

The only solutions are $f(x) = x^{2}$ and $f(x) = 0$. It's easy to see that these both work. Let $P(x,y)$ denote the expression. By $P(0,0)$, we have $f(0) = 0$. By $P(0,y)$, we get $f(f(y)) = f(y^{2})$, so now let $P(x,y)$ be the expression $f(x^{2} + f(y)) = f(x^{2}) + f(y^{2}) + 2f(xy)$. By symmetry, $f(x^{2}+f(y)) = f(y^{2} + f(x))$. Furthermore, $P(-x, y)$ results in $f(-x) = f(x)$. I claim that $f$ is injective for the positives (that is, if $f(a) = f(b), a, b\geq 0$, then $a=b$). If $f(a) = f(b)$, by $P(1,a)$ and $P(1,b)$, we have $f(a^{2}) = f(b^{2})$. WLOG, $b\leq a$. Then, considering $P(x,a), P(x,b)$, we have \[f(x^{2} + f(a)) = f(x^{2} + f(b)) \Rightarrow f(x^{2}) + f(a^{2}) + 2f(ax) = f(x^{2}) + f(b^{2}) + 2f(bx)\]This means $f(ax) = f(bx)$ for all $x$. Consider $P(\frac{a}{b}x, y)$. We have \[f\left(\frac{a^{2}}{b^{2}}x^{2} + f(y)\right) = f\left(\frac{a^{2}}{b^{2}}x\right) + f(y^{2}) + 2f\left(\frac{a}{b}xy\right)\]\[= f(x^{2}) + f(y^{2}) + 2f(xy) = f(x^{2} + f(y)) \Longrightarrow f\left(x^{2} + \frac{b^{2}}{a^{2}}f(y)\right) = f(x^{2} + f(y))\]Because if $f(a) = f(b)$, then $f(a^{2}) = f(b^{2})$, which also means $f(a^{2}x) = f(b^{2}x)$, this also means \[f\left(x^{2} + \frac{b^{2^{k}}}{a^{2^{k}}} f(y)\right) = f(x^{2} + f(y))\]Since $b < a$, we can fix some $k$ and fix $y$, then this will mean that for all real numbers $t$ at least $\left(\frac{b}{a}\right)^{2^{k}}f(y)$, then \[f\left(t + \left(1 - \frac{b^{2^{k}}}{a^{2^{k}}}\right)f(y)\right) = f(t)\]As $k$ goes to infinity, then $\left(\frac{b}{a}\right)^{2^{k}}f(y)$ goes to $0$, so as $k$ goes to infinity, we can say that $f(x) = f(x+r)$, for some fixed number $r$ and $x > 0$. Now, since $f(x) = f(x+r)$, this means for all $y$, we will have $f(xy) = f((x+r)y)$. Since $\frac{x+r}{x}$ spans from $1$ to $\infty$, this means that $f(1) = f(y)$ for all $y > 1$. It also means that $f(1) = f(y)$ for all $0 < y < 1$, so $f(x) = c$ for all $c\neq 0$. Unless $c$ is $0$ (which we've already covered), this gives a contradiction since $c\neq c + c + 2c$. Therefore, if $a,b \geq 0$ and $f(a) = f(b)$, then $a = b$. We can extend this by using $f(a) = f(-b)$ to say that if $f(a) = f(b)$, then $|a| = |b|$. Because $f(x^{2} + f(y)) = f(y^{2} + f(x))$, this means $|x^{2} + f(y)| = |y^{2} + f(x)|$. I claim that $x^{2} + f(y) = y^{2} + f(x)$. If $x^{2} + f(y) = -(y^{2} + f(x))$, then $x^{2} + f(x) = -y^{2} - f(y)$. Clearly $|x|\neq |y|$. By $P(x,x)$, we have $f(x^{2} + f(x)) = 4f(x^{2})$. If $x^{2} + f(x) = -y^{2} - f(y) = c$, then \[4f(x^{2}) = f(x^{2} + (c-x^{2})) = f(c) = f(-c) = f(y^{2} + f(y)) = 4f(y^{2})\]By injectivity, this means $|x| = |y|$, a contradiction. Therefore, for all $x,y$, we have $x^{2} + f(y) = y^{2} + f(x)$. Rearranging, we must have $f(x) - x^{2} = c$ for some constant $c$, so $f(x) = x^{2} + c$. Since $f(0) = 0$, we must have $c = 0$, so $f(x) = x^{2}$ and $f(0)$ are our only two solutions.

IndoMathXdZ wrote: Determine all the functions $f : \mathbb{R} \to \mathbb{R}$ such that \[ f(x^2 + f(y)) = f(f(x)) + f(y^2) + 2f(xy) \]for all real numbers $x$ and $y$. Solved with MathLuis. Let \(P(x,y)\) assert the above functional equation. \(P(0,0)\) tells us that \(f(0)=0\) and \(P(x,0)\) tells us that \(f(f(x))=f(x^2)\). We now prove some claims. Claim 1. \(f(a)=0\iff a=0\) or \(f\) is identically zero. Proof. \(P(x,a)\) implies that \[f(a^2)+2f(ax)=0\]but \(f(a^2)=f(f(a))=f(0)=0\) so \(f(ax)=0\) for all \(x\) implying that \(f\) is identically zero or \(a=0\) as wanted. Claim 2. \(f(a)=f(b)\) implies that \(f(ax)=f(bx)\) for all \(x\). Proof. Comparing \(P(x,a)\) and \(P(x,b)\) does the job. Claim 3. The image of \(f\) is nonnegative. Proof. Otherwise, there exist \(s\) and \(t\) for which \(f(s)=-t^2\). Since \[0=f(f(s)+t^2)=f(f(t)+s^2)\]by claim 1 we can say that \(f(t)=-s^2\). Now, comparing \(P(\sqrt{s^2+t^2},s)\) and \(P(\sqrt{s^2+t^2},t)\) gives us that \[f(s\sqrt{s^2+t^2})=f(t\sqrt{s^2+t^2})\]and by claim 2, we see that \(f(s)=f(t)\), so \(s=\pm t\) (since \(s^2=t^2\)). Then \(P(t,t)\) gives us that \[0=f(t^2)+f(t^2)+2f(t^2)\]implying that \(t=0\) and we are done. Claim 4. \(f\) is injective over positive reals. Proof. Assume that \(f(a)=f(b)\) is non-zero. Clearly, \(f\left(\frac{a}{b}\right)=f(1)\). Now we will take \(P\left(\sqrt{\frac{a}{b}-f(1)},1\right)\) to get that \[f\left(f\left(\frac{a}{b}-f(1)\right)\right)+2f\left(\sqrt{\frac{a}{b}-f(1)}\right)=0\]and since \(f\) is nonnegative, we get that both terms have to be zero, and by claim 1, we see that \(a=b\) as desired. Since \[f(x^2+f(y))=f(y^2+f(x))\]we see that \(f(x)-x^2=f(y)-y^2=c\), and with \(f(0)=0\) this gives \(f(x)=x^2\) which indeed fits.

The main idea of this solution is very similar to that of RMM 2019/5. Anyways, here goes the solution. The answer is $f(x) \equiv x^2$ and $$f(x) \equiv \begin{cases} 0 \qquad \text{if } x=0 \\ t \qquad \text{otherwise} \end{cases}$$for some constant $t \in \mathbb R$. These clearly work. Now we show these are the only solutions. Let $P(x,y)$ be the given assertion. $P(0,0)$ gives $f(0) = 0$. $P(x,0)$ gives $$f(f(x)) = f(x^2) \qquad{(1)}$$Now comparing $P(x,y)$ and $P(-x,y)$ we obtain $f$ is even. Claim (Key Claim): If $f(a) = f(b)$ for some $a,b \in \mathbb R$ with $a^2 \ne b^2$, then $f$ must be of the second form. proof: WLOG $a \ne 0$. Comparing $P(x,a)$ and $P(x,b)$ yields $f(x) = f(\alpha x) ~ \forall ~ x \in \mathbb R$ where $\alpha = b/a$. Then comparing $P(x,y)$ and $P(\alpha x,y)$ gives $$f(\alpha^2 x^2 + f(y)) = f(x^2 + f(y)) = f(\alpha^2 x^2 + \alpha^2 f(y))$$As $\alpha^2 \ne 1$ and $f(y) \ne 0$ for some $y$, so we conclude that there are constants $c,M > 0$ such that $$f(x) = f(x+kc) ~ \forall ~ x \ge M , k \in \mathbb N$$Now for $y \ge M$ and $k \in \mathbb N$ comparing $P(x,y)$ and $P(x,y+kc)$ we obtain $$f(xy) = f(x(y+kc)) ~ \forall ~ x \in \mathbb R, y \ge M, k \in \mathbb N$$Now it is not hard to see that $\Big(xy,x(y+kc) \Big)$ can become any pair $(a,b)$ with $a,b \in \mathbb R^+$ and $a < b$. Since $f$ is even, so this yields $$f(x) \equiv \begin{cases} 0 \qquad \text{if } x=0 \\ t \qquad \text{otherwise} \end{cases}$$for some constant $t \in \mathbb R$. $\square$ So now assume $f$ is not of the second form. Then $f(a) = f(b)$ implies $a^2 = b^2$. Using this in $(1)$ yields $$f(x)^2 = x^4 ~ \forall ~ x \in \mathbb R \qquad{(2)}$$FTSOC, let $f(\beta) = -\beta^2$ for some $ \beta \ne 0$. $P(\beta,\beta)$ gives $f(\beta^2) = 0$, but this then contradicts $(2)$. $\blacksquare$

Let $P(x,y)$ denote the given assertion. $P(0,0): f(f(0))=f(f(0))+3f(0)\implies f(0)=0$. $P(0,x): f(f(x))=f(x^2)$. Definition: Partially injective means if $f(a)=f(b)$ then $a^2=b^2$. Claim: If $f$ is partially injective, then $\boxed{f(x)=x^2}$, which works. Proof: Suppose $f$ is partially injective. Then $|f(x)|=|x^2|$ for all $x$. If $f(x)=-x^2$ for some $x\ne 0$, then $P(x,x): f(f(x))+3f(x^2)=0\implies 4f(x^2)=0$. Since $f$ is partially injective and $f(0)=0$, we have $f(x^2)=0\implies x^2=0\implies x=0$, a contradiction. Thus, $f(x)=x^2$ for all $x$ $\blacksquare$. Claim: If $f(a)=0$ for some $a=0$, then $\boxed{f\equiv 0}$, which works. Proof: Note that $f(a^2)=f(f(a))=0$. So $P(x,a): f(x^2)=f(f(x))+2f(xa)$. Thus, $f(xa)=0\implies f\equiv 0$. $\blacksquare$ Henceforth assume if $f(x)=0$, then $x=0$. Claim: $f$ is even. Proof: We have $f(f(x))=f(x^2)=f((-x)^2)=f(f(-x))$. Now $P(-x,y): f(x^2+f(y))=f(f(x))+f(y^2)+2f(-xy)$. Now $f(xy)=f(-xy)$, so $f$ is even. $\blacksquare$ $P(y,x): f(y^2+f(x))=f(f(y))+f(x^2)+2f(xy)=f(f(x))+f(y^2)+2f(xy)$. Comparing with the original FE gives \[f(x^2+f(y))=f(y^2+f(x))=f(x^2)+f(y^2)+2f(xy)=f(f(x))+f(f(y))+2f(xy)\]Call this new assertion $Q(x,y)$. Claim: $f$ is partially injective. Proof: Suppose not and $f(a)=f(b)$, with $a^2\ne b^2$. By evenness, we can WLOG $a>b>0$. Subclaim 1: $f(ax)=f(bx)$ for all reals $x$. Proof of subclaim: Note that $f(x^2+f(a))=f(f(x))+f(f(a))+2f(xa)$ and $f(x^2+f(a))=f(f(x))+f(f(a))+2f(xb)$ from $Q(x,a)$ and $Q(x,b)$. Thus, $2f(xa)=2f(xb)\implies f(ax)=f(bx)$. $\square$ Now we have $f(x)=f\left(\frac{a}{b}x\right)$. Let $\frac{a}{b}=k$. Then $f(x)=f(kx)$. Subclaim 2: $f(x)\ge 0$ for all reals $x$. Proof of subclaim: Suppose not and $f(m)=-n^2$. From $Q(n,m)$ we have $0=f(m^2+f(n))\implies f(n)=-m^2$. Now from $Q(n,km)$, we have $0=f(k^2m^2+f(n))=0\implies f(n)=-k^2m^2$. But $-m^2\ne -k^2m^2$ $\square$. Let $z=\sqrt{kf(x)-f(x)}$. Note that $z>0$. $Q\left(x,z\right): f(kf(x))=f(f(x))+f(f(z))+2f(xz)$. Since $f(kf(x))=f(f(x))$, we have \[f(f(z))+2f(xz)=0\implies f(f(z))=2f(xz)=0\implies z=0,\]a contradiction. $\blacksquare$

$P(0,0) : f(0) = 0$. $P(x,0) : f(x^2) = f(f(x))$. $P(x,y) , P(y,x) : f(x^2+f(y)) = f(y^2+f(x))$. Let $a,b$ be such that $f(a) = f(b)$ then $P(x,a) , P(x,b) : f(xa) = f(xb)$. Case $1 : f$ is constant. Note that we have $f(0) = 0$ so $f(x) = 0$. Case $2 : f$ isn't constant. Claim $: f(x) \ge 0$. Proof $:$ Assume not then there exists $k$ such that $f(k) = -t^2$. we had $f(x^2+f(y)) = f(y^2+f(x))$ so $P(t,k) : f(k^2 + f(t^2)) = 0 \implies f(t^2) = -k^2$. $P(\sqrt{t^2+k^2},t^2) , P(\sqrt{t^2+k^2},k^2) : f(s\sqrt{s^2+t^2}) = f(t\sqrt{s^2+t^2}) \implies f(k) = f(t)$. $P(k,k) : 4f(k^2) = 0 \implies k = 0$ but $f(0) = 0$ so contradiction. Claim $: f(a) = f(b) \implies f(a^{2n}) = f(b^{2n})$. Proof : $f(a) = f(b) \implies f(f(a)) = f(f(b)) \implies f(a^2) = f(b^2) \implies f(f(a^2)) = f(f(b^2)) \implies ... \implies f(a^{2n}) = f(b^{2n})$. Claim $: f$ is injective. Proof $:$ Assume not then there exists $a,b$ such that $f(a) = f(b)$ and $a \neq b$. $f(a) = f(b) \implies f(\frac{a}{b}) = f(1)$ and WLOG assume $a > b$ Also from last claim we can assume $\frac{a}{b} > f(1)$. $P(\sqrt{\frac{a}{b}-f(1)},1) : f(\frac{a}{b}) = f(f(\sqrt{\frac{a}{b}-f(1)})) + f(1) + 2f(\sqrt{\frac{a}{b}-f(1)}) \implies f(f(\sqrt{\frac{a}{b}-f(1)})) + 2f(\sqrt{\frac{a}{b}-f(1)}) = 0 \implies 2f(\sqrt{\frac{a}{b}-f(1)}) = 0 \implies \frac{a}{b} = f(1)$ but we had $a>b$ so contradiction. so $f$ is injective now note that we had $f(x^2+f(y)) = f(y^2+f(x))$ so $x^2 + f(y) = y^2 + f(x) \implies f(y) = y^2$. Answers $: f(x) = 0, f(x) = x^2$.

Let $P(x,y)$ be the given assertion. $P(0,0)$ gives $f(0)=0.$ $P(x,0)$ gives $f(x^2)=f(f(x)).$ Comparing $P(x,1)$ and $P(-x,1)$ gives $f$ is even. $P(z,x)$ and $P(z,y)$ gives $f(xz)=f(yz)$ when $f(x)=f(y).$ Note that if $f(k)=0$ for some $k\neq 0$ then $f\equiv 0.$ Otherwise if we prove injectivity, $f(x)=x^2,$ both work. Comparing $P(x,y)$ and $P(y,x)$ gives $f(x^2+f(y))=f(y^2+f(x)).$ Claim 1: $f(x)\geq 0~~\forall x.$ Proof. Let $f(a)=-b^2$ for some $a,b\geq 0.$ It follows that $f(b)=-a^2.$ Now comparing $P(\sqrt{a^2+b^2},x_{a,b})$ forces $a=b.$ Then $P(a,a)$ gives $a=0.$ $\blacksquare$ Claim 2: $f(a)=f(b)\implies a=b.$ Proof. Otherwise, $f(a/b)=f(1).$ Induction will lead to $(a/b)^k>f(1).$ But $P(\sqrt{(a/b)^k-f(1)},1)$ gives $f(1)=(a/b)^k,$ absurd. $\blacksquare$

Interesting The answer is $f \equiv 0$ and $f(x)=x^2$, which both clearly work, so we prove that these are the only solutions. Let $P(x,y)$ denote the assertion. From $P(0,0)$ we obtain $f(0)=0$, so $P(0,x)$ gives $f(f(x))=f(x^2)$—call this fact $(\spadesuit)$. Now by using $(\spadesuit)$ and comparing $P(x,1)$ with $P(-x,1)$, we obtain $f(x)=f(-x)$, hence $f$ is even. If $f$ is injective, then $(\spadesuit)$ immediately gives $f(x)=x^2$, which is a solution, so suppose it's not. Then if $f(a)=f(b) \neq 0$ (so $a,b \neq 0$), from $P(x,a)$ and $P(x,b)$ as well as $(\spadesuit)$ we conclude that $f(ax)=f(bx)$. Since this holds for all $x \in \mathbb{R}$, we have that $f(rx)=f(x)$ for all $x$, where $r=a/b$ or $r=b/a$. Thus define $$S=\{r \mid f(rx)=f(x)~\forall x \in \mathbb{R}\}.$$Note that because $f$ is even, $r \in S \implies -r \in S$. Further, it is clear that $r \in S \implies 1/r \in S$. Finally, from $P(x,ry)$ we obtain $r \in S \implies r^2 \in S$. Suppose that $f \not \equiv 0$—if $f \equiv 0$, we are done. I will prove that $[1,\infty) \subseteq S$. Because we supposed $f$ was also not injective, we can find some $t>1 \in S$, and thus by repeatedly applying $r \in S \implies r^2 \in S$ we can find arbitrarily large elements of $S$. Let $r$ be one of these massive elements. Pick some $y$ such that $f(y) \neq 0$; by comparing $P(x,y)$ and $P(rx,y)$ we find that $f(r^2x^2+f(y))=f(x^2+f(y))$. Thus $$S \ni \frac{r^2x^2+f(y)}{x^2+y}=r^2-\frac{(r^2-1)f(y)}{x^2+f(y)}.$$By varying $x$ within $[0, \infty)$, it follows that every real in $[1,r^2)$ lies in $S$, hence $[1,\infty) \in S$ as $r^2$ is unbounded. Then by using $r \in S \implies -r,1/r \in S$, it follows that $R \setminus \{0\} \subseteq S$, which implies that $f$ is constant over $R \setminus \{0\}$, say equalling some constant $C$. Then $P(x,1)$ for $x$ large (so $x^2+f(1)\neq 0$) yields $C=4C \implies C=0$, hence we actually have $f \equiv 0$, so we're done. $\blacksquare$ EDIT: ok I just realized $f$ is never actually injective; it should be partial injectivity instead. I think this is trivial to fix.

Let $P(x,y)$ be the given F.E. $P(0,0) \implies f(0)=0$ $P(x,0) \implies f(x^{2})=f(f(x))$. Let $f(x^2+f(y))=f(x^{2})+f(f(y))+2f(xy)$ be $Q(x,y)$. Obviously, $f \equiv 0$ is a solution. So let us look at a non-constant solution. Let $t \neq 0$ be a real such that $f(t)=0$ (it might not exist). Then $f(t^{2})=f(f(t))=f(0)=0$, so $Q(x,t)$ gives us $f(xt)=0$. The last one means that $f \equiv 0$ - contradiction. So, if $x \neq 0$, then $f(x) \neq 0$. We'll prove that if $f(x_{1})=f(x_{2})$, then $|x_{1}|=|x_{2}|$. Assume the contrary and let $a$ and $b$ be such reals that $f(a)=f(b)$ and $|a|\neq |b|$. Apparently, $a$ and $b$ are not zeros, so WLOG let $k=\frac{a}{b}$ and $|k|<1, k\neq 0$. Comparing $Q(x,a)$ and $Q(x,b)$, we are given that $f(ax)=f(bx)$, so $f(x)=f(kx)$ and, more generally $f(x)=f(xk^{n})$ for $\forall x \in \mathbb{R}, n\in \mathbb{N}$. Looking at $P(x,y)$ and $P(kx,y)$, we get that $f(k^{2}x^{2}+f(y))=f(x^{2}+f(y))=f(k^{2}x^{2}+k^{2}f(y))$. In the last one we take $y=1$ and $x=k^{2}x^{2}+f(1)$ to get that $f(x)=f(x+D)$, where $D=(k^{2}-1)f(1) \neq 0$ (the same it true even when we swap $k$ with $1/k$, so we can guarantee $D<0$) and $x\geq c\geq f(1)$, so $c$ can be selected arbitrarily large, so let's take it to make sure that $c+D>0$. Now in $Q(x,y)$ we take $y=y+D$ for $y\geq c$ to get that $f(xy)=f(xy+xD)$. Let's look at the system $xy=u, xy+xD=v$. We get that $x=\frac{v-u}{D}, y=\frac{uD}{v-u}$. Let $v$ be an arbitrary positive and $u>v$. We must have $y\geq c \iff u(c+D)\leq cv \iff u\leq \frac{cv}{c+D}$, so for $u \in (v;\frac{cv}{c+D}]$ $f(u)=f(v)$. It is obviously true when $u=v$, so we can extend it to the following: if $v>0$, $u \in [v;Tv]$, then $f(u)=f(v)$, where $T=\frac{c}{c+D}>1$.The last one means that $f$ is constant in $\bigcup_{i=0}^{\infty}[vT^{i};vT^{i+1}]=[v;+\infty)$, because $\lim_{i\to \infty}T^{i}=\infty$. Now, in $P(x,y)$ with $f(f(x))=f(x^{2})$ for large enough $x,y$ we get that $f(v)=4f(v)\implies f(v)=0$, contradiction. So, we have that $|a|=|b|$. From there, we get that $|f(x)|=x^{2}$. Assume there are non-zero $X$ and $Y$ such that $f(X)=X^2$, $f(Y)=-Y^2$. Looking at $P(X,Y)$ and $P(Y,X)$, we find that $f(X^2+f(Y))=f(f(X)+Y^2)\implies f(X^2-Y^2)=f(X^2+Y^2)$. From the claim before we get that $XY=0$, contradiction. A direct check shows us that the only non-constant solution is $f(x)=x^2$.

$\bullet$ $P(0,0) \Longrightarrow f(0) = 0$ $\bullet$ $P(x,0) \Longrightarrow f(x^2) = f(f(x))$ $\bullet$ $P(1,x) - P(-1,x) \Longrightarrow f(x) = f(-x)$ $\bullet$ $P(x,y) \Longrightarrow f(x^2+f(y)) = f(y^2+f(x)) = f(x^2) + f(y^2) + 2f(xy) = f(f(x)) + f(f(y)) + 2f(xy).$ Let's say $f(a) = f(b) \Longrightarrow P(x,a) - P(x,b) \Longrightarrow f(ax) = f(bx)$ From this we can figure out that if $f$ has any nonzero roots we get that its a constant function. $\underline{f(x) = 0}$ So from now on we can assume $f(x) = 0 \Longrightarrow x = 0$ And now we wish to prove that $f(x) = x^2$ as its the only other solution. Now since $f(ax) = f(bx) \Longrightarrow f(x) = f(\frac{b}{a} x)$ let $\omega = \frac{b}{a}$ Assume there exists $t$ such that $f(t) < 0$ $P(\omega x, y) \Longrightarrow f(\omega^2 x^2 + f(y)) = f(\omega^2 x^2) + f(y^2) + 2f(\omega x y) = f(x^2) + f(y^2) + 2f(xy) = f(x^2 + f(y)) $ $P(\omega \sqrt{-f(t)}, t) \Longrightarrow f(- \omega^2 f(t) + f(t)) = f(-f(t) + f(t)) = f(0)$ After which we obtain $f(t)(1-\omega^2) = 0 \Longrightarrow f(t) = 0 \Longrightarrow t = 0$ Contradiction. $f(x) \ge 0 \hspace{5mm} \forall x$ Claim: $f$ is injective since $f(\omega x) = f(x)$ then $P(1) \Longrightarrow f(\omega) = f(1)$ and by repeated squaring we get $\omega > f(1)$ let $\alpha = \sqrt{\omega - f(1)}$ $P(\alpha, 1) \Longrightarrow f(\omega) = f(f(\alpha)) + f(1) + 2f(\alpha) \Longrightarrow f(f(\alpha)) + 2f(\alpha) = 0$ but since $f$ is a non negative function we have $f(f(\alpha)) = f(\alpha) = 0 = \alpha \Longrightarrow \omega = f(1)$ a contradiction. since we have $f(x^2) = f(f(x))$ and $f$ is injective we get our second solution $\underline{f(x) = x^2}$

Denote the assertion with $P(x, y)$. $P(0,0)$ implies that $f(f(0)) = f(f(0)) + 3f(0)$ or $f(0) = 0$. Then, $P(x,0)$ implies that $f(x^2) = f(f(x))$ and thus $P(x,1)$ means that \[ f(x^2 + f(1)) = f(x^2) + f(1) + 2f(x) \]which implies that $f$ is even. Claim: If $f(a) = f(b)$, then $f(ax) = f(bx)$ for all $x \in {\mathbb R}$ Proof. $P(x,a)$ and $P(x,b)$ imply respectively that \[ f(x^2 + f(a)) = f(f(x)) + f(a^2) + 2f(ax) \]and that \[ f(x^2 + f(b)) = f(f(x)) + f(b^2) + 2f(bx) \]Since $f(a^2) = f(f(a)) = f(f(b)) = f(b^2)$ $f(ax) = f(bx)$ holds. $\blacksquare$ Define the set \[ S = \{r \mid f(x) = f(rx) \forall x\} \]and note that $1 \in S$. Suppose there exists a nonzero $z$ for which $f(z) = 0$. Note that $f(z^2) = f(f(z)) = f(0) = 0$. Then, $P(x,z)$ implies that $f(x^2) = f(f(x)) + f(z^2) + 2f(zx)$ which simplifies as $f(zx) = 0$. Since $x$ is arbitrary, this implies that $f$ is always $0$. Else, assume that no such $z$ exists. Claim: $f$ is injective Proof. For the sake of contradiction assume not. Then $S$ contains an element $a$ such $a > 1$. Note that $f(a^2) = f(f(a)) = f(f(1)) = f(1)$ Then, $P(ax,1)$ means that \begin{align*} f(a^2x^2 + f(1)) &= f(f(ax)) + f(1) + 2f(ax) = f(f(x)) + f(1) + 2f(x) \\ & = f(x^2 + f(1)) = f(a^2x^2 + a^2f(1)) \end{align*}so $f$ is eventually periodic with period $c = (a^2 - 1)f(1)$. Then, $P(x,y+kc)$ for integer $k$ and sufficiently large $y$ implies that \[ f(x^2 + f(y)) = f(f(x)) + f(y^2) + 2f(x(y + kc)) \]or \[ f(xy) = f(x(y + kc)) \]By varying $y$ and $(y + kc)$, we get that \[ 1 + \frac{kc}{y} \in S \]This implies that $f$ is constant on the positive reals, however this implies that $f(x) = 0$, contradiction. $\blacksquare$ Thus, $f$ is injective on the positive reals and since $f(f(x)) = f(x^2)$, it follows that $f(x) = x^2$ for positive reals, and since $f$ is even, it holds for all reals. Thus, the solution set is $f(x) = 0$ and $f(x) = x^2$.

IndoMathXdZ wrote: Determine all the functions $f : \mathbb{R} \to \mathbb{R}$ such that \[ f(x^2 + f(y)) = f(f(x)) + f(y^2) + 2f(xy) \]for all real numbers $x$ and $y$. Let $P(x,y)$ be the assertion of the problem. $P(0,0)$ gives that $f(0)=0$. We will Prove that $f$ is either injective or constant. From $P(x,0)$ we have $f(x^2)=f(f(x))$. If $f$ is an injective function then $f(x)=x^2$ and we are done. Assuming $f$ is not injective there are $a>b$ such that $f(a)=f(b)$. Using $f(y^2)=f(f(y))$ we can write $P(x,y)$ as \[ f(x^2 + f(y)) = f(f(x)) + f(f(y)) + 2f(xy).\]Here, $P(x,a)$ and $P(x,b)$ gives us $f(ax)=f(bx)$ for all $x>0$. Define $A= \big\{\dfrac{a}{b} : f(a)=f(b)\big\}$. From above we know that if $c\in A$ then $f(x) = f(cx)$ for all $x>0$. Indeed $f(x)=f(cx)=f(c^2x)=\cdots$. So if $c\in A$ then $c^n\in A$ for all $n\in \mathbb Z$. On the other hand, if $c\in A$ from assertions $P(cx,1)$ and $P(x,1)$ we conclude that $f(c^2x^2+l)= f(x^2+l)$ where $l= f(1)$. So $$Q_c(x) : = \dfrac{c^2x^2+l}{x^2+l}\in A\;\; \forall x>0 \; \text{and}\; c>1, c\in A.$$$Q_c(x)$ is a continuous function with $Q_c(0) = 1$ and $Q_c(+\infty) = c$. Therefore, by intermediate value theorem $Q_c(x)$ covers the entire of interval $(1,c)$. So $(1,c)\subset A$. Now it is just enough to observe that $c^n$ is also a member of $A$ and therefore $(1,c^n)\subset A$ for all $n\in \mathbb N$. We infer that$(1,+\infty) \subset A$. So for all $y>x>0$ we have $\dfrac{y}{x}\in A$ and then $f(x) = f(\dfrac{y}{x}\cdot x) = f(y)$. Thus $f$ is constant and this finishes the proof.

IndoMathXdZ wrote: Determine all the functions $f : \mathbb{R} \to \mathbb{R}$ such that \[ f(x^2 + f(y)) = f(f(x)) + f(y^2) + 2f(xy) \]for all real numbers $x$ and $y$. Hint : try to construct $q(x)$ to be a continuous function such as $f(1)=f(q(x))$ as $x$ covers $\mathbb{R}$. Rest is pretty much P1-P2 level.

ltf0501 wrote: Let $P(x,y)$ be the assertion. $P(0,0)\to f(0)=0$ $P(0,x)\to f(f(x))=f(x^2)$ $\therefore P(x,y):\ f(x^2+f(y))=f(x^2)+f(y^2)+2f(xy)$ $P(1,x)-P(-1,x)\to f(x)=f(-x)$, so we just consider the case in $\mathbb{R}_{\ge 0}$ If $f(a)=f(b)$, $P(t,a),P(t,b)$ imply $f(ta)=f(tb),\ \forall t\in \mathbb{R}$. Suppose $f(a)=0$ for some $a \neq 0$, then $f(ta)=0\ \forall t\in \mathbb{R}$, i.e $f(x)\equiv 0$. Now suppose that $f(x)=0 \Leftrightarrow x=0$. $P(\sqrt{|x^2+f(z)|},y)\to f(x^2+f(y)+f(z))=f(x^2+f(z))+f(y^2)+2f(\sqrt{|x^2+f(z)|}y)=f(x^2)+f(y^2)+f(z^2)+2f(xz)+2f(\sqrt{|x^2+f(z)|}y)$. Change the variables $y,z$ we have $$f(xz)+f(\sqrt{|x^2+f(z)|}y)=f(xy)+f(\sqrt{|x^2+f(y)|}z) $$Supose there exist $y>z>0$ s.t. $f(z)y^2<f(y)z^2$. Let $x^2=\frac{z^2f(y)-y^2f(z)}{y^2-z^2}$, then $\sqrt{|x^2+f(z)|}y=\sqrt{|x^2+f(y)|}z$, implying $f(xz)=f(yz)\Rightarrow f(y)=f(z)$. However, since $f(z)y^2<f(y)z^2\Rightarrow y^2<z^2$, which is a contradiction. Hence $\forall y>z>0,\ \frac{y^2}{f(y)}\ge \frac{z^2}{f(z)}$. Now we prove that $f$ is injective. Suppose not, then $\exists t>1$ s.t. $f(x)=f(tx),\ \forall x\in\mathbb{R}_{\ge 0}$. Hence $\frac{t^2x^2}{f(tx)}=\frac{x^2}{f(x)}\Rightarrow f(x)>0,\ \forall x>0$. Choose $n\in\mathbb{N}$ sufficiently large such that there exists $c>0$ satisfying $c^2+f(y)=t^ny^2$. $P(c,y)\to f(x^2)+2f(xy)=0$, which is a contradiction. Hence $f$ is injective. Since $f(f(x))=f(x^2)\Rightarrow f(x)=x^2$, as desired.

Why does $f(xy)=f(xz)$ imply $f(y)=f(z)$ shouldn't that be $f(x)=f(y)$??

We will show that the only solutions are $f\equiv 0$ and $f(x)=x^2$. Let $P(x,y)$ denote the given assertion. We begin with some simple observations. $P(0,0)\Rightarrow f(0)=0$. $P(1,0)\Rightarrow f(x^2)=f(f(x))$. So the FE is symmetric in $x$ and $y$, hence $f(x^2+f(y))=f(y^2+f(x))$. Suppose there exists $a\neq 0$ such that $f(a)=0$, $P(x,a)$ gives $f(ax)=0$ for all $x$ so $f\equiv 0$. Otherwise, we have $f(a)=0\iff a=0$. We will try to prove that $f$ is semi injective (I made this name up) so suppose that there exists $a\neq \pm b$ such that $f(a)=f(b)$. From $P(x,a)$ and $P(x,b)$ we get that $f(ax)=f(bx)$ for all $x\in \mathbb R$. Also, from $P(1,x)$ and $P(-1,x)$ we get $f(x)=f(-x)$. Now we are ready to devour the problem Let $S$ denote the set of reals $r$ such that $f(x)=f(xr)$ for all $x$. Note that $x\in S\Rightarrow -x\in S$, $x\in S\Rightarrow \frac{1}{x}\in S$, $x,y\in S\Rightarrow xy\in S$. We have $\left(\frac{b}{a}\right)^n\in S$ so there exists $\alpha>0$ arbitrarily close to $0$ in $S$. Now from $P(\alpha x,y)$ and $P(x,y)$ we get that $f(\alpha^2x^2+f(y))=f(x^2+f(y))$. Note that $\alpha^2\in S$ so $$f(\alpha^2x^2+f(y))=f(x^2+f(y))=f(\alpha^2x^2+\alpha^2f(y))$$ We will show that this implies that $f(x)=f(x+c)$, $\forall x>0$. Firstly $x\mapsto \frac{x}{\alpha}$ gives $f(x^2+f(y))=f(x^2+\alpha^2f(y))$. Now if there exists $y$ such that $f(y)>0$, then setting $c=(1-\alpha^2)f(y)$ will give the desired conclusion. If $f(y) < 0$ for all $y\neq 0$ let $x=\sqrt{x^2-f(y)}$ and set $c=(\alpha^2-1)f(y)$ to get the desired conclusion. Now we have that $f(x)=f(x+c)$, $\forall x>0$. From $P(x,y+c)$ and $P(x,y)$, we get $f(xy)=f(x(y+c))$ so $1+\frac{c}{y}\in S$ for all $y>0$. Hence $[1,t]\subset S$ for some $t$. Notice that by multiplying this interval by $t\in S$ we get that $[t,t^2]\subset S$, hence $[1,t^2]\subset S$ so by repeating this process we get that $[1,\infty)\subset S$ and since $x\in S\Rightarrow \frac{1}{x}\in S$ and $x\in S\Rightarrow -x\in S$ we deduce that $S=\mathbb R^*$. So $f$ is constant (besides $f(0)$) and this also implies $f\equiv 0$. Now we know that $f(a)=f(b)\iff a=\pm b$. From $f(x^2)=f(f(x))$ we get $f(x)=\pm x^2$. Notice that $f(x)=x^2$ works so suppose there exists $x\neq 0$ such that $f(x)=-x^2$. Case 1. $f(x)=-x^2$, $\forall x\in \mathbb R$ By simply substituting back into the FE we get a contradiction. Case 2. There exists $y\neq 0$ such that $f(y)=y^2$. From $P(y,x)$ we get $f(y^2-x^2)=y^4-x^4+2f(xy)$ and this can only happen if $|y|=|x|\sqrt{2}$ so this implies that there are at most $2$ values of $x$ such that $f(x)=-x^2$ and at most $2$ values of $y$ such that $f(y)=y^2$, hence $f(z)=0$ for all other $z$ and this is clearly a contradiction. $\square$

The answers are $f(x)=0$ and $f(x)=x^2$, both of which clearly work. Let $P(x,y)$ denote the assertion. Claim 1: $f(f(x))=f(x^2)$. Note that $f(0)=0$ follows from $P(0,0)$ and $P(0,y)$ immediately gives you $f(f(y))=f(y^2)$ as desired. Now we can modify our equation to read $f(x^2+f(y))=f(f(x))+f(f(y))+2f(xy)$. The right hand side is symmetric, so we have \[f(x^2+f(y))=f(y^2+f(x))\] Claim 2: If $f(x)=f(y)$ then $f(x^2+f(z))=f(y^2+f(z))$ and $f(xz)=f(yz)$ for all $z$. The first one can be shown by noting that \[f(x^2+f(z))=f(z^2+f(x))=f(z^2+f(y))=f(y^2+f(z))\]and for the second one, take $P(z,x)$ and $P(z,y)$ and compare them. Note that this implies that if $f(x)=0$ for any $x\neq 0$ then $f(x)$ is identically zero. Suppose $f(a)=f(b)$ for some $a^2\neq b^2$ and $a$, $b$ nonzero. Suppose from now on that $f$ is not identically zero. Then let $c\neq 0$ be in the range of $f$. We have $f(xa)=f(xb)$ and therefore $f(x^2a^2+c)=f(x^2b^2+c)$ for all $x$, and finally \[f(yx^2a^2+cy)=f(yx^2b^2+cy)\]for all $x$, $y$. Claim 3: For all $m$, $n$ such that $b^2m\neq a^2n$, there exists values $x$ and $y$ such that $yx^2a^2+cy=m$ and $yx^2b^2+cy=n$. Letting \[x=\sqrt{\left|\frac{c(n-m)}{b^2m-a^2n}\right|}\text{ and }y=\frac{b^2m-a^2n}{c(b^2-a^2)}\]does the trick. This proves that $f(m)=f(n)$ for most pairs $(m,n)$. Claim 4: For all $m$, $n$, $f(m)=f(n)$. We have already covered all cases except $b^2m=a^2n$. However, in this case note that $f(a^2+f(0))=f(b^2+f(0))$ implies $f(a^2)=f(b^2)$ so \[f(m)=f\left(a^2\cdot \frac{m}{a^2}\right)=f\left(b^2\cdot \frac{n}{b^2}\right)=f(n)\]so this is true as well. However, this implies $f$ is constant, which can only means constantly zero, contradiction. Therefore, $f(a)=f(b)\implies a^2=b^2$. Looking back to Claim 1, we see that $f(x^2)=f(f(x))$. This implies $f(x)^2=x^4$ for all $x$. Note that using $f(x^2+f(y))=f(y^2+f(x))$ implying \[x^4+y^4+2x^2f(y)=x^4+y^4+2y^2f(x)\]we get that $2x^2f(y)=2y^2f(x)$. So if $f(x)=x^2$ while $f(y)=-y^2$ where $x$, $y$ are nonzero then $-2x^2y^2=2x^2y^2$ is a contradiction. Therefore, we either have $f(x)=x^2$ for all $x$ or $f(x)=-x^2$ for all $x$, of which clearly only the first one works.