Please guys, you should learn how to $LATEX$ Let's $LATEX$ it so it can become clearer : Find all the functions $f: \mathbb R \rightarrow \mathbb R$ satisfying : $f(x^2+y^2+2f(xy))=(f(x+y))^2$ for all real numbers $x,y$

Take $x=y=0$ and put $k=f(0)$ we get $f(2k)=k^2$ take $y=0$ we get $f(x^2+2k)=(f(x))^2=(f(-x))^2$ for all $x \in \mathbb R$ So for $x=k^2$ we get $f((-k)^2)=f(k^2)$ or $f((-k)^2)=-f(k^2)$ Suppose $f((-k)^2)=f(k^2)$, then take $x=y=k$ we get $f(2k^2+2f(k^2))=k^4$ now take $x=-y=k$ we get $f(2k^2+2f((-k)^2))=k^2$ so $k^4=k^2$ which means that $k \in \{0,-1,1\}$ we can check that $k=-1$ and $k=1$ are not possible cases. So $f(0)=0$ Take $y=0$ in the original functional equation we get $f(x^2)=(f(x))^2$ for all $x \in \mathbb R$ and this is has been solved before, we find that $f(x)=x$ for all $x \in \mathbb R$ Have I made a mistake somewhere ?

erdos wrote: Take $x=y=0$ and put $k=f(0)$ we get $f(2k)=k^2$ take $y=0$ we get $f(x^2+2k)=(f(x))^2=(f(-x))^2$ for all $x \in \mathbb R$ So for $x=k^2$ we get $f((-k)^2)=f(k^2)$ or $f((-k)^2)=-f(k^2)$ Suppose $f((-k)^2)=f(k^2)$, then take $x=y=k$ we get $f(2k^2+2f(k^2))=k^4$ now take $x=-y=k$ we get $f(2k^2+2f((-k)^2))=k^2$ so $k^4=k^2$ which means that $k \in \{0,-1,1\}$ we can check that $k=-1$ and $k=1$ are not possible cases. So $f(0)=0$ Take $y=0$ in the original functional equation we get $f(x^2)=(f(x))^2$ for all $x \in \mathbb R$ and this is has been solved before, we find that $f(x)=x$ for all $x \in \mathbb R$ Have I made a mistake somewhere ? for one thing, you're writing $ f((-k)^2) = f(k^2) $ or $ f((-k)^2) = -f(k^2) $ where you really mean $ f(-k^2) = f(k^2) $ or $ f(-k^2) = -f(k^2) $. and you're missing the case where $ f(-k^2) = -f(k^2) $

Oh yeeh , I have totally forgotten about it So my solution posted above is not totally correct. This sounds a hard problem. Does anybody have an idea about it ?

there's also the fact that $ f $ can be identically $ 0 $ or identically $ 1 $. heh, this turned out to be a harder problem than i thought... (or maybe it's actually very easy and i'm just blind?)

f(x^2)=f(x)^2 It was claimed above that this equation can be solved here! I don't think so. Not unless some sort of assumption on continuty is given. [/quote]

I think that even the Most experienced people on this forum haven't found a solution yet. Please can anyone help ??

Could you please let the people think? I really do not think it is the end of the world if we do not solve it in 4 days.

OK. I am really sorry.

Thank you harazi, I spend 3 hours to find this solution. Put f(0)=a we have f(x ² +y ² +2f(xy))= [f(x+y)] ² Put now X=x+y and Y=xy then the equation becomes : f(X ² -2Y+2f(Y))=f(X) ² Put X=0 Then f(2f(Y)-2Y)=f(0) ² =a ² Put now g(x)= 2f(x)-2x then f(x)= 1/2g(x)+x Composing we have : f(g(x))=1/2g(g(x))+g(x)=a ² (*) Linear equation with constant coefficient: So g(x) must be in the forme : g(x)=Ax+B Reporting this to (*) and computing we mill have: A(A+2)=0 and B(A+2)=2a ² 1°case: A=0 then B=a ² and f(x)=x+1/2a ² or [f(x)] ² =[f(-x)] ² so a=0 And f(x)=x. 2°case: A=-2 then a=0 and g(x)=-2x+B then f(x)=B/2 or f(0)=a=0 Then f(x) =0. Correct me if not

you're also missing the possibility $ f(x) \equiv 1$. and there are a couple of other points in your solution i'm unclear about...

First of all X=x+y and Y=xy implies that X^2>=4Y. So if you let X=0 then Y has to be negative. Furthermore How can you possibly say that g(g(x)) + g(x) is linear and thus.... This again brings me to a point in one of the last posts: f(x^2)=f(x)^2 was said to imply f(x)=x^2, where as it is completely wrong. One can construct many funmctions satisfying that equation. The solution needs to be more detalied...

Just to be explicit: take the equation g(g(x)) + 2g(x) =2a^2 that you derived. x has to be negative.(I explained this above) So you can't really use the above equation, Because you don't know how g(x) behaves when x>0.

for Ali IF WE HAVE g(g(x))+2g(x)=2a ² then g(x)=Ax+B For this take x(0)=x and x(n+1)=g(x(n)) then x(n+1)+2x(n)=a ² take y(n)=x(n)-2/3a ² then y(n+1)+2y(n)=0 and y(0)=x-2/3a ² The caracteristical equation is : x+2=0 and go on

It seems like you didn't get my point: x>0 is not allowed in that equation. Anyways I can only say that the solution is completely wrong. Problem: Solve g(g(x)) + 2 g(x) = 0. Solution: Well you would say that g(x)=Ax +B right??? Now here is my solution: Define f(x)=g(x)+2x So f(f(x) -2x) =0. take f(1)=f(-1)=0 and f(x)=2x+1 for x>0. and f(x)=2x-1 for x<0. I am seriously tired of this argument, so you don't need to send me back some other meaningless post.

Oh, I recognize it. It's my problem, but nobody in Russia have inform me that it was proposed to IMO and short-listed. Lucky guys. As for the solution, it is not so much easy. I may post in PM, if anybody wants. The only hint is that the cardinality of the set of such functions is hypercontinuum.

hahaha!!

Fedor, just a question: Does the solution use any kind of calculus or advanced mathematics? You really frightened me with your hypercontinuum... Darij

2Darij: No, it does not. Hypercontinuum is not so mad: an equation $(f(x))^2=1$ has the same cardinality of the set of solutions.

, ,sorry i didnt see sol ofn bojan basic

This is an interesting f.e. problem. I am sorry I did not notice there are three pages following this post until I submitted my message. Pls forget about this proof if you are familiar with this post. Let $ a = f(0)$. $ P(x,0): f(x^2 + 2a) = f(x)^2$.(*) Suppose $ a < 0$, we let $ x = \sqrt { - 2a}$ and (*) gives $ a = f(x)^2\ge0$. Contradiction! Hence $ a\ge0$. Plug this into the RHS of the original formula, $ f(x^2 + y^2 + 2f(xy)) = f(x^2 + y^2 + 2xy + 2a)$. (**) If $ f(x) = x + a,\forall x$, it is easy to verify $ a = 0$ and $ f(x) = x$ is one solution. From now on, we assume there exists $ b_0$ with $ c_0 = f(b_0)\ne b_0 + a$. 1) Let $ c = f(b)$ for arbitrary $ b\ne0$, then $ f(u + 2c - 2b) = f(u + 2a),\forall u\ge 2|b| + 2b$ When $ u\ge 2|b| + 2b$, we can always solve $ (x + y)^2 = u$ and $ xy = b$ Then we can plug in (**) and show the claim. Hence, $ 2c - 2b - 2a$ is a period for $ f(x)$ when $ x\ge 2|b| + 2b + 2a$. In particular, we see $ T_0 = |2c_0 - 2b_0 - 2a_0| > 0$ is a period when $ x\ge N_0 = 4|b_0| + 2a$. 2) $ f(x) = C$ when $ x\ge2a$, where $ C = 0$ or $ C = 1$. Let $ b\in [N_0,N_0 + 1]$ be arbitrary with $ c = f(b)$. Then we have $ f(b^2 + 2a) = c^2$ and $ f((b + T_0)^2 + 2a) = f(b + T_0)^2 = c^2$. By 1), when $ x\ge 4b + 4T_0 + 2a$, both $ 2c^2 - 2b^2 - 6a$ and $ 2c^2 - 2(b + T_0)^2 - 6a$ are periods. So $ [2c^2 - 2b^2 - 6a] - [ 2c^2 - 2(b + T_0)^2 - 6a] = 2T_0^2 + 4bT_0$ is also a period. Now let $ b_0 > b_1$ be arbitrary on $ [N_0,N_0 + 1]$, we see for $ x \ge N_1 = 4N_0 + 4 + 4T_0 + 2a$, $ [ 2T_0^2 + 4b_0T_0] - [ 2T_0^2 + 4b_1T_0] = 4(b_0 - b_1)T_0$ is a period. In other words, any number in between $ 0$ and $ 4T_0$ is a period of $ f(x)$ when $ x \ge N_1$. It is easy to see that $ f(x) = C$ for some constant $ C$ when $ x\ge N_1$. Then use (*), we have $ C^2 = C$. So $ C = 0$ or $ C = 1$. Again from (*), we know $ f( - x) = \pm f(x)$. For any $ x\ge0$, we can find $ b\le - N_1$ and $ x + 2c - 2b\ge N_1$ because $ c = 0,\pm1$. By 1), $ f(x + 2a) = f(x + 2c - 2b) = C$. This shows the claim. 3) If $ a = 0$, $ f(x) = 0$. By 2), $ f(x) = C,\forall x\ge0$. Since $ f(0) = 0$, $ f(x) = 0$ for all $ x\ge0$. (*) implies $ f( - x) = \pm f(x) = 0$. 4) If $ a > 0$, $ f(x) = 1,\forall x\ge - \frac23;f(x) = \pm1,\forall x < - \frac23$. By (*), $ f(2a) = a^2$. By 2), $ a^2 = C$. Since $ a > 0$, we have $ f(x) = 1$ when $ x\ge2$. Now LHS of (*) is 1, $ f(x) = \pm 1,\forall x$. Now we look back the original equation. If $ f(b) = - 1$ for some $ b = xy$, we have $ f(x^2 + y^2 - 2) = 1$. In other words, $ f(u) = 1,\forall u\ge2|b| - 2$. Once $ 2|b| - 2\le b$, we will get contradiction. Hence, $ b < - \frac23$. It is not hard to see that is sufficient for the equation to hold. Conclusion: 1) $ f(x) = x$. 2) $ f(x) = 0$. 3) $ f(x) = 1,\forall x\ge - \frac23;f(x) = \pm1,\forall x < - \frac23$.

See here for official solution.

Substitute $y=0$ gives us $f(x^2+2f(0))=f(x)^2$ for all real number $x$. This also gives $f(x)^2=f(-x)^2$ for all real number $x$. Note that for any real number $a,b$ that $a^2\geq 4b$, there exists real numbers $x,y$ that $x+y=a$ and $xy=b$. From $f(x^2+y^2+2f(xy)) =f(x+y)^2$ for all real numbers $x,y$, we get that $$f(a^2-2b+2f(b))=f(a)^2=f(a^2+2f(0))$$for all real numbers $a,b$ that $a^2\geq 4b$. This can be simplify to $$P(x,y): f(x-2y+2f(y))=f(x+2f(0))$$for all real numbers $x,y$ that $x\geq \max \{ 0,4y\}$. Also, we've $$Q(x,y): f(x^2-y^2+2f(-xy))=f(x-y)^2$$for all real numbers $x,y$. If $f(y)=y+f(0)$ for all real number $y$, substitute in $f(x^2+y^2+2f(xy))=f(x+y)^2$ gives us $$x^2+y^2+2xy+3f(0)=(x+y+f(0))^2$$for all real numbers $x,y$. This easily gives $f(x)=x$ for all real number $x$, which clearly is one of the solution. Now, we’ll consider only the case when there exists real number $k$ that $f(k)\neq k+f(0)$. Let $M=2f(k)-2k-2f(0)$. Note that $M\neq 0$. From $P(x-2f(0),k)$, we’ve $f(x+M)=f(x)$ for all real number $x\geq 2f(0)+\max \{ 0,4k\}$. If $M>0$, we get $f(x+|M|)=f(x)$ for all real number $x\geq 2f(0)+\max \{ 0,4k\}$. If $M<0$, we get $f(x)=f(x-M)=f(x+|M|)$ for all real number $x-M\geq 2f(0)+\max \{ 0,4k\}\iff x\geq M+2f(0)+\max \{ 0,4k\}$. In both cases, we have $f(x)=f(x+|M|)$ for all real number $x\geq 2f(0)+\max \{ 0,4k\}$. We can prove by induction that $f(x)=f(x+v|M|)$ for all positive integer $v$ and real number $x\geq 2f(0)+\max \{ 0,4k\}$. For any real numbers $c,d$ that $f(c)=f(c+d)$, we’ve the following results: 1) $f(c^2+2f(0))=f(c^2+2cd+d^2+2f(0))$ This follows from $f(c^2+2f(0))= f(c)^2=f(c+d)^2=f(c^2+2cd+d^2+2f(0))$. 2) $f(x)=f(x-2d)$ for all real number $x\geq 2f(c)-2c+\max \{ 0,4c,4c+4d\}$ This follows from $P(x,c),P(x,c+d)$, which gives us $f(x-2c+2f(c))=f(x-2c-2d+2f(c))$ for all real number $x\geq \max \{ 0,4c,4c+4d\}$. 3) $f(x)=f(x-2(2cd+d^2))$ for all real number $x\geq 2f(c)^2-2(c^2+2f(0))+\max \{ 0,4c^2+8f(0),4c^2+8f(0)+8cd+4d^2\}$. This is true by applying the result 1) combine with 2) and note that $f(c^2+2f(0))=f(c)^2$. Back to the problem. In this paragraph, we'll show that $f(\ell )=f(\ell -\Delta)$ for all real numbers $\Delta \in (0,|M|)$ and $\ell \geq 2f(0)+\max \{ 0, 4k\} +|M|$. From $f(x)=f(x+|M|)$ for all real number $x\geq 2f(0)+\max \{ 0,4k\}$, using result 3), we get that $$f( t)=f(t-2(2x|M|+M^2))$$for all real number $t\geq 2f(x)^2-2(x^2+2f(0)) +\max \{ 0,4x^2+8f(0), 4x^2+8f(0) +8x|M|+4M^2 \}$ where $x\geq 2f(0)+\max \{ 0,4k\}$. There exists real number $h$ that $2(2h|M|+M^2) =\Delta $. There also exists positive integer $n$ that $h+n\geq 2f(0)+\max \{ 0,4k\}$. So, we’ve $$f(t)=f(t-2(2(h+n)|M|+M^2))=f(t-\Delta -4n|M|)$$for all real number $t\geq 2f(h+n)^2-2((h+n)^2+2f(0))+ \max \{ 0,4(h+n)^2+8f(0),4(h+n)^2+8f(0)+8(h+n)|M|+4M^2\}$. For any real number $\ell \geq 2f(0)+\max \{ 0,4k\}+|M|$, there exists positive integer $m>4n$ that $$\ell +m|M|\geq 2f(h+n)^2-2((h+n)^2+2f(0))+ \max \{ 0,4(h+n)^2+8f(0),4(h+n)^2+8f(0)+8(h+n)|M|+4M^2\}.$$So, $f(\ell +m|M|)=f(\ell -\Delta -4n|M|+m|M|)=f(\ell -\Delta +(m-4n)|M|)$. Since $\ell -\Delta >\ell -|M|\geq 2f(0)+\max \{ 0,4k\}$ and $\ell \geq 2f(0)+\max \{ 0,4k\}+|M|\geq 2f(0) +\max \{ 0,4k\}$, we get $$f(\ell +m|M|)=f(\ell)\text{ and } f(\ell -\Delta +(m-4n)|M|)=f(\ell -\Delta).$$Hence, $f(\ell )=f(\ell -\Delta)$, which we aim at the beginning of this paragraph. Using the result in preceding paragraph, together with $f(x)=f(x+|M|)$ for all real number $x\geq 2f(0)+\max \{ 0,4k\}$, we get that $f(x)$ must be constant for all real number $x\geq 2f(0)+\max \{ 0,4k\}$. For convenient, we write it as $f(x)=C$ for all real number $x\geq D$. Note that we can force $D$ to be positive. Since there exists real number $r$ that $r$ and $r^2+2f(0))$ are greater than $D$, we get $C=f(r^2+2f(0))=f(r)^2=C^2$. Hence, $C\in \{ 0,1\}$. So we've two following cases: 1) $C=0$. Since $f(x)=0$ for all real number $x\geq D$ and $f(x)^2=f(-x)^2$ for all real number $x$, we get that $f(x)=0$ for all real number $x$ that $|x|\geq D$. For any real number $e$, there exists positive real number $u$ that $u(u+e)>D$ and $(u+e)^2+u^2>D$. The former gives $f(-u(u+e))=0$ and so $f((u+e)^2+u^2+2f(-u(u+e)))=f((u+e)^2+u^2)=0$. From $Q(u+e,u)$, we get that $0=f(e)^2\implies f(e)=0$. Hence, $f(x)=0$ for all real number $x$, which clearly is one of the solution. 2) $C=1$. Since $f(x)=1$ for all real number $x\geq D$ and $f(x)^2=f(-x)^2$ for all real number $x$, we get that $f(x)\in \{ -1,1\}$ for all real number $x$ that $|x|\geq D$. For any real number $e$, there exists positive real number $u$ that $u(u+e)>D$ and $(u+e)^2+u^2-2>D$. The former gives $f(-u(u+e))\in \{ -1,1\} $ and so $(u+e)^2+u^2+2f(-u(u+e)) \geq (u+e)^2+u^2-2>D$. Hence, $f((u+e)^2+u^2+2f(-u(u+e))) =1$. From $Q(u+e,u)$, we get that $1=f(e)^2\implies f(e)\in \{ -1,1\}$. In other words, $f(x)\in \{ -1,1\}$ for all real number $x$. Let $A=\{ x\in \mathbb{R}\mid f(x)=1\}$ and $B=\{ x\in \mathbb{R}\mid f(x)=-1\}$. Note that they form a partition of $\mathbb{R}$. From $f(x^2+y^2+2f(xy)) =f(x+y)^2$ for all real numbers $x,y$, we get that $$f(x^2-2y+2f(y))=f(x)^2=1$$for all real numbers $x,y$ that $x^2\geq 4y$. This gives us $f(x-2y+2f(y))=1$ for all real number $x\geq \max \{ 0,4y\}$. Using this equation, we get that $$\begin{cases} x-2-2y\in A \ \text{for all} \ x,y\in \mathbb{R}\text{ that }x\geq \max \{ 0,4y\} ,y\in B\\ x+2-2y \in A \ \text{for all} \ x,y\in \mathbb{R}\text{ that }x\geq \max \{ 0,4y\} ,y\in A \end{cases} $$In case that $y>0$, we may (and will) substitute $x$ by $4y+z$ where $z$ is real number that $z\geq 0$. So, we can rewrite the condition in the following four cases \begin{align*} 1) & \ x-2y+2\in A \ \text{for all} \ x,y\in \mathbb{R}\text{ that }x\geq 0,y\leq 0,y\in A \\ 2) & \ x+2y+2\in A \ \text{for all} \ x,y\in \mathbb{R}\text{ that }x\geq 0,y\geq 0,y\in A \\ 3) & \ x-2y-2\in A \ \text{for all} \ x,y\in \mathbb{R}\text{ that }x\geq 0,y\leq 0,y\in B \\ 4) & \ x+2y-2\in A \ \text{for all} \ x,y\in \mathbb{R}\text{ that }x\geq 0,y\geq 0,y\in B. \end{align*}If there exists real number $0\leq w\leq 2$ that $w\in B$, using condition 4), we get that $w<2w-2 \implies w>2$, contradiction. So, $x\in A$ for all real number $x$ that $0\leq x\leq 2$. Using condition 2), we can extend it to $x\in A$ for all real number $x$ that $x\geq 0$. If there exists real number $\frac{-2}{3} \leq w\leq 0$ that $w\in B$, using condition 3), we get that $w<-2w-2\implies w<\frac{-2}{3}$, contradiction. So, $x\in A$ for all real number $x$ that $\frac{-2}{3}\leq x\leq 0$. Hence, now we’ve $f(x)=1$ for all $x\geq \frac{-2}{3}$. This means there exists subset $\mathcal{S}$ of $\{ x\in \mathbb{R}\mid x<\frac{-2}{3} \}$ that $$f(x)=\begin{cases} -1& , \text{if} \ x\in \mathcal{S}\\ 1& , \text{otherwise} \end{cases}.$$Not hard to verify that this function is one of the solution for all subset $\mathcal{S}$ of $\{ x\in \mathbb{R}\mid x<\frac{-2}{3} \}$. Finally, we have three type of solutions to the problem: 1. $f(x)=0$ for all $x\in \mathbb{R}$, 2. $f(x)=x$ for all $x\in \mathbb{R}$, and 3. $f(x)=\begin{cases} -1& , \text{if} \ x\in \mathcal{S}\\ 1& , \text{otherwise} \end{cases}$ for any subset $\mathcal{S}$ of $\{ x\in \mathbb{R}\mid x<\frac{-2}{3} \}$.

Amazing practice problem. The answer is $f(x) = 0$, $f(x) = x$, and $f(x) = -1$, for any subset of $(-\infty, -\frac{2}{3})$ and $f(x) = 1$ otherwise. It's easy to check that these are the only solutions. We'll prove that these are the only ones. Let $x + y = a$ and $xy = b$. Therefore we can rewrite the functional equation as \[ P(a,b) : f(a^2 - 2b + 2f(b)) = f(a)^2 \ \forall a^2 \ge 4b \]Now, define $2f(b) - 2b = g(b)$. \[ P(a,b) : f(a^2 + g(b)) = f(a)^2 \]It is quite immediate to notice that $f(-a)^2 = f(a)^2$ and $f$ is eventually non-negative (by fixing one value of $\ell$ and take $a \ge \max \{0, 4 \ell \}$ if $g(\ell) \ge 0$ and $a \ge \max \{ 4 \ell, \sqrt{-g(\ell)} \} $ if $g(\ell) < 0$. $\textbf{Case 01.} \ \text{If } g \ \text{is constant}$ Then $f$ is linear. We can check that $f(x) = x$ is the only solution. $\textbf{Case 02.} \ \text{If } g \ \text{is non constant}$ Suppose not, we will prove that $f$ is eventually constant. Firstly, notice that if $g(b)$ is non-constant, there exists two distinct real numbers $\alpha$ and $\beta$ such that $g(\alpha) - g(\beta) = c > 0$. Notice that \[ f(u)^2 = f(u^2 + g(\alpha) ) = f(u^2 + g(\beta) + c) = f((\sqrt{u^2 + c})^2 + g(\beta)) = f(\sqrt{u^2 + c})^2 \]Denote $\sqrt{u^2 + c} = v$. Now, we have that if $v^2 - u^2 = c$, then $f(u) = f(v)$. Now, we have \[ g(v) - g(u) = (2f(v) - 2v) - (2f(u) - 2u) = 2(f(v) - f(u)) + 2(u - v) = 2(u - v) = \frac{2c}{u + v} \]Now, we have \[ f(s^2 + g(u)) = f(s^2 + g(v)) = f(s)^2 \]Let $u,v \in [A,3A]$ where $A$ is large. Let $\varepsilon = g(v) - g(u) = \frac{k}{u + v} \in [\delta, 2 \delta]$. Therefore, \[ f(s^2 + g(u) + \varepsilon) = f(s^2 + g(v)) = f(s)^2 = f(s^2 + g(u)) \]This forces that for large $a,b > 12A$, we then have $f(a) = f(b)$ if $a^2 - b^2 \in [\delta, 2 \delta]$. Repeating this process by changing the interval, we get that $f$ is eventually constant. Now, take two large integers $a,b > M$ such that $b^2 - a^2 = g(u)$ for some integer $u$. Suppose the constant is $c$. However, \[ c^2 = f(b)^2 = f(a)^2 = f(a^2 + g(u)) = f(b^2) = c \]This gives us $c = 0$ or $c = 1$. Take a large negative number $-P$. We know that $|f(-P)| = |f(P)| = c$ which is a constant. Therefore, $f(x) \in \{ -c, c \}$ as well for $x \le -P$. Hence, \[ f(s^2 + 2f(-P) -2(-P)) = f(s^2 + 2P - c_0) = f(s)^2 \]for all large integer $P$ and some constant $c_0$. This forces $f(s)^2 = c$ for all integer $s$. Hence, if $c = 0$, we get $f \equiv 0$ which satisfy the equation. Now, assume that $c = 1$. \[ R(x,y): f(x^2 + y^2 + 2f(xy)) = 1 \]Notice that if $f(0) = -1$. Then $R(\sqrt{2},0)$ gives us $f(0) = 1$, a contradiction. Therefore, $f(0) = 1$ which therefore gives $f(x) = 1$ for all $x \ge 2$. Now, suppose $f(t) = -1$. Suppose $-\frac{2}{3} \le t \le 2$. Then there exists $x,y \in \mathbb{R}$ such that \[ x^2 + y^2 - 2 = xy = t \]as $2 + t = x^2 + y^2 \ge 2|xy| = 2|t|$. Therefore, this gives \[ 1 = f(x^2 + y^2 + 2f(xy)) = f(t) \]a contradiction. Therefore, $x^2 + y^2 - 2 = xy = t$ must gives no solution. Hence, if $f(t) = -1$, then $t < -\frac{2}{3}$. This gives a solution $f(x) = -1$, for any subset of $(-\infty, -\frac{2}{3})$ and $f(x) = 1$ otherwise.

The solutions are $f(x)=x$, $f(x)=0$, and for any set $A\subseteq \left(-\infty, -\tfrac 23\right)$ $$f(x) = \begin{cases} -1 & x\in A \\ 1 & x\notin A; \end{cases}$$the first two are easy to verify. To verify the last solution, it's helpful to reparametrize the functional equation as $$f(a^2+2f(b)-2b) = f(a)^2 \text{ whenever } a^2\geq 4b \qquad (*)$$ If $b<0$, then it's easy to check that $2f(b)-2b \geq -\tfrac 23$ for any $b$, which makes the equation visibly true; if $b\geq 0$, then $a^2+2f(b)-2b \geq 2b+2 >0$, which again makes the equation visibly true. This completes the verification. Now we prove that these are all solutions. Claim 1: If $f$ is injective, then $f$ must be an identity function. Proof: For any $r,s$, take $x$ large enough so that $$f(x^2+2f(r)-2r) = f(x)^2 = f(x^2+2f(s)-2s),$$or $f(r)-r=f(s)-s$ after injectivity. This gives $f(x)=x+c$ for some constant $c$; it's easy to check that $c=0$. $\blacksquare$ Claim 2: Assume that $f(r)=f(s)$ such that $r> s$, then there exists $K$ which $f$ is constant on $[K,\infty)$. Proof: First of all, we note that $f(x)\geq 0$ for all sufficiently large $x>M_0$. This can be easily seen by fixing $b$ and taking $a$ sufficiently large. Now, consider \begin{align*} f(\sqrt{x+2r})^2 &= f(x+2r + 2f(r)-2r) \\ &= f(x+2f(r)) \\ &= f(x+2f(s)) \\ &= f(\sqrt{x+2s})^2, \end{align*}for all $x$ large enough (say $>100|r|$) Thus, $f(\sqrt{x+2r}) = f(\sqrt{x+2s})$ for all sufficiently large $x$, say $x>M_1$. We also take $\varepsilon \in (0, \sqrt{M_1+2r} - \sqrt{M_1+2s})$. Now, fix $u,v$ such that $\tfrac{\varepsilon}{2} < u^2-v^2 < \varepsilon$, and $u,v$ are larger than $M$ (which will be told later). Then, we can find $B$ so that $\sqrt{B+2r} - \sqrt{B+2s} = u^2-v^2 \quad (\spadesuit)$. This choice of $B$ must be bounded above by a constant depending only on $r,s,f$, since we have a fixed lower bound of $u^2-v^2$. So, let's suppose that it's bounded above by $M_2$. By repeating the argument above, we find that $$f\left(\sqrt{A+\sqrt{B+2r}}\right) = f\left(\sqrt{A+\sqrt{B+2s}}\right)$$for any $A > \max\{100\sqrt{B+2r},M_0\}$. Thus, let $M_3 = \max\{100\sqrt{M_2+2r},M_0\}$. The above equations hold for $A>M_3$. Hence, if $M > \sqrt{M_3+\sqrt{M_2+2r}}$, we can find a suitable $A$ which $A+\sqrt{B+2r} = u^2$. From $(\spadesuit)$, we get that $A+\sqrt{B+2s}=v^2$. Therefore $f(u)=f(v)$ for any $u,v>M$ and $u^2-v^2 \in (0.5 \varepsilon, \varepsilon)$. By applying this fact repeatedly, it's easy to see that $f$ is constant for a large enough value. $\blacksquare$ Thus, we see that $f$ is a constant $c$ over $[K,\infty)$. By fixing $b$ and taking $a$ very large in the equation, we deduce that $c^2=c$ or $c\in\{0,1\}$. Since $f(a)^2 = f(-a)^2$, we get that $f$ must be $\pm c$ in $(-\infty, -K]$ too. Now, we take $b$ very negative, so that $2f(b)-2b > K$. This gives $f(a)^2 = c$ or $f(a)=\pm c$ for any $a$. In particular, if $c=0$, then we have forced $f$ to be a zero function. The only case left is $f(x)=\pm 1$ for all $x$. The equation is then reduced to $$f(a^2+2f(b)-2b) = 1 \text{ whenever } a^2\geq 4b.$$It remains to show that if $f(t)=-1$ for some $t$, then $t\leq -\tfrac 23$. We note that the equation gives $$f(a^2 - 2 - 2t)=1 \text{ whenever } a^2\geq 4t.$$ If $t\geq 0$, then the above translates to $f(k)=1$ for any $k\geq 2t-2$. Thus, $t\leq 2t-2$ or $t\geq 2$. However, since $f(a^2+2f(0))=1$ for any nonnegative $a$, we get that $f(x)=1$ for any $x\geq 2$, contradiction. If $t<0$, then the above translates to $f(k)=1$ for any $k\geq -2t-2$. Thus, $t<-2t-2$ or $t< -\tfrac 23$ as desired.

We claim the solutions are $f(x)\equiv 0,x,f_S(x)$, where $S\subseteq(-\infty,-2/3)$ is arbitrary, and $f_S(x)$ is defined to be $1$ outside of $S$ and $-1$ on $S$. Let $P(x,y)$ be the given FE. Note that \[P(x,0)\implies f(x^2+2f(0)) = f(x)^2,\]so we can rewrite our FE as \[Q(x,y): f(x^2+y^2+2f(xy)) = f(x^2+y^2+2xy+2f(0)).\] We have the following key lemma. Lemma: If $f(a)=f(b)$, then \[f(z+2a)=f(z+2b)\]for all $z\ge 2f(0)+2\cdot\max\{|a|,|b|\}$. Proof: Fix $z\ge 2f(0)+2\cdot\max\{|a|,|b|\}$. By the inequality condition on $z$, we may pick reals $x$ and $y$ such that $xy=a$ and $x^2+y^2 = z-2f(0)$, and similarly we may pick reals $x'$ and $y'$ such that $x'y'=b$ and $x'^2+y'^2=z-2f(0)$. Comparing $Q(x,y)$ detail $Q(x',y')$, we see that $f(z+2a)=f(z+2b)$, as desired. $\blacksquare$ We also have the following easy lemma. Lemma: There exists a positive constant $A$ such that $f(x)\ge 0$ for all $x\ge A$. Proof: Note that \[P(x,1/x)\implies f(x^2+1/x^2+2f(1))=f(x+1/x)^2\ge 0,\]so $A=\max\{2+2f(1),0\}$ suffices. $\blacksquare$ We can use these two lemmas to get the following killer claim. Claim: If $f(a)=f(b)$ for some $a\ne b$, then there exists constants $B$ and $k$ such that $x\ge B\implies f(x)=k$. Proof: Using the lemma, we may shift up the values of $a$ and $b$ arbitrarily, so assume that they are greater than $A$. Now, fix some $T\ge 2\cdot\max\{a,b\}$. Let $x$ and $y$ be given by $x+y=\sqrt{T+2a}$, $x-y=\sqrt{T-2a}$, and let $x'$ and $y'$ be given by $x'+y'=\sqrt{T+2b}$, $x'-y'=\sqrt{T-2b}$. We see that $xy=a$, $x'y'=b$, and $x^2+y^2=x'^2+y'^2=T$, so comparing $P(x,y)$ and $P(x',y')$ tells us that \[f(\sqrt{T+2a}) = f(\sqrt{T+2b})\]for $T\ge 2a$ (assuming that $a>b$ WLOG). Focusing our attention to $T\in[2a,3a]$, we see that by the lemma, there exists a constant $B$ such that for $z\ge B$, \[f(z+2(\sqrt{T+2a}-\sqrt{T+2b})) = f(z)\]for all $T\in[2a,3a]$. In this interval, the expression $\sqrt{T+2a}-\sqrt{T+2b}$ can take all the values in some interval $[\alpha,\beta]$ with $0<\alpha<\beta$, so we have the following statement. There is a positive constant $B$ and an interval $[\alpha,\beta]$ with $0<\alpha<\beta$ such that \[f(z+\gamma)=f(z)\]for all $z\ge B$ and $\gamma\in[\alpha,\beta]$. Let $0<\delta<\beta-\alpha$. Then, \[f(z)=f(z+\beta)=f((z+\delta) + (\beta-\delta))=f(z+\delta)\]for all $z\ge B$, since $\beta-\delta\in[\alpha,\beta]$. So $f(z)=f(z+\delta)$ for all $z\ge B$ and $\delta\in(0,\beta-\alpha)$, so an easy induction implies that $f$ is constant beyond $B$, as desired. $\blacksquare$ If $f$ is injective, then $Q(x,y)$ implies that $f(xy)=xy+f(0)$ for all real $x,y$, so $f(x)=x+f(0)$. Plugging this into the FE shows that only $f(x)=x$ works, so in this case, we the solution $\boxed{f(x)\equiv x}$. Now, suppose that $f$ is not injective. By the above claim, we have that $f(x)=k$ for all $x\ge B>0$. From $P(x,0)$, we see that $f(-x)\in\{f(x),-f(x)\}$ for all $x$. Fixing any real $z$, we have \[P(z+x,-x)\implies f(x^2+(z+x)^2+2f(-x(z+x))) = f(z)^2\]for all real $x$. By making $x$ sufficiently large, we see that $f(-x(z+x))\in\{k,-k\}$ and $x^2+(z+x)^2-2|k|\ge B$, so in fact $f(z)^2=k$ for all real $z$. This implies that $k\in\{0,1\}$. If $k=0$, then this implies that $\boxed{f(x)\equiv 0}$. Now assume $k=1$, so this implies $f(x)\in\{1,-1\}$ for all real $x$. Our FE now becomes \[f(x^2+t^2/x^2+2f(t)) = 1.\]If $f(t)=-1$ for some real $t$ (potentially zero), then we see that \[f(x^2+t^2/x^2-2) = 1\]for all real $x\ne 0$, so $t$ must be outside of the image of the function \[x\mapsto x^2+t^2/x^2-2,\]where $x$ varies over all nonzero reals. The image of this function is $[2|t|-2,\infty)$, so $t<2|t|-2$, or $t\in(-\infty,-2/3)\cup(2,\infty)$. This means that $f(x)=1$ for all $x\in[-2/3,2]$. Suppose we know that $f(x)=1$ for all $x\in[0,C]$ where $C>0$. Then, \[f(x^2+y^2+2)=1\]for all nonnegative real $x$ and $y$ such that $xy\le C$. Thus, we have $f(x)=1$ for all $x\in[0,2C+2]$. We know the statement is true for $C=2$, and since iterating the map $C\mapsto 2C+2$ sends the value to infinity, we see that $f(x)=1$ for all $x\ge -2/3$. This gives $\boxed{f(x)\equiv f_S(x)}$ for any set $S\subseteq(-\infty,-2/3)$. This covers all the cases, so we're done. >>>

Can someone please check the below solution (it seems very different from the above ones, so maybe I messed up somewhere) The solutions are $f(x) \equiv x$ ; $f(x) \equiv 0$ and $$f(x) = \begin{cases} -1 \qquad & \text{if } x \in T \\ 1 \qquad & \text{otherwise} \end{cases}$$for any set $T \subseteq \left( - \infty , \tfrac{-2}{3} \right)$. The first two functions trivially work. For the third function: RHS is always $1$. Now just consider two cases according to the value of $f(xy)$ and use the upper bound on elements of $T$. Now we prove the above are the only solutions. Let $P(x,y)$ be the given assertion. $P(x,0)$ gives $$f(x)^2 = f(x^2 + 2f(0)) \qquad{(1)}$$Let $2f(0) = 2c$. Then the original equation becomes $$f(x^2 + y^2 + 2f(xy)) = f(x^2 + y^2 + 2xy + 2c) \qquad{(2)}$$Lets first assume $f$ is injective. Then $(2)$ directly gives us $f(x) \equiv x+c$, and it is not hard to note that only $c = 0$ works ; i.e. $f(x) \equiv x$ is the only solution here. Now we consider the crucial case when $f$ is not injective. Let $f(a) = f(b)$ with $a \ne b$. Claim: There are constants $M,r > 0$ such that $f(x+r) = f(x) ~ \forall ~ x \ge M$ Proof: It is not hard to see that $(2)$ implies $$f(x + 2f(y)) = f(x + y + 2c) ~ \forall ~ x,y \in \mathbb R \text{ with } x \ge 4|y|$$Now taking $y=a,b$ in the above implies our claim. $\square$ Claim: For some $d \in \{0,1\}$ we have $f(x) = d ~ \forall ~ x \ge M$. Proof: So replacing $x$ by $x+r$ in $(1)$ gives $$f((x+r)^2 + 2c) = f(x^2 + 2c) ~ \forall ~ x \ge M \qquad{(3)}$$Now any fix any $p > q \ge M$. Note that a $x \ge M$ can be chosen such that $$(x+r)^2 + 2c,x^2 + 2c \ge M ~~ \text{and} ~~ \frac{\Big(((x+r)^2 + 2c) - (x^2 + 2c) \Big) - (p-q)}{r} \in \mathbb Z$$Then that along with $(3)$ and our previous claim would give that $f(p) = f(q)$. Hence there is a constant $d$ such that $f(x) = d ~ \forall ~ x \ge M$. Taking $x > 0$ sufficiently large in $(1)$ yields $d^2 = d$, i.e. $d \in \{0,1\}$. $\square$ Claim: $f(x) \in \{d,-d\} ~ \forall ~ x \in \mathbb R$. Proof: Vary $x,y$ in $P(x,y)$ such that both of $x,y$ are negative and $xy = M$. We obtain there is a constant $M_0 < 0$ such that $$f(x)^2 = d ~ \forall ~ x \le M_0 \qquad{(4)}$$Now consider any real number $u$. Take $n$ to be a sufficiently large positive number. Consider $P(n+u,-n)$. We obtain $f(u)^2 = d$, which proves our claim. $\square$ So now our $P(x,y)$ becomes $$Q(x,y) ~:~ f(x^2 + y^2 + 2f(xy)) = 1 $$If $d=0$, then we directly have $f(x) \equiv 0$. Now assume $d=1$. Let $S := \{x : f(x) = 1\}$ and $T := \{x : f(x) = -1 \}$. Claim: $\left[ \tfrac{-2}{3},\infty \right] \subseteq S$, and consequently $T \subseteq \left( - \infty , \tfrac{-2}{3} \right)$. Proof: Consider any $t \in T$. Fixing $xy = t$ in $Q(x,y)$ and varying $x,y$ we obtain all real numbers $\ge 2|t| - 2$ must be in set $S$ . Thus $t < 2|t| - 2$. This forces $t \notin \left[\frac{-2}{3},2 \right] $. Thus $\left[\frac{-2}{3},2 \right] \subseteq S$. Then $Q(x,0)$ gives $[2,\infty] \subseteq S$. This proves our claim. $\square$ This completes the proof of the problem. $\blacksquare$

The answer is $f(x)\equiv x, f(x)\equiv 0$ and $f(x)=1\forall x\ge -\frac 23$ and $f(x)\in \{-1,1\}$ for all $x<-\frac 23$. Let $P(x,y)$ denote the assertion in the question. Case 1: $f$ is injective. In this case, let $s=x+y$. Then $s^2-2xy+2f(xy)$ is constant if $s$ stays the same, so $f$ is linear for the interval $(-\infty, \frac s2)$. Picking $s$ large enough implies $f(x)\equiv x+C$. A simple substitution suggests $C=0$ so $f(x)\equiv x$. Case 2: $f$ is not injective. Suppose $f(a)=f(b)$. WLOG $a<b$. Pick $x,y,u,v$ such that $x^2+y^2=u^2+v^2=K$ for some $K>2\max\{|a|,|b|\}$, $xy=a$ and $uv=b$. Now, $P(x,y)$ and $P(u,v)$ gives $f(u+v)^2 = f(K+2f(b))=f(K+2f(a))=f(x+y)^2$. We have $u+v=\sqrt{K+2b}$ and $x+y=\sqrt{K+2a}$. It follows that $|f(\sqrt{K+2a})| = |f(\sqrt{K+2b})|$ for all $K>\max\{2|a|, 2|b|\}$. Claim: There exists a constant $C$ such that $f(x)$ is constant for $x\ge C$. First, I prove that $f(x)\ge 0$ for a constant $x$ independent of $a,b$: if $t>2+2f(1)$ then there exists $x^2+y^2=t-2f(1), xy=1$ and thus $f(t)=f(x^2+y^2+2f(xy))=f(x+y)^2\ge 0$. Picking $K$ such that $\sqrt{K+2a} > 2+2f(1)$ we obtain that $f(\sqrt{K+2a})=f(\sqrt{K+2b})$, or $f(\sqrt{K}) = f(\sqrt{K+nd})$ where $d=b-a, n\in \mathbb{N}$. Suppose $\sqrt{K+d}-\sqrt{K}=c$ and $\sqrt{K+\epsilon+d}-\sqrt{K+\epsilon}=c-\delta$. Then, for all $x>2(K+\epsilon+d)$, we have $f(\sqrt{x}) = f(\sqrt{x+c}) = f(\sqrt{x+c-\delta})$ for some arbitrarily small $\delta$ that is not affected by other external variables other than $\epsilon$. By continuity, any sufficiently small $\delta$ is constructible, so we can carve a "path" from $f(m)$ to $f(n)$ where $n-m=\delta_1+\delta_2+\cdots+\delta_t$ and have $f(m)=f(m+\delta_1)=f(m+\delta_1+\delta_2)=\cdots=f(m+\delta_1+\cdots+\delta_t)$. Say $f(x)=c$ for all $x>C$. Then $P(C+5, C+5)$ implies $c=c^2$ so $c\in \{0,1\}$. Also, $P(x,y)$ and $P(-x,-y)$ suggests $f(x+y)^2 = f(-x-y)^2$ so it follows that $f(x)=\pm c$ for all $x<-C$. Case 1: $c=0$. Now, $P(x,-x-d)$ for $x$ large suggests $f(x^2+(x+d)^2+2f(-x(x+d))=f(d)^2$. Taking $x(x+d)>C$ gives $|f(-x(x+d))|=c$ so LHS is c, so $f(x)\equiv 0$. Case 2: $c=1$. Again we can prove that $f(x)\in \{1,-1\}$ for all $x$. Thus, $f(x^2+y^2+2f(xy))=1$. Suppose $f(x)$ is not identically 1. Then if $f(t)=-1$, then we have 3 cases. If $f(0)=-1$ then $P(0,y)$ suggests $f(y^2-2)=1$. Substituting $y=\sqrt{2}$ gives a contradiction. If $f(t)=-1$ for some $t>0$ then for any $xy=t$ we have $x^2+y^2\ge 2t$. Since the range of $x^2+y^2-2$ is $\mathbb{R}_{\ge 2t-2}$ it follows $t<2t-2$ i.e. $t>2$. However, this implies that if $x^2+y^2=t-2$ then $f(xy)=-1$, and we can pick $0<xy<2$ to get a contradiction. Thus, $f(t)=1$ for all $t\ge 0$. If $f(-t)=-1$ for some $t>0$ then take $x<0, y>0$ and then $x^2+y^2-2 \ne t$. Since the range of $x^2+y^2$ is $\mathbb{R}_{\ge 2t-2}$ it follows that $2t-2 > -t$ i.e. $t>\frac 23$. Thus, $f(x)=1$ for all $x\ge -\frac 23$. Observe that $f(x)=1\forall x\ge -\frac 23$ and $f(x)\in \{1,-1\}$ for all $x<-\frac 23$ is a solution because if $x^2+y^2+2f(xy)<0$ then $xy\le -\frac 23$ and thus $x^2+y^2 \ge \frac 43$.

Nice Problem, very brilliant!

Critical.

Solved with lrjr24 Note that this is equivalent to \[ f(s^2+2(f(p)-p)) = f(s)^2 \]for $s^2 \ge 4p$. Denote this assertion with $P(s, p)$. By $P(s, p)$ and $P(-s, p)$ for $p < 0$ it follows that $|f(s)| = |f(-s)|$. If $f$ is injective, then $f(x) - x$ is constant so $f(x) = x + c$. Checking gives that only $c = 0$ works. Else, suppose that there are two distinct $a, b$ such $f(a) - a > f(b) - b$. Then, $f$ is periodic with period $t = (f(a) - a) - (f(b) - b)$ on some interval $I = (c, \infty)$. Since sufficiently large $f$ is nonnegative, $f$ must be nonnegative on $I$. Claim: $f$ is eventually constant. Proof. Note that for $\sqrt{x} \in I$ \[ f\left(\sqrt{x}\right)^2 = f(x + 2(f(p) - p)) = f(x + 2(f(p) - p) + t) = f\left(\sqrt{x + t}\right)^2. \]Then, by $P(x, a)$ for $x \ge 2a$ it follows that $f\left(\sqrt{x + 2a}\right) = f(x + 2f(a))$ As such, for $y \ge 2\sqrt{x+t}$, \[ f\left(\sqrt{y + 2\sqrt{x}}\right) = f\left(\sqrt{y + 2\sqrt{x + t}}\right) \]Then, take sufficiently large $a > b$. It follows that for $r^2, s^2 > 4\sqrt{a + t}$ and \[ 2\sqrt{b + t} - 2\sqrt{b} < r^2 - s^2 < 2\sqrt{a + t} - 2\sqrt{a}, \]$f(r) = f(s)$. Fix a $r$. By repeating the above we have that $f(r) = f(s)$ if \[ s^2 \in \left[r^2 + n \cdot \left(2\sqrt{b + t} - 2\sqrt{b}\right), r^2 + n \cdot \left(2\sqrt{a + t} - 2\sqrt{a}\right)\right] \]for some $n$. From taking a sufficiently large $n$ it follows that $f$ is constant on some interval with length larger than $t$ in $I$. By periodicity it follows that $f$ is constant on $I$. $\blacksquare$ Let this constant on sufficiently large inputs be $d$. Since $f$ is constant for sufficiently large, taking $P(s, p)$ for large enough $s$ means that $d^2 = d$ so either $d = 0$ or $d = 1$. Thus, by $P(s, p)$ for sufficiently negative $p$ it follows that \[ f(s^2 + k) = f(s)^2 \]for arbitrarily large $k > 0$. As such, $f(x)^2 = d$ for all $x$. If $d = 0$, then $f(x) = 0$. From here on, suppose $d = 1$. Claim: $f$ is $1$ on $\left(-\frac23, \infty\right)$. Proof. By $P(s, 0)$ it follows that \[ f(s^2 + 2f(0)) = 1 \]Since $f(0) = \pm 1$, it follows that $f$ is constant on $[2, \infty)$. For the sake of contradiction suppose that for some $x \in \left[0, 2\right)$ that $f(x) = -1$. Then, by $P\left(s, x\right)$ it follows that \[ f(s^2 - 2 - 2x) = 1 \]for $s^2 \ge 4x$. Since $2x - 2 \le x$ this leads to a contradiction. Now, suppose that for some $x \in \left(-\frac23, 0\right)$ that $f(x) = -1$. Then, by $P\left(s, x\right)$ \[ f(s^2 - 2 - 2x) = 1 \]However, $-2 - 2x \le x$ holds, contradiction. $\blacksquare$ As such, either $f(x) = x$, $f(x) = 0$, or \[ f(x) = \begin{cases} 1 & x \ge -\frac23 \\ \pm 1 & x < -\frac23 \end{cases} \]pointwise.

Took long time for solving Let $z=z+y, t=xy$, with $z^2 \ge 4t$. Define $g(x) = 2(f(x)-x)$. Then, we have: $f(z^2+g(t)) = f(z)^2 (*)$ for all $z^2 \ge 4t$. Let $c=2f(0) = g(0)$. Thus: $t=0$ in (*) results to: $f(z^2+c)=f(z)^2(\#)$. Note that: $c \ge 0$ (else: $c<0$ implies $z^2+c=0$ for some $z$, which leads to: $c/2 = f(z)^2$, which is false). Also: $f(x) \ge 0$ for all $x>c$. Consider cases: Case-1: $g$ is constant: Clearly we would have $c=0$ with $f(x)=x$. Case-2: $g$ is non-constant. Consider some $a, b \in \mathbb R$, such that $f(a)-f(b)=d > 0$ for some $d$. For some sufficiently large constant $M$, with $p, q \ge M$, we have: $q^2-p^2=d$, and thus: $p^2+f(a)=q^2+f(b)$. Due to (*), we have: $f(p)=f(q)$. Also $g(p)-g(q) = 2(p-q) = \frac{d}{p+q} = \frac{d}{p+\sqrt{p^2+s}}$. Thus, there exists some interval $[\epsilon, 2 \epsilon]$ such that, every value is of the form $g(p)-g(q)$. Here, we will start the another hunt: Consider $y>x\ge 2 \sqrt{M}$ and $\epsilon < y^2-x^2 < 2 \epsilon$. Due to the above work, there exists some $p, q \le M$ with $x^2+g(p)+y^2+g(q)$ which implies $f(x)^2=f(y)^2$ (taking $x^2 \ge 4p, y^2\ge 4q$). Taking $M$ large-enough such that: $4M \ge c^2$, we have $f(x)=f(y)$. Therefore, $f$ is eventually constant, i.e. $f(x)=k$ for some constant $k$ for all $x \ge 2 \sqrt{M}$. Taking $z > 2 sqrt{M}$ in (#), we have: $k=0, 1$. Note that: $f(-z)=|f(z)|$ for all $z < - 2 \sqrt{M}$ with $|f(z)| \le 1$. Thus: $g(u)=2f(u)-2u \ge -2 - 2u$ for all $u \le - 2 sqrt{M}$, which gives $g$ is unbounded function. For every value of $z$, there exists some value of $t$ such that: $z^2+g(t)>2 \sqrt{M}$ which leads to $f(z)^2=f(z^2+g(t)) = k = k^2$. Thus: $f(z) = \pm k$. Case-1: $k=0$ which evidently leads to $f \equiv 0$. Case-2: $k=1$. Then: $c=2f(0)= \pm 2$. Since $c >0$, $c=2$. Thus: $f(x)=1$ for all $x >2$. Let $f(t)=-1$ for some $t<2$. Then: $t-g(t) = 3t+2$. If $t-g(t) \ge 0$, then: taking $z^2=t-g(t)$ leads to: $f(z)^2=f(z^2+g(t)) = f(t) = -1$, contradiction. Thus, we have: $t-g(t) < 0$ or $t < -\tfrac{2}{3}$. Thus, note that: for any subset $S$ of $(-\infty, -\tfrac{2}{3})$, the function $f$: $f(x) = -1$ for all $x \in X$, $f(x)=1$ else-wise, works...! Thus, we have: $f(x)\equiv x, 0$ along with: For any subset $S$ of $(-\infty, -\tfrac{2}{3})$, the function $f$ such that: $f(x) = -1$ for all $x \in X$ and $f(x)=1$ else-wise and we are done.