See here http://www.mathlinks.ro/Forum/viewtopic.php?p=18289#18289

Here's my Ugly Proof... Let given functional eq as T-eq We know $ f$ is surjective cuz for given $ x$, $ 2y$ moves all in Real number. Let $ f(a) = 0$ for some $ a$, take $ y = 0$ in T-eq then $ f(x^2 + f(0)) = f(x)^2$, Thus $ f(x)^2 = f(x^2 )$, so $ f( - a) = 0$ Take $ x = 0, y = a, - a$, we get $ f(a) = 2a + f(0)^2$, $ f( - a) = - 2a + f(0)^2$ so $ a = 0$ Here, We get $ f(x) = 0 \Longleftrightarrow x = 0$ Now put $ y = - \frac { f(x)^2}{2}$, we get $ f(x^2 - \frac { f(x)^2}{2} + f( - \frac { f(x)^2}{2})) = 0$ which means $ x^2 - \frac { f(x)^2}{2} + f( - \frac { f(x)^2}{2}) = 0$ Thus $ f(p) = f(q) \Longrightarrow p^2 = q^2$ If $ f(p) = f( - p) = r$, where $ p > 0$ put $ y = 0$ in T-eq, $ f(x^2 ) = f(x)^2$ Thus $ f(x) \ge 0$ when $ x \ge 0$, Thus $ r \ge 0$ put $ y = p, - p$ in T-eq, we get $ f(x^2 + p + r ) = 2p + f(x^2)$, $ f(x^2 - p + r) = - 2p + f(x^2 )$ Put $ x = 0$ in first one, $ f(p + r) = 2p$, put $ x = \sqrt {p + r}$ in Second one, we get $ f(2r) = - 2p + f(p + r) = 0$ so $ r = 0$ means $ p = 0$ Therefore, $ f(p) = - f( - p)$ Thus $ f(p) = f(q)$ means $ p^2 = q^2$ so $ p = q$ which means $ f$ is injective Now, take $ x = 0$ in T-eq, we get $ f(y + f(y)) = 2y$. We know $ f$ is bijective. Thus $ y + f(y)$ moves all of real numbers. So we can set $ y + f(y) = t$ with $ \forall t \in R$ Let's reconsider about T-eq, we get $ f(x + t) = f(x) + f(t)$ with $ x\ge 0$ Then $ f(x + y) = f(|x| + x + y - |x|) = f(|x| - x) + f(y - |x|) = f(|x|) + f(x) - f(|x| - y)$ $ = f(x) + f(y)$ Thus $ f(x + y) = f(x) + f(y)$ Thus $ f(q) = kq$ for $ q \in Q$ Then $ \forall \alpha \ge 0$, $ f(x + \alpha) = f(x) + f(\sqrt {\alpha})^2 \ge f(x)$ which means $ f$ in increasing function. $ \forall x \in R$, let two rational sequence $ (a_n), (b_n)$ which are converge to $ x$ with $ a_n < x$, $ b_n > x$ for all $ n in N$ Then $ f(a_n ) = k a_n \le f(x) \le f(b_n ) = k b_n$, so $ f(x) = kx$ for all $ x \in R$ cuz $ f(x^2 ) = f(x)^2$, $ f(1) = 1$. so we get $ f(x) = x$ for all $ x \in R$

we have $f$ is surjective and $f(0)=0$, $f(y+f(y))=2y$ and $f(x^2)=f(x)^2 $. We can $f(a)=0$ $ \Rightarrow $ $a=0$. So $g(x)=f(x)+x$ is surjective in $ g:(-\infty ,0]\rightarrow (-\infty, 0] $. For all $z\le 0$, $x\in R$, we have $f(x+z)=f(x)+f(z)$ and $f(x)=-f(-x)$ - odd function. So $g $ is surjective in all $R$. For all $x,y\ge 0$ we get that $f(x+y)=f(x)+f(y)$ and $f$ is monotone increasing. Answer: $f(x)=x$.

$P(x,y)$ and $P(-x,y)$ give us $f(x)^{2}=f(-x)^{2}$ So $f(x)=f(-x)$ or $f(x)=-f(-x)$ Suppose that $f(0)\neq 0$, If there exist a $t$ satisfy $f(t)=-f(-t)$ then $P(0,t)$ and $P(0,-t)$ give us $f(0)=0$, a condradiction. Since $f(x)^{2}=f(-x)^{2}$, $f$ is even. If there exist $p$ s.t. $f(p)\leq 0$ and $p\neq 0$ We can get that $p=0$ from $P(\sqrt{-f(p)},p)$ and $P(\sqrt{-f(p)},-p)$ So $f(x)>0$ for all non-zero real number $x$ a contracdition with $P(x,-y)$ for a sufficient large $y$ So $f(0)=0$, and it's easy to get $f(x^{2})=f(x)^{2}$ and $f(x+y)=f(x)+f(y)$ Hence $f(x)=x$

Here is a solution by my friend and teammate RAMİL JABİYEV (İ like his fe solutions that is why I post here) Claim 1:$f (a)=0$ $\Longrightarrow$ $a=0$ Proof:Just see $P (0,a) $ and $P (0,-a) $ (Note:Here we used that $f (x)^2=f (-x)^2$) Then done... Claim 2:$f (f (x))+f (x)=2x $ Proof:Firstly lets look at $P (x,-\frac {f (x)^2}{2}) $ which gives us $x^2-\frac {f (x)^2}{2}-f (-\frac {f (x)^2}{2})=0$.Then $P (0,-\frac {f (x)^2}{2}) $ yields that $f $ is odd.Now $P (x,-x^2) $ gives the desired relation . Claim 3:$f(2x)=2f (x) $ Proof:Using the above claim it is nothing Claim 3:$f (x)=x $ is a solution Proof: $P(x,f (y)) $ and using the fact that $f $ is odd we conclude that $f $ is additive and since $f $ is bounded in some interval we get $f (x)=x $ for any $x\in R $.

tenplusten wrote: ... and since $f $ is bounded we get $f (x)=x $ for any $x\in R $. Quite funny "since $f(x)$ is bounded ...", we get ... an unbounded function .... Do you think this sentence is quite consistant ?

$f (x^2)=f (x)^2\ge 0$ .So f is bounded in some interval

tenplusten wrote: $f (x^2)=f (x)^2\ge 0$ .So f is bounded in some interval According to me, each word is important in maths statements. And "f is bounded" is not equivalent to "f is lowerbounded over $\mathbb R^+$" With this last sentence, I do agree.

Alternative ending: After we establish that $ 2x = f(x) + f(f(x)) $ we set $a_n = n_{th}$ iteration of $f$. Then, we get that $a_n + a_{n-1} = a_{n-2}$ and solving the recurrence we find that $a_n = l + m(-2)^n$. But $f$ is lowerbounded in positive reals and therefore we must have $m=0$ and $a_n = l$ which is constant. Thus, $f(f(x))=f(x)$ and due to surjectivity we have $f(x)=x$ in positive reals. Now choose $x$ large enough s.t. $x^2 + y + f(y) > 0$ and from the initial relation we get that $f(y)=y, \forall y \in \mathbb{R}$