xeroxia wrote: [*] $f(x^2) = f(x)^2 -2xf(x)$ [*] $f(x) = f(x-1)$ [*] $1<x<y \Longrightarrow f(x) < f(y).$ Isn't condition (ii) and (iii) contradictory? By (iii)$1<x -1 < x \Longrightarrow f(x-1) < f(x)$. Please edit or clarify.

ii should be $f(-x)=f(1-x)$

Mahi wrote: xeroxia wrote: [*] $f(x^2) = f(x)^2 -2xf(x)$ [*] $f(x) = f(x-1)$ [*] $1<x<y \Longrightarrow f(x) < f(y).$ Isn't condition (ii) and (iii) contradictory? By (iii)$1<x -1 < x \Longrightarrow f(x-1) < f(x)$. Please edit or clarify. I have edited the problem statement to $f(-x) = f(x-1)$. Sorry for the typo.

Put $x$ and $-x$ in i) to get that $f(-x)=f(x)-2x$ and hence $f(x)>2x$ From ii) we get $f(x)=f(-(x+1))=f(x+1)-2(x+1)$, hence $f(x+1)=f(x)+2x+2$.(*) Consider the sequence $(a_n)$ such that $a_0=f(0)$ and $a_{n+1}=a_n+2n+2$. Solving we get that $a_n=n^2+n+c$, where $c$ is some constant. It is clear that for $n\in\mathbb{N}_0$ we have $f(n)=a_n$. From i) we get $n^4+n^2+c=(n^2+n+c)^2-2n(n^2+n+c)$ which is equivalent with $2n^2(c-1)+(c^2-1)=0$, therefore we must have $c=1$, hence $f(n)=n^2+n+1$ for all integers $n\geq 0$. Use induction and (*) to prove that for $x>1$ and $n\in\mathbb{N}$ we have $f(x+n)=f(x)+2xn+n^2+n$.(**) Let $x=\frac{p}{q}$. Then use i) and (**) in $f((x+q)^2)=f(x+q)^2-2(x+q)f(x+q)$ to get $f(x)=x^2+x+1$ for all $x\in\mathbb{Q}$. Now from iii) we get that $f(x)=x^2+x+1$ for all $x>1$ and so $f(x)=x^2+x+1$ for all $x<-1$. From ii) if $x>0$ we get $f(x)=f(-x-1)=(-x-1)^2-(x-1)+1$ hence $f(x)=x^2+x+1$ for all $x\geq 0$ and finally: $f(x)=x^2+x+1, \forall x\in\mathbb{R}$.$\blacksquare$

$ f(x)-f(-x)=2x $ $ \Rightarrow $ $f(x)-f(x-1)=2x$(*) and $ f(x)>2x $(**). So, $f(n)=n^2+n+1$ for any $n \in Z$. Since (*), we solve at the $ f:(1, +\infty) \rightarrow R^+ $ . We have $f$ is strictly monotone and $g(x^2)=f(x)-x-1$ is strictly monotone too. (all increasing) So, we easily get $g(x)=x$ for all $x\in (1, +\infty)$ and $f(x)=x^2+x+1$ for all $x\in R$. Answer: $ f(x)=x^2+x+1 $ for all $x\in R$.

Property 1. $f(x)=x+\sqrt{x^{2}+f(x^{2})}$ We can get this easily from i). Property 2. $f(x)=f(x-1)+2x$ Put $-x$ in property 1, we get $f(-x)=-x+\sqrt{x^{2}+f(x^{2})}=f(x-1)$, subtracting with property one we get desired result. Property 3. $f(n)=n^{2}+n+1$ where $n$ is any integer. Put $x=0$ in i) we get $f(0)=f(0)^{2}$, $f(0)>0 \Rightarrow f(0)=1$. For positive integer $n$ we have $f(n)=f(n-1)+2n=...=f(0)+2(n+n-1+...+1)=n^{2}+n+1$ $f(-n)=f(n-1)=n^{2}-n+1$. Property 4. $f(x)=x^{2}+x+1$ Let $x_{0}>1, f(x_{0})=(x_{0})^{2}+x_{0}+1+\alpha$. We will show that $\alpha=0$. Consider $\alpha \neq 0$ We have from i) $f(x_{0}^{2})=f(x_{0})^{2}-2x_{0}f(x_{0})=x_{0}^{4}+x_{0}^{2}+1+\alpha (\alpha+2x_{0}^{2}+2)$ Let's denote $\alpha_{1}=\alpha (\alpha+2x_{0}^{2}+2)$. We'll construct numbers $\alpha_{2}$, $\alpha_{3}$,... in the same way. Case 1. $\alpha>0$. As $\alpha_{i}>2\alpha_{i-1}$ There exist a number $m$ such that $\alpha_{m}>1$. $f(a)=a^{2}+a+1+\alpha_{m}$ $f(a^{2})=a^{4}+a^{2}+1+\alpha_{m} (\alpha_{m}+2a^{2}+2)>a^{4}+a^{2}+1+2a^{2}+2 =(a^{2}+1)^{2}+(a^{2}+1)+1>([a^2]+1)^2+([a^2]+1)+1=f([a^{2}]+1) \Rightarrow f(a^{2})>f([a^{2}]+1)$ Contradiction to iii. Case 2. $\alpha<0$, we have $f(x)>2x \Rightarrow \alpha>-x_{0}^{2}+x_{0}-1 \Rightarrow \alpha+2x_{0}^{2}+2>x_{0}^{2}+x_{0}+1>2 \Rightarrow \alpha_{i}>2\alpha_{i-1} \Rightarrow$ there exist a number $m$ such that $\alpha_{m}<-2$, this implies analogously to case 1 $f(a^{2})<f([a^{2}])$ which is impossible.

Why we don't have $f(x)=f(-x)$ after changing $x \rightarrow -x $ to $i.$ ?