# A famous functional equation

That is following problem: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that
$f(x^2 + y + f(y)) = (f(x))^2 + 2\cdot y\; \forall x,y\in\mathbb{R}.(*)$
It is famous! Why? Because, it is from AMM(problem 10908, posted by Wu Wei Chao) and it is one in problems from Bulgarian TST 2003, Vietnam TST 2004 and Iran TST 2007. However, in Vietnam TST 2004 it has form:

Find all real values of $a$, for which there exists one and only one function $f: \mathbb{R} \to \mathbb{R}$ and satisfying the equation
$f(x^2 + y + f(y)) = (f(x))^2 + a\cdot y\;\forall x,y\in\mathbb{R}.(**)$
We easy see that, if $a$ is an answer then $a=2$. In fact,
If $a= 0$. We have at least two functions satisfying, therefore $a= 0$ not satisfy.
If $a\not = 0$. Suppose $f$ is function satisfy, because $\{f^{2}(x)+ay|x,y\in\mathbb{R}\} = \mathbb{R}$(note that : $a\not = 0$) we have $f$ is surjection (1).
By (1) exist $b$ such that $f(b) = 0$, from (**) we have $f(x^{2} + b) = f^{2}(x) + a b\forall x\in\mathbb{R}$ and  $f(x^{2} - b) = f^{2}(x) - ab\forall x\in\mathbb{R}$ , therefore $2ab = f(x^{2} + b) - f(x^{2} - b)\forall x\in\mathbb{R}$, choose $x = 0$ here we have $b = 0$(note that :$f( - b) = 0$). So $f(x) = 0$ iff $x = 0$ (2).
In (**) choose $y = 0$ we have $f(x^{2}) = f^{2}(x)\forall x\in\mathbb{R}$ (3).
In (3) choose $x = 1$ and by (2) we have $f(1) = 1$ (4).
In (**) choose $y = 1$ we have $f(x^{2} + 2) = f(x^{2}) + a\forall x\in\mathbb{R}$ (5). In (5) choose $x = 0$ then $a= f(2)$. Therefore $a^{2} = f^{2}(2) = f(4) = f(2) + a= 2a$, so $a= 2$.
So, we’ll back to equation (*). These are solutions of this one:
1. Solution of harazi, a member of the site mathlinks.ro.
In (*) put $y=-\dfrac{f^2(x)}{2}$ we have $f\left(x^2-\dfrac{f^2(x)}{2} +f\left( -\dfrac{f^2(x)}{2}\right)\right)=0\forall x\in\mathbb{R}$, from here and by above we have $f\left( -\dfrac{f^2(x)}{2}\right )=$

$-x^2+\dfrac{f^2(x)}{2}\forall x\in\mathbb{R} (1)$
From (1), in (*) Put $y$ by $-\dfrac{f^2(y)}{2}$ we have $f(x^2-y^2)=f^2(x)-f^2(y)\forall x,y\in\mathbb{R}$
From here we have $f(x+y)=f(x)+f(y)\; \forall x,y\in\mathbb{R} (2)$
Now, by $f^2(x)=f(x^2)\forall x\in\mathbb{R}$ and (2) we see that $f^2(x+y)=f((x+y)^2)\forall x,y\in\mathbb{R}$

$\Rightarrow (f(x)+f(y))^2=f(x^2+2xy+y^2)\forall x,y\in\mathbb{R}$

$\Rightarrow f^2(x)+2f(x)f(y)+f^2(y)=f(x^2)+2f(xy)+f(y^2)\forall x,y\in\mathbb{R}$

$\Rightarrow f(xy)=f(x)f(y)\forall x,y\in\mathbb{R} \; (3)$
Final, by (2),(3) and Proposition 2.7 ([1]) we have $f(x)=x\forall x\in\mathbb{R}$ or $f(x)=0\forall x\in\mathbb{R}$. Checking condition (*) we obtain

$f(x)=x\forall x\in\mathbb{R}.$

2. Solution of N.T.TUAN, a member of the site mathlinks.ro.
In (*) choose $x=0$ we have $f(x+f(x))=2x\forall x\in\mathbb{R}\; (1)$ .
By above: $|f(x)|=|f(-x)|\forall x\in\mathbb{R}$, if $b=f(a)=f(-a)(a>0)$ then $b>0$ (because $f(x)>0\forall x>0$). Setting $y=a$ and $y=-a$ in (*) we have $f(x^2+a+b)=f^2(x)+2a$

$\forall x\in\mathbb{R}\Rightarrow f(a+b)=2a$

$f(x^2-a+b)=f^2(x)-2a\forall x\in\mathbb{R}$

$\Rightarrow f(2b)=0(\text{by above})$
Therefore $2b=0\Rightarrow b=0$, contradiction! So, $f(x)=-f(-x)\forall x\in\mathbb{R}$, form that, see

$f(x)>0\forall x>0(2)$ and $f(x)<0\forall x<0(3)$.
Now, for all $y<0$, in (*) setting $x=\sqrt{-y}$ we have $f(f(y))+f(y)-2y=0\forall y Final,from (3),(4) and idea of Example 1.10 ([1]) we see that$latex f(x)=x\forall x<0\$, but

$f(x)=-f(-x)\forall x\in\mathbb{R}$, we have
$f(x)=x\forall x\in\mathbb{R}.$