Find all functions $f(x)$ such that $f\left(x^2+f(y)\right)$=$y+(f(x))^2$

Question

Let $\mathbb R$ denote the set of all real numbers. Find all function $f: R\to \ R$ such that $$f\left(x^2+f(y)\right)=y+(f(x))^2$$ It is the problem. I tried to it by putting many at the place of $x$ and $y$ but I can't proceed. Please somebody help me.

Answer 1

Let me summarize the proposed ideas and give a self-contained answer: Let $f : \Bbb{R} \to \Bbb{R}$ satisfy the functional equation

$$ f(x^2 + f(y)) = y + f(x)^2 \tag{*}$$

for all $x, y \in \Bbb{R}$.

Step 1. $f(0) = 0$.

(This solution is due to @Leo163.) Let $a \in \Bbb{R}$ be such that $f(a) = 0$. For instance, plugging $y = -f(x)^2$ to $\text{(*)}$ confirms the existence of such $a$. Then plugging $(x, y) = (a, a)$ to $\text{(*)}$ gives

$$ f(a^2) = f(a^2 + f(a)) = a + f(a)^2 = a.$$

Finally, plugging $(x, y) = (0, a^2)$ to $\text{(*)}$ gives

$$ 0 = f(a) = f(f(a^2)) = a^2 + f(0)^2. $$

This shows that $f(0) = 0$.

Step 2. $f$ satisfies various identities:

1. $f(x^2) = f(x)^2$.
2. $f(f(x)) = x$. In particular, $f$ is bijective.
3. $f(-x) = -f(x)$.
4. $f(x+y) = f(x)+f(y)$.

The identity 1 (resp. 2) follow by plugging $y = 0$ (resp. $x = 0$) to $\text{(*)}$. For 3, we may assume that $x \geq 0$. Then replacing $(x, y)$ in $\text{(*)}$ by $(x^{1/2}, f(-x))$, we have

$$ 0 = f(x - x) = f(-x) + f(x^{1/2})^2 = f(-x) + f(x) $$

and 3 follows. Finally, replacing $(x, y)$ in $\text{(*)}$ by $(x^{1/2}, f(y))$ gives

$$ f(x + y) = f(x) + f(y), \qquad \forall x \geq 0, \ y \in \Bbb{R}. $$

The restriction that $x \geq 0$ can be removed by combining this with 3: if $x < 0$, then

$$ f(x+y) = -f(-x-y) = -(f(-x) + f(-y)) = f(x) + f(y). $$

Step 3. $f(x) = x$.

If $x > 0$ then by the first identity of the previous step,

$$ f(x) = f((x^{1/2})^2) = f(x^{1/2})^2 > 0. $$

(We can exclude the equality because $f$ is bijective and $f(0) = 0$.) Combining this with the fact that $f$ is an odd function, we have:

$$ f(x) \geq 0 \quad \Longleftrightarrow \quad x \geq 0.$$

Now for any $x \in \Bbb{R}$,

$$ f(x) -x = f(x) - f(f(x)) = f(x - f(x)). $$

Therefore $f(x) - x \geq 0$ if and only if $x - f(x) \geq 0$, which implies that $f(x) = x$.

---

Remark. The Cauchy functional equation has pathological solutions which are also involution, and we need to use the extra structure given by the equation $\text{(*)}$ to exclude such possibilities. The first identity of Step 2 was essential in our solution.

Answer 2

Suppose $a$ is such that $f(a)=0$ (we know there is such $a$, it is enough to consider $y=-f(x)^2$ for some $x$), then substituting $x=a$ and $y=a$, we have $$f(a^2)=a.$$ Then, set $x=0$ and $y=a^2$, to obtain $$0=f(f(a^2))=a^2+f(0)^2.$$ Since we are only dealing with real numbers, it means that $f(0)=0$, and no other number has $0$ as image.

As consequences, we have $f(f(y))=y$ and $f(y^2)=f(y)^2$ for every $y\in\mathbb{R}$.

Answer 3

Observe

1. $f(x^2+f(0)) = [f(x)]^2$
2. $f(f(x)) = x+[f(0)]^2$.

Plugging $0$ into expression 1 yields \begin{align} f_2(0):=f(f(0)) = [f(0)]^2 \end{align} and plugging $f(0)$ into expression 1 yields \begin{align} f(f(0)^2+f(0)) =[f_2(0)]^2 = [f(0)]^4. \end{align} Moreover, by expression 2, we have \begin{align} f(f(0)^2+f(0))= f([f(0)]^2)=f(f_2(0)) = f_3(0). \end{align} Hence we have \begin{align} f_3(0) = [f(0)]^4. \end{align}

Next, using both 1 and 2, we have \begin{align} f(f(x^2+f(0)))=x^2+f(0)+[f(0)]^2. \ \ (\ast) \end{align} Plugging $0$ into $(\ast)$ yields \begin{align} f_3(0) = f(0)+[f(0)]^2. \end{align} Combining everything yields \begin{align} [f(0)]^4= f_3(0) = f(0)+[f(0)]^2 \ \ \Rightarrow \ \ f(0)[f(0)^3-f(0)-1] =0. \end{align} Solving for $f(0)$ yields that either $f(0) = 0$ or $f(0)$ is a root of the polynomial $p(x) = x^3-x-1$, which only has one real root.

Moreover, observe \begin{align} f_4(0) = [f(0)]^4 \end{align} and \begin{align} f_4(0) = f(f(f_2(0))) = f_2(0) + [f(0)]^2 = 2[f(0)]^2 \end{align} which means \begin{align} [f(0)]^4-2[f(0)]^2 = 0. \end{align} Thus, $f(0)$ is also a root of $x^4-2x^2 = x^2(x^2-2)$.

In conclusion, $f(0)$ has to be $0$. Thus, $f(f(x)) = x$ and $f(x^2) = [f(x)]^2$.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

\begin{equation} \mrm{f}\pars{x^{2} + \mrm{f}\pars{y}} =y + \bracks{\mrm{f}\pars{x}}^{2} \label{1}\tag{1} \end{equation}

Deriving both members of \eqref{1} respect of $\ds{x}$ and $\ds{y}$ yield, respectively: \begin{equation} \left\{\begin{array}{rcl} \ds{\mrm{f}'\pars{x^{2} + \mrm{f}\pars{y}}\pars{2x}} & \ds{=} & \ds{2\mrm{f}\pars{x}\mrm{f}'\pars{x}} \\[2mm] \ds{\mrm{f}'\pars{x^{2} + \mrm{f}\pars{y}}\mrm{f}'\pars{y}} & \ds{=} & \ds{1} \end{array}\right. \label{2}\tag{2} \end{equation}

---

By comparing both expressions in \eqref{2}: \begin{equation} {\mrm{f}\pars{x}\mrm{f}'\pars{x} \over x} = {1 \over \mrm{f}'\pars{y}} = \alpha\,, \qquad \pars{~\alpha\ \mbox{is independent of}\ x\ \mbox{and}\ y~} \label{2.a}\tag{2.a} \end{equation}

---

The second one is quite trivial: \begin{equation} {1 \over \mrm{f}'\pars{y}} = \alpha \implies \mrm{f}\pars{y} = {1 \over \alpha}\, y + \beta\,,\qquad \pars{~\beta\ \mbox{is a constant}~}\label{3}\tag{3} \end{equation}

---

\eqref{2.a} and \eqref{3} yield: \begin{align} &{\mrm{f}\pars{x}\mrm{f}'\pars{x} \over x} = \alpha \,\,\,\stackrel{\mrm{see}\ \eqref{3}}{\implies}\,\,\, {\bracks{x/\alpha + \beta}/\alpha \over x} = \alpha \implies {1 \over \alpha^{2}}\, x + {\beta \over \alpha} = \alpha x \implies \left\{\begin{array}{rcl} \ds{1 \over \alpha^{2}} & \ds{=} & \ds{\alpha} \\[2mm] \ds{\beta \over \alpha} & \ds{=} & \ds{0} \end{array}\right. \\[5mm] &\ \stackrel{\mrm{see}\ \eqref{3}}{\implies}\,\,\, \bbx{\ds{\mrm{f}\pars{x} = {x \over \alpha}\,,\qquad \alpha^{3} = 1}} \label{4}\tag{4} \end{align}

---

By inserting \eqref{4} into \eqref{1} we conclude that $\ds{\alpha = 1}$: $$ \bbx{\ds{\mrm{f}\pars{x} = x}} $$

Answer 5

Here is a short approach without Cauchy equation. Starting from $f(0)=0$ by @Leo163, we obtain $f(f(x))=x$ by plugging $y=0$. This shows $f$ is a bijection. Now plug $y=f^{-1}(\zeta)$, we see $$f(x^2+\zeta)=f^{-1}(\zeta)+(f(x))^2\geq f^{-1}(\zeta)=f(\zeta)$$ Hence $f$ is monotone increasing. Now suppose for some $x_0$ we have $f(x_0)> x_0$, then $x_0=f(f(x_0))\geq f(x_0)> x_0$, a contradiction. Thus we have $f(x)\leq x$. Similarly we have $f(x)\geq x$, so $f(x)=x$.