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How to solve the following functional equation:

Find all $f:\mathbb{R}\to\mathbb{R} $ such that: $$ f\left(x^2+f(y)\right)=y+f(x)^2 $$ Holds for every $x,y\in\mathbb{R}$.

A friend gave it to me, probably its an olympiad question.

I started with the equation $f(f(y))=y+f(0)^2$ which seems to be quite helpful, but I couldn't do it effectively. How to solve it properly?

Redundant Aunt
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2 Answers2

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This problem is from the IMO 92.

Let $P(x,y)$ be the assertion $f\left(x^2+f(y)\right)=y+f(x)^2$. Then: $$ P(0,y):\space f(f(y))=y+f(0)^2 $$ Thus we have $f\left(f\left(x^2+f(y)\right)\right)=f\left(y+f(x)^2\right) \iff x^2+f(y)+f(0)^2=f\left(y+f(x)^2\right)$. Let $Q(x,y)$ be this assertion. Furthermore, $P(0,y)$ implies, that $f$ is bijective. Therefore, let $a\in\mathbb R$ be, such that $f(a)=0$. Consequently: $$ Q(a,a):\space a^2+f(0)^2=0\iff a=f(0)=0 $$ Thus, $f(f(y))=y$. Now we have: $$ P(x,0):\space f(x^2)=f(x)^2\implies f(x)≥0\space\forall x\in\mathbb{R_{≥0}}\\ P(-x,0):\space f(x^2)=f(-x)^2\implies f(-x)^2=f(x)^2\implies f(-x)=-f(x)\space\left(\text{$f(x)=f(-x)$ is impossible, because $f$ is bijective.}\right)\\ P\left(x,f(y)\right):\space f\left(x^2+y\right)=f(y)+f(x)^2=f(y)+f(x^2)\implies f(y+z)=f(y)+f(z)\forall y\in\mathbb R, z\in\mathbb{R_{≥0}} $$ Now, we calculate $f(y-z)$ with $y\in\mathbb R, z\in\mathbb{R_{≥0}}$: $$ f(y-z)=-f(-y+z)=-f(-y)-f(z)=f(y)+f(-z) $$ Thus, we have $f(y+z)=f(y)+f(z)\space\forall y,z\in\mathbb R$. But we have $f(x)≥0\space\forall x\in\mathbb{R_{≥0}}$, so $f$ doesn't lay dense, and by Cauchy, we have that $f(x)=cx$ for some real constant $c$. Substituting this in the original equation, we can see, that $c=1$ and thus $f(x)=x$ is the only solution.

Redundant Aunt
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5

Here is an elementary way to show that $ f=1_\mathbb R$

Claim 1: $f$ is an odd function.

If we substitute $x=y=0$ in the given equation, then we get $$f(f(0))=(f(0))^{2}.$$ We then substitute $x=0$ in the given equation and find out that $\forall y \in \Bbb R$, $$f(f(y))=y+(f(0))^{2}.$$ Now, we observe that, for all $x,y \in \Bbb R$, $$y+(f(x))^{2}=f(x^{2}+f(y))=f((-x)^{2}+f(y))=y+(f(-x))^{2}.$$ Hence $\forall x \in \Bbb R,$ $$f(-x)=f(x) \space \space or \space \space f(-x)=-f(x).$$ But, if for some $x \in \Bbb R$, $f(-x)=f(x)$, then we would have, $$x+(f(y))^{2}=f(y^{2}+f(x))=f(y^{2}+f(-x))=-x+(f(y))^{2}$$ for any $y \in \Bbb R$ implying $x=0$. So, for any $x \neq 0$, $f(-x)=-f(x).$
.Hence $f$ is an odd function.

Claim 2: $f(x)>0$ if $x>0$ and $f$ is increasing.

Note that $f(x)\neq 0$ if $x\neq 0$. Indeed, if $f(x)=0$, then $$0=f(0)=f(f(x))=x.$$ Furthermore, if $x>0$, then $f(x)>0$. Indeed, $f(x)=f(\sqrt{x}^2)=f(\sqrt{x})^2>0$.

Then in fact, $f$ is increasing. Since $f$ is odd, it suffices to show it is increasing on $(0,\infty)$. Well, if $x>y>0$, $$f(x)-f(y)=f(\sqrt{x}^2)+f(-y)=f(\sqrt x)^2+f(-y)=f(x+f(f(-y)))=f(x-y)>0.$$

But then, if $f(x) >x$, $x>f(f(x))>x$. Likewise, if $f(x)<x$, $x=f(f(x))<f(x)<x$. As these are impossible, we must have $f(x)=x$ for all $x$

Arpit Kansal
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