# Functions satisfying: $f(f(x)^2+f(y))=xf(x)+y$

The problem is to find all the continuous functions $$f:\mathbb{R}\to \mathbb{R}$$ defined by :$$f(f(x)^2+f(y))=xf(x)+y$$

I'm trying my best to figure out a way to find the expression of this unknown function by plugging some numbers, but I could not. Thanks in advance for your help.

• Hint: Taking $x=y=0$ we see that $o$ is in the range of $f$. Let $f(x)=0$. The the equation gives $f(f(y))=y$ for all $y$. Commented Jun 17, 2019 at 0:03

The answers are $$f(x) = \pm x$$.

First, plug in $$x = 0$$ and $$y = 0$$, to get that $$f(f(0)^2 + f(0)) = 0$$. Therefore, let $$x = f(0)^2 + f(0)$$. Then, we get $$f(f(y)) = y$$. Then, plug in $$y = f(0)^2 + f(0)$$. We get that $$f(0) = f(0)^2 + f(0)$$, or that $$f(0) = 0$$.

Now, let $$f(a) =1$$. Clearly, $$f(1) =a$$ as well. Now, plugging $$x = a, y = 1$$ in to the original equation, we get that $$f(a+1) = a+1$$. However, plugging in $$x = 1, y = 1$$ gives us that $$f(a^2 + a) = a + 1$$, showing that $$a^2 + a = a + 1$$, which means that $$a = \pm 1$$.

Case 1: $$f(1) = 1$$. Let $$x = 1$$. Then, we get $$f(1+f(y)) = 1+y$$. Now, let $$y = 1$$. Then, we get that $$f(f(x)^2 + 1) = xf(x)+1$$. However, if, in the first equation, we let $$y = xf(x)$$, we get that $$f(1+xf(x)) = 1 + xf(x)$$, which means that $$f(x)^2 = x(f(x))$$ or that $$f(x) = x$$.

Proceed similarly in the second case.

Assuming $$f$$ is differentiable:

\begin{align} f(f(x)^2+f(y)) &= xf(x)+y \\ \frac{\partial{}}{\partial x}\left[f(f(x)^2+f(y))\right] &= f'(f(x)^2+f(y))\cdot 2f(x)f'(x) \\ \frac{\partial{}}{\partial x}\left[xf(x)+y\right] &= xf'(x)+f(x) \\ \Rightarrow f'(f(x)^2+f(y)) &= \frac{xf'(x)+f(x)}{2f(x)f'(x)} \tag{1}\\ \frac{\partial{}}{\partial y}\left[f(f(x)^2+f(y))\right] &= f'(f(x)^2+f(y))\cdot f'(y) \\ \frac{\partial{}}{\partial y}\left[xf(x)+y\right] &= 1 \\ \Rightarrow f'(f(x)^2+f(y)) &= \frac{1}{f'(y)} \tag{2} \end{align} From $$(1)$$ and $$(2)$$ we can see that $$f'(y)$$ is independent of $$y$$. So $$f(x) = ax + b$$. Solving this for $$f(f(x)^2+f(y))=xf(x)+y$$, we get $$a=\pm 1$$ and $$b=0$$. So $$f(x)=\pm x$$.

Obs.: we can discard $$f(x) = 0$$ and $$f'(x) = 0$$ by inspection, so $$(1)$$ is well defined.

By plugging in $$x=0$$, we see that

$$f(f(y)+f(0)^2)=y,$$

so $$f$$ is surjective (as $$y$$ can be any real), injective (as we may find $$y$$ from $$f(y)$$), and thus bijective. In addition, letting $$x$$ be so that $$f(x)=0$$, we see that

$$f(f(y))=y.$$

Now plug in $$x=f(t)$$ to see

$$f(f(f(t))^2+f(y))=f(t)f(f(t))+y$$

$$f(t^2+f(y))=tf(t)+y=f(f(t)^2+f(y)).$$

Now, as $$f$$ is injective, we may apply $$f^{-1}$$ to both sides to see

$$t^2+f(y) = f(t)^2+f(y) \implies f(t)\in\{-t,t\}.$$

It is easily seen that $$f(x)=x$$ for all real $$x$$ and $$f(x)=-x$$ for all real $$x$$ are solutions; we now show that they are the only ones. Assume for the sake of contradiction that for some real $$x\neq 0$$, $$f(x)=x$$, and for some real $$y\neq 0$$, $$f(y)=-y$$. Then

$$f(f(x)^2+f(y))=xf(x)+y \implies f(x^2-y)=x^2+y.$$

This means that $$x^2-y=x^2+y$$ or $$y-x^2=y+x^2$$, the first of which implies that $$y=0$$ and the second of which implies that $$x=0$$, each a contradiction, finishing the proof.

Note that this doesn't use continuity at all -- so these are the only solutions regardless of whether $$f$$ is required to be continuous.