Determine all $f : \mathbb R \rightarrow \mathbb R$ such that $f(f(x) + y) = f(x^2 - y) + 4yf(x)$.

Question

Determine all $f : \mathbb R \rightarrow \mathbb R$ such that $$f\big(f(x) + y\big) = f(x^2 - y) + 4yf(x)$$ for all $x,y\in\mathbb{R}$.

$f(x) = 0$ is obvious , but how can we find all other functions?

Thanks in advance!

Answer 1

Let $f:\mathbb{R}\to\mathbb{R}$ satisfy $$f\big(f(x) + y\big) = f(x^2 - y) + 4y\,f(x)$$ for all $x,y\in\mathbb{R}$. By setting $y:=\dfrac{x^2-f(x)}{2}$ in the given functional equation, we get $$\begin{align}f\left(\frac{x^2+f(x)}{2}\right)&=f\left(f(x)+\frac{x^2-f(x)}{2}\right)\\&=f\left(x^2-\frac{x^2-f(x)}{2}\right)+4\,\left(\frac{x^2-f(x)}{2}\right)\,f(x)\\&=f\left(\frac{x^2+f(x)}{2}\right)+4\,\left(\frac{x^2-f(x)}{2}\right)\,f(x)\end{align}$$ for all $x\in\mathbb{R}$. Therefore, $4\,\left(\dfrac{x^2-f(x)}{2}\right)\,f(x)=0$ for each $x\in \mathbb{R}$. Hence, we can conclude that, for each real number $x$, either $$f(x)=0\text{ or }f(x)=x^2\,.\tag{*}$$ In particular, this means $f(0)=0$.

Suppose that there are two nonzero real numbers $s$ and $t$ such that $f(s)=0$ and $f(t)=t^2$. Then, by plugging in $x:=s$ and $y:=t$ in the given functional equation, we have $$t^2=f(t)=f\big(f(s)+t\big)=f(s^2-t)+4t\,f(s)=f(s^2-t)\,.$$ By (*), $f(s^2-t)=0$ or $f(s^2-t)=(s^2-t)^2$. Thus, either $$t^2=0\text{ or }t^2=(s^2-t)^2\,.$$ This means $t=0$, $s=0$, or $s^2=2t$. Since $s$ and $t$ are nonzero, we obtain $$s^2=2t\,.\tag{#}$$ However, (#) holds for any values of $s\neq 0$ and $t\neq 0$ such that $f(s)=0$ and $f(t)=t^2$. It can then be easily seen (why?) that, unless

• $f(x)=0$ for all $x\in\mathbb{R}$ or
• $f(x)=x^2$ for all $x\in\mathbb{R}$,

the equation (#) yields a contradiction.

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I shall also solve the old (now changed) version. Let $f:\mathbb{R}\to\mathbb{R}$ satisfy $$f\big(f(x) + y\big) = f(x^2 - y) + 4x\,f(x)$$ for all $x,y\in\mathbb{R}$. We proceed as before by setting $y:=\dfrac{x^2-f(x)}{2}$. However, this time we get $$4x\,f(x)=0$$ for all $x\in\mathbb{R}$. Thus, if $x\neq 0$, then $f(x)=0$. Now, by setting $x$ and $y$ to be $0$, we obtain $$f\big(f(0)\big)=f(0)\,.$$ If $f(0)\neq 0$, then $f\big(f(0)\big)=0$, whence $f(0)=0$, which is a contradiction. Therefore, $f(0)$ must be $0$, and so $f(x)=0$ for each $x\in\mathbb{R}$.

Answer 2

Here's another solution: $$P(x,y)\implies f(f(x)+y)=f(x^2−y)+4yf(x)$$ $$P(x,-f(x))\implies f(0)=f(x^2+f(x))-4f(x)^2 \iff f(x^2+f(x))=f(0)+4f(x)^2$$ $$P(x,x^2)\implies f(x^2+f(x))=f(0)+4x^2f(x)$$ Combining the above 2 statements:$$ \implies f(0)+4f(x)^2=f(0)+4x^2f(x) \iff f(x)^2=x^2f(x)$$ Putting $x=0$ we get that $f(0)=0$, and it gives two cases:

Case 1: $f(x)=0 \ \ \ \forall \ x \in \mathbb{R}$

Case 2: there exists a real number $a$ such that $f(a) \ne 0$

Then $$f(a)^2=a^2f(a) \implies f(a)=a^2$$ $$f(a) \ne 0 \implies a^2 \ne 0 \implies a \ne0$$ So $f(a)=a^2$ for all $a \ne 0$, this and the fact that $f(0)=0$ gives $f(x)=x^2$ for all $x$

Checking back in the original equation, we get that both solutions work, the only solutions $\Box.$