# Find all functions $f:\Bbb{R} \to \Bbb{R}$ such that for all $x,y,z \in \Bbb{R}$ , $f(xf(x)+f(y))=x^2+y$

Find all functions $$f:\Bbb{R} \to \Bbb{R}$$ such that for all $$x,y, \in \Bbb{R}$$ , $$f(xf(x)+f(y))=x^2+y$$

We can easily get a strong condition $$f(f(y))=y$$ by setting $$x=0$$ . By this equation we know $$f$$ is injective and surjective. I got lost from there. By observation I know $$f(x)=x$$ and $$f(x)=-x$$ are solution. So I was trying to make $$x^2+y=f(xf(x)+f(y))$$ close to $$f(x)^2+y$$ or $$x^2+f(y)$$. Any hints would be helpful.

• If you substitute $x$ by $f(x)$, then using $f(f(x))=x$ you get $f(x)^2=x^2$. – Redundant Aunt Apr 1 at 12:01
• Let $f$ be such a function. For all $x,y\in\mathbb{R}$ we have $$f(xf(x)+f(y))=x^{2}+y=f(-xf(-x)+f(y))$$ As $f$ is bijective it follows that $$xf(x)+f(y)=-xf(-x)+f(y)\Leftrightarrow f(x)=-f(-x).$$ From this it immediately follows that $f(0)=0$. Also let $x\in\mathbb{R}$ such that $f(x)=1$, then $x^{2}=f(xf(x)+f(0))=f(x)=1$, so $x=\pm1$. For all $x\in\mathbb{R}$ it now also follows that $$f(x(f(x))-f(x^{2}))=x^{2}-x^{2}=0\Leftrightarrow xf(x)=f(x^{2})$$ – Floris Claassens Apr 1 at 12:14
• artofproblemsolving.com/community/q1h1675275p10669235 – Sil Apr 6 at 11:07

## 3 Answers

Let $$f:\mathbb{R}\to\mathbb{R}$$ and for $$x,y\in\mathbb{R}$$ denote by $$P(x,y)$$ the assertion $$f(xf(x)+f(y))=x^2+y$$. Assume that $$f$$ satisfies $$P(x,y)$$ for all $$x,y\in\mathbb{R}$$. Then $$P(0,y):\quad f(f(y))=y\implies f\text{ bijective}\\ P(f(x),y):\quad f(f(x)f(f(x))+f(y))=f(x)^2+y\implies x^2+y=f(xf(x)+f(y))=f(f(x)f(f(x))+f(y))=f(x)^2+y\\ \implies f(x)^2=x^2$$ We are no left to prove that either $$f(x)=x$$ for all $$x$$, or $$f(x)=-x$$ for all $$x$$, i.e. that $$f$$ doesn't jump around between $$x\mapsto x$$ and $$x\mapsto -x$$. For this, assume that there are $$a,b\in\mathbb{R}\backslash\{0\}$$ with $$f(a)=a$$ and $$f(b)=-b$$. Then $$P(a,b):\quad f(a^2-b)=a^2+b$$ Now if $$f(a^2-b)=a^2-b$$ then $$a^2-b=a^2+b$$ and thus $$b=0$$, contradiction. But if $$f(a^2-b)=-(a^2-b)$$ then $$-(a^2-b)=a^2+b$$ and thus $$a=0$$, so again contradiction. Therefore, such $$a,b$$ don't exist, and thus either $$f(x)=x\ \forall x$$ or $$f(x)=-x\ \forall x$$, and one can verify easily that this are indeed solutions to the equation.

Plugging in $$x=y=-1$$ shows that $$f(0)=0$$ and plugging in $$x=0$$ then shows that $$f(f(y))=y$$. This implies that $$f$$ is invertible, and plugging in $$y=0$$ shows that $$f(xf(x))=x^2=f(f(x^2)),$$ and hence $$xf(x)=f(x^2)$$ for all $$x\in\Bbb{R}$$. This shows that $$f(-x)=-f(x)$$ for all $$x\in\Bbb{R}$$, and that $$f(x^2+y)=xf(x)+f(y)=f(x^2)+f(y),\tag{1}$$ for all $$x,y\in\Bbb{R}$$, from which it follows that $$f$$ satisfies Cauchy's functional equation $$f(x+y)=f(x)+f(y).$$ Much has been said about this functional equation, which has many pathological solutions. Note that this means $$f$$ is $$\Bbb{Q}$$-linear, and if $$f$$ is either continuous at a point, bounded on an interval or monotonic on an interval, then $$f$$ is $$\Bbb{R}$$-linear and so $$f(x)=cx$$ for some $$c\in\Bbb{R}$$.

In an earlier version of this answer I rushed to the conclusion that $$f(x)=cx$$, which quickly implies that $$c=\pm1$$ and indeed both functions $$f(x)=\pm x$$ satisfy the functional equation.

• I fail to see how you can conclude that $f(x)=\pm x$ at the end. – Floris Claassens Apr 1 at 14:02
• @FlorisClaassens So do I, unfortunately I rushed to my conclusion. – Servaes Apr 2 at 0:10

\begin{align} &\text{As already noted, we have }\qquad\qquad\qquad\qquad\qquad f(f(y)) = y \implies f(x) = f^{-1}(x) \\ &\text {Also, using the substitution } x\to y \text{ we get}\,\,\quad f((y+1)f(y)) = y^2+y \to f(0) = 0\\ &\text{using } f= f^{-1}:\quad f(x*f(x) + f(y)) = x^2+y \implies x*f(x) + f(y) = f(x^2+y)\\ &\text{from above line we then get}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad f(x) = \frac{f(x^2+y) -f(y)}{x} \end{align}

From $$f(x) = \frac{f(x^2+y) -f(y)}{x}$$ we now obtain $$xf(x) = f(x^2)$$ by setting $$y=0$$.

A substitution $$x\mapsto f(x)$$ now yields $$f(x)x = f(f(x)^2)$$.

Combining those two equations, we get $$f(x^2) = f(f(x)^2) \implies x^2 = f(x)^2 \implies f(x) = \pm x$$