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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.

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    $\begingroup$ If you substitute $x$ by $f(x)$, then using $f(f(x))=x$ you get $f(x)^2=x^2$. $\endgroup$ – Redundant Aunt Apr 1 at 12:01
  • $\begingroup$ 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 $$f(x(f(x))-f(x^{2}))=x^{2}-x^{2}=0\Leftrightarrow xf(x)=f(x^{2})$$ $\endgroup$ – Floris Claassens Apr 1 at 12:14
  • $\begingroup$ artofproblemsolving.com/community/q1h1675275p10669235 $\endgroup$ – Sil Apr 6 at 11:07
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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.

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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.

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    $\begingroup$ I fail to see how you can conclude that $f(x)=\pm x$ at the end. $\endgroup$ – Floris Claassens Apr 1 at 14:02
  • $\begingroup$ @FlorisClaassens So do I, unfortunately I rushed to my conclusion. $\endgroup$ – Servaes Apr 2 at 0:10
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$$ \begin{align} &\text{As already noted, we have }\qquad\qquad\qquad\qquad\qquad f(f(y)) = y \implies f(x) = f^{-1}(x) \\ &\text {Also, using the substitution } x\to y \text{ we get}\,\,\quad f((y+1)f(y)) = y^2+y \to f(0) = 0\\ &\text{using } f= f^{-1}:\quad f(x*f(x) + f(y)) = x^2+y \implies x*f(x) + f(y) = f(x^2+y)\\ &\text{from above line we then get}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad f(x) = \frac{f(x^2+y) -f(y)}{x} \end{align}$$

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 $f(x^2) = f(f(x)^2) \implies x^2 = f(x)^2 \implies f(x) = \pm x$

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