Find all function $f:\mathbb{R}\rightarrow \mathbb{R}$ so that $f(xf(y))+f(yf(x)=2xy$.

By putting $x=y=0$ we get $f(0)=0$ and by putting $x=y=1$ we get $f(f(1))=1$. Let $y=f(1)\Rightarrow f(x)+f(f(x)f(1))=2x$, which tells us that $f$ is an injective function. The only solutions I came up with so far are $f(x)=x$ and $f(x)=-x$.

  • 3
    $\begingroup$ $f(xf(x))=x^2=f(-xf(-x))$ and therefore $xf(x)=-xf(-x)$. Hence $f(x)=-f(-x)$. $\endgroup$
    – Pp..
    Commented Jan 6, 2015 at 19:47
  • $\begingroup$ @Pp : That is definition of all $ odd functions $ $\endgroup$
    – Narasimham
    Commented Jan 6, 2015 at 20:14
  • 2
    $\begingroup$ @Narasimham Yes, it is a property that $f$ must have. $\endgroup$
    – Pp..
    Commented Jan 6, 2015 at 20:15
  • $\begingroup$ See artofproblemsolving.com/community/q1h619883p3702109 $\endgroup$
    – Sil
    Commented Oct 13, 2019 at 7:05
  • $\begingroup$ "Let $y=f(1)\Rightarrow f(x)+f(f(x)f(1))=2x$", no, $y=f(1)$ gives $f(x)+f(f(1)f(x))=2xf(1)$. (It still implies injectivity and $f(1)^2=1$ by choosing $x=f(1)$) $\endgroup$
    – Sil
    Commented Oct 13, 2019 at 7:57

2 Answers 2


Putting $x=1$ in $f(x)+f(f(x)f(1))=2xf(1)$ we get that $f(1)=\pm 1$

Let $f(1)=1$.

$x\rightarrow xf(x), y\rightarrow \frac{1}{x} \Rightarrow f(xf(x)f(\frac{1}{x}))+f(x)=2f(x) \rightarrow f(x)f(\frac{1}{x})=1$.

Putting now $x\rightarrow \frac{1}{x}, y \rightarrow 1$ we get $\frac{1}{f(x)}+ \frac{1}{f(f(x))}=\frac{2}{x}$. Using $f(f(x))=2x-f(x)$ we get $(f(x)-x)^2=0$, and there we get one of our two solutions, $f(x)=x$. We get $f(x)=-x$ as the other solution by putting $f(1)=-1$.


This is all I have found so far:

Just keep feeding the snake with its own tail. As you noted $f(f(1))=1$. So with $x=y=f(1)$ we then see that $1=f(f(1))=(f(1))^2$ so that $f(1)=\pm 1$. As you almost correctly noted, we have for $y=f(1)$ that $$ f(x)+f(f(x)f(1))=2xf(1) $$ which indeed shows that $f$ is injective. With that observation, the comment by Pp.. works to conclude that $f$ is odd since $f(xf(x))=x^2=f(-xf(-x))$ then implies $f(x)=-f(-x)$.

Now, if we knew that $f$ was a polynomial it would be easy to conlude that $f(xf(x))=x^2$ implies $\deg f(\deg f+1)=2$ so $f$ has to have degree 1 in that case. Since $f(0)=0$ and $f(1)=\pm 1$ that would then imply $f(x)=\pm x$.

But if $f$ is not a polynomial, all I can tell this far is that $f$ is an injective, odd function with $f(0)=0$ and $f(1)=\pm 1$ and a bucnh of functional equations attached to it. Maybe someone else can elaborate further ...


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