Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $f\left( f(x)^2+f(y) \right)=xf(x)+y$ for all real numbers $x$ and $y$.

Clearly $f(x)=x$ is a solution, check by substitution.

I'm at a loss as to how to show this is the only one or find others.

Any help would be very much appreciated.

  • $\begingroup$ Where is this from? $\endgroup$
    – zyx
    Apr 10, 2013 at 5:51
  • 1
    $\begingroup$ It's a problem for the training of high school math Olympians aiming to go to the IMO. $\endgroup$
    – John Marty
    Apr 10, 2013 at 5:53
  • $\begingroup$ I mean what competition was it originally from? $\endgroup$
    – zyx
    Apr 10, 2013 at 5:54
  • $\begingroup$ I'm not sure. I was just given it as a problem to think about, not as homework, just if you're wondering. $\endgroup$
    – John Marty
    Apr 10, 2013 at 5:56
  • $\begingroup$ I am the last person on this site to care about homework. We are having many conversations on the meta about those issues. Thanks for the nice puzzle, it is sort of therapeutic to make a long series of mindless deductions and sooner or later the thing cracks open. $\endgroup$
    – zyx
    Apr 10, 2013 at 5:59

1 Answer 1


lemmas in order they were noticed, all easy:

$f$ is injective

$f$ is surjective

$f(f(y)=y$ (let $f(x)=0$)

$f(-x)=-f(x)$ for nonzero $f(x)$ ($f$ is odd)

$f(0)=0$ by bijectivity and oddness

$f(f(x)^2)=xf(x)$ taking $y=0$

$f(u^2)=uf(u)$ by taking $x=f(u)$

$f(f(x)^2 + xf(x)) = xf(x) + x^2$ , (let $y=x^2$)

$f(x)^2=x^2$ (apply $f$ to both sides of preceding +injectivity)

$f(x) = \pm x$

We see now that $f(x)=-x$ is also a solution. Let $f(x) = s(x)x$ with $|s|=1$. Note $s(-x)=s(x)$

Taking $x$ with $x^2 > |y|$, $s(x)s(y)=1$ ($x$ sets the sign of each side, then compare the $y$ terms) so that $s(x)=s(y)$. Also $s(x^2)=s(x)$ for nonzero $x$ from $f(x^2)=xf(x)$.

Iterating this, the sign is constant on $|x|>1$ and on $|x|<1$. You can check by calculation whether there is a solution with the sign chosen differently on these intervals.

  • 2
    $\begingroup$ An alternative: After the 3rd point, $f(x^2+f(y))=f(f(f(x))^2+f(y))=f(x)f(f(x))+f(y)=xf(x)+f(y)=f(f(x)^2+f(y))$ so $x^2=f(x)^2$. Now if $f(a)=a, f(b)=-b, ab \not =0$, then $(a^2+b)^2=f(a^2+f(f(b)))^2=(af(a)+f(b))^2=(a^2-b)^2$, so $a^2b=0$, a contradiction. Thus $f(x)=x \, \forall x \in \mathbb{R}$ or $f(x)=-x \, \forall x \in \mathbb{R}$. $\endgroup$
    – Ivan Loh
    Apr 10, 2013 at 6:15
  • $\begingroup$ @Ivan (thanks), I did a quick search and this problem is very similar to IMO 1992 problem 2. As a contest person, do you know where this problem appeared? $\endgroup$
    – zyx
    Apr 10, 2013 at 6:17
  • $\begingroup$ Well a lot of these functional equations are similar and can be solved using the same approach. I do not know the source of this problem. $\endgroup$
    – Ivan Loh
    Apr 10, 2013 at 6:35
  • $\begingroup$ How do you prove surjectivity? I get stuck in trying to prove it. I understand the rest. $\endgroup$ Apr 10, 2013 at 6:47
  • $\begingroup$ @IshanBanerjee, adjust $y$ so as to hit any desired value. $\endgroup$
    – zyx
    Apr 10, 2013 at 6:53

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