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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.

Caleb Stanford
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John Marty
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  • Where is this from? – zyx Apr 10 '13 at 05:51
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    It's a problem for the training of high school math Olympians aiming to go to the IMO. – John Marty Apr 10 '13 at 05:53
  • I mean what competition was it originally from? – zyx Apr 10 '13 at 05:54
  • I'm not sure. I was just given it as a problem to think about, not as homework, just if you're wondering. – John Marty Apr 10 '13 at 05:56
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    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. – zyx Apr 10 '13 at 05:59
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    A source has been found. The first search hit at this answer, http://math.stackexchange.com/questions/735448/1992-imo-functional-equation-problem , is a PDF file of contest problems of which no. 6 is IMO 1992.2, and no.7 is this question, listed as "Balkan 2000/1". That is Balkan Mathematical Olympiad, 2000, problem 1. – zyx Apr 01 '14 at 18:24

1 Answers1

6

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.

user26857
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zyx
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    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 $f(x)=-x , \forall x \in \mathbb{R}$. – Ivan Loh Apr 10 '13 at 06:15
  • @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? – zyx Apr 10 '13 at 06:17
  • 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. – Ivan Loh Apr 10 '13 at 06:35
  • How do you prove surjectivity? I get stuck in trying to prove it. I understand the rest. – Ishan Banerjee Apr 10 '13 at 06:47
  • @IshanBanerjee, adjust $y$ so as to hit any desired value. – zyx Apr 10 '13 at 06:53
  • The proof of injectivity, requested in the other answer, is to compare the functional equation written using $(x,y)$ and $(x,z)$ as the variables. If $f(y)=f(z)$ for fixed $x$ then the left sides of the equations are equal, which compels $y=z$ by looking at the right hand sides. – zyx Apr 18 '13 at 22:50
  • @Ivan, in case you are interested, a source has been located, as explained in the comments under the question. – zyx Apr 01 '14 at 18:26