Finding all real valued functions that satisfy $f(f(y) + xf(x)) = y + (f(x))^2$

Question

I would like some help with finding all real valued functions that satisfy this equation:

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

I tried the usual substitutions like $x = y = 0$, but my experience with this kind of problem is very limited.

EDIT: I'm an idiot and copied the wrong right side. Updated it now.

$f(x) = x$ is a solution, I think, but this was just by guessing. — flawr, Sep 17 '14 at 18:19
Let us try $x=0$. then $f^2(y) = y + f^2(0)$. This suggest that $f(x) = x+c$ is a solution.
For $y=0$ we then get $f(f(0) +xf(x)) = f^2(x) \implies f(c+x(x+c)) = x+2c \implies 2c+x(x+c) = x+2c \implies x(x+c)=x \implies x+c = 0 \text{ for } x \neq 0 \implies x=c$ which is a contradiction. So $c=0$ if $f$ has the form $f(x) = x+c$ — flawr, Sep 17 '14 at 18:28
@RobinGoodfellow so sorry, i copied that exponent wrong :\ it's actually $(h(x))^2$. — Lucas Virgili, Sep 17 '14 at 19:18
This is a BMO (Balkan Math Olympiad) problem. See here for a clear solution. — user26486, Sep 17 '14 at 19:27

score 2 · Accepted Answer · answered Sep 17 '14 at 19:16

2

$$f(f(y) + xf(x)) = y + f^2(x)$$ Let $x=0$, then : $$f(f(y))=y+f^2(0)$$ So as $y$ is one-one and onto and invertible, that said that $f$ is onto, so there's no harm supposing that $f(x)=0$ or $x=f^{-1}(0)$: $$f(f(y))=y\implies f(y)=f^{-1}y$$ Does that give you a hint[$f$ is a one-one onto invertible self-inverse function]. Try $f(x)=x$ or $f(x)=-x$.

answered Sep 17 '14 at 19:16

RE60K

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sorry for the trouble, i copied the problem incorrectly :\ – Lucas Virgili Sep 17 '14 at 19:20
@LucasVirgili you have copied wrong question, but I have given correct answer already because I have met this question quite a few times before – RE60K Sep 17 '14 at 19:26
I shall try it, thanks :) – Lucas Virgili Sep 17 '14 at 19:28

Finding all real valued functions that satisfy $f(f(y) + xf(x)) = y + (f(x))^2$

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