here is the problem enter image description here

Here is my solution :

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

$x=0; y=f(0)$ gives $f(f(f(0)=0=f(0)$ (because $f(f(0))=f(0) \iff f(f(f(0)))=f(f(0))=f(0)$)

$x=0$ gives $f(f(y))=-y$

$x=0 ; y=f(y)$ gives $f(-y)=-f(y)\iff f(-f(y))=y$

so $y=-f(y)$ gives $f(2x)=2f(x)$

now we can prove by induction that $f(nx)=nf(x)$

it is true for $n=2$

let's suppose that it is true for $n$

we have $f(nx+f(-f(x))=f(nx)+f(x)\iff f((n+1)x)=(n+1)x$

so it is true now we have $f(x)=xf(1)$

plugging this in the first equation gives $f(1)^2=-1$ which is impossible

I just want to know if my solution is right

In the official solution they prove that $f$ is bijective (which is true because $f(f(y))=-y$ ;) ) then they use cauchy function , they find that $f(x)=cx$ , plugging this in the first equation gives an impossible result.

  • $\begingroup$ @vadim123 Because from the first line, $f(f(0)) = f(0)$. $\endgroup$ – rogerl Mar 8 at 21:59
  • $\begingroup$ Looks right; you meant "which is impossible" near the bottom :) $\endgroup$ – rogerl Mar 8 at 22:01
  • $\begingroup$ Ok updated ..... $\endgroup$ – user600785 Mar 8 at 22:14
  • 2
    $\begingroup$ Possible duplicate of Functional equation. $\endgroup$ – Sil Apr 7 at 9:22

On the 5th line, I assume you meant to set $y=-f(x)$, which indeed yields $f(2x)=2f(x)$. The rest of the proof is fundamentally correct, though unclear at times (e.g. it was not immediately obvious to me why $f(x)=xf(1)\implies f(1)^2=-1$.) In an actual contest, you should justify and detail all of these processes.


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