IMO 2002 A1

[mutasimmim]

Find all functions $R\rightarrow R$ such that $f(f(x)+y)=2x+f(f(y)-x)$

[*Mahi*]

IMO Shortlist 2002 Algebra 1

Hint:


Prove that $f$ is surjective.
Use $f^{-1}(0)$

[mutasimmim]

Let $P(x,y)$ be the given assertion. Then,
[1]:
$P(x,0)\Longrightarrow f(f(x))=2x+f(f(0)-x)$.
[2]:
$P(0,y)\Longrightarrow f(f(y))=f(y+f(0))$
[3]
Combining $[1] $and $[2]$ we get $f(y+f(0))=2y+f(f(0)-y))$ that is, $f(y+c)-f(c)=y$ for any choice of $y$ and $c$.
Now $P(f(y),x)\Longrightarrow f(f(f(y)+y))=2f(y)+f(0)\Longrightarrow f(f(y+f(0))+y)=2f(y)+f(0)$ [Using [2]]
$\Longrightarrow f(f(y+f(0))+y)-f(y)=f(y)+f(0)$ Using [2] again, $f(y+f(0))=f(y)+f(0)\Longrightarrow f(y+f(0))-f(0)=f(y)\Longrightarrow y=f(y)$

[Nirjhor]

Show that $f$ is bijective and plug in $x=0$ to get $f(x)=x+c$ for any $c\in\mathbb{R}$.