math_mat wrote: ... Write $x+f(0)=A$ and $f(0)-x=B$ and we get : $f(A) - f(B) = A - B$ for all $A, B \in \mathbb{R}$. ... Unfortunately, this is wrong : you got $f(A)-f(B)=A-B$ only for those $A,B$ such that $A+B=2f(0)$, and not $\forall A,B$

pco wrote: math_mat wrote: ... Write $x+f(0)=A$ and $f(0)-x=B$ and we get : $f(A) - f(B) = A - B$ for all $A, B \in \mathbb{R}$. ... Unfortunately, this is wrong : you got $f(A)-f(B)=A-B$ only for those $A,B$ such that $A+B=2f(0)$, and not $\forall A,B$ Yes, I also saw my mistake and I was trying to fix it, without success apparently. I deleted the previous post.

it is problem A:1 IMO shortlist-2002. You can see IMO page.

djb86 wrote: Find all functions $f : \mathbb{R} \to \mathbb{R}$ that satisfy the condition \[f(f(x) + y) = 2x + f(f(y) - x)\quad \text{ for all } x, y \in\mathbb{R}.\] Let $P(x,y)$ be the assertion $f(f(x)+y)=2x+f(f(y)-x)$ Let $a=f(0)$ $P(\frac{a-x}2,-f(\frac{a-x}2))$ $\implies$ $x=f(f(-f(\frac{a-x}2))-\frac{a-x}2)$ and so $f(x)$ is surjective. $P(x,y)$ $\implies$ $f(f(x)+y)=2x+f(f(y)-x)$ $P(y,-x)$ $\implies$ $f(f(y)-x)=2y+f(f(-x)-y)$ And so, adding these two lines : $f(f(x)+y)=2x+2y+f(f(-x)-y)$ setting $y=\frac{f(-x)-f(x)}2$ in this last equation, we get new assertion $Q(x)$ : $f(-x)=f(x)-2x$ $\forall x$ $P(x+a,0)$ $\implies$ $f(f(x+a))=2x+2a+f(-x)$ $P(0,x+a)$ $\implies$ $f(x+2a)=f(f(x+a))$ $Q(x)$ $\implies$ $f(-x)=f(x)-2x$ Adding these three lines, we get $f(x+2a)=f(x)+2a$ $P(0,x)$ $\implies$ $f(f(x))=f(x+a)$ and so $f(f(f(x)))=f(f(x+a))=f(x+2a)=f(x)+2a$ $P(0,f(x))$ $\implies$ $f(f(f(x)))=f(f(x)+a)$ So $f(f(x)+a)=f(x)+2a$ And surjectivity implies then $\boxed{f(x)=x+a}$ $\forall x$ which indeed is a solution

Let $P(x,y)$ be the assertion $ f(f(x)+y)=2x+f(f(y)-x). $ $P(x,f(x))$ $ \Rightarrow $ $f(0)-2x=f(f(-f(x))-x) $ and so $f(x)$ is surjective. Let for some $c$, such that $f(c)=0$. From $x=c$ we have $ f(y)=2c+f(f(y)-c) $ $ \Rightarrow $ $f(f(y)-c)=(f(y)-c)-c$. Hence, $f$ is surjective then $f(x)=x+a$ for all real $x$.

Let $P(x,y)$ be that assertion. Now $P(x,-f(x))$ implies $f$ is onto so there exist $a$ such that $f(a)=0$ and now $P(a,x)$ implies $f(f(x)-a)=f(x)-2a$ also as $f$ is onto so we get $f(x)=x+a$.