WakeUp wrote: Determine all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy \[f(f(x)+y)=2x+f(f(y)-x)\quad\text{for all real}\ x,y. \] See http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=303332

Let P（x,y） be the assertion f（f（x）＋y）=2x＋f（f（y）－x） 1.f is surjective P（x,－f（x））,f（f（－f（x））－x）=f（0）－2x.So f is surjective. 2.f is injective Suppose that f（a）=f（b）=t.By surjective, w s.t. f（w）=a＋b. P（a,w）,f（t＋w）=2a＋f（b）=2a＋t. P（b,w）,f（t＋w）=2b＋f（a）=2b＋t. So a=b. P（0,y）,f（f（0）＋y）=f（f（y）） By injective, f（y）=y＋f（0）. So c∈R s.t. f（x）=x＋c.It is easy to check that this function satisfies the condition.So the answer is c∈R s.t. f（x）=x＋c（x∈R）

Claim: $f$ is surjective Proof: $P(x,-f(x)): f(0)-2x=f(f(-f(x))-x)$, which implies $f$ is surjective. $P(0,y): f(f(0)+y)=f(f(y))$. Claim: It suffices to show $f$ is injective. Proof: If $f$ is injective, then $f(0)+y=f(y)\implies \boxed{f(x)=x+c}$ for all $x\in\mathbb{R}$ and constants $c\in\mathbb{R}$, which works. Claim: $f$ is injective. Proof: Suppose $f(a)=f(b)$. Suppose $f(k)=a+b$. $P(a,k): f(f(a)+k)=2a+f(a)$. $P(b,k): f(f(a)+k)=2b+f(a)$. Subtracting the two equations gives $2a=2b\implies a=b$.

$P(f(x),x)\Rightarrow f(f(f(x))+x)=2x+f(0)\Rightarrow f$ is surjective Assume $f(a)=f(b)$ for some $a\ne b$ and set $f(k)=a+b$. $P(a,k)\Rightarrow f(f(a)+k)=2a+f(f(k)-a)=2a+f(b)=2a+f(a)$ $P(b,k)\Rightarrow f(f(a)+k)=2b+f(a)$ Contradiction. Hence $f$ is bijective. $P(0,x)\Rightarrow\boxed{f(x)=x+c},c\in\mathbb R$ which are solutions.

Let $P(x,y):=f(f(x)+y)=2x+f(f(y)-x)$ Claim: $f$ is surjective Proof: $P(f(x),x)$ yields $f(f(f(x))+x)=2x+f(0)$ thus $f$ is surjective $\square$. Claim: $f$ is injective Proof: Let $f(a)=f(b)$, furthermore since $f$ is surjective let $f(c)=a+b$ $P(a,c)$ yields $f(f(a)+c)=2a+f(f(c)-a)=2a+f(a+b-a)=2a+f(b)\Longrightarrow f(f(a)+c)-f(b)=2a$ $P(b,c)$ yields $f(f(b)+c)=2b+f(f(c)-b)=2a+f(a+b-c)=2a+f(a)\Longrightarrow f(f(b)+c)-f(a)=2b$ Thus from the last two results we obtain that $2a=2b\Longrightarrow a=b$ and thus $f$ is also injective $\square$. $P(0,x)$ yields $f(f(0)+x)=f(f(x))\overset{\text{injectivity}}{\Longrightarrow}f(x)=x+c$ where $c=f(0)$ So to sum up $\boxed{f(x)=x+c, \forall x\in\mathbb{R}\text{ and some constant }c}$ $\blacksquare$.