Keeping $x$ fixed and varying $y$, we see that the RHS spans overs $\mathbb{Z}$, meaning $f$ is surjective. Let $f(a)=0$. Setting $x=y=a$ gives $f(a)=-a$ so $-a=0$ i.e. $a=0$. Setting $x=0$ we obtain $f(f(y))=-y$. Now $f(x+f(y))=f(x)-y\implies f(f(x+f(y)))=f(f(x)-y)$. Now since $f(f(y))=-y$ for all $y$, clearly $f(f(x+f(y)))=x+f(y)$. And note that setting $x=-y,y=x$ in the original equation implies that $f(f(x)-y)=f(-y)-x$. And so we have that $x+f(y)=f(-y)-x$ i.e. $2x=f(-y)-f(y)$. This is clearly impossible if we vary $x$ and keep $y$ fixed. So no $f$ exists. This seems a bit too easy, is there anything wrong with the solution?

Alternate solution: From the previous solution, we can conclude that $f(0)=0$. Plugging $x=-f(y)$ we see that $f(-f(y))=y$ Plugging $y=-f(y)$ we get $f(x+y)=f(x)+f(y)$ which gives $f(x)=cx$ But, plugging it into the original equation we see that it doesn't work, so no solutions.

easy to prove that f is bijective; let $y=0$,$f(x+f(0))=f(x)$hence $f(0)=0$,$f(f(y))=-y$ so$f(x+f(y))=f(x)+f(f(y))$ by bijectiveness,f satisfies Cauchy's funtion hence $f(x)=cx$ then$c^2=-1$contradiction!

WakeUp wrote: Setting $x=0$ we obtain $f(f(y))=-y$. Now $f(x+f(y))=f(x)-y\implies f(f(x+f(y)))=f(f(x)-y)$. Now since $f(f(y))=-y$ for all $y$, clearly $f(f(x+f(y)))=x+f(y)$. Am I missing something, or should $f(f(x+f(y))) = -x-f(y)$ instead of $x+f(y)$? filipbitola's solution seems correct though.

indeed - hence my scepticism.. Instead we have that $-x-f(y)=f(-y)-x\implies -f(y)=f(-y)$, then proceed as filitbola's solution..

filipbitola wrote: Alternate solution: ... Plugging $y=-f(y)$ we get $f(x+y)=f(x)+f(y)$ which gives $f(x)=cx$ ... Are you sure about this? I know for conditions like $f$ have to be bounded or continuous or monotonic.. I'm not sure about bijectiveness Is it always true?

$f(x+y) = f(x) + f(y)$ gives the only solution $f(x) = cx$ if $f$'s domain is integers. (It also applies if $f$'s domain is rationals.) Just try: $f(0) = 0$ by plugging $x=y=0$. Now, letting $f(1) = c$ and plugging $y = 1$, we get $f(x+1) = f(x) + f(1) = f(x) + c$. By induction we can get that $f(x) = cx$ for all $x \in \mathbb{Z}^+$. Plugging $y = -x$ gives $f(-x) = -f(x)$, so $f(x) = cx$ for all integers $x$.

Assume there exists $a,b$ such that $f(a) = f(b)$. $P(a,b) : f(a+f(b)) = f(a) - b$ $(1)$. $P(a,a) : f(a+f(a)) = f(a) - a$ $(2)$. From $(1),(2)$ we have $a = b$ so $f$ is bijective. $P(0,0) : f(f(0)) = f(0) \implies f(0) = 0$. $P(0,x) : f(f(x)) = -x \implies f(x+f(y)) = f(x) + f(f(y))$. So $f$ is a Cauchy funtion and it's $\mathbb{Z}\to\mathbb{Z}$ so $f(x) = cx$ which gives contradiction with main equation. Answers : None.

$\boxed{\text{ no }}$. Clearly $f$ is surjective. If $f(a)=0$, then $P(x,a): f(x)=f(x)-a\implies a=0$. So $f(0)=0$. $P(-f(x),x): 0=f(-f(x))-x\implies f(-f(x))=x$. $P(x,-f(y)): f(x+y)=f(x)+f(y)$, since $f$ is over integers, $f(x)=cx$ for some constant $c$, which clearly doesn't work.