Find all the functions $f : \Bbb Z \rightarrow \Bbb Z$ with the property that for all integers $(x,y)$ the equation holds: $$f(x+f(f(y)))=y+f(f(x))$$
My solution: $f(x+f(f(y)))=y+f(f(x))$ means that also is possible
$$f(y+f(f(x)))=x+f(f(y))$$
Putting this in the first equation we get:
$$f[f(y+f(f(x)))]=y+f(f(x))$$
This is equal to $f(f(K))=K$ with $K=y + f(f(x))$, in other words, any number.
Replacing: $f(f(x))=x$ and $f(f(y))=y$ in $f(x+f(f(y)))=y+f(f(x))$:
$$f(x+y)=y+x$$
That's equal to $f(x)=x$.
It's this correct or not?, If it's wrong, any hints?.
Looks good. To be a bit more precise, you could explain that $y+f(f(x))$ can be any integer $n$ by choosing $x$ arbitrarily and then setting $y=n-f(f(x))$, and similarly for $x+y$ at the end of your argument.