Find all functions $f: \mathbb{N}^*\to \mathbb{Z}$ satisfies $$f(x+|f(y)|)=x+f(y), \forall x, y\in\mathbb{N}^*$$ My current progress:
- For $f(y)\geq 0$, it's obvious that $f(x)=x$ satisfies.
- For $f(y)<0$, then $f(x-f(y))=x+f(y)$. Because of the condition, $x>0>f(y)$.
... And that's it. I don't know what to do next, because I'm only familiar with $f: \mathbb{R}\to \mathbb{R}$.
First, we show that $f(y)\leq 0 , \forall y $ is impossible. Suppose this is the case then $y=1, x=|f(1)|+1 $ gives $ f(2|f(1)|+1)=1>0 $ so there is some $x_0 = 2|f(1)|+1 $ such that $f(x_0)>0$.
Now let inf$\{ x\in \mathbb{N} | f(x)>0 \}=c$ and let $f(c)=y$ then we claim $f(x)=x , \forall x > y$ as any $x>y$ can be written as $x=y+t $ then putting this in the first equation we get $f(x)=f(t+y)=f(t+f(c))=t+f(c)=t+y=x$. Now since c is infimum we get $f(k)\leq 0 , k<c$. Now for some $k<c$ if $|f(k)|\geq 1$ then $f(x+|f(k)|)=x+f(k)<x+|f(k)| $ Now we can choose x large enough so that $x+|f(k)|>y$ which gives contradiction.
So all possible solutions are $f(x)=0 , x<k$ and $ f(x)=x , x\geq k$, where $k \in \mathbb{N}.$ Thanks for pointing out the mistake in the previous proof. I hope this is correct.