Find all functions $f: \mathbb{N}^\to \mathbb{Z}$ satisfies $f(x+|f(y)|)=x+f(y), \forall x, y\in\mathbb{N}^$

Question

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 $f: \mathbb{R}\to \mathbb{R}$.

" it's obvious that f(x)=x satisfies." But is that the only one? You've been asked to find them all. — fleablood, Sep 20 '21 at 06:09
It's obvious that $f(x)=x$ satisfies, but it not obvious it is the only one that does. So one must show that it is the only one. — fleablood, Sep 20 '21 at 16:27
Also you $f(y)$ need not be either $\ge 0$ nor $< 0$ for all values of $y$. — fleablood, Sep 20 '21 at 16:29

Subhajit Ghosh · Answer 1 · 2021-09-20T10:23:09.843

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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.

edited Sep 20 '21 at 10:23

answered Sep 20 '21 at 04:42

Subhajit Ghosh

221

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There's still a problem: if there is some $ x _ 0 \in \mathbb N ^ * $ with $ f ( x _ 0 ) = 0 $, then setting $ x = y = x _ 0 $ in the functional equation, we get $ x _ 0 = 0 $, which is a contradiction. So all of the functions you mentioned, other that identity itself, are ruled out. – Mohsen Shahriari Sep 20 '21 at 16:14
Nice observation @ Mohsen. I completely missed it !! Thanks. – Subhajit Ghosh Sep 20 '21 at 16:17

Find all functions $f: \mathbb{N}^*\to \mathbb{Z}$ satisfies $f(x+|f(y)|)=x+f(y), \forall x, y\in\mathbb{N}^*$

1 Answers1

Find all functions $f: \mathbb{N}^\to \mathbb{Z}$ satisfies $f(x+|f(y)|)=x+f(y), \forall x, y\in\mathbb{N}^$