Functional equation $f(n+f(n))=f(n)$

Question

Solve my functional equation:

$$ f(n+f(n))=f(n)$$

if $f:\mathbb{N}\rightarrow \mathbb{M}$ and $f(k)=1$ for some $k$. Do you have any hints? I'm sorry I won't show much work, but I really struggle with this problem. If we take this $k$, we have:

$$f(k+f(k))=f(k+1)=f(k)=1 $$

but what do next?

Answer 1

As you have already found $f$ is constant for $n\ge k$. Now let $n\le k$ such that $f(n)\neq 1$ but $f(n)=f(n+f(n))=f(n+2f(n))...=f(n+kf(n))=1$ . Thus we have proved that $f(n)=1$ for all n.

Answer 2

Assuming zero is not a natural number:

1) Let $f (c)=d $. $f (c+d)=f (c+f (c))=f (c)=d $. $f (c+2d)=f (c+d +d)=f (c+d+f (c+d)=f (c+d)=d $ and by induction $f (c+md)=d $

2) Let $f(k)=1$. $f (1+k)=f (k+f (k))=f (k)$ and by induction $f (n)=1$ for $n\ge k$.

3) $f (c+kd)=d $. But $kd+c \ge k+c \ge k $ so $f (c+kd)=1$. So $f (c)=1$ and therefore $f (n)=1$ for all natural numbers.

If $0$ is considered a natural number 1) and 2) are still valid. 3) is valid only if $d >0$ but if $f (c)=0$ we have no inconsistancy provided $c <k $.

Thus we can conclude: there is some $k $ so that $f (n)=0 \forall n <k$ and $f (n)=1\forall n\ge k $. ($k $ could still be $0$ and $f (n)=1$ but $k $ could be any natural number.)