A functional equation with no term outside functions

Question

Find all $f:\mathbb{N} \rightarrow \mathbb{N}$ satisfying $$f(m-n+f(n))=f(m)+f(n)$$ for all $m,n \in \mathbb{N}.$

I have no idea about how to find them, because there are no terms outside of the function.

Answer 1

Pick $n\in\mathbb N$, let $a=f(n)-n$, $b=f(n)$. Then we obtain $$ f(m+a)=f(m)+b$$ for all $m\in\mathbb N$. Specifically, $a=0$ is only possible if $b=0$. By induction $$f(m+ka)=f(m)+kb $$ Assume $f$ is not the constant zero map. Then for suitable $n$ we have $a\ne 0$. The quotient $\frac ba$ must be the same for all such $n$ because for $a',b'$ obtained from $n'$ we have $f(m+aa')=f(m)+ab'=f(m)+a'b$.

With $m=f(n)$ we find $f(f(n))=2f(n)$, hence with $f(n)\ne 0$ for $n$ we obtain $a=f(f(n))-f(n)=f(n)$ and $b=f(f(n))=2f(n)$, so $\frac ba=2$. We conclude that for all $n$ with $f(n)\ne0$ we have $f(n)=2\cdot(f(n)-n)$, i.e., $$f(n)\in\{0,2n\}.$$ If $n,m$ are nonzero and $f(n)=2n$ and $f(m)=0$ then $$f(m+n)=f(m-n+f(n))=f(m)+f(n) =2n\notin\{0,2(m+n)\}$$ so that this constellation is not possible. We conclude that either $f(n)=2n$ for all $n$ or that $f$ is identically zero.