# Seeking a more direct proof for: $m+n\mid f(m)+f(n)\implies m-n\mid f(m)-f(n)$

If $f:\mathbb N\to\mathbb Z$ satisfies:

$$\forall n,m\in\mathbb N\,, n+m\mid f(n)+f(m)$$

How to show that this implies:

$$\forall n,m\in\mathbb N,\,n-m\mid f(n)-f(m)?$$

I was almost incidentally able to prove this by classifying such functions, but that seems circuitous for such a result. Is there a proof that is (more) direct?

This is actually pretty easy. Let $n>m$, and take an $N$ such that $N(n-m)>m$. Set $a=N(n-m)-m$. Then $$m+a=N(n-m),\qquad n+a=m+a+n-m=(N+1)(n-m).$$ Now $$f(n)-f(m)=f(n)+f(a)-(f(m)+f(a)),$$ but by assumption $n-m\mid f(m)+f(a)$ and $n-m\mid f(n)+f(a)$, and we are done.
• i.e. $\ \ \ \begin{eqnarray} && n\!-\!m\!\!\!&&\mid \ n\!+\!a\!\!\!&&\mid \ \color{#0a0}{f_{\large n}\!+f_{\large a}}\\ \Rightarrow &&n\!-\!m\!\!\!&&\mid m\!+\!a\!\!\!&&\mid \color{#c00}{f_{\large m}\!+f_{\large a}}\end{eqnarray}\Bigg\}\,\Rightarrow\ n\!-\!m\mid (\color{#0a0}{f_{\large n}\!+f_{\large a}})-(\color{#c00}{f_{\large m}\!+f_{\large a}}) = f_{\large n}\!-f_{\large m}\$ – Bill Dubuque Jun 4 '14 at 22:57