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?

Solution 1:

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.