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.
