All functions $g:\mathbb{N}\to\mathbb{Z}$ such that $m+n~|~g(m)+g(n)$

Question

I just thought about the following question: Find all functions $g:\mathbb{N}\to\mathbb{Z}$ such that $m+n~|~g(m)+g(n)$ for all $m,n\in\mathbb{N}$.

Clearly every polynomial $g(X)\in\mathbb{Z}[X]$ in which $X$ only occurs to an odd power is a solution.

Taking $m=n$ gives $m~|~g(m)$ and hence, introducing $h(m)=m\cdot g(m)$, we can rewrite the condition as $m+n~|~n(h(m)-h(n))$ for all naturals $m$ and $n$. This in turn is equivalent to $m+n~|~h(tm)-h(tn)$ for all naturals $m,n,t$.

But now I am stuck. Do you have any ideas or solutions? Do you think/spot/know that there are more solutions than the mentioned polynomials?

Answer 1

First, note that the set of functions satisfying your condition has the property that it is closed under integer linear combinations. Give two such functions, $p,q,$ and integers $a,b$, the function $r(n)=ap(n)+bq(n)$ has the above property.

Let $p_k(x)=x(x^2-1)\dots(x^2-k^2)$. This is a polynomial with only odd degrees.

Then for any sequence of integers $a_k$, you can define:

$$p(n)=\sum_{k=0}^\infty a_kp_k(n)$$

This is well defined because for $k>n$, $p_k(n)=0$, so it is, locally, a finite sum for any $n$. (Left to the reader: show $p(n)$ has your property. It does.)

This means that there are uncountably many such functions.

This is related to an old question of mine: finding all $f:\mathbb N\to\mathbb Z$ with $m-n\mid f(m)-f(n)$. There, I answered my own question, but I think your question might be slightly different. Still, the above is "like" what I did in that older question, so possibly worth checking out.

In particular, if $g:\mathbb Z\to\mathbb Z$ is an odd function with $\forall m,n:m-n\mid f(m)-f(n)$, then $g$ restricted to the natural numbers is in your class. These functions are classified by a sequence of integers, $a_k$ using the functions:

$$\sum_{k=0}^\infty a_k\frac{\mathrm{lcm}\{1,\dots,2k+1\}}{(2k+1)!}p_k(x)$$

But your condition is really that such odd function satisfy $m-n\mid f(m)-f(n)$ only for $n<0$ and $m>0$. So that might allow for more functions.

But I think I can outline a proof much like my proof for my question. Namely, if we have the values $f(0),\dots,f(n)$, that determines the value of $f(n+1)$ modulo ${\mathrm{lcm}\{n+1,\dots,2n+1\}}$ since we know:

$$f(n+1)\equiv -f(d)\pmod{n+1+d}$$ for $d=0,\dots,n$. (The case $d=n+1$ follows from the case $d=0$.)

But $\mathrm{lcm}\{n+1,\dots,2n+1\} = \mathrm{lcm}\{1,\dots,2n+1\}$.

So we can inductively find $a_k$ so that:

$$f(n)=\sum_{k=0}^\infty a_k\frac{\mathrm{lcm}\{1,\dots,2k+1\}}{(2k+1)!}p_k(n)$$

This also means that your condition implies $\forall m,n.m-n\mid g(m)-g(n)$ too. I wondered if there was a direct proof of this, and I asked here and got a quick easy proof.

That easy proof lets you re-use my result directly when classifying your functions, since it means that your $g$ can be extended to a function $g_1:\mathbb Z\to\mathbb Z$ with $m-n\mid g_1(m)-g_1(n)$ and $g_1(-m)=-g_1(m)$. So that lets us use my classification result.