What is the solution, $f(n)$, of the following functional equation: $mf(m)+nf(n)=(m+n+xmn)f(m+n+xmn)$?

Question

What is the solution, $f(n)$, of the following functional equation:

$$mf(m)+nf(n)=(m+n+xmn)f(m+n+xmn) ,$$

where $f$ takes on integer values, $m$ and $n$ are integers, and $x$ is an indeterminate? It is a fundamental step in the proof of a famous theorem of Weierstrass that a non-rational meromorphic function which admits an algebraic addition theorem is necessarily periodic. The equation, due to A.R. Forsyth, is "solved" by him according to his following description: "Since the left-hand side is the sum of two functions of distinct and independent magnitudes, the form of the equation shows that it can be satisfied only if $x= 0$,so that..."

I am unable to follow this proof that necessarily $x=0$. If one can show it, then it is easy to show that the only solution of the functional equation is $f(n)= a$ constant.

Answer 1

Let $\;g(n):=nf(n).\;$ If $x=0$ then the functional equation is $\;g(n)+g(m)=g(n+m)\;$ for all $n,m\in\mathbb{Z},\;$ which is Cauchy's functional equation for integers and the general solution is $g(n)=cn$ which implies $\;c=f(n)\;$ for all $\;n\ne 0, n\in\mathbb{Z},\;$ while $f(0)$ is arbitrary.

Now assume $x\ne 0,\;\phi(n):=1+xn,\;h(\phi(n))=g(n),\;$ and let $n\oplus m:=n+m+xnm\;$ where $\phi(n\oplus m)=\phi(n)\phi(m).\;$ Then the functional equation is $\;g(n)+g(m)=g(n\oplus m)\;$ for all $n,m\in\mathbb{Z},\;$ but now rewriting it as $\; h(\phi(n))+h(\phi(m))=h(\phi(n)\phi(m))\;$ leads to $\;k\:h(\phi(n))=h(\phi(n)^k)\;$ for all $k,n,m\in\mathbb{Z}\;$ with $k>0$.

Given $n\ne0,\;$let $\;t:=(\phi(n)^k-1)/x.\;$ Now $\;g((n-1)/x)=h(n),\;$ thus $\;k\:n\:f(n)=t\:f(t).$ Now, assuming that $\;|f(t)|\ge b>0\;$ for $k$ big enough, then the right side grows exponentially and the left linearly which is a contradiction. Thus eventually $f(t)=0$ and hence $f(n)=0.$ Note that $|f(t)|\ge b\;$ is implied by $f(t)$ being a nonzero integer.

My proof is very similar to the one in the other answer, but has a few more details. I think Forsyth's remark may be similar to the one in that too narrow margin of Fermat.

Added note: I implicitly define $h$ by $h(n)=g((n-1)/x)$ only for $n=1\pmod{x}$ and this is equivalent to $h(\phi(n))=g(n).$

Answer 2

Perhaps Forsyth meant the remark as a sort of heuristic. The paper linked in the comments by GH from MO mentions that his book is non-rigorous and has been criticized for this.

Nevertheless, we can give a rigorous proof that pretty much follows Forsyth's description. Suppose that there is such a function $f:\mathbb Z\to \mathbb Z$ and $x\neq 0$. Let $a> 1$ be some integer that is $1\pmod x$. Then our functional equation gives us for any natural $u,v$. $$\frac{a^u-1}{x}f\left(\frac{a^u-1}{x}\right)+\frac{a^v-1}{x}f\left(\frac{a^v-1}{x}\right)=\frac{a^{u+v}-1}{x}f\left(\frac{a^{u+v}-1}{x}\right)$$ which in particular shows that the left hand side only depends on $u+v$. This means that there are integers $b,c$ such that $$\frac{a^u-1}{x}f\left(\frac{a^u-1}{x}\right)=b+cu$$ and so for any large enough $u$ we must have $f\left(\frac{a^u-1}{x}\right)=b+cu=0$ since otherwise the left side is much larger in magnitude than the right side. This in turn means that $f\left(\frac{a^u-1}{x}\right)=0$ for any $u$. To conclude notice that for any $m\neq 0$ $$f(m)=f\left(\frac{(mx+1)^1-1}{x}\right)=0$$ and of course, $f(0)$ can be arbitrary.

Side remark: As a minor correction, the solution when $x=0$ is $f(m)=\text{constant}$ for all nonzero $m$ and an arbitrary value for $f(0)$, since our equation does not give any information about $f(0)$.