Fermat's last theorem for entire functions

Question

Let $f,g,h$ be entire functions, i.e., holomorphic functions on $\mathbb{C}$. Suppose $f^n+g^n=h^n$ for some integer $n\geq2$. What can we say about $f,g,h$?

Clearly this is Fermat's last theorem for entire functions. I did a little search on the internet but, somewhat surprisingly, I found nothing relevant. Where can I find the answer? Thanks in advance. :)

Edit: In particular, I would like to know why there are no nontrivial solutions for $n\geq4$. Here a trivial solution is a solution of the form $f=ap,g=bp,h=cp$ where $a,b,c\in\mathbb{C}$ satisfy $a^n+b^n=c^n$ and $p$ is entire.

Answer 1

I found a solution for $n\geq4$ after browsing similar questions on this site. Here is a sketch:

The equation $X^n+Y^n=Z^n$ defines a smooth projective algebraic plane curve $X\subset\mathbb{P}^2$ of genus $(n-1)(n-2)/2\geq2$ if $n\geq4$. By the uniformization theorem, its universal cover is biholomorphic to the unit disk $\mathbb{D}$. Factoring out a common entire factor if necessary, we may assume $f,g,h$ have no common zero. Then the map $[f:g:h]:\mathbb{C}\to X\subset\mathbb{P}^2$ lifts to a holomorphic map $\mathbb{C}\to\mathbb{D}$, which is necessarily constant by Liouville's theorem. Thus $[f:g:h]$ is constant. This shows $(f,g,h)$ is a trivial solution.

Answer 2

Ribenboim discusses this in Chapter XIII, Section 2, of 13 Lectures on Fermat's Last Theorem. First, he proves,

Theorem. If $n\ge3$, if $p(z)$ is a nonzero polynomial of degree at most $n-2$, if $f(z)$ and $g(z)$ are entire functions such that $(f(z))^n+(g(z))^n=p(z)$, then $f(z)$, $g(z)$, and $p(z)$ are constants.

From this, he deduces the corollary:

If $n\ge3$, if $f(z)$, $g(z)$, $h(z)$ are nonzero entire functions such that $h(z)$ never vanishes, and if $(f(z))^n+(g(z))^n=(h(z))^n$, then there exist nonzero complex numbers $a$, $b$ such that $f(z)=ah(z)$, $g(z)=bh(z)$, $a^n+b^n=1$.

He gives the proof of the Theorem under the stronger hypothesis that $p(z)$ has degree at most $n-3$ – he says the proof for $n-2$ is somewhat more technical.

Let $\zeta$ be a primitive $2n$th root of $1$ (e.g., $\zeta=e^{\pi i/n}$). Then $${p(z)\over(g(z))^n}=\prod_{j=1}^n\left({f(z)\over g(z)}+\zeta^{2j-1}\right)$$
The meromorphic function on the left has at most $n-3$ zeros. So, at least three factors on the right side never vanish. Thus, the meromorphic function $f(z)/g(z)$ misses three values. By Picard's Theorem, $f(z)/g(z)$ is a constant. The Theorem follows.

Ribenboim cites F. Gross, On the functional equation $f^n+g^n=h^n$, Amer. Math. Monthly 73 (1966) 1093-1096.

Answer 3

Your question is equivalent to asking for non-constant meromorphic functions $f, g$ in $\Bbb C$ satisfying $$ f^n + g^n = 1 $$ for some integer $n \ge 2$. For $n=2$ and $n=3$ all solutions of the equation are known, see for example

• Baker, I. N. “On a Class of Meromorphic Functions.” Proceedings of the American Mathematical Society, vol. 17, no. 4, 1966, pp. 819–822. JSTOR, https://www.jstor.org/stable/2036259.

or Fermat like equation for meromorphic functions. on this site.

There are no solutions for $n \ge 4$, this goes back to

• Gross, Fred. On the equation $f^n + g^n = 1$. Bull. Amer. Math. Soc. 72 (1966), 86--88. https://projecteuclid.org/euclid.bams/1183527448

There is also a short proof using the “Second fundamental theorem” of the Nevanlinna theory: Write the equation $f^n + g^n = 1$ as $$ \prod_{k=1}^n \left( \frac f g - a_k \right) = \frac{1}{g^n} $$ where $a_1, \ldots a_n$ are the $n$-th roots of $(-1)$. We can assume that $F=f/g$ is not constant (otherwise we have a “trivial” solution). Zeros of $F- a_k$ can only occur at poles of $g$, and are therefore of multiplicity $n$ or more.

Using the notation of the Nevanlinna theory this implies $$ \overline{N}(r, a_k, F) \le \frac 1n N(r, a_k, F) \, . $$ Substituting this in the second fundamental theorem gives $$ \begin{align} (n-2) T(r, F) &\le \sum_{k=1}^n \overline{N}(r, a_k, F) + S(r, F) \\ &\le \frac 1n \sum_{k=1}^n N(r, a_k, F) + S(r, F) \\ &\le T(r, F) + S(r, F) \end{align} $$ or $(n-3) T(r, F) \le S(r, F)$. This is a contradiction because the “error term” $S(r, F)$ is small compared to $T(r, F)$.