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.

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    $\begingroup$ As there are now several valid answers providing different proofs, I have decided not to accept any of them. Thanks for all the inspring answers! $\endgroup$
    – Yuxiao Xie
    Jun 2, 2020 at 14:33
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    $\begingroup$ Of course, you are quite free not to accept any answer. In my opinion, it's very much the wrong response to the situation. If there are two or more posts that answer your question, I think you should just pick one – whichever one you want – and accept it, and write nice comments on the others. I assure you that no one will be offended by that action. $\endgroup$ Jun 3, 2020 at 0:35
  • $\begingroup$ For $n = 6$ this is equivalent to this question: math.stackexchange.com/questions/179282 $\endgroup$ Jun 3, 2020 at 7:59

3 Answers 3


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.


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

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 $T(r, F)$.

  • $\begingroup$ This is essentially the proof Ribenboim gives, with attribution to Gross, as reported in my answer, only appealing to Nevanlinna theory instead of Picard's Theorem. $\endgroup$ Jun 3, 2020 at 0:38
  • $\begingroup$ I don't think that it is is essentially the same proof if it uses a complete different theory. Also it works for arbitrary meromorphic functions, not only for the case that $h$ has no or finitely many zeros. Note also that it uses that $f/g$ takes values with higher multiplicities, not that $f/g$ omits values, so Picard's theorem cannot be applied here. $\endgroup$
    – Martin R
    Jun 3, 2020 at 4:32
  • $\begingroup$ Fair enough. It's just that your second display looks an awful lot like my first display. $\endgroup$ Jun 3, 2020 at 5:51
  • $\begingroup$ @GerryMyerson: Yes, both solutions start with a factorization of $(f/g)^n + 1$. The remaining parts are different though. – I do not claim that this solution is new, I would have added a citation if I had found one. $\endgroup$
    – Martin R
    Jun 3, 2020 at 6:46

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.


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