# Entire functions satisfying identity is constant

I am trying to solve the following problem:

Let $f$ and $g$ be entire holomorphic functions satisfying the identity $$f(z)^2=g(z)^6-1$$ for all $z\in\mathbb C$. Show that $f$ and $g$ are constant. Would the same conclusion be valid if $f$ and $g$ were assumed to be entire meromorphic functions?

I think I need to show that $f$ and $g$ are bounded and then use Liouville's theorem, but I am not sure how.

Entire case

From the equation we have that $f^2-g^6=(f+g^3)(f-g^3)=-1$.

Therefore, $f+g^3$, and $f-g^3$ don't vanish.

This means that there is an entire $h$ such that \begin{align}f+g^3&=-e^{ih}\\f-g^3&=e^{-ih}\end{align}

It follows that \begin{align}f&=\frac{-e^{ih}+e^{-ih}}{2}\\g^3&=\frac{-e^{ih}-e^{-ih}}{2}\end{align}

But then $-e^{2ih}=2e^{ih}g^3+1=(2^{1/3}e^{ih/3}g)^3+1$. Since $e^{ih}$ doesn't take the value $0$, then $2^{1/3}e^{ih/3}g$ cannot take any of the three cubic roots of $1$. Since $2^{1/3}e^{ih/3}g$ is entire, by Picard's theorem it follows that $2^{1/3}e^{ih/3}g$ is constant.

Therefore, $h$ is constant and so must be $f$ and $g$.

Meromorphic case

Let's start with a meromorphic solution of $A^3+B^3=1$. It is know that this has such solutions. Even more, all meromorphic solutions are of the form \begin{align}A&=\frac{1+3^{-1/2}\mathcal{P}'(a(z))}{2\mathcal{P}(a(z))}\\B&=\omega\frac{1-3^{-1/2}\mathcal{P}'(a(z))}{2\mathcal{P}(a(z))}\end{align} for $a$ entire and $\omega$ a cube root of unity. See I.N. Baker, On a class of meromorphic functions. Proc. Amer. Math. Soc. 17 (1966),819–822.

Observe that $A,B$ don't have common zeros and by multiplying the equation by the cube of an entire function, we can assume that they don't have poles, although now the equation looks like $$A^3+B^3=C^3$$

We can multiply the whole equation by the cube of an entire function such that, after multiplication, the zeros of $A$ and $B$ have even multiplicity. Therefore, $A,B$ are the square of two entire functions $A_0,B_0$, respectively.

The equation now looks like $A_0^6+B_0^6=C^3$, for a new $C$. Dividing by $A_0^3$ the equation becomes $1+[(B_0/A_0)^3]^2=(C/A_0)^3$.

Therefore, $g=C/A_0$, $f=(B_0/A_0)^3$ is a meromorphic solution to the original equation.

• Of what theorem is used for the fact that $A(z)\ne 0$ for any $z$ for an entire function, then there is an entire $B$ such that $A(z)=e^{iB(z)}$? I think I must have forgotten something. – user284331 Apr 2 '18 at 2:56
• @user284331 You can prove it directly by integrating $\int_{z_0}^{z}A'/A$. The fact that $A$ is never zero implies that $A'/A$ is entire and therefore the integral is independent on the path. It is also a consequence of Weierstrass theorem although that is really circular because that theorem is proven using that little lemma. – user547557 Apr 2 '18 at 3:02
• Okay, I have checked that. Ah... just missed a simple fact, I have learnt something very important today from your answer, thanks. – user284331 Apr 2 '18 at 3:22