Picard's theorem applied to $f^n + g^n =1$

1.7k Views Asked by At

So I have the following problem. Part 1 is just to state Picard's theorem, so for that we have that any entire holomorphic function takes on every value with possibly one exception.

Part 2 is to show that for $n\geq 2$ there are no nowhere vanishing and nonconstant entire functions $f, g$ such that $f^n+g^n=1$ The proof of that goes as follows. $f^n$ is entire, and $f^n=1-g^n$. Since $g\neq 0$, $g^n\neq 0$ hence $f^n \neq 1$ hence $f^n$ is an entire function omitting both $0$ and $1$, contradiction.

The final part is assume that $n>2$. Find all the solutions $f$ and $g$ to $f^n+g^n=1$ in the ring of entire functions. The hint is to transform the problem to the setting of meromorophic functions on the complex line, but I don't know how to proceed.

Thanks for any help!

1

There are 1 best solutions below

3
On BEST ANSWER

Assume $n \ge 3$ and $f^n+g^n=1$ for two entire functions $f, g$. Then $$ \bigl(\frac fg \bigr)^n + 1 = \frac 1 {g^n} \quad . $$ The right-hand side is never zero, which means that the meromorphic function $h = f/g$ does not take any of the values $w_1, \ldots, w_n$ which are defined as the solutions of $w^n = -1$.

So $h$ is meromorphic in $\mathbb C$ and omits $n \ge 3$ values and therefore must be constant.

Or, if you prefer, $1/(h-w_1)$ is an entire function and omits the $ n - 1 \ge 2$ values $1/(w_k - w_1)$, $k = 2, \ldots, n$.