Basically the [Fermat curve](https://en.wikipedia.org/wiki/Fermat_curve)
is $x^n+y^n=1$.

For $n > 2$, can we parametrize it by $x=g(t),y=g'(t)$ for already named function $g(t)$ of complex argument?

i.e., $g(t)^n+g'(t)^n=1$.

---

Some of what I know:

For $n=2$, solution is $g(t)=\sin(t)$.

For $n=4$, [Wolfram alpha](http://www.wolframalpha.com/input/?i=solve+f%28x%29%5E4%2Bf%27%28x%29%5E4%3D1) appears to give functional equation.

For $n \ge 3$, SAGE gives functional equation with a lot of constants:

```
#n = 3
x == 1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(-t^3 + 1)^(1/3)/t - 1)) + _C + 1/3*log((-t^3 + 1)^(1/3)/t + 1) - 1/6*log(-(-t^3 + 1)^(1/3)/t + (-t^3 + 1)^(2/3)/t^2 + 1),
g(x) == (-t^3 + 1)^(1/3)
#n = 5
x == 1/5*sqrt(5)*(sqrt(5) + 1)*arctan((sqrt(5) + 4*(-t^5 + 1)^(1/5)/t - 1)/sqrt(2*sqrt(5) + 10))/sqrt(2*sqrt(5) + 10) + 1/5*sqrt(5)*(sqrt(5) - 1)*arctan(-(sqrt(5) - 4*(-t^5 + 1)^(1/5)/t + 1)/sqrt(-2*sqrt(5) + 10))/sqrt(-2*sqrt(5) + 10) + _C - 1/10*(sqrt(5) + 3)*log(-(-t^5 + 1)^(1/5)*(sqrt(5) + 1)/t + 2*(-t^5 + 1)^(2/5)/t^2 + 2)/(sqrt(5) + 1) - 1/10*(sqrt(5) - 3)*log((-t^5 + 1)^(1/5)*(sqrt(5) - 1)/t + 2*(-t^5 + 1)^(2/5)/t^2 + 2)/(sqrt(5) - 1) + 1/5*log((-t^5 + 1)^(1/5)/t + 1),
g(x) == (-t^5 + 1)^(1/5)

```