# Transformation which takes Fermat curve $x^n+y^n=1$ to a hyperelliptic curve?

Motivated by this where it is possible to take certain Fermat curves like $$x^3+y^3=1$$ into Elliptic curves.

I was wondering if it is always possible to transform any Fermat curve $$x^n+y^n=1$$ birationally into some hyperelliptic curve?

1. According to "The Group of Automorphisms of the Fermat Curve" (Tzermias 1995), the automorphism group of the Fermat curve with $$n \ge 4$$ in characteristic $$0$$ is the semidirect product $$\Sigma_3 \ltimes (\Bbb{Z}/n)^2$$ which has order $$6n^2$$.
2. The genus of the Fermat curve is $$g=(n-1)(n-2)/2$$.
3. According to "Automorphism Groups of Hyperelliptic Riemann Surfaces" (Bujulance, Etayo, Martinez 1987), a hyperelliptic Riemann surface of genus $$g>15$$ has at most $$8(g+1)$$ automorphisms.
4. If $$n \ge 8$$ then $$g = (n-1)(n-2)/2 \gt 15$$, $$8(g+1) = 4(n^2-3n+4) \lt 6n^2$$ and so the Fermat curve is not hyperelliptic.
5. EDIT added: Evidently from Theorem p.175 in the paper cited (3) the bound $$8(g+1)$$ applies if $$g>9$$ and so the $$n \ge 6$$ Fermat curve is not hyperelliptic.