# Is there a substitution which transforms every Fermat curve into an elliptic curve?

A Fermat Curve of degree $$n$$ is the set of solutions to $$x^n+y^n=z^n$$, $$x,y,z\in \mathbb R$$. In this question, the OP provides a substitution which relates a Fermat Curve of degree $$n=3,4$$ to two different elliptic curves. To transform the Fermat Curve of degree $$3$$, the substitutions $$a=\frac{12z}{x+y},\quad b=\frac{36(x-y)}{x+y}$$ produce $$b^2=a^3-432$$, an elliptic curve. Similarly for the Fermat Curve of degree $$4$$, the substitutions $$a=\frac{2(y^2+z^2)}{x^2},\quad b=\frac{4y(y^2+z^2)}{x^3}$$ give $$b^2=a^3-4a$$. However, the substitutions used are not at all obvious, which leads me to wonder,

Is there a similar substitution which can relate a Fermat curve of arbitrary degree to an elliptic curve?

How can we even begin to prove this? I suspect the proof or disproof of this statement will be way above my level; I truly have no clue where to begin. Can someone help out? If this is somehow an open problem, then any links to literature is also appreciated!