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Let $\omega$ be a primitive cube root of unity, let $\lambda = 1 - \omega$, and let $R = \mathbb Z[\omega]$, the ring of Eisenstein integers. In Alon Amit's proof here, the author concludes (in an incorrect fashion, but I have fixed that error) that if $(x,y,z)$ is a primitive solution over $R$ to the Fermat equation $x^3 + y^3 = uz^3$, where $u$ is one of the six units $\pm 1$, $\pm \omega$, $\pm(1 + \omega)$ in $R$, then $\lambda$ must divide exactly one of $x$, $y$, or $z$. He then assumes that $\lambda$ divides $z$, and proceeds with the proof as if that is the general case. It would seem we need a different argument if $\lambda$ divides, say, $x$, since the unit $u$ moves to a different term if we make the obvious rearrangement/sign absorptions in an attempt to mimic the $z$-case.

How do we make the necessary modification(s)?

Johnny Apple
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2 Answers2

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The proofs I know start as follows. Let $(x,y,z)$ be a nontrivial integer solution to $x^3+y^3=z^3$ with coprime $x,y,z$. Then we have $3\mid xyz$, because otherwise $x^3+y^3\equiv -2,0,2\bmod 9$ and $z^3\equiv 1,-1$, so that $x^3+y^3\neq z^3$. We may assume that $3\mid z$ and $3\nmid xy$ by rearranging. And now we use that $3=(1-\omega)(1-\omega^2)$. This yields $\lambda=1-\omega \mid z$ in $R=\Bbb Z[\omega]$. Then we have $$ x^3+y^3=(x+y)(x+\omega y)(x+\omega^2y)=z^3 $$ over $R=\Bbb Z[\omega]$.

For a full proof see here, for example:

The equation $x^3 + y^3 = z^3$ has no integer solutions - A short proof

Dietrich Burde
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I believe I have concocted an answer to my question. Suppose instead, without loss of generality, that $\lambda$ divides $x$. Then write

$$y^3 + u(-z)^3 = (-x)^3$$

By an exercise from the post, this forces $\pm 1 + u \equiv 0 \mod \lambda^2$, and, again from the same post, this forces $u = \pm 1$. But then the Fermat equation reads

$$y^3 + (\pm z)^3 = (-x)^3$$

and we are now in the same situation (in fact, better) than if we had assumed $\lambda$ divides $z$ at the outset.

Johnny Apple
  • 4,571