Reference request: Clean proof of Fermat's last theorem for $n=3$.

Solution 1:

Recently I have found two simple proofs of FLT for n=3. They use the ring $\mathbb{Z}[\sqrt[3]{2}]$. If $x^3+y^3+z^3=0$ then

(1) $x^6-4 (yz)^3=a^2$,

where $a=y^3-z^3$.

The idea is to deduce from (1) that the algebraic number $x^2-\sqrt[3]{4} yz$ is a square, more precisely:

(2) $x^2-\sqrt[3]{4} yz=u \beta^2$,

where $\beta \in \mathbb{Z}[\sqrt[3]{2}]$, and $u$ is a unit.

Once this is established, then it is not difficult to get a contradiction via infinite descent. It is similar to Euler's proof, but another number field is used.

The (2) follows from the unique factorization in the Euclidean ring $\mathbb{Z}[\sqrt[3]{2}]$

My second proof deduces (2) without using the unique factorization.

The idea is simple: find a number $\alpha \in \mathbb{Z}[\sqrt[3]{2}]$, such that $\alpha^2$ is divided by the $x^2-\sqrt[3]{4} yz$ and $\alpha^2/(x^2-\sqrt[3]{4} yz)$ has a small norm. It turns out that the last number may have small norm indeed: less than 9. Then the norm of it can be either 1 or 4, and (2) easily follows. The method can be used to prove FLT for other concrete $n$ as well.

It is easy to get such $\alpha$ because $\alpha$ belongs to $I$, where $I$ is the ideal of norm $a$, generated by $a$ and $x^2-\sqrt[3]{4} yz$. We know from the theory of algebraic numbers, that there is an $\alpha \in I$, such that $N(\alpha)<m N(I)$, where $m$ is easily calculated constant, that turns to be less than 3. Then $N(\alpha^2/(x^2-\sqrt[3]{4} yz))<(3 a)^2/a^2=9$.

This proof is shorter and easier than that of Euler and Kummer.

Now I'll explain how to get contradiction from (2). We may assume, that $x$ is odd, otherwize take $y$ to play the role of $x$.

Units in the ring $\mathbb{Z}[\sqrt[3]{2}]$ have the form: $u=\pm(\sqrt[3]{2}-1)^k$, where $k$ is integer. Therefore, from (2) follows, that either

(3) $x^2−\sqrt[3]{4} yz=\beta_1^2$

or

(4) $(\sqrt[3]{2}-1)(x^2−\sqrt[3]{4} yz)=\beta_2^2$,

where $\beta_1$ and $\beta_2$ belong to the ring $\mathbb{Z}[\sqrt[3]{2}]$.

However (4) is impossible, because the coefficient at $\sqrt[3]{2}$ turns out to be odd in the left part and even in the right part of (4). Therefore (3) takes place, that is:

(5) $x^2−\sqrt[3]{4} yz=(a_0+a_1 \sqrt[3]{2}+a_2 (\sqrt[3]{2})^2)^2$, where $a_0$, $a_1$ and $a_2$ are integers.

From (5) we get:

(6) $a_0^2+4 a_1 a_2=x^2$

(7) $2 a_2^2+2 a_0 a_1=0$

(8) $a_1^2+2 a_0 a_2=-yz$

It follows, that all three numbers $a_0$, $a_1$ and $a_2$ are not zero. We may assume, that $a_2>0$, otherwise change sign of all three numbers.

It follows from (7) that $a_0=\pm m b_0^2$, $a_1=\mp m b_1^2$, $a_2=m b_0 b_1$, where $b_0$ and $b_1$ are relatively prime positive integers and $m$ is a positive integer.

Now it follows from (6) that:

(9) $m^2 b_0 (b_0^3 \mp 4 b_1^3)=x^2$

From (9) it follows that $b_0$ is a square, that is $b_0=x_1^2$, and $(b_0^3 \mp 4 b_1^3)$ is a square, that is:

(10) $x_1^6 \mp 4 b_1^3$ is a square

It can be shown, that $x_1^6 \mp 4 b_1^3 \neq 1$ (a separate lemma). Then from (9) it follows, that $b_0<x^2$, that is $x_1<x$ and we've got the infinite descent.