We can now begin the proof for n=3. The proof for n=3 within the realm of rational integers appears quite complex and seems to share little in common with the proof for n=4. However, if we consider the proof that x^3+y^3+z^3=0 has no solutions in \mathbb{Z}[\omega], we find many similarities to the case n=4. In fact, the proof is arguably simpler than that for n=4 (note that in the rational integers \mathbb{Z}, the proof for n=4 is simpler than for n=3; Fermat completed the proof for n=4, but not for n=3). I believe it was these similarities that gave the mathematician Lamé the confidence that he could complete the proof of Fermat’s Last Theorem using this path, even before the proof was finished. (However, this confidence was misplaced: Lamé’s path could not be fully completed. Kummer went much further along this path, but even then, Kummer did not prove Fermat’s Last Theorem.)
The proof is similar to the second proof for n=4. We first add a unit to the equation and then prove that regardless of what that unit is, the equation has no solutions in \mathbb{Z}[\omega]. This is a clever trick that allows us to prove that more equations have no solutions while using fewer steps. In fact, there are proofs that only show x^3+y^3+z^3=0 has no solutions, but they require a very careful analysis of the units involved, which is quite tedious. This proof is simplified by referencing the proof on the Fermat’s Last Theorem Blogspot and combining it with the second proof for n=4 in this series, primarily by reducing the detailed analysis of units.
Lemma
In this article, let \varepsilon_1, \varepsilon_2, \varepsilon_3, \varepsilon denote units in \mathbb{Z}[\omega], and let \xi = 1-\omega. If the equation \varepsilon_1 x^3 + \varepsilon_2 y^3 + \varepsilon_3 z^3 = 0 with \xi|x and \xi \nmid yz has a solution in \mathbb{Z}[\omega], then the equation can be transformed into: \varepsilon x^3 + y^3 + z^3 = 0
The proof is simple. Divide all terms of the equation by \varepsilon_3 to get (\varepsilon_1/\varepsilon_3) x^3 + (\varepsilon_2/\varepsilon_3) y^3 + z^3 = 0, and then consider the equation modulo \xi. Since \xi \nmid yz, we have y^3 \equiv \pm 1 \pmod{9} and z^3 \equiv \pm 1 \pmod{9}. Note that 9 = \xi^4 \omega. Considering the equation modulo \xi^3, we have: 0 \pm (\varepsilon_2/\varepsilon_3) \pm 1 \equiv 0 \pmod{\xi^3} Therefore, \varepsilon_2/\varepsilon_3 = \pm 1. We can incorporate the factor of -1 (if it is negative) into y (since (-1)^3 = -1). By setting \varepsilon_1/\varepsilon_3 = \varepsilon, we obtain \varepsilon x^3 + y^3 + z^3 = 0.
Proof
Now we can begin the main proof.
The equation x^3+y^3+z^3=0 has no solutions in \mathbb{Z}[\omega] such that xyz \neq 0.
If the equation x^3+y^3+z^3=0 has a solution in \mathbb{Z}[\omega] with xyz \neq 0, then \xi must divide xyz. Otherwise, if \xi \nmid x, \xi \nmid y, \xi \nmid z, we would have: \begin{aligned} x^3 &\equiv \pm 1 \pmod{9} \\ y^3 &\equiv \pm 1 \pmod{9} \\ z^3 &\equiv \pm 1 \pmod{9} \end{aligned} This would lead to \pm 1 \pm 1 \pm 1 \equiv 0 \pmod{9}, which is impossible regardless of the signs. Thus, \xi | xyz. Since the roles of x, y, z are symmetric, we can assume without loss of generality that \xi | x. The purpose of this step is to show that for n=3, Fermat’s Last Theorem also belongs to the class of equations \varepsilon x^3 + y^3 + z^3 = 0 with \xi | x, allowing for further congruence analysis.
The following steps are almost identical to those for n=4, and even simpler. Suppose an equation of the type \varepsilon x^3 + y^3 + z^3 = 0 with \xi | x and \xi \nmid yz has a solution. Let (x, y, z) be a set of pairwise coprime solutions with the minimal norm N(x). Note that N(x) must be minimized across all possible units \varepsilon (the six units in \mathbb{Z}[\omega]) and all solutions for a fixed \varepsilon. We pick any one such solution with minimal N(x).
First, we determine the power of \xi in x. Since -\varepsilon x^3 = y^3 + z^3, let y^3 \equiv e \pmod{9} and z^3 \equiv f \pmod{9}, where e, f \in \{-1, 1\}. Considering both sides modulo \xi^3, we have e + f \equiv 0 \pmod{\xi^3}, which implies e + f = 0. Consequently, y^3 + z^3 is at least a multiple of 9. Since 9 = \xi^4 \omega, the power of \xi in x must be at least 2, i.e., \xi^2 | x.
The core part is the factorization: -\varepsilon x^3 = (y+z)(y+z\omega)(y+z\omega^2) The three terms on the right have the following relationships: \begin{aligned} (y+z) - (y+z\omega) &= (1-\omega)z = \xi z \\ (y+z)\omega - (y+z\omega) &= (\omega-1)y = -\xi y \\ (y+z\omega) - (y+z\omega^2) &= \omega(1-\omega)z = \omega\xi z \\ (y+z\omega)\omega - (y+z\omega^2) &= (\omega-1)y = -\xi y \end{aligned} Since y and z are coprime, the greatest common divisor of (y+z) and (y+z\omega), or (y+z\omega) and (y+z\omega^2), can be at most \xi. Since the left side is divisible by \xi^6, at least one term on the right must be divisible by \xi. If one term is divisible by \xi, the others must also be divisible by \xi. However, since their greatest common divisor is at most \xi, the power of \xi in two of the terms must be exactly 1, while the remaining term "absorbs" all the remaining powers of \xi (at least power 4). Since y+z, y+z\omega, y+z\omega^2 are symmetric (we can multiply z by powers of \omega to permute them), we can assume without loss of generality that \xi^4 | y+z. Thus, we can set: \begin{aligned} x &= \xi^2 \chi \\ y+z &= \xi^4 r' \\ y+z\omega &= \xi s' \\ y+z\omega^2 &= \xi t' \end{aligned} Then -\varepsilon \chi^3 = r's't'. Since r', s', t' are pairwise coprime, they must each be a cube up to a unit factor (associates of cubes). We can set: r' = \varepsilon_1 r^3, \quad s' = \varepsilon_2 s^3, \quad t' = \varepsilon_3 t^3
Note that: (y+z) + (y+z\omega)\omega + (y+z\omega^2)\omega^2 = 0 This yields: \varepsilon_1 \xi^3 r^3 + (\varepsilon_2 \omega) s^3 + (\varepsilon_3 \omega^2) t^3 = 0 Where \varepsilon_1, \varepsilon_2 \omega, \varepsilon_3 \omega^2 are all units. According to the Lemma, this equation can be written in the form: \varepsilon' \xi^3 r^3 + s^3 + t^3 = 0 Then (\xi r, s, t) is a solution for the unit \varepsilon', and clearly N(\xi r) < N(x) (because \xi^4 r^3 is a factor of x^3 and N(\xi) > 1). This contradicts our assumption of minimality. Therefore, the equation \varepsilon x^3 + y^3 + z^3 = 0 with \xi | x and \xi \nmid yz has no solutions in \mathbb{Z}[\omega], and consequently x^3 + y^3 + z^3 = 0 has no solutions in \mathbb{Z}[\omega]. Q.E.D.
Original Address: https://kexue.fm/archives/2910
For more details on reprinting, please refer to: Scientific Space FAQ