From Fermat's Last Theorem, Part 9: n=3

从费马大定理谈起（九）：n=3 苏剑林 · 2014-09-01 · Mathematics number theory Diophantine equation fermat's last theorem

Now we can begin the proof of $n=3$. In the realm of the ordinary integers, the proof for n=3 looks fairly complicated, and it doesn't seem to have much 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 within $\mathbb{Z}[\omega]$, it turns out to have a lot in common with the n=4 case—and in fact the proof is simpler than for n=4 (note that, in the ordinary integers, the proof for n=4 is easier than for n=3; Fermat completed the proof for n=4, but not for n=3). I think it's precisely this kind of similarity that gave Lamé—who had not yet completed the proof at the time—the confidence that this path would lead him to a proof of Fermat's Last Theorem. (Unfortunately, this confidence turned out to be misplaced: Lamé's path could not be pushed all the way through. Kummer went further along this path than anyone, but even he did not manage to prove Fermat's Last Theorem.)

The proof here is similar to the second proof for $n=4$. We first adjoin a unit to the equation, and then show that no matter what the unit is, the equation has no solutions in $\mathbb{Z}[\omega]$. This is a rather clever trick: it lets us rule out more equations while using fewer steps. In fact, there does exist a proof that only shows $x^3+y^3+z^3=0$ has no solutions, but it requires very careful case analysis of the units involved, which is quite a hassle. The proof presented here is one I adapted from the proof on Fermat's Last Theorem blogspot, combined with the second proof for n=4 in this series, simplified so as to reduce the amount of careful analysis of units needed. more

Lemma

In this post, let $\varepsilon_1,\varepsilon_2,\varepsilon_3,\varepsilon$ denote a unit in $\mathbb{Z}[\omega]$, and write $\xi=1-\omega$. Then if the equation $\varepsilon_1 x^3+\varepsilon_2 y^3 +\varepsilon_3 z^3=0,\,\xi|x,\xi\nmid yz$ has a solution in $\mathbb{Z}[\omega]$, the equation can be rewritten as
$$\varepsilon x^3+ y^3 +z^3=0$$

The proof is simple: divide every term 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 it modulo $\xi$. Since $\xi\nmid yz$, we have $y^3\equiv\pm 1(\bmod\,9),z^3\equiv \pm 1(\bmod\,9)$; note that $9=\xi^4 \omega$, so working modulo $\xi^3$ gives

$$0\pm (\varepsilon_2/\varepsilon_3)\pm 1\equiv 0(\bmod\,\xi^3)$$

Hence $\varepsilon_2/\varepsilon_3=\pm 1$. Absorbing the factor of $-1$ in front of $y^3$ (if the sign is negative) into $y$ ($(-1)^3=-1$), and setting $\varepsilon_1/\varepsilon_3=\varepsilon$, we obtain $\varepsilon x^3+ y^3 +z^3=0$.

Proof

Now we can start the main proof.

The equation $x^3+y^3+z^3=0$ has no solution with $xyz\neq 0$ in $\mathbb{Z}[\omega]$.

Suppose the equation $x^3+y^3+z^3=0$ has a solution with $xyz\neq 0$ in $\mathbb{Z}[\omega]$. Then we must have $\xi|xyz$—otherwise $\xi\nmid x,\xi\nmid y,\xi\nmid z$, which would give

$$\begin{aligned}x^3\equiv \pm 1(\bmod\,9)\\ y^3\equiv \pm 1(\bmod\,9)\\ z^3\equiv \pm 1(\bmod\,9)\end{aligned}$$

From this we get $\pm 1\pm 1\pm 1(\bmod\,9)$, and no matter which sign is chosen, this cannot hold. Hence we must have $\xi|xyz$. Since $x,y,z$ all play symmetric roles here, without loss of generality assume $\xi|x$. The purpose of this step is to show that Fermat's Last Theorem for n=3 also reduces to an equation of type $\varepsilon x^3+y^3 +z^3=0,\xi|x$, so that we can proceed with the corresponding congruence analysis.

The steps below are almost identical to the n=4 case, and in fact even simpler. Suppose some equation of type $\varepsilon x^3+y^3 +z^3=0,\xi|x,\xi\nmid yz$ has a solution. Let $(x,y,z)$ be a pairwise coprime solution with the smallest possible $N(x)$. Again, note that here $N(x)$ needs to range over all $\varepsilon$ (i.e., all six units), and also over all solutions for a fixed $\varepsilon$, from which we then pick the solution with the smallest $N(x)$. There may be more than one such solution, but we just need to pick any one of them.

First, let's determine the power of $\xi$ dividing $x$. Since $-\varepsilon x^3=y^3 +z^3$, write $y^3\equiv e(\bmod\,9),z^3\equiv f(\bmod\,9)$, $e,f\in\{-1,1\}$; then considering both sides modulo $\xi^3$, we get $e+f\equiv 0(\bmod\,\xi^3)$, that is, $e+f=0$. This means $y^3+z^3$ must be a multiple of at least 9, but since $9=\xi^4\omega$, the power of $\xi$ dividing $x$ must be at least 2, i.e., $\xi^2|x$.

The core of the argument is now the factorization:

$$-\varepsilon x^3=(y+z)(y+z\omega)(y+z\omega^2)$$

The three terms on the right satisfy the relation

$$\begin{aligned}(y+z)-(y+z\omega)=(1-\omega)z=\xi z\\ (y+z)\omega-(y+z\omega)=(\omega-1)y=-\xi y\\ (y+z\omega)-(y+z\omega^2)=\omega(1-\omega)z=\omega\xi z\\ (y+z\omega)\omega-(y+z\omega^2)=(\omega-1)y=-\xi y \end{aligned}$$

Since $y,z$ are coprime, $y+z$ and $y+z\omega$, as well as $y+z\omega$ and $y+z\omega^2$, can share at most the common factor $\xi$. But the left-hand side has the factor $\xi^6$, so at least one term on the right must have the factor $\xi$. Once one term has the factor $\xi$, the other two must also have the factor $\xi$, but the three terms can share at most the common factor $\xi$. Therefore, two of the terms can only carry a power of $\xi$ equal to 1, while the remaining term "absorbs" all of the rest of the power of $\xi$ (which is at least 4). However, these three terms $y+z,y+z\omega,y+z\omega^2$ actually play symmetric roles, since one can multiply $z$ by suitable powers of $\omega$ to permute them. So, without loss of generality, assume $\xi^4|y+z$; then we can write

$$\begin{aligned}x=\xi^2 \chi\\ y+z=\xi^4 r'\\ y+z\omega=\xi s'\\ y+z\omega^2=\xi t' \end{aligned}$$

Then $-\varepsilon\chi^3=r's't'$ and $r',s',t'$ are pairwise coprime, so up to a unit factor each of them is a perfect cube, i.e., an associate of a cube. Hence we can write

$$r'=\varepsilon_1 r^3,\ s'=\varepsilon_2 s^3,\ t'=\varepsilon_3 t^3$$

Notice that

$$(y+z)+(y+z\omega)\omega+(y+z\omega^2)\omega^2=0$$

from which we obtain

$$\varepsilon_1 \xi^3 r^3+(\varepsilon_2 \omega) s^3+(\varepsilon_3 \omega^2) t^3=0$$

where $\varepsilon_1,\varepsilon_2 \omega,\varepsilon_3 \omega^2$ are all units. By the lemma, this equation must take the form

$$\varepsilon' \xi^3 r^3+s^3+t^3=0$$

Then $(\xi r,s,t)$ is a solution for some choice of unit $\varepsilon'$, and clearly $N(\xi r) < N(x)$ (since $\xi^4 r^3$ is a factor of $x^3$, $N(\xi) > 1$), which contradicts our assumption of minimality. Hence the equation $\varepsilon x^3+y^3 +z^3=0,\xi|x,\xi\nmid yz$ has no solution in $\mathbb{Z}[\omega]$, and therefore $x^3+y^3+z^3=0$ has no solution in $\mathbb{Z}[\omega]$. $\blacksquare$