Starting from Fermat's Last Theorem (VI): n=4 (2)

从费马大定理谈起（六）：n=4（2） 苏剑林 · 2014-08-19 · Mathematics number theory Diophantine equation fermat's last theorem

In the previous post I mentioned that, in order to prove the case n=4, it seemed necessary to prove that $x^4+y^4=z^2$ has no solutions, rather than merely proving that $x^4+y^4=z^4$ has none. However, while working on this at noon today, I found another proof of the n=4 case. It is likewise carried out in $\mathbb{Z}[i]$, but this time the exponents are all 4 in a slightly different sense — not simply of the form $x^4+y^4=z^4$, but rather $\varepsilon x^4+y^4=z^4$, where $\varepsilon$ is a unit. I feel this proof is closer in spirit to the proof for general odd prime n, so I've added this supplementary post for readers' reference. You may want to compare it with the previous post.

Lemma

Let $\varepsilon_1,\varepsilon_2,\varepsilon_3,\varepsilon$ denote a unit in $\mathbb{Z}[i]$. We first prove the following:

If the equation $\varepsilon_1 x'^4 +\varepsilon_2 y'^4+\varepsilon_3 z'^4=0$ has a solution in $\mathbb{Z}[i]$ with all components nonzero, then, after suitable simplification and rearrangement, the equation must take the form $\varepsilon x^4+y^4=z^4$, where $(x,y,z)$ is some permutation of $(x',y',z')$, and $\xi^2|x$.

The proof proceeds similarly to the previous post. First we show $\xi|x'y'z'$; if not, then we would have

$$\begin{aligned}x'^4 &\equiv 1(\bmod\, 8)\\ y'^4 &\equiv 1(\bmod\, 8)\\ z'^4 &\equiv 1(\bmod\, 8) \end{aligned}$$

which gives $8|(\varepsilon_1+\varepsilon_2+\varepsilon_3)$, and this cannot hold no matter what value $\varepsilon_1,\varepsilon_2,\varepsilon_3$ takes.

So suppose $\xi|x'$, and further suppose $\xi\nmid y'z'$; then

$$\begin{aligned}x'^4 &\equiv 0(\bmod\, \xi^4)\\ y'^4 &\equiv 1(\bmod\, 8)\\ z'^4 &\equiv 1(\bmod\, 8) \end{aligned}$$

Rearranging the original equation gives

$$(\varepsilon_1/\varepsilon_2) x'^4 +y'^4+(\varepsilon_3/\varepsilon_2) z'^4=0$$

Taking each term modulo $\xi^4$ ($-8i=\xi^6$), we obtain

$$\xi^4|(1+\varepsilon_3/\varepsilon_2)$$

Hence $\varepsilon_3/\varepsilon_2=-1$; taking $\varepsilon_1/\varepsilon_2=\varepsilon,(x',y',z')=(x,y,z)$ then gives

$$\varepsilon x^4+y^4=z^4$$

Since $z^4-y^4\equiv x^4\equiv 0(\bmod\,8),\ \xi^6=-8i$, at least $\xi^2|x$ must hold.

Proof

Now suppose $(x,y,z)$ is a pairwise coprime solution of $\varepsilon x^4+y^4=z^4,\ \xi^2|x$, and suppose that among all such solutions this one has the smallest $\xi$-degree contained in $x$. Note that here $\varepsilon$ is an arbitrary, unspecified unit, so "smallest" here means: among all possible choices of $\varepsilon$, take the solution for which the $\xi$-degree is smallest. Write $\xi^m|x,\xi^{m+1}\nmid x,m\geq 2$.

In $\mathbb{Z}[i]$, we can factor completely:

$$\varepsilon x^4=(z+y)(z-y)(z+yi)(z-yi)$$

Observe that

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

The above computation shows that $(z+y),(z-y)$ has at most the common divisor $\xi^2$, $(z-y),(z+yi)$ has at most the common divisor $\xi$, and $(z+yi),(z-yi)$ has at most the common divisor $\xi^2$. Meanwhile, the left-hand side has at least the divisor $\xi^8$, so at least one of the four terms on the right must have the divisor $\xi^2$. However, it's impossible for all the terms to have the divisor $\xi^2$, since that would contradict the fact that $(z-y),(z+yi)$ has at most the common divisor $\xi$. Thus at least two of the terms must have $\xi$-degree equal to 1, since having $z+y$ with divisor $\xi^2$ necessarily forces $z-y$ to have divisor $\xi^2$, and having $z+yi$ with divisor $\xi^2$ necessarily forces $z-yi$ to have divisor $\xi^2$, and vice versa; if only one term had $\xi$-degree equal to 1, the other three would all have to be at least 2, a contradiction. So among the four terms, one must have degree at least 4, from which it follows that the degrees of $\xi$ on the right-hand side are respectively $\geq 4,2,1,1$. Without loss of generality, set

$$\begin{aligned}x&=\xi \chi\\ z+y&=\xi^{4m-4} u'\\ z-y&=\xi^2 v'\\ z+yi&=\xi s'\\ z-yi&=\xi t' \end{aligned}$$

Then

$$\varepsilon\chi^4=u'v's't'$$

and since $u',v',s',t'$ are pairwise coprime, each of them must be an associate of some fourth power, and none of them can have the divisor $\xi$, or else this would contradict $\xi^{m+1}\nmid x$. We can therefore write

$$\begin{aligned}u'=\varepsilon_1 u^4,\ v'=\varepsilon_2 v^4\\ s'=\varepsilon_3 s^4,\ t'=\varepsilon_4 t^4 \end{aligned}$$

At this point

$$\begin{aligned}2z&=\xi^{4m-4} u^4 +\xi^2 v^4=\xi s^4+\xi t^4\\ 2y&=\xi^{4m-4} u^4 -\xi^2 v^4=(\xi s^4-\xi t^4)(-i) \end{aligned}$$

that is,

$$\begin{aligned}\varepsilon_1 \xi^{4m-5} u^4 +\varepsilon_2 \xi v^4&=\varepsilon_3 s^4+\varepsilon_4 t^4\\ \varepsilon_1\xi^{4m-5} u^4 -\varepsilon_2\xi v^4&=(\varepsilon_3 s^4-\varepsilon_4 t^4)(-i) \end{aligned}$$

Adding the two equations together term by term,

$$2\varepsilon_1 \xi^{4m-5} u^4=(1-i)\varepsilon_3 s^4+(1+i)\varepsilon_4 t^4$$

Dividing each term by $\xi$ and simplifying gives

$$(-i)(\varepsilon_1/\varepsilon_4) \xi^{4m-4} u^4 +i (\varepsilon_3/\varepsilon_4) s^4=t^4$$

By the lemma,

$$i (\varepsilon_3/\varepsilon_4)=1$$

Setting $(-i)(\varepsilon_1/\varepsilon_4)=\varepsilon'$, we get

$$\varepsilon' \xi^{4m-4} u^4 +s^4=t^4$$

This shows that $(\xi^{m-1} u,s,t)$ is a solution of the equation with the unit taken to be $\varepsilon'$, and $\xi\nmid u$, but $m-1 < m$, which contradicts our assumption. Therefore the equation $\varepsilon x^4+y^4=z^4$ has no solution in $\mathbb{Z}[i]$.

Remarks

The argument above winds back and forth in a way that may seem hard to grasp, but this is exactly what reveals the general pattern of proof for arbitrary n. The most crucial step is the complete factorization of $z^n-y^n$ within the extended number field. If unique factorization holds there, a contradiction follows directly. At the same time, basic congruence analysis turns out to be essential — its results "conveniently" arrange the coefficients just right, allowing us to apply Fermat's method of infinite descent to derive the contradiction. All of this seems almost too coincidental to be true, and yet also strangely inevitable...