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

从费马大定理谈起（五）：n=4 苏剑林 · 2014-08-19 · Mathematics number theory Diophantine equation fermat's last theorem

The time has come!

The previous articles laid the groundwork for the proof of Fermat's Last Theorem — of course, compared with the full proof of Fermat's Last Theorem, these articles are but a drop in the ocean. Still, they are already enough to complete the proof of Fermat's Last Theorem for the case n=4. We will soon see the elegant proof that the Gaussian integers give us for n=4, and this will convince us that this road can be traveled much further.

The indeterminate equation $x^4+y^4=z^2$ has no solutions in $\mathbb{Z}[i]$ with all entries nonzero.

The reader will notice that we are considering $x^4+y^4=z^2$ rather than $x^4+y^4=z^4$; the former is a strengthening of the latter. But this strengthening was not made in order to prove a more general statement — rather, our proof simply does not apply to $x^4+y^4=z^4$ at all! That is to say, by the method of this article, we can prove that $x^4+y^4=z^2$ has no solutions, but we cannot "merely prove" that $x^4+y^4=z^4$ has no solutions. This is indeed a curious phenomenon: some statements are only easier to prove once they have been strengthened, much as with mathematical induction on inequalities — if you don't strengthen the statement first, the induction fails.

The tool we need is very simple: it is the "analysis modulo 1+i" within $\mathbb{Z}[i]$ that we developed in the third article. For convenience, let us write $\xi=1+i$ below.

Step One

The first step is to show that if $(x,y,z)$ is a Gaussian integer solution of $x^4+y^4=z^2$, then $\xi|xyz$.

Suppose $\xi\nmid xyz$, so $\xi\nmid x,\xi\nmid y,\xi\nmid z$, then

$$\begin{aligned}x^4 &\equiv 1(\bmod\, 8)\\ y^4 &\equiv 1(\bmod\, 8)\\ z^2 &\equiv \pm 1(\bmod\, 4) \end{aligned}$$

Hence from $x^4+y^4-z^2\equiv 0(\bmod\,4)$ we get $1+1-\pm 1\equiv 0(\bmod\,4)$, a contradiction, so $\xi|xyz$.

Step Two

The second step: suppose $(x,y,z)$ is a solution of $x^4+y^4=z^2$ with all entries nonzero and pairwise coprime. Then $\xi$ can divide only one of $x,y,z$. But it cannot divide $z$, because if it divided $z$, then it would not divide $x,y$, and we would have ($\xi|z\Rightarrow \xi^4|z^4$)

$$\begin{aligned}x^4 &\equiv 1(\bmod\, 8)\\ y^4 &\equiv 1(\bmod\, 8)\\ z^2 &\equiv 0(\bmod\, \xi^2) \end{aligned}$$

Note that $\xi^2=2i,\xi^4=-4,\xi^6=-8i$. If $\xi^2|z$, then $\xi^4|z^2$, so the three equations above amount to saying $x^4+y^4-z^2\equiv 1+1-0\equiv 2(\bmod\,\xi^4)$, i.e. $4|2$, a contradiction; if $\xi\nmid\left(\frac{z}{\xi}\right)$, then $\left(\frac{z}{\xi}\right)^2\equiv \pm 1(\bmod\,4)$, i.e. $z^2\equiv \pm \xi^2(\bmod\,4\xi^2)$, which is equivalent to $z^2\equiv \pm \xi^2(\bmod\,8)$, so $x^4+y^4-z^2\equiv 1+1-\pm\xi^2\equiv 2(1\mp i)(\bmod\,\xi^6)$, i.e. $\xi^6|\xi^3$, again a contradiction.

So $\xi$ divides either $x$ or $y$; without loss of generality say $\xi|x$. It then follows necessarily that $z^2\equiv 1(\bmod\, 4)$, otherwise a contradiction results.

Step Three

The third step: let us list the results we have obtained so far:

$$\begin{aligned}x^4 &\equiv 0(\bmod\, \xi^4)\\ y^4 &\equiv 1(\bmod\, 8)\\ z^2 &\equiv 1(\bmod\, 4) \end{aligned}$$

Step Four

The fourth step is our core step. Suppose a solution exists; then there is a pairwise-coprime solution $(x,y,z),\ \xi|x$, and among all such solutions we pick the one for which $N(x)$ is smallest. Within $\mathbb{Z}[i]$ we can factor

$$x^4=(z-y^2)(z+y^2)$$

Let $u=z+y^2,v=z-y^2$; then

$$u+v=2z=-i\xi^2 z,\ u-v=2y^2=-i\xi^2 y^2$$

Any common divisor of $u,v$ must also be a common divisor of $u+v,u-v$, and $(u+v,u-v)=(-i\xi^2 z,-i\xi^2 y^2)=\xi^2$, so $u,v$ can have at most the common divisor $\xi^2$, whereas the left-hand side $x^4$ has at least the divisor $\xi^4$. Hence one of $u,v$ must have at least the divisor $\xi^2$, and consequently the greatest common divisor of $u,v$ is exactly $\xi^2$. Write $x=\xi \eta,u=\xi^2 \mu,v=\xi^2 \nu$, with $\mu,\nu$ coprime; then

$$\eta^4=\mu\nu$$

Since $\mu,\nu$ are coprime, $\mu,\nu$ must each be a fourth power up to a unit — that is, each is an associate of a fourth power. Let $\varepsilon_1,\varepsilon_2$ denote a unit, and set $\mu=\varepsilon_1 \kappa^4,\nu=\varepsilon_2 \iota^4$, i.e.

$$\eta^4=(\varepsilon_1 \kappa^4)(\varepsilon_2 \iota^4)=(\varepsilon_1\varepsilon_2) (\kappa\iota)^4$$

Then $\varepsilon_1\varepsilon_2$ is also a fourth power, but the only fourth power among the units is 1, so $\varepsilon_1\varepsilon_2=1$. Then from $u-v=2y^2=-i\xi^2 y^2$ we get

$$-i y^2=\varepsilon_1 \kappa^4-\varepsilon_2 \iota^4$$

Step Five

The fifth step: we enumerate the cases of $\varepsilon_1,\varepsilon_2$.

5.1. Suppose $\varepsilon_1=i,\varepsilon_2=-i$; then

$$-i y^2=i \kappa^4+i\iota^4$$

that is,

$$(iy)^2=\kappa^4+\iota^4$$

This shows that $(\kappa,\iota,iy)$ is also a solution. However, $N(\kappa),N(\iota)$ are all smaller than $N(x)$ (since $\varepsilon_1 \varepsilon_2\kappa^4\iota^4\xi^4=x^4$), and one of $\kappa,\iota$ must have the divisor $\xi$ — contradicting the minimality of $N(x)$.

5.2. Suppose $\varepsilon_1=-i,\varepsilon_2=i$; then

$$-i y^2=-i \kappa^4-i\iota^4$$

that is,

$$y^2=\kappa^4+\iota^4$$

This shows that $(\kappa,\iota,y)$ is also a solution. However, $N(\kappa),N(\iota)$ are all smaller than $N(x)$ (since $\varepsilon_1 \varepsilon_2\kappa^4\iota^4\xi^4=x^4$), and one of $\kappa,\iota$ must have the divisor $\xi$ — contradicting the minimality of $N(x)$.

5.3. Suppose $\varepsilon_1=-1,\varepsilon_2=-1$; then

$$-i y^2=-\kappa^4+\iota^4$$

If neither of $\kappa,\iota$ is a multiple of $\xi$, then $\kappa^4 \equiv\iota^4 \equiv 1(\bmod\,\xi^4)$, so $\xi^4|(-\kappa^4+\iota^4)$, giving $\xi|y$, which contradicts the fact that $x,y$ are coprime. So one of $\kappa,\iota$ (and only one) must be a multiple of $\xi$, and hence $-\kappa^4+\iota^4\equiv \pm 1(\bmod\,\xi^4)$. But the left-hand side $\xi\nmid y$, so $y^2\equiv \pm 1(\bmod\,\xi^4)$, hence $-iy^2\equiv \pm i(\bmod\,\xi^4)$. The congruences on the two sides disagree — a contradiction.

5.4. Suppose $\varepsilon_1=1,\varepsilon_2=1$; then

$$-i y^2=\kappa^4-\iota^4$$

The analysis is essentially the same as in 5.3, and also leads to a contradiction.

All cases have now been ruled out, so the original assumption fails: there is no solution of $x^4+y^4=z^2$ in Gaussian integers with all entries nonzero.

Commentary

The proof may look long when written out, but it is actually quite simple. In essence, we have only used analysis modulo $1+i$, which is the exact analogue of parity analysis over the integers. If the reader is unfamiliar with Gaussian integers, these congruences will seem murky and hard to grasp, but once one is familiar with Gaussian integers, everything becomes quite obvious. After all, is parity analysis over the ordinary integers not equally straightforward? If the reader has already gone through the proof for n=4 over the ordinary integers, it's worth comparing the two — you'll find similar structures, though the proof here seems even a bit shorter. (Length doesn't necessarily mean complexity — what matters is whether each step is obvious.)

What is it that forces us to split into four cases in the last step? The units! There are four distinct units in the Gaussian integers, hence the four cases; in more general rings of integers there may be even more units, which is one reason why the proof becomes harder for larger n. But this difficulty is not the core one — the core difficulty is the failure of unique factorization, which is a story for later.

Finally, let me say a bit more about why it is $x^4+y^4=z^2$ rather than $x^4+y^4=z^4$. The latter genuinely cannot be shown to lead to a contradiction using the techniques of this article. So a natural question arises: is there a proof that establishes the latter directly, on its own? I don't know. I spent several days trying to find a proof of just the latter statement, but without success. Perhaps this is simply a peculiarity of the case where $n$ is composite.

Addendum (Aug. 20) #

One can directly prove that $\varepsilon x^4+y^4=z^2,\ \xi|x$ has no solutions, which simplifies the proof — see the next article. Just modify the steps of the next article slightly.