It is time!
In the previous articles, we have paved the way for the proof of Fermat’s Last Theorem. Of course, compared to the complete proof of Fermat’s Last Theorem, these few articles are just a drop in the ocean. However, they are already sufficient to complete the proof of Fermat’s Last Theorem for the case n=4. We will soon see the concise proof brought by Gaussian integers for n=4, which strengthens our belief that this path can lead much further.
The Diophantine equation x^4 + y^4 = z^2 has no solutions in \mathbb{Z}[i] where x, y, z are all non-zero.
As the reader can see, we are considering x^4 + y^4 = z^2 rather than x^4 + y^4 = z^4. The former is a stronger version of the latter. However, we do not strengthen it just to prove a more general proposition; rather, our proof for x^4 + y^4 = z^4 simply does not work on its own! That is to say, according to the method in this article, we can prove that x^4 + y^4 = z^2 has no solutions, but we cannot "only prove" that x^4 + y^4 = z^4 has no solutions. This is indeed a very curious phenomenon. Some propositions are easier to prove only after they have been strengthened, much like using mathematical induction to prove certain inequalities—if they are not strengthened, the induction fails.
The tools we use are very simple: the "analysis modulo 1+i" within \mathbb{Z}[i] mentioned in the third article. For convenience, we denote \xi = 1+i below.
Step One
In the first step, we prove that if (x, y, z) is a set of Gaussian integer solutions to x^4 + y^4 = z^2, then \xi | xyz.
Assume \xi \nmid xyz. Then \xi \nmid x, \xi \nmid y, \xi \nmid z. Thus: \begin{aligned} x^4 &\equiv 1 \pmod{8} \\ y^4 &\equiv 1 \pmod{8} \\ z^2 &\equiv \pm 1 \pmod{4} \end{aligned} Therefore, x^4 + y^4 - z^2 \equiv 0 \pmod{4} leads to 1 + 1 - (\pm 1) \equiv 0 \pmod{4}, which is a contradiction. Thus, \xi | xyz.
Step Two
In the second step, assume (x, y, z) is a set of non-zero, pairwise coprime Gaussian integer solutions to x^4 + y^4 = z^2. Then \xi can only divide exactly one of x, y, z. However, it is impossible for \xi to divide z. If \xi | z, then \xi does not divide x or y. Then we have (since \xi | z \Rightarrow \xi^4 | z^4): \begin{aligned} x^4 &\equiv 1 \pmod{8} \\ y^4 &\equiv 1 \pmod{8} \\ z^2 &\equiv 0 \pmod{\xi^2} \end{aligned} Note that \xi^2 = 2i, \xi^4 = -4, \xi^6 = -8i. If \xi^2 | z, then \xi^4 | z^2. Thus, the above three equations imply x^4 + y^4 - z^2 \equiv 1 + 1 - 0 \equiv 2 \pmod{\xi^4}, which means 4 | 2, a contradiction. If \xi \nmid (z/\xi), then (z/\xi)^2 \equiv \pm 1 \pmod{4}, which means z^2 \equiv \pm \xi^2 \pmod{4\xi^2}, equivalent to z^2 \equiv \pm \xi^2 \pmod{8}. Therefore, x^4 + y^4 - z^2 \equiv 1 + 1 - (\pm \xi^2) \equiv 2(1 \mp i) \pmod{\xi^6}, which means \xi^6 | \xi^3, a contradiction.
So \xi divides either x or y. Without loss of generality, let \xi | x. Consequently, we must have z^2 \equiv 1 \pmod{4}, otherwise a contradiction arises.
Step Three
In the third step, we list the results we have obtained: \begin{aligned} x^4 &\equiv 0 \pmod{\xi^4} \\ y^4 &\equiv 1 \pmod{8} \\ z^2 &\equiv 1 \pmod{4} \end{aligned}
Step Four
The fourth step is our core step. Suppose a solution exists; then there exists a set of pairwise coprime solutions (x, y, z) with \xi | x. Among all such solutions, choose the one where the norm N(x) is minimal. In \mathbb{Z}[i], we can factor: x^4 = (z - y^2)(z + y^2) Let u = z + y^2 and v = z - y^2. Then: u + v = 2z = -i\xi^2 z, \quad u - v = 2y^2 = -i\xi^2 y^2 Any common divisor of u and v must also be a common divisor of u+v and u-v. Thus (u+v, u-v) = (-i\xi^2 z, -i\xi^2 y^2) = \xi^2. Therefore, u and v have at most \xi^2 as a common divisor. Since the left side x^4 has at least a factor of \xi^4, one of u or v must have at least a factor of \xi^2. Consequently, the greatest common divisor of u and v is exactly \xi^2. Let x = \xi \eta, u = \xi^2 \mu, v = \xi^2 \nu, where \mu and \nu are coprime. Thus: \eta^4 = \mu\nu Since \mu and \nu are coprime, they must each be a fourth power up to a unit factor. That is, they are associates of fourth powers. Let \varepsilon_1, \varepsilon_2 be units, and let \mu = \varepsilon_1 \kappa^4, \nu = \varepsilon_2 \iota^4. Then: \eta^4 = (\varepsilon_1 \kappa^4)(\varepsilon_2 \iota^4) = (\varepsilon_1 \varepsilon_2) (\kappa\iota)^4 Thus \varepsilon_1 \varepsilon_2 is also a fourth power. Among the units in \mathbb{Z}[i], the only fourth power 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
In the fifth step, we enumerate the possibilities for \varepsilon_1 and \varepsilon_2.
5.1. If \varepsilon_1 = i, \varepsilon_2 = -i, then: -i y^2 = i \kappa^4 + i \iota^4 which is: (iy)^2 = \kappa^4 + \iota^4 This indicates that (\kappa, \iota, iy) is also a solution. However, N(\kappa) and N(\iota) are both smaller than N(x) (since \varepsilon_1 \varepsilon_2 \kappa^4 \iota^4 \xi^4 = x^4), and one of \kappa, \iota must be divisible by \xi. This contradicts the minimality of N(x).
5.2. If \varepsilon_1 = -i, \varepsilon_2 = i, then: -i y^2 = -i \kappa^4 - i \iota^4 which is: y^2 = \kappa^4 + \iota^4 This indicates that (\kappa, \iota, y) is also a solution. However, N(\kappa) and N(\iota) are both smaller than N(x) (since \varepsilon_1 \varepsilon_2 \kappa^4 \iota^4 \xi^4 = x^4), and one of \kappa, \iota must be divisible by \xi. This contradicts the minimality of N(x).
5.3. If \varepsilon_1 = -1, \varepsilon_2 = -1, then: -i y^2 = -\kappa^4 + \iota^4 If neither \kappa nor \iota is a multiple of \xi, then \kappa^4 \equiv \iota^4 \equiv 1 \pmod{\xi^4}, which implies \xi^4 | (-\kappa^4 + \iota^4), leading to \xi | y. This contradicts the fact that x and y are coprime. Thus, exactly one of \kappa, \iota is a multiple of \xi, which implies -\kappa^4 + \iota^4 \equiv \pm 1 \pmod{\xi^4}. However, on the left side \xi \nmid y, so y^2 \equiv \pm 1 \pmod{\xi^4}, which means -iy^2 \equiv \pm i \pmod{\xi^4}. The two sides are not congruent, a contradiction.
5.4. If \varepsilon_1 = 1, \varepsilon_2 = 1, then: -i y^2 = \kappa^4 - \iota^4 The analysis is essentially the same as in 5.3, also leading to a contradiction.
All cases have been refuted. Therefore, the original assumption is false, and there are no non-zero Gaussian integer solutions to x^4 + y^4 = z^2.
Commentary
The proof may seem long, but it is actually quite simple. Overall, we only used analysis modulo 1+i, which is equivalent to parity analysis in real integers. If the reader is not familiar with Gaussian integers, this congruence might seem confusing, but for those familiar with them, everything is quite obvious. Imagine, isn’t parity analysis in real integers easy? If the reader has already read the proof for n=4 in real integers, they might compare the two and find similarities, though the proof in this article seems even shorter. (Length does not necessarily imply complexity; the key is whether each step is self-evident.)
What caused us to split into four cases in the final step? Units! There are four different units in Gaussian integers, hence the four cases. In more general number rings, there may be even more units, which is one reason why proofs for larger n are more difficult. However, this difficulty is not the most fundamental one; the core difficulty is the failure of the Unique Factorization Theorem, but that is a story for another time.
Finally, I should mention again: why x^4 + y^4 = z^2 instead of x^4 + y^4 = z^4? The techniques in this article indeed cannot derive a contradiction for the latter directly. A natural question is whether there is a proof that only proves the latter. I do not know; I tried for several days to find a process that only proves the latter, but to no avail. Perhaps for composite n, the proofs naturally have such characteristics.
Supplement (Aug 20)
One can directly prove that \varepsilon x^4 + y^4 = z^2, \xi | x has no solutions, thereby simplifying the proof process. Refer to the next article. The steps in the next article can be slightly modified for this purpose.
Original Address: https://kexue.fm/archives/2831
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