In the previous article, I mentioned that to prove the case n=4, it seems necessary to prove that x^4+y^4=z^2 has no solutions, rather than just proving x^4+y^4=z^4 has no solutions. However, while researching this afternoon, I discovered another proof for n=4. This proof is also situated in \mathbb{Z}[i], but it deals with the form where all exponents are 4. Furthermore, it does not merely address x^4+y^4=z^4, but rather the form \varepsilon x^4+y^4=z^4, where \varepsilon is a unit. I believe this proof process is closer to the proofs for other odd prime values of n. Therefore, I have supplemented this article for your reference. Readers may find it useful to compare this with the previous article.
Lemma
Let \varepsilon_1, \varepsilon_2, \varepsilon_3, \varepsilon denote units 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] where x', y', z' are all non-zero, then after appropriate 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 process is similar to the one in the previous article. First, we prove that \xi | x'y'z'. If not, then: \begin{aligned} x'^4 &\equiv 1 \pmod 8 \\ y'^4 &\equiv 1 \pmod 8 \\ z'^4 &\equiv 1 \pmod 8 \end{aligned} This would imply 8 | (\varepsilon_1 + \varepsilon_2 + \varepsilon_3), which is impossible regardless of the values of the units \varepsilon_1, \varepsilon_2, \varepsilon_3.
Now, assume \xi | x' and \xi \nmid y'z'. Then: \begin{aligned} x'^4 &\equiv 0 \pmod{\xi^4} \\ y'^4 &\equiv 1 \pmod 8 \\ z'^4 &\equiv 1 \pmod 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 (noting that -8i = \xi^6), we get: \xi^4 | (1 + \varepsilon_3/\varepsilon_2) Thus, \varepsilon_3/\varepsilon_2 = -1. Setting \varepsilon_1/\varepsilon_2 = \varepsilon and (x', y', z') = (x, y, z), we obtain: \varepsilon x^4 + y^4 = z^4 Since z^4 - y^4 \equiv x^4 \equiv 0 \pmod 8 and \xi^6 = -8i, it follows that at least \xi^2 | x.
Proof
Assume (x, y, z) is a set of pairwise coprime solutions to \varepsilon x^4 + y^4 = z^4 with \xi^2 | x, and that this solution has the minimal power of \xi contained in x. It is important to note that \varepsilon is an arbitrary unit and is not fixed; therefore, "minimal" here refers to the solution with the smallest \xi-valuation across all possible units \varepsilon. Let \xi^m | x and \xi^{m+1} \nmid x, where m \geq 2.
In \mathbb{Z}[i], we can factorize the equation completely: \varepsilon x^4 = (z+y)(z-y)(z+yi)(z-yi) Note 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 calculations show that the greatest common divisor of (z+y) and (z-y) is at most \xi^2, the gcd of (z-y) and (z+yi) is at most \xi, and the gcd of (z+yi) and (z-yi) is at most \xi^2. Since the left side is divisible by at least \xi^8, at least one term on the right must be divisible by \xi^2. However, it is impossible for all terms to be divisible by \xi^2, as this would contradict the fact that the gcd of (z-y) and (z+yi) is at most \xi.
Furthermore, at least two terms must have a \xi-valuation of exactly 1. This is because if (z+y) is divisible by \xi^2, then (z-y) must also be divisible by \xi^2; similarly, if (z+yi) is divisible by \xi^2, then (z-yi) must be as well. If only one term had a \xi-valuation of 1, it would imply the other three terms have valuations of at least 2, leading to a contradiction. Thus, among the four terms, the valuations of \xi are \geq 4m-4, 2, 1, 1. Without loss of generality, let: \begin{aligned} x &= \xi^m \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' Since u', v', s', t' are pairwise coprime, they are each associates of a fourth power. Furthermore, none of them can be divisible by \xi, otherwise it would contradict \xi^{m+1} \nmid x. Thus, we can set: \begin{aligned} u' = \varepsilon_1 u^4, \quad v' = \varepsilon_2 v^4 \\ s' = \varepsilon_3 s^4, \quad t' = \varepsilon_4 t^4 \end{aligned} From this, we have: \begin{aligned} 2z &= \varepsilon_1 \xi^{4m-4} u^4 + \varepsilon_2 \xi^2 v^4 = \varepsilon_3 \xi s^4 + \varepsilon_4 \xi t^4 \\ 2y &= \varepsilon_1 \xi^{4m-4} u^4 - \varepsilon_2 \xi^2 v^4 = (\varepsilon_3 \xi s^4 - \varepsilon_4 \xi t^4)(-i) \end{aligned} Which can be rewritten as: \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: 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: (-i)(\varepsilon_1/\varepsilon_4) \xi^{4m-4} u^4 + i (\varepsilon_3/\varepsilon_4) s^4 = t^4 By the Lemma, we must have: i (\varepsilon_3/\varepsilon_4) = 1 Letting (-i)(\varepsilon_1/\varepsilon_4) = \varepsilon', we obtain: \varepsilon' (\xi^{m-1} u)^4 + s^4 = t^4 This shows that (\xi^{m-1} u, s, t) is a solution for the unit \varepsilon'. Since \xi \nmid u and m-1 < m, this contradicts the assumption of minimality. Therefore, the equation \varepsilon x^4 + y^4 = z^4 has no non-zero solutions in \mathbb{Z}[i].
Commentary
The process above may seem circuitous, but it reveals the general proof patterns for Fermat’s Last Theorem for general n. The most critical step is utilizing the extended number field to completely factorize z^n - y^n. If the Unique Factorization Theorem holds, a contradiction can be derived. At the same time, basic congruence analysis is essential; the results of the congruence analysis "perfectly" help us align the coefficients so that we can "just" use Fermat’s method of infinite descent to derive a contradiction. All of this seems like a series of coincidences, yet it also feels inevitable...
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