The Proof for \(n = 3\)¶

Summary

Euler's proof of Fermat's Last Theorem for the case \(n = 3\) – and why it forces us to leave the ordinary integers behind. The entry into algebraic number theory.

Prerequisites¶

1. Why \(n = 3\) Is Harder Than \(n = 4\)¶

In the proof for \(n = 4\), we could treat the equation \(x^4 + y^4 = z^2\) entirely within \(\mathbb{Z}\). The key was the parametrisation of Pythagorean triples – a formula describing all solutions of \(x^2 + y^2 = z^2\).

For \(n = 3\), no analogue exists. The equation \(x^3 + y^3 = z^3\) cannot be factorised in \(\mathbb{Z}\) in a way that makes a descent possible. The reason: the factorisation

\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \]

yields two factors whose coprimality is hard to control. One would need information about \(\gcd(x + y, \, x^2 - xy + y^2)\), which is not easy to obtain in \(\mathbb{Z}\).

Euler's brilliant idea: instead of computing in \(\mathbb{Z}\), we extend the number domain.

2. The Eisenstein Integers¶

Let \(\omega = e^{2\pi i/3} = \frac{-1 + \sqrt{-3}}{2}\) be a primitive cube root of unity. The Eisenstein integers are the ring:

\[ \mathbb{Z}[\omega] = \{ a + b\omega \mid a, b \in \mathbb{Z} \} \]

Geometrically, the Eisenstein integers form a regular triangular lattice in the complex plane. Every element has the form \(a + b\omega\) with integer coordinates.

Basic Properties¶

Norm. For \(\alpha = a + b\omega\) we define the norm as:

\[ N(\alpha) = \alpha \cdot \bar{\alpha} = a^2 - ab + b^2 \]

where \(\bar{\alpha} = a + b\bar{\omega} = a + b\omega^2\) is the conjugate (since \(\omega^2 = \bar{\omega}\)). The norm is always a non-negative integer and is multiplicative: \(N(\alpha\beta) = N(\alpha) \cdot N(\beta)\).

Units. The units (invertible elements) of \(\mathbb{Z}[\omega]\) are precisely the elements with norm \(1\):

\[ \mathbb{Z}[\omega]^\times = \{ \pm 1, \pm \omega, \pm \omega^2 \} \]

That is six units – more than the two units \(\pm 1\) in \(\mathbb{Z}\).

Prime elements. An Eisenstein element \(\pi\) is called prime if it is not a unit and \(\pi \mid \alpha\beta\) implies \(\pi \mid \alpha\) or \(\pi \mid \beta\). The prime structure of \(\mathbb{Z}[\omega]\) differs from that of \(\mathbb{Z}\):

The prime \(3\) decomposes specially: \(3 = -\omega^2 (1 - \omega)^2\), so \(\lambda := 1 - \omega\) is a prime element with \(N(\lambda) = 3\).
Primes \(p \equiv 2 \pmod{3}\) remain prime in \(\mathbb{Z}[\omega]\).
Primes \(p \equiv 1 \pmod{3}\) split: \(p = \pi \bar{\pi}\) for some prime element \(\pi\).

The Decisive Advantage¶

In \(\mathbb{Z}[\omega]\) we can factorise \(x^3 + y^3\) completely:

\[ x^3 + y^3 = (x + y)(x + \omega y)(x + \omega^2 y) \]

Three linear factors instead of two! This finer factorisation makes the descent possible.

3. Unique Factorisation¶

The proof works only if \(\mathbb{Z}[\omega]\) is a principal ideal domain (PID) – that is, if every element has an essentially unique decomposition into prime elements.

Theorem. \(\mathbb{Z}[\omega]\) is a Euclidean domain (with the norm function as Euclidean function) and therefore in particular a PID.

Proof sketch. For \(\alpha, \beta \in \mathbb{Z}[\omega]\) with \(\beta \neq 0\), consider \(\alpha/\beta \in \mathbb{Q}(\omega)\). This element can be approximated by a lattice element \(\gamma \in \mathbb{Z}[\omega]\) with \(N(\alpha/\beta - \gamma) < 1\) (because the triangular lattice is dense enough). Then \(\alpha = \beta\gamma + \rho\) with \(N(\rho) < N(\beta)\) – exactly the division with remainder that is needed. \(\square\)

Not self-evident!

For \(p = 3\), \(\mathbb{Z}[\omega]\) is a PID – but \(\mathbb{Z}[\zeta_p]\) is not a PID for general \(p\). Already for \(p = 23\), unique factorisation fails. This was the point where Lamé failed and Kummer invented ideal theory.

4. The Proof: Step by Step¶

We prove: \(x^3 + y^3 = z^3\) has no solution in positive integers.

Equivalently (and more technically convenient), we prove the more general statement in \(\mathbb{Z}[\omega]\):

\[ \alpha^3 + \beta^3 + \gamma^3 = 0 \quad \text{has no solution with } \alpha, \beta, \gamma \in \mathbb{Z}[\omega] \setminus \{0\} \text{ and } \lambda \nmid \alpha\beta\gamma \]

where \(\lambda = 1 - \omega\) is the prime element above \(3\). (The symmetric form \(\alpha^3 + \beta^3 + \gamma^3 = 0\) is equivalent to \(x^3 + y^3 = z^3\) with the sign of \(z\) reversed.)

In fact, we prove an even stronger version by infinite descent. The proof proceeds in several stages.

Preparation: Cubic residues modulo \(\lambda\)¶

Since \(N(\lambda) = 3\), we have \(\mathbb{Z}[\omega]/(\lambda) \cong \mathbb{Z}/3\mathbb{Z} = \{0, 1, 2\}\). Every element of \(\mathbb{Z}[\omega]\) not divisible by \(\lambda\) is congruent to \(\pm 1 \pmod{\lambda}\). It follows that every cube of such an element is likewise congruent to \(\pm 1 \pmod{\lambda}\).

More precisely, for every \(\alpha\) with \(\lambda \nmid \alpha\):

\[ \alpha^3 \equiv \pm 1 \pmod{\lambda^4} \]

This is the analogue of the statement "every square is \(\equiv 0\) or \(1 \pmod{4}\)" in \(\mathbb{Z}\) – but one level more intricate.

The Descent¶

Assumption. Let \((\alpha, \beta, \gamma)\) be a solution of \(\alpha^3 + \beta^3 + \gamma^3 = 0\) with \(\lambda \nmid \alpha\beta\gamma\) and with minimal \(\lambda\)-valuation in one of the terms. More precisely: we assume that \(\lambda \mid \gamma\) (after reordering), and write \(\gamma = \lambda^n \delta\) with \(\lambda \nmid \delta\) and minimal \(n \geq 1\).

We show that from this solution a new solution with smaller \(n\) can be constructed – contradiction.

Step 1: Factorisation. In \(\mathbb{Z}[\omega]\):

\[ \alpha^3 + \beta^3 = -\gamma^3 = -\lambda^{3n} \delta^3 \]

\[ (\alpha + \beta)(\alpha + \omega\beta)(\alpha + \omega^2\beta) = -\lambda^{3n} \delta^3 \]

Step 2: Coprimality of the factors. One shows that the three factors \(\alpha + \beta\), \(\alpha + \omega\beta\), \(\alpha + \omega^2\beta\) can be separated pairwise by \(\lambda\): their pairwise differences are \((1 - \omega)\beta = \lambda\beta\) and \((1 - \omega^2)\beta\), so \(\lambda\) is the only common factor. After careful analysis of the \(\lambda\)-valuation, one can show that exactly one of the three factors is divisible by \(\lambda^{3n-2}\) and the other two are not divisible by \(\lambda\).

Step 3: Force cubic structure. Since \(\mathbb{Z}[\omega]\) is a PID and the three factors (up to \(\lambda\)-parts) are coprime, each factor (up to units and powers of \(\lambda\)) must be a cube. In particular, there exist \(\alpha_1, \beta_1, \gamma_1 \in \mathbb{Z}[\omega]\) with:

\[ \alpha + \beta = \varepsilon_1 \lambda^{3n-2} \gamma_1^3, \quad \alpha + \omega\beta = \varepsilon_2 \alpha_1^3, \quad \alpha + \omega^2\beta = \varepsilon_3 \beta_1^3 \]

where \(\varepsilon_1, \varepsilon_2, \varepsilon_3\) are units.

Step 4: Derive a new equation. From the three equations, one can (by skilful combination) derive an equation of the form

\[ \alpha_1^3 + \beta_1^3 + \varepsilon \lambda^{n'} \gamma_1^3 = 0 \]

where \(n' < n\). The unit \(\varepsilon\) can be absorbed by choosing suitable associates.

Step 5: Contradiction. We have found a solution with \(\lambda\)-valuation \(n' < n\). Since \(n\) was minimal, this is a contradiction. \(\blacksquare\)

5. The Gap in Euler's Original¶

Euler's proof, as it appeared in his 1770 Algebra, contained a subtle gap. In the decisive step of the descent, he used the fact that certain elements in \(\mathbb{Z}[\omega]\) must be cubes – and thereby implicitly assumed unique prime factorisation in \(\mathbb{Z}[\omega]\).

This assumption happens to be correct: \(\mathbb{Z}[\omega]\) is indeed a PID. But Euler did not prove this; he took it for granted. The fact that UPF holds in \(\mathbb{Z}[\omega]\) was rigorously established only later – among others by Gauss.

The gap is repairable: one can make Euler's proof completely correct by establishing the PID property of \(\mathbb{Z}[\omega]\) first. But the gap reveals a deep conceptual problem: for general \(p\), \(\mathbb{Z}[\zeta_p]\) is not a PID, and Euler's strategy breaks down.

6. A Taste of Where the Method Fails¶

For \(p = 5\), we consider \(\mathbb{Z}[\zeta_5]\) with \(\zeta_5 = e^{2\pi i/5}\). Here UPF still holds – the proof for \(n = 5\) (Dirichlet/Legendre, 1825) uses this, albeit with considerably more effort.

For \(p = 23\), the situation is dramatically different: \(\mathbb{Z}[\zeta_{23}]\) has class number \(h_{23} = 3 \neq 1\). UPF fails, and a naïve factorisation approach yields false results. This is precisely where Kummer's ideal theory stepped in, as we shall explore further in the article on rings and fields.

The moral: The proof for \(n = 3\) works because we are lucky. The Eisenstein integers are "good enough" – they have unique factorisation. For general \(p\), an entirely different approach is needed.

7. From Euler to Kummer – and Beyond¶

The proof for \(n = 3\) marks a turning point in the history of FLT:

Aspect	\(n = 4\) (Fermat)	\(n = 3\) (Euler)
Number domain	\(\mathbb{Z}\)	\(\mathbb{Z}[\omega]\)
Factorisation	Pythagorean triples	cubic factorisation
Descent via	size of \(z\)	\(\lambda\)-valuation
New mathematics	Infinite Descent	Algebraic Number Theory

From here the path branches:

Foundational tools: To understand the proof for general \(p\), we need groups, rings, fields, Galois theory – the "language" of modern algebra.
Specialised tools: Elliptic curves and modular forms – the objects that Wiles' proof connects.
The proof itself: Galois representations, deformation theory, the \(R = T\) theorem – the heart of the matter.

In the following tool topics, we build these foundations piece by piece.