The Proof for \(n = 4\)¶
Summary
Fermat's own proof – the only case of FLT that he himself proved. We introduce the method of infinite descent and show: \(x^4 + y^4 = z^2\) has no solution in positive integers.
Prerequisites¶
1. The Method of Infinite Descent¶
Infinite descent (descente infinie) is a proof technique that Fermat himself invented – and which he called his favourite method. The basic idea is strikingly simple:
Assume an equation has a solution in positive integers. Then show that from this solution a smaller solution can be constructed – one in which the numbers involved are strictly smaller. From this smaller solution, an even smaller one could be constructed, and so on – ad infinitum.
But: positive integers cannot become infinitely small. There is no infinite descent in \(\mathbb{Z}^+\). Therefore the assumption is false, and the equation has no solution.
Formally, the descent uses the well-ordering principle: every non-empty subset of \(\mathbb{N}\) has a smallest element. An infinitely descending sequence of positive integers therefore cannot exist.
Descent vs. Contradiction
Infinite descent is a special proof by contradiction: one assumes a solution exists and derives an impossible consequence (the infinite descending sequence). In modern formulation, equivalently: one considers a minimal solution and shows that an even smaller one would have to exist – contradiction.
2. Pythagorean Triples¶
Before we can carry out the proof for \(n = 4\), we need a classical result: the complete description of all solutions of \(x^2 + y^2 = z^2\).
Theorem (Parametrisation of Pythagorean triples). All primitive Pythagorean triples \((x, y, z)\) with \(\gcd(x, y) = 1\) and \(x\) even have the form:
where \(s > t > 0\), \(\gcd(s, t) = 1\), and \(s \not\equiv t \pmod{2}\) (i.e., \(s\) and \(t\) have different parity).
Proof sketch. We write \(x^2 = z^2 - y^2 = (z-y)(z+y)\). Since \((x, y, z)\) is primitive, \(z - y\) and \(z + y\) are coprime (up to a factor of 2). Since \(x\) is even, \(z\) and \(y\) are both odd, so \(z - y\) and \(z + y\) are both even. Set \(z - y = 2u\) and \(z + y = 2v\), then \(x^2 = 4uv\) with \(\gcd(u, v) = 1\). Since the product \(uv\) is a perfect square and the factors are coprime, \(u\) and \(v\) must themselves be perfect squares: \(u = t^2\), \(v = s^2\). Substituting yields the parametrisation. \(\square\)
Examples:
| \(s\) | \(t\) | \(x = 2st\) | \(y = s^2 - t^2\) | \(z = s^2 + t^2\) |
|---|---|---|---|---|
| 2 | 1 | 4 | 3 | 5 |
| 3 | 2 | 12 | 5 | 13 |
| 4 | 1 | 8 | 15 | 17 |
| 4 | 3 | 24 | 7 | 25 |
3. From FLT to a Stronger Statement¶
Fermat's proof for \(n = 4\) does not directly show that \(x^4 + y^4 = z^4\) has no solution, but rather the stronger statement:
Theorem (Fermat)
The equation \(x^4 + y^4 = z^2\) has no solution in positive integers.
Why is this stronger? Because \(z^4 = (z^2)^2\) is a special perfect square. If there is no solution with \(z^2\) on the right-hand side, then certainly not with \(z^4\).
Hence: \(x^4 + y^4 = z^2\) has no solution \(\implies\) \(x^4 + y^4 = z^4\) has no solution.
4. The Proof in Detail¶
We prove: \(x^4 + y^4 = z^2\) has no solution in \(x, y, z \in \mathbb{Z}^+\).
Assumption for contradiction. Let \((x, y, z)\) be a solution with minimal \(z\). Without loss of generality, let \(\gcd(x, y) = 1\) (otherwise we cancel the common factor and obtain a smaller solution).
Step 1: Apply Pythagorean triples.
The equation \(x^4 + y^4 = z^2\) can be read as \((x^2)^2 + (y^2)^2 = z^2\) – a Pythagorean triple! Since \(\gcd(x, y) = 1\), the triple is primitive, and we can apply the parametrisation. Without loss of generality, let \(x\) be even (otherwise swap \(x\) and \(y\)). Then there exist \(s, t\) with \(s > t > 0\), \(\gcd(s, t) = 1\), \(s \not\equiv t \pmod{2}\), such that:
Step 2: Another Pythagorean triple.
From \(y^2 = s^2 - t^2\) it follows that \(y^2 + t^2 = s^2\) – again a Pythagorean triple! Since \(\gcd(s, t) = 1\) and \(s \not\equiv t \pmod{2}\), this triple is also primitive. Now \(y\) is odd (because \(x\) is even and \(\gcd(x, y) = 1\)), so \(t\) is even. The parametrisation yields \(u, v\) with \(u > v > 0\), \(\gcd(u, v) = 1\), \(u \not\equiv v \pmod{2}\):
Step 3: Analyse \(x^2\) as a product.
Substituting \(s = u^2 + v^2\) and \(t = 2uv\) into \(x^2 = 2st\):
Hence \((x/2)^2 = uv(u^2 + v^2)\). Since \(\gcd(u, v) = 1\), the three factors \(u\), \(v\), and \(u^2 + v^2\) are pairwise coprime. Their product is a perfect square, so each individual factor must be a perfect square:
for certain positive integers \(a, b, c\).
Step 4: The descent.
From \(u^2 + v^2 = c^2\) and \(u = a^2\), \(v = b^2\) it follows:
This is the same equation as our original one! And we have:
Hence \(c < z\) – we have found a smaller solution.
Contradiction. We had chosen \((x, y, z)\) as a solution with minimal \(z\), but \((a, b, c)\) is a solution with \(c < z\). Contradiction! \(\blacksquare\)
5. Why the Descent Works¶
The proof has an elegant structure:
Solution (x, y, z) with z minimal
→ Parametrisation as Pyth. triple → (s, t)
→ Second Pyth. triple → (u, v)
→ Three coprime squares → (a², b², c²)
→ New solution (a, b, c) with c < z
→ CONTRADICTION
The key is that each step makes the numbers smaller. From \(z\) via \(s\) (which is smaller than \(z\)) via \(u\) and \(v\) (which are smaller than \(s\)) to \(c\) (which is smaller than \(z\)). The well-ordering of \(\mathbb{N}\) guarantees that this process cannot continue indefinitely.
Why the stronger statement?
Fermat's trick of considering \(z^2\) instead of \(z^4\) is no coincidence. In the descent, an equation \(a^4 + b^4 = c^2\) arises – only if the right-hand side is allowed to be a general perfect square (not just a fourth power) does the inductive loop close. Had we considered only \(x^4 + y^4 = z^4\), the equation arising in the descent, \(a^4 + b^4 = c^2\), would not have the same form – and the proof would break down.
6. Historical Context¶
This proof is the only case of FLT for which Fermat himself left a verifiable proof. It appears in his Observationes (appendix to the 1670 edition of the Arithmetica) and there proves the statement that the area of a right triangle with integer sides cannot be a perfect square – which is equivalent to \(x^4 + y^4 = z^2\).
What the proof shows – and what it doesn't:
- ✅ FLT is true for \(n = 4\) (and hence for all \(n\) divisible by \(4\): \(n = 8, 12, 16, \ldots\))
- ❌ The method cannot be directly transferred to \(n = 3\) – the simple parametrisation of "cubic triples" is lacking
- ❌ For general primes \(p\), the elementary descent fails
The case \(n = 3\) requires, as we shall see in Article 4, a decisive conceptual leap: one must extend the number domain from \(\mathbb{Z}\) to \(\mathbb{Z}[\omega]\). Here begins the path to algebraic number theory.
Further Reading¶
- Fermat: Observationes ad Diophantum (1670) – the original proof
- Nigel Boston: The Proof of Fermat's Last Theorem, Ch. 1
- Harold Edwards: Fermat's Last Theorem – historically detailed exposition