fermats-last-theorem-n3

Fermat’s Last Theorem at $n=3$

Summary: The case $n=3$ of Fermat’s Last Theorem: $x^3+y^3=z^3$ has no nontrivial integer solutions. Stronger, no two cubes can sum or differ by a cube, nor can they sum or differ to twice a cube. Euler gave the first widely-accepted proof (1770) in Vollständige Anleitung zur Algebra §243 — reproduced in ch2.0.15-questions-cubes — using infinite descent and arithmetic in $\mathbb{Z}[\sqrt{-3}]$.

Sources: chapter-2.0.15

Last updated: 2026-05-09

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Statement

For positive integers $x,y,z$:

$\boxed{x^3+y^3\ne z^3.}$

Three companion results in Elements of Algebra:

1. §243: $x^3+y^3\ne z^3$ and $x^3-y^3\ne z^3$ (the case Fermat proves).
2. §247: $x^3+y^3\ne 2z^3$ and $x^3-y^3\ne 2z^3$ (no sum/difference of two cubes equals twice a cube), except trivially when $y=x$.
3. Implicit (§246): $2a^3+x^3\ne y^3$ as a special case of (2).

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Euler’s Proof Skeleton (§243)

Below is the structural outline; full algebraic detail is in ch2.0.15-questions-cubes.

Setup. Assume coprime $x,y$ both odd (the case treats sum; difference is symmetric). Set $p=(x+y)/2$, $q=(x-y)/2$. Then

$x^3+y^3=2p(p^2+3q^2).$

We need $2p(p^2+3q^2)$ a cube.

Branch on whether $3\mid p$.

Case 1: $3\nmid p$

The two factors $p$ and $p^2+3q^2$ are coprime (modulo a factor of 4 absorbed elsewhere), so each must independently be a cube. To represent $p^2+3q^2$ as a cube, use the identity in $\mathbb{Z}[\sqrt{-3}]$:

$(t+u\sqrt{-3}{)}^3=t^3-9t{u}^2+3u(t^2-u^2)\sqrt{-3},$

which gives

$p=t(t-3u)(t+3u),q=3u(t^2-u^2),p^2+3q^2=(t^2+3u^2{)}^3.$

Constraint that $p/4$ also be a cube reduces (via parity and coprimality) to three pairwise coprime factors $2t,t+3u,t-3u$ each independently being a cube. Set $t+3u=f^3$, $t-3u=g^3$. Then

$2t=f^3+g^3.$

Since $2t$ must be a cube too, $f^3+g^3$ is a cube — but $f,g$ are smaller than $x,y$.

Case 2: $3\mid p$

Set $p=3r$. The formula becomes $\frac{9}{4}r(3r^2+q^2)$. The same argument on $q^2+3r^2=(t^2+3u^2{)}^3$ produces $q=t(t^2-9u^2)$, $r=3u(t^2-u^2)$. Constraint reduces to coprime factors $t+u=f^3$, $t-u=g^3$, $2u=f^3-g^3$. Hence $f^3-g^3$ is a cube with $f,g$ smaller than $x,y$.

Descent Closure

In both cases, an assumed coprime solution forces a strictly smaller solution of the same kind ($a^3+b^3=$ cube, or $a^3-b^3=$ cube). By infinite-descent, no solution exists. $\square$

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The $\mathbb{Z}[\sqrt{-3}]$ Subtlety

Euler’s proof implicitly assumes that if $p^2+3q^2=$ cube and $\gcd (p,q)=1$, then $p+q\sqrt{-3}=(t+u\sqrt{-3}{)}^3$ for some integers $t,u$.

This is not literally correct in $\mathbb{Z}[\sqrt{-3}]$, which is not a unique factorization domain. The classic counterexample: $4=2\cdot 2=(1+\sqrt{-3})(1-\sqrt{-3})$.

The repair: pass to the maximal order $\mathbb{Z}[\omega ]$ where $\omega =(-1+\sqrt{-3})/2$ is a primitive cube root of unity. This ring (the Eisenstein integers) is a UFD, and Euler’s argument goes through cleanly there. Modern textbooks present FLT-3 in $\mathbb{Z}[\omega ]$.

The descent skeleton itself is unaffected by the gap; only the cube-extraction step needs care.

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Why $n=3$ Is Harder Than $n=4$

Aspect	$n=4$ (fermats-last-theorem-n4)	$n=3$
Tools	Pythagorean parametrization (in $\mathbb{Z}$)	Cube-power expansion in $\mathbb{Z}[\sqrt{-3}]$
Coprime factorization	Three coprime squares	Three coprime cubes after Brahmagupta-Fibonacci factorization
Subtlety	None — fully elementary	Unique factorization required
Original proof	Fermat himself (descent)	Euler 1770 (gap discovered later)
Modern repair	Trivial	Pass to $\mathbb{Z}[\omega ]$

For $n=5$, Dirichlet and Legendre (1825); for $n=7$, Lamé (1839); each requires its own ad-hoc cyclotomic ring. Kummer’s regular-prime theory (1847) handled all “regular” primes uniformly. Wiles (1995) closed the rest via modularity.

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Use in Elements of Algebra

• Cited (footnote 81 of ch2.0.10-cubic-formula-as-cube) to bound the bootstrap for ”$1+x^3=$ cube” at $x=0,-1,2$ — terminating the cascade that the surd-as-cube method otherwise iterates indefinitely.
• Used (§247) to derive the sister theorem $x^3\pm y^3\ne 2z^3$.
• Referenced (§246) to explain why the formula for ”$a^3+b^3+x^3$ a cube” diverges to $x=\infty$ when $a=b$.

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Historical Notes

• Fermat (c. 1637) asserted FLT in the margin of his Diophantus but published no proof for $n=3$.
• Euler (1753, 1770) provided the first proof, with a gap noted later.
• Gauss (~1801) independently provided a clean proof in $\mathbb{Z}[\omega ]$ in his unpublished work.
• Andrew Wiles (1995) finally proved FLT for all $n\ge 3$.

The descent argument Euler uses is the same shape as Fermat’s $n=4$ argument — testimony to descent’s power as a uniform proof scheme.

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Related pages

• ch2.0.15-questions-cubes
• sum-of-two-cubes
• fermats-last-theorem-n4
• infinite-descent
• brahmagupta-fibonacci-identity
• ch2.0.13-impossibility-biquadrate-sums
• three-cubes-as-cube
• indeterminate-analysis
• imaginary-numbers