Theorem 01 · Solved · Wiles, 1995
Fermat's Last Theorem
No cube is the sum of two cubes — nor any higher power.
In 1637 Pierre de Fermat scribbled a note in the margin of a book: the equation \(x^n + y^n = z^n\) has no solutions in whole numbers once \(n\) is bigger than 2 — and, he added, "I have discovered a truly marvellous proof of this, which this margin is too narrow to contain." He never wrote it down.
For \(n = 2\) there are endless solutions — the Pythagorean triples like \(3^2 + 4^2 = 5^2\) that every schoolchild meets. But for cubes, fourth powers, and beyond, Fermat claimed you can search forever and never find three whole numbers that fit. It took the world 358 years to prove him right.
Precisely: there are no positive integers \(x, y, z\) with \(x^n + y^n = z^n\) for any \(n > 2\). The case \(n = 2\) is the exception that proves the rule — its solutions, the Pythagorean triples \((3,4,5)\), \((5,12,13)\), \((8,15,17)\), …, are infinite in number and easy to generate.
It is enough to rule out \(n = 4\) and every odd prime \(n = p\), since any larger exponent contains one of these as a factor. Fermat himself disposed of \(n = 4\) by his method of infinite descent — showing any solution would force a strictly smaller one, impossibly forever. Euler did \(n = 3\), Dirichlet and Legendre \(n = 5\), and Sophie Germain made sweeping progress on a whole family of primes at once.
Kummer (1840s) then cracked all "regular" primes together by inventing ideal numbers to repair unique factorisation among the cyclotomic integers — the deepest nineteenth-century advance in the subject. But the irregular primes resisted, and after three centuries the elementary tricks had run dry.
Why so hard? An equation in three unknowns secretly describes a curve, and the modern insight is that this innocent-looking statement is really a question about the geometry of such curves and the arithmetic objects attached to them — a world away from the clever factorisations where it began.
Statement: for \(n \ge 3\) the only integer solutions of \(x^n + y^n = z^n\) have \(xyz = 0\). By the factor argument it reduces to \(n = 4\) (Fermat) and odd primes \(n = \ell\). The classical dividing line is whether \(\ell\) is regular — that is, \(\ell \nmid h\), the class number of \(\mathbb{Q}(\zeta_\ell)\); Kummer settled the regular case, and the Herbrand–Ribet theorem sharpened the criterion.
The route that finally worked is geometric. To a hypothetical solution \(a^p + b^p = c^p\), Frey (1984) attached the elliptic curve \(y^2 = x(x - a^p)(x + b^p)\), whose discriminant \((abc)^{2p}\) and conductor are so divisible by \(p\)th powers that the curve would be far "too unramified" to be modular.
Serre and Ribet (1986, the epsilon conjecture / level-lowering) made this precise: such a Frey curve could not be modular. But the Taniyama–Shimura–Weil conjecture predicted that every elliptic curve over \(\mathbb{Q}\) is modular. A solution would therefore manufacture a non-modular elliptic curve — a contradiction — once one could prove modularity for the (semistable) Frey curves.
Status: solved (1995). Andrew Wiles, with Richard Taylor, proved the modularity of all semistable elliptic curves over \(\mathbb{Q}\), completing the proof; the full modularity theorem followed (Breuil–Conrad–Diamond–Taylor, 2001).
A live search over pairs \((a,b)\), each cell shaded by how close \(a^n + b^n\) comes to a perfect \(n\)th power — rose for an exact hit, gold for a near-miss (watch \(9^3+10^3=1729\) flare). The exponent \(n\) cycles 2 → 8; the exact solutions vanish past 2. The Fermat curve \(x^n + y^n = 1\): a circle at \(n=2\), squaring off as \(n\) grows. Only \(n=2\) carries rational points. \(n\) cycles on its own. The Frey curve a solution would build — the bridge to modularity. \(n\) cycles through the exponents.
The case n = 4 · Proved by Fermat himself
Infinite Descent
You can't fall down whole numbers forever.
Here is the one piece of the puzzle Fermat really did prove — for fourth powers — and the idea is beautifully simple. Suppose a solution in whole numbers existed. Fermat showed you could always use it to build another solution made of strictly smaller whole numbers.
But then that smaller solution builds an even smaller one, and that one a smaller one still — forever. And there's the catch: whole numbers can't shrink forever. Counting downward you must eventually hit bottom and stop. An endless staircase going down, with no lowest step, simply cannot exist — so the solution you assumed could never have been there to begin with.
This is the argument for \(n = 4\). Fermat showed that any solution of \(x^4 + y^4 = z^4\) is bound up with a Pythagorean triple, and from the structure of such triples one can construct a second, genuinely smaller solution of the very same form.
Repeat the construction and you get an infinite, strictly decreasing sequence of positive integers — which is impossible, because they have a floor. This is the single case of his Last Theorem that Fermat actually wrote down a full proof for: for \(n = 4\), at least, the margin note was no bluff.
The method is infinite descent, and it rests on the well-ordering of \(\mathbb{N}\): every non-empty set of positive integers has a least element, so no strictly decreasing sequence of them can run forever. It is a complete, rigorous proof for \(n = 4\), and Euler adapted the same idea to settle \(n = 3\).
But descent does not scale. For a general exponent the natural construction lives in rings of algebraic integers that lose unique factorisation, and the clean "build a strictly smaller solution" step breaks down. That failure is exactly why the elementary attack stalled for three centuries — and why the eventual proof had to leave elementary number theory altogether, for the Frey curve, Ribet's theorem and Wiles's modularity machinery.
Each solution forces a strictly smaller one — a staircase marching down with no bottom step. The descent marches toward the floor on a loop. For \(n = 4\): from one solution build a smaller one of the same kind, forever — but positive integers have a floor at 1. It steps down on its own. Infinite descent on the well-ordered \(\mathbb{N}\): a real proof for \(n=4\), but it does not scale to general \(n\). It steps down on its own.
The full proof 03 · The bridge · Modularity
The Proof: Modularity
Two worlds, secretly the same — and Fermat falls out.
How do you prove something is impossible forever? Not by searching — no search can ever finish. Wiles found the answer in a place no one expected: a deep, surprising bridge between two utterly different kinds of mathematical object.
Picture two worlds of mathematics that look nothing alike — as different as music is from architecture. The astonishing discovery is that they are secretly the same: every object in the first world is really an object in the second, wearing a disguise. A whole-number solution to Fermat's equation could be used to build one particular object in the first world — but that object is exactly the kind the second world forbids. Something cannot live in a world and be banned from it at the same time, so the solution cannot exist.
A modular form is a function on the upper half-plane left unchanged by the modular group \(\mathrm{SL}(2,\mathbb{Z})\) — the symmetries \(z \mapsto \frac{az+b}{cz+d}\) — and so completely determined by its values on a single tile of an infinitely repeated tessellation. Each form has a \(q\)-expansion \(\sum a_n q^n\) whose coefficients are arithmetic gold.
The Taniyama–Shimura–Weil conjecture says every elliptic curve \(E/\mathbb{Q}\) is matched by such a form: their \(L\)-functions — one built from counting points on \(E\), the other from the coefficients \(a_n\) — coincide. Frey's curve, built from a Fermat solution, would have to be modular too.
But Ribet's theorem (1986) showed that, by "level-lowering," any such form would have to drop to level 2 — where no modular forms exist at all. So a Fermat solution forces a non-modular elliptic curve, turning Fermat into a special case of modularity. Wiles then spent seven years, in secret, proving enough of it: that all semistable elliptic curves — the class the Frey curve belongs to — are modular.
Modularity is proved through Galois representations. To an elliptic curve \(E\) one attaches the action of \(\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\) on its \(p\)-torsion and Tate module; \(E\) is modular exactly when this representation arises from a modular form. Wiles' engine is a modularity lifting theorem: if the mod-\(p\) representation is modular (and suitably irreducible), so is the \(p\)-adic one.
The technical heart is a numerical criterion equating the size of a deformation ring with that of a Hecke algebra — the slogan "\(R = T\)." The base case uses Langlands–Tunnell at \(p = 3\) (where \(\mathrm{GL}_2(\mathbb{F}_3)\) is solvable), and a clever 3–5 switch covers the curves that escape it.
Wiles' 1993 announcement had a gap in the Euler-system bound; he and Taylor repaired it in 1994 with the Taylor–Wiles patching argument. The circle of ideas — deformations of Galois representations, \(R = T\) — now drives much of modern number theory.
Status: solved. Wiles (with Taylor) proved semistable modularity, enough for Fermat; Breuil, Conrad, Diamond and Taylor (2001) proved the full modularity theorem for every elliptic curve over \(\mathbb{Q}\).
Two utterly different worlds of mathematics — secretly the same object in disguise. A Fermat solution would need a thing one world forbids. Two worlds, one bridge. Ford circles: one tangent circle at every rational, nested by denominator — the symmetry of \(\mathrm{SL}(2,\mathbb{Z})\). The tessellation deepens on a loop. The proof by contradiction, step by step. Each step reveals in turn, looping.