The cube that will not double

The story is that the oracle at Delos, asked how to end a plague, said the cubical altar of Apollo should be doubled. The islanders built one with edges twice as long, produced eight times the altar, and the plague continued.

Whether or not any of that happened, the mathematics it names is real and it stayed open for two thousand two hundred years.

The problem, stated as a number #

A cube of edge $1$ has volume $1$. A cube of double the volume has edge $x$ with $x^3 = 2$, so $x = \sqrt[3]{2}$, which is about $1.2599$.

The question is not whether that number exists — it plainly does, as the side of a cube anyone could carve — but whether it can be constructed: reached from a unit segment by the two operations, in finitely many steps.

By the previous rung of this ladder, a constructible number has degree a power of two over the rationals. So the whole problem reduces to one question: what is the degree of $\sqrt[3]{2}$?

The islanders’ mistake is worth a sentence, because it is the same mistake in every dimension. Doubling every edge multiplies a length by $2$, an area by $4$ and a volume by $8$; the factor is the doubling raised to the number of dimensions. To double a volume the edge must go up by $2^{1/3}$, which is a little over a quarter rather than a doubling, and the resulting altar looks disappointingly similar to the old one. The scaling exponent is what the oracle’s instruction was really about, and it is the same exponent that makes a circle’s area grow with the square of its radius rather than with the radius.

The degrees a construction can land on #

Before hunting for the degree of a particular number it is worth looking at the target.

Every step multiplies the dimension by two or by one. Starting from $1$, the reachable dimensions are therefore

$$1,\quad 2,\quad 4,\quad 8,\quad 16,\quad 32,\ \dots$$

and nothing else. The list has enormous gaps in it and the first gap is at three.

That is the whole of the argument’s target. The question “can this be constructed?” has become “is this number’s degree on that list?”, and a number of degree three is disqualified before anything geometric is examined.

Degree three, established by exhaustion #

$\sqrt[3]2$ satisfies $x^3 - 2 = 0$, which has degree three. To know that three is the degree rather than merely a degree, the polynomial must be shown to have no factorisation over the rationals.

For a cubic that is unusually easy. A factorisation of a cubic into smaller pieces must include a linear factor, because the degrees have to add to three and the only ways to split three are $1+2$ and $1+1+1$. A linear factor $qx - p$ means a rational root $p/q$. So:

A cubic with whole-number coefficients and no rational root is irreducible over the rationals.

And rational roots are a finite search. The rational root theorem says that if $p/q$ is a root in lowest terms then $p$ divides the constant term and $q$ divides the leading coefficient. For $x^3 - 2$ the constant term is $-2$ and the leading coefficient is $1$, so the candidates are $±1$ and $±2$ — four of them, listed in the figure at the top of this page with the value of the polynomial beside each.

None is zero. The nearest miss is $-1$, at $x = 1$. So $x^3-2$ has no rational root, is irreducible, and $[\mathbb{Q}(\sqrt[3]2):\mathbb{Q}] = 3$.

The theorem behind that finite list is itself a consequence of unique factorisation, and the argument is one line: substituting $p/q$ in lowest terms and clearing denominators gives $p^3 = 2q^3$, so $p$ divides $2q^3$ while sharing no factor with $q$, which forces $p$ to divide $2$. The same manoeuvre run to its end is the classical proof that $\sqrt2$ is irrational, and it is not a coincidence that the two look alike: both are statements that a certain equation has no solution in whole numbers, and both are settled by counting prime factors on each side.

Three is not a power of two. There is no $k$ with $2^k = 3$, and three does not divide $2^k$ for any $k$, since $2^k$ has no odd factor above one. The cube cannot be doubled.

Why the exhaustion is the honest picture #

The figure at the top of this page is a table of failures, and that is deliberate.

An impossibility has no picture. There is no diagram of a construction that does not exist, and any drawing purporting to show one would be showing something else. What can be drawn is the search that would have found it, run to the end, with what turned up instead.

The rational root theorem is what makes the search finite, and finiteness is what makes the drawing a proof rather than a gesture. Four candidates, four divisions, four non-zero answers. Nothing is being taken on trust and nothing is being approximated: the arithmetic in the table is done over the whole numbers, by clearing denominators before adding anything up, so a value is zero or it is not and there is no tolerance anywhere in the judgement.

That second figure is the control, and it is there for the reason every control is there. A test that returns “no root” whatever it is handed proves nothing by returning “no root”. Given $x^3-8$, whose root is the perfectly ordinary number $2$, the same search finds it. So the failure on $x^3-2$ is a property of $x^3-2$.

What the construction would have needed #

It is worth seeing the shape of the thing that is missing, because the near-misses are instructive.

Hippocrates of Chios reduced the problem to a cleaner one around 430 BC: to double the cube, it suffices to find two lengths $u$ and $v$ with

$$\frac{1}{u} = \frac{u}{v} = \frac{v}{2},$$

two mean proportionals between $1$ and $2$. Chaining those equalities gives $u^2 = v$ and $v^2 = 2u$, so $u^4 = 2u$ and $u^3 = 2$. The first mean proportional is the edge wanted.

That reduction is why so many ancient solutions look like machines for finding two mean proportionals — Archytas’ solution using the intersection of a cone, a cylinder and a torus; Eratosthenes’ sliding frames; Nicomedes’ conchoid. Every one of them works. None of them uses only a compass and a straightedge.

The contrast in that figure is the whole subject in miniature. One mean proportional between $1$ and $n$ is exactly the square-root construction — the perpendicular into a semicircle — and it costs one circle. Two mean proportionals is the cube root, and no amount of circles delivers it, because circles only ever adjoin square roots and the degrees only ever double.

Where the two mean proportionals come from #

The reduction deserves a second look, because it explains why so much ancient effort went into curves rather than into circles.

Between two numbers $a$ and $b$ there is one mean proportional $m$ with $a : m = m : b$, and it is the geometric mean $\sqrt{ab}$. Between them there are two mean proportionals $u, v$ with $a : u = u : v = v : b$, and those are $\sqrt[3]{a^2b}$ and $\sqrt[3]{ab^2}$.

The pattern is exact: one mean needs a square root, two means need a cube root, $k$ means need a $(k+1)$-th root. So the Delian problem is the $k=2$ case of a family whose $k=1$ case is drawn on this page in one circle. Nothing about the statement of the problem suggests it is harder than its neighbour; the difficulty is entirely in what the instruments happen to be able to do.

There is a pleasing consequence. If a single instrument could produce two mean proportionals in one action, every one of the classical problems except squaring the circle would fall to it, and several ancient devices were exactly that instrument. What none of them is, is a compass.

The neighbours who could #

Menaechmus, in the fourth century BC, solved the problem with conic sections. The two mean proportional equations $u^2 = v$ and $v^2 = 2u$ are a parabola and a parabola; taken as $u^2 = v$ and $uv = 2$ they are a parabola and a hyperbola. Where the curves cross is the answer.

Nothing is wrong with that solution. It is a complete and correct answer to the geometric question, and it is why the conics were studied in the first place: they were invented for this problem.

What it is not is a straightedge-and-compass construction, because a compass draws circles and a parabola is not a circle. The impossibility result does not say the cube cannot be doubled. It says the cube cannot be doubled with those two instruments, and the qualifier carries the entire content.

A marked ruler does it too. Slide a ruler with two marks a unit apart until the marks land on two given lines with the edge passing through a given point, and the construction produces cube roots; the technique is called neusis and Archimedes used it freely. Paper folding does it as well, by a single crease that brings two points onto two lines simultaneously. Both reach every number satisfying an equation of degree three or four.

So the boundary being mapped here is not a boundary of what is geometrically possible. It is the boundary of one stated operation set, and that is exactly what makes it a result about computation rather than about drawing.

Why cubics are the easy case #

The step that makes this proof three lines long instead of a chapter is the one about factorisation, and it is worth being clear that it does not generalise.

For a polynomial of degree three or four, having no rational root settles irreducibility for degree three and almost settles it for degree four. A cubic that factors must shed a linear piece, so no rational root means no factorisation. A quartic that factors might instead split into two quadratics with no rational root anywhere — $x^4 + 4 = (x^2-2x+2)(x^2+2x+2)$ is the standard example, and its rational-root table is as empty as any irreducible quartic’s.

From degree five upward the rational root theorem says almost nothing, and irreducibility becomes a real subject with real techniques: Eisenstein’s criterion, reduction modulo a prime, and factorisation algorithms that are the working tools of computer algebra. Working modulo a prime is the cheapest of them — a polynomial that stays irreducible when its coefficients are reduced mod some prime was irreducible to begin with, and the reduced problem is finite.

None of that is needed here. Every classical impossibility turns on a cubic, and every cubic surrenders to a list of divisors. That is a considerable stroke of luck and it explains why these three problems, and not others, are the ones with famous short answers.

Where the two thousand years went #

The gap between the question and the answer is worth accounting for, because it is not a story of people being slow.

The Greeks had no algebra. The statement “$\sqrt[3]2$ has degree three over $\mathbb{Q}$” has no translation into their language, because it needs polynomials with coefficients, a notion of a field, and a notion of dimension — three ideas that arrived in the sixteenth, eighteenth and nineteenth centuries respectively. What the Greeks had was a very good sense that the problem was hard and a growing collection of solutions using other tools.

The proof is Pierre Wantzel’s, published in 1837, and it is short: the case analysis of intersections, the doubling, and the degree count — three pages, all of it reproduced across this essay and the previous one. Wantzel was twenty-three. The same paper settled the trisection of the angle and characterised the constructible polygons, which is to say it closed two of the three classical problems and answered a fourth question Gauss had raised.

His result was barely noticed for a century, partly because it is a negative result and partly because Galois’ much deeper theory arrived at almost the same moment and absorbed the attention. That is a recurring pattern in this field: an impossibility proof is short, decisive, and much less celebrated than the machinery built to prove harder things.

The shape of every argument in this field #

The proof above has a shape worth naming, because the next three essays repeat it exactly.

1. Turn the construction into a number.
2. Find a polynomial with whole-number coefficients that the number satisfies.
3. Show the polynomial is the smallest, usually by running the rational root theorem to the end of a finite list.
4. Observe that the degree is not a power of two.

Step 3 is the only one with any content, and it is the only one a figure can carry. Steps 1 and 2 are algebra done in the prose; step 4 is arithmetic anybody can check.

That last figure will matter later. It is a blind search — no theorem consulted, just every integer polynomial up to a degree and a coefficient bound, evaluated at the number — and for $\sqrt[3]2$ it succeeds. Keeping it in view here means that when the same search is run at $\pi$ and finds nothing, the nothing is a report about $\pi$ and not about the search.

Where this ladder goes #

The next rung applies exactly the same four steps to trisecting an angle, and the interesting part is not the failure but its scope: some angles trisect perfectly well, and which ones is a question with a measured answer rather than a slogan.

After that, the polygons, where the degree count survives but the arithmetic gets more interesting — the relevant degree is not obvious from $n$ and turns out to be Euler’s totient of it. And then the circle, which breaks the pattern entirely, because $\pi$ has no minimal polynomial to find.