A box with every face diagonal whole

Two right angles and a box found the diagonal of a box by using Pythagoras twice: across a face, then from the face’s diagonal up to the far corner, so that a box with edges $a$, $b$, $c$ has a long diagonal of $\sqrt{a^2 + b^2 + c^2}$. That essay was about lengths. This one asks about whole numbers. A rectangle with whole sides and a whole diagonal is a Pythagorean triple, and every one of them lies on a single circle and hangs from a single tree. What about a box?

The question has three levels. A box has three kinds of face, so it has three face diagonals, $\sqrt{a^2 + b^2}$, $\sqrt{a^2 + c^2}$ and $\sqrt{b^2 + c^2}$, and one space diagonal through its interior. A box with whole edges and all three face diagonals whole is an Euler brick. If the space diagonal is whole as well, it is a perfect cuboid. Euler bricks exist, infinitely many of them. Perfect cuboids are the subject of one of the oldest open questions about whole numbers that anybody can state with a ruler.

Three triples sharing their legs #

An Euler brick is three Pythagorean triples welded together: the pairs $(a, b)$, $(a, c)$ and $(b, c)$ must each be the legs of a right triangle with whole hypotenuse. Each edge serves as a leg in two of the triangles. The smallest such box, by its longest edge, is $44 \times 117 \times 240$, found by the German mathematician Paul Halcke in 1719. Its face diagonals are

$$\sqrt{117^2 + 240^2} = 267, \qquad \sqrt{44^2 + 240^2} = 244, \qquad \sqrt{44^2 + 117^2} = 125,$$

and its space diagonal is $\sqrt{73{,}225} = 270.60\ldots$, not whole. Each face is a familiar triple in disguise: 117, 240, 267 is three times 39, 80, 89; 44, 240, 244 is four times 11, 60, 61; and 44, 117, 125 is primitive. Three unrelated triples, chosen so that their legs coincide in pairs — and finding such a coincidence by hand, as Halcke did, meant comparing lists of triples until two of them happened to share a leg in the right way.

The way to find all the bricks below a bound follows that description directly. Generate every Pythagorean pair of legs up to the bound, primitive triples by Euclid’s formula $m^2 - n^2$, $2mn$ — the formula that the rope that squares a corner used for its knots — and all their multiples. For every number that occurs as a leg, list its partners — the other legs it makes a whole hypotenuse with. Two partners $b$ and $c$ of a common edge $a$ already give two whole face diagonals; the box is an Euler brick exactly when $b^2 + c^2$ is a square as well. Below 300,000 there are a few million Pythagorean pairs, and checking every pair of partners of every edge takes under a second.

The smallest bricks #

A brick multiplied by any whole number is still a brick, so the essential ones are the primitive bricks, whose three edges share no common factor. The ten smallest are 44 × 117 × 240, 240 × 252 × 275, 140 × 480 × 693, 85 × 132 × 720, 160 × 231 × 792, 1008 × 1100 × 1155, 187 × 1020 × 1584, 429 × 880 × 2340, 832 × 855 × 2640 and 780 × 2475 × 2992. Below 300,000 there are 113 primitive bricks in all. In every one, the squared space diagonal falls between two consecutive squares: $73{,}225$ for Halcke’s brick lies between $270^2 = 72{,}900$ and $271^2 = 73{,}441$.

The list has no visible pattern. The edges of a brick are not close to one another in any systematic way — 240 × 252 × 275 is nearly a cube, 85 × 132 × 720 is a long plank — and nothing in one row predicts the next. That is typical of a Diophantine problem with more equations than a single formula can solve: there are families, but they do not exhaust the solutions.

A brick from every triple #

One family is old and generates infinitely many bricks. Nicholas Saunderson, the blind Lucasian professor at Cambridge, showed around 1740 that any Pythagorean triple $u^2 + v^2 = w^2$ produces a brick with edges

$$a = u\,|4v^2 - w^2|, \qquad b = v\,|4u^2 - w^2|, \qquad c = 4uvw.$$

The triple 3, 4, 5 gives $3 \times 39$, $4 \times 11$ and $240$ — Halcke’s brick exactly. The triple 5, 12, 13 gives $828 \times 2035 \times 3120$, and every further triple gives another. The figure checks the formula on every primitive triple with hypotenuse up to 61, and each output passes: three whole face diagonals. The algebra behind it is an identity in $u$, $v$, $w$: each of the three sums of squares factors into a perfect square once $u^2 + v^2 = w^2$ is used.

So there are infinitely many Euler bricks, and Saunderson’s formula proves it. But it does not produce them all — most of the 113 primitive bricks below 300,000 are not of this form — and it never produces a perfect one, because the space diagonal it gives, $\sqrt{a^2 + b^2 + c^2}$, is never whole for any triple. Euler himself found more general families in the 1770s, and later writers found parametrisations with two free parameters; none covers everything, and none comes near the space diagonal.

Which numbers can be edges at all #

An edge of a brick must be a leg of two different right triangles at once, and that already restricts it. A number is a leg of some Pythagorean triple whenever it is at least 3, but to be a leg twice over, with two different partners, it needs room in its factorisation: the number of ways $a$ is a leg is governed by the divisors of $a^2$, through the identity $a^2 = (c - b)(c + b)$, and a prime $a$ is a leg of exactly one primitive triple. So prime numbers are never the edges of bricks in the way that 44, 117 and 240 are; the edges in the list are all products of several small primes, and the partners of a heavily composite number are numerous. The hypotenuses obey a different rule, the one two squares and a lattice found: a whole number is a sum of two squares exactly when its primes of the form $4k + 3$ appear to even powers. A brick’s three face diagonals must all pass that test, while its edges must each be legs twice — two kinds of arithmetic constraint on the same six numbers.

That is the practical shape of the search. Most numbers are partners of very few others, and the work concentrates on the composite numbers with many partners, among which the rare triples of partners closing up into a third square are found. The perfect cuboid asks, in addition, that the three squared edges sum to a square — a condition on the edges alone, unrelated to how they pair up — and nothing about the partner structure makes it more likely.

Points at whole distances #

The perfect cuboid has a cousin in the plane that is settled. Ask for a set of points in the plane, not all on one line, such that every pair is a whole distance apart: finite sets of any size exist — points on a circle at the right angles give arbitrarily many — but Norman Anning and Paul Erdős proved in 1945 that an infinite such set must lie on a line, by an argument that turns whole distances into hyperbolas with too few lattice points, and the diameter that measures every side built some of the finite ones. A perfect cuboid would be eight points in space — the corners of the box — with every edge, face diagonal and space diagonal whole: twenty-eight distances, all whole. Its non-existence would be a statement of the same kind as Anning and Erdős’s, but for a fixed small configuration, and for small configurations no general argument is known.

Euler’s own interest came from a different direction. In the 1770s he studied the problem of finding three numbers whose pairwise sums of squares are all squares as part of a programme on Diophantine equations — equations to be solved in whole or rational numbers — and he gave families of bricks from two parameters, more general than Saunderson’s one-parameter construction, together with the remark that he could not make the fourth square appear. That remark has been repeated, in effect, by everyone who has worked on the problem since. The name Euler brick honours the families, and the question he left is the perfect cuboid.

How many bricks there are #

Counting the bricks below a bound shows how sparse they are. Up to a longest edge of a thousand there are five primitive bricks; up to ten thousand, nineteen; up to a hundred thousand, sixty-five; up to a million, a few hundred, with several thousand counting multiples. The primitive count grows like a small power of the bound — a fraction of a power, far slower than the bound itself — so bricks thin out as the edges grow, and a box chosen at random among those with whole face diagonals on two faces is less and less likely to have the third.

That thinning is the heuristic reason perfect cuboids are thought to be rare or absent. A perfect cuboid needs a fourth square on top of the three, and if the four conditions behaved like independent events, the expected number of perfect cuboids with edges below $L$ would grow so slowly, if at all, that a search up to $10^{12}$ finding none is unsurprising. The heuristic does not decide whether the expected number is finite or infinite, and so it does not decide the question.

Factors every brick must have #

What can be proved about bricks are necessary conditions, and they come from remainders. A square leaves only a few remainders on division by a small number — modulo 4, only 0 and 1; modulo 16, only 0, 1, 4 and 9 — and three sums of squares that must all be squares constrain the edges’ remainders sharply. Working the cases shows that in every primitive Euler brick some edge is divisible by 4 and some edge by 16, some edge by 3 and some by 9, some by 5 and some by 11. The figure checks each statement on all 113 primitive bricks below 300,000 and finds no exception, as there cannot be.

For a perfect cuboid the constraints multiply. Each additional square — the space diagonal — adds conditions modulo every small prime, and the known results force a perfect cuboid’s edges to be divisible by a long list of small primes and its smallest edge to be larger than any searched so far. Such conditions are how searches are made efficient: they prune the candidates that cannot work before any square root is taken. They do not, by themselves, rule perfect cuboids out, because a large enough number can satisfy any finite list of divisibility conditions.

Missing by one diagonal #

A perfect cuboid has four diagonal conditions and three edges. Drop any one condition and solutions are easy to find — which is what makes the problem so tantalising. Euler bricks drop the space diagonal, and are for that reason also called body cuboids. Face cuboids keep the space diagonal and two face diagonals and drop the third face. The search finds 154 primitive ones with every edge below 300,000, the smallest $104 \times 153 \times 672$: its face diagonals over $104 \times 153$ and $104 \times 672$ are 185 and 680, its space diagonal is 697, and only the face $153 \times 672$ fails, with diagonal $\sqrt{474{,}993} \approx 689.20$. Other near misses keep all four diagonals and let one edge be irrational, or keep everything but let one face diagonal miss; all of them exist in abundance.

The third kind of near miss is the edge cuboid, which has every diagonal whole and every edge whole except one, which is irrational — the square root of a whole number — so it is a perfect cuboid that has given up on an edge instead of a diagonal. All three kinds exist in abundance, and so do boxes with rational rather than whole edges, which are the same problem after scaling. Every relaxation that removes one of the seven conditions — three edges whole, three face diagonals whole, one space diagonal whole — has infinitely many solutions.

The search finds no perfect cuboid with every edge below 300,000. Far larger searches agree, pushing the bounds on a perfect cuboid’s edges into the trillions over several decades of computation, and every one of the four conditions, and every three of them together, have been met many times. All four together have never been.

Why this is harder than a triple #

A Pythagorean triple is a rational point on a circle, and the circle has a rational parametrisation — every triple on one circle is the statement that a line of rational slope through one rational point meets the circle again at another — which is why triples can be listed by a formula. An Euler brick is a rational point on a surface cut out by three equations in six unknowns, and that surface has families of rational curves on it, which is why Saunderson and Euler could write families of bricks down. A perfect cuboid is a rational point on a surface with one more equation, and that surface is of a kind — general type, in the language of algebraic geometry — whose rational points are expected, by a conjecture of Enrico Bombieri and Serge Lang, to lie on finitely many curves, so that perfect cuboids, if they exist, would be organised by finitely many patterns. No pattern is known.

The progression is the same one the areas a right triangle can have followed: one triangle with a given area is a point on an elliptic curve, and whether there is one comes down to deep questions about that curve. Each extra condition moves a question about whole numbers into a harder kind of geometry, and the perfect cuboid sits where the geometry is least understood.

What the search shows and assumes #

The search is complete below its bound: every Pythagorean pair with both legs at most 300,000, or a million for the counts, is generated, and every pair of partners of every edge is checked, so every brick and every body cuboid with all edges below the bound is found. Squareness is tested exactly on whole numbers, which in this range fit comfortably in double precision. The divisibility checks confirm theorems on the bricks found; they do not prove the theorems, which need the remainder arguments. And nothing below 300,000, or below $10^{12}$, says anything about larger boxes.

Still open: the perfect cuboid #

Is there a box with whole edges, whole face diagonals and a whole space diagonal? The question goes back at least to Euler and is listed among the classic unsolved problems in number theory. It is known that a perfect cuboid’s edges would have to be divisible by many small primes, that its smallest edge would exceed any bound searched so far, and that no parametric family of bricks yet found contains one. Whether one exists is not known; neither is whether there are finitely many, which is what the Bombieri–Lang conjecture would imply. It is one of the cleanest problems in mathematics that a child can state and nobody can answer.

Three squares easily, four never #

An Euler brick is three Pythagorean triples sharing their legs: the smallest is Halcke’s 44 × 117 × 240, Saunderson’s formula turns every triple into another, and there are 113 primitive ones with every edge below 300,000, every one with edges divisible by 16, 9, 5 and 11. Ask for the space diagonal to be whole too and there are 154 boxes below 300,000 that miss by a single face diagonal, the smallest 104 × 153 × 672 — and none that misses by nothing. Each condition is easy, any three can be met together, and whether all four can is one of the oldest open questions about whole numbers.