Euler brick

In mathematics, an Euler brick, named after Leonhard Euler, is a cuboid whose edges and face diagonals all have integer lengths. A primitive Euler brick is an Euler brick whose edge lengths are relatively prime.

Definition and properties

The definition of an Euler brick in geometric terms is equivalent to a solution to the following system of Diophantine equations:

$\begin{cases} a^2 + b^2 = d^2\\ a^2 + c^2 = e^2\\ b^2 + c^2 = f^2\end{cases}$

where a, b, c are the edges and d, e, f are the diagonals. Euler found at least two parametric solutions to the problem, but neither gives all solutions.[1]

If (a, b, c) is a solution, then (ka, kb, kc) is also a solution for any k. Consequently, the solutions in rational numbers are all rescalings of integer solutions. Given an Euler brick with edge-lengths (a, b, c), the triple (bc, ac, ab) constitutes an Euler brick as well.

Examples

The smallest Euler brick, discovered by Paul Halcke in 1719, has edges $(a, b, c) = (44, 117, 240)$ and face diagonals 125, 244, and 267.

Some other small primitive solutions, given as edges (a, b, c) — face diagonals (d, e, f), are below:

• (85, 132, 720) — (157, 725, 732);
• (140, 480, 693) — (500, 707, 843);
• (160, 231, 792) — (281, 808, 825);
• (240, 252, 275) — (348, 365, 373).

Perfect cuboid

 Does a perfect cuboid exist?

A perfect cuboid (also called a perfect box) is an Euler brick whose space diagonal also has integer length.

In other words, the following equation is added to the system of Diophantine equations defining an Euler brick:

$a^2 + b^2 + c^2 = g^2,\,$

where g is the space diagonal. Thus (a, b, c, g) must be a Pythagorean quadruple. As of November 2012, no example of a perfect cuboid had been found and no one has proven that none exist. Exhaustive computer searches show that, if a perfect cuboid exists, one of its edges must be greater than 3·1012.[2][3] Furthermore, its smallest edge must be longer than 1010.[4]

Some facts are known about properties that must be satisfied by a primitive perfect cuboid, if one exists, based on modular arithmetic:[citation needed]

• One edge, two face diagonals and the body diagonal must be odd, one edge and the remaining face diagonal must be divisible by 4, and the remaining edge must be divisible by 16
• 2 edges must have length divisible by 3 and at least 1 of those edges must have length divisible by 9
• 1 edge must have length divisible by 5.
• 1 edge must have length divisible by 7.
• 1 edge must have length divisible by 11.
• 1 edge must have length divisible by 19.
• 1 edge or space diagonal must be divisible by 13.
• 1 edge, face diagonal or space diagonal must be divisible by 17.
• 1 edge, face diagonal or space diagonal must be divisible by 29.
• 1 edge, face diagonal or space diagonal must be divisible by 37.

Solutions have been found where the space diagonal and two of the three face diagonals are integers, such as:

$(a, b, c) = (672, 153, 104).\,$

Solutions are also known where all four diagonals but only two of the three edges are integers, such as:

$(a, b, c) = (18720, \sqrt{211773121}, 7800)$

and

$(a, b, c) = (520, 576, \sqrt{618849}).$

Perfect parallelepiped

A perfect parallelepiped is a parallelepiped with integer-length edges, face diagonals, and body diagonals, but not necessarily with all right angles; a perfect cuboid is a special case of a perfect parallelepiped. In 2009, a perfect parallelepiped was shown to exist,[5] answering an open question of Richard Guy. Solutions with only a single oblique angle have been found.[clarification needed]

Notes

1. ^
2. ^ Durango Bill. The “Integer Brick” Problem
3. ^
4. ^ Randall Rathbun, Perfect Cuboid search to 1e10 completed - none found. NMBRTHRY maillist, November 28, 2010.
5. ^ Sawyer, Jorge F.; Reiter, Clifford A. (2011). "Perfect parallelepipeds exist". Mathematics of Computation 80: 1037–1040. arXiv:0907.0220. doi:10.1090/s0025-5718-2010-02400-7..