Megan Grywalski



An Euler brick is a cuboid whose side lengths and face
diagonals are integers.
For example, a cuboid with dimensions (a, b, c) = (44, 117,
240) is an Euler brick. It is actually the smallest Euler brick,
which was found by Paul Halcke in 1719. This Euler brick has
face diagonals (d, e, f) = (125, 244, 267).
An Euler brick satisfies the following equations:
a2 + b2 = d2
a2 + c2 = e2
b2 + c2 = f2
d = dab
a2  b2
a2  c2
Must be
integers
b2  c2
e = dac
f = dbc

A perfect cuboid, or perfect Euler brick, is an Euler brick
whose space diagonal is also an integer.

A perfect Euler brick must satisfy the equations for sides (a,
b, c), face diagonals (d, e, f), and space diagonal g with a, b,
c, d, e, f, and g ∈ ℤ:
a2
a2  b2
+
=
a2 + c2 = e2
b2 + c2 = f2
a2 + b2 + c2 = g2
d = dab
b2
e = dac
d2
f = dbc
a2  c2
b2  c2
a2  b2  c2
g = dabc
It is unknown if a perfect Euler
brick exists, nor has anyone
proved that one does not exist.


In 1740, Nicholas Saunderson, a blind mathematician, came
up with a parameterization that produced Euler bricks.

If (u, v, w) is a Pythagorean triple u2 + v2 = w2 then
(a, b, c) = (|u(4v2 - w2)|, |v(4u2 - w2)|, |4uvw|)
if u = 2st, v = s2 - t2, w = s2 +t2
a =6ts5 − 20t3s3 + 6t5s, b = −s6 + 15t2s4 − 15t4s2 + t6, c =
8s5t − 8st5.
The surface r(s, t) = <6ts5 − 20t3s3 + 6t5s,−s6 + 15t2s4 −
15t4s2 + t6, 8s5t − 8st5> is an Euler brick.
a2 +b2 +c2 = f(t, s)(s2 +t2)2
f(t, s) = s8 + 68s6t2 - 122s4t4 + 68s2t6 + t8,
if you found s, t for which f(t, s) is a square then this would be a
perfect Euler brick.
Using these parameters with the help of a
computer it was found that there exists an (a,
b, c) with a having 68162 digits, b with 56802
digits, and c with 56803 digits so that the
space diagonal √(a2 + b2 + c2) is 10-60589 close
to an integer.



Exhaustive computer searches show that, if a
perfect cuboid exists, one of its sides must
be greater than 1012.
Solutions have been found where two of the
three face diagonals and the space diagonal
are integers, such as:
(a, b, c) = (672, 153, 104)
Some solutions have been found where all
four diagonals but only two of the three sides
are integers, such as:
(a, b, c) = (18720, √211773121, 7800) and
(a, b, c) = (520, 576, √618849)