megan grywalski. an euler brick is a cuboid whose side lengths and face diagonals are integers. ...
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Finding the perfect Euler brick
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 e = dac f = dbc
Euler brick
22
22
22
cb
ca
ba
Must be integers
A perfect cuboid, or perfect Euler brick, is an Euler brick whose space diagonal is also an integer.
A perfect Euler brick
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 + b2 = d2
a2 + c2 = e2
b2 + c2 = f2 a2 + b2 + c2 = g2
d = dab e = dac f = dbc g = dabc
It is unknown if a perfect Euler brick exists, nor has anyone proved that one does not exist.
222
22
22
22
cba
cb
ca
ba
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)