# Finding the perfect Euler brick - PowerPoint PPT Presentation

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Finding the perfect Euler brick. Megan Grywalski. Euler brick. An Euler brick is a cuboid whose side lengths and face diagonals are integers.

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Finding the perfect Euler brick

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## Finding the perfect Euler brick

Megan Grywalski

### Euler brick

• 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

Must be

integers

### A perfect Euler brick

• 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+ 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.

• 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 = w2then

(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, bwith 56802 digits, and cwith 56803 digits so that the space diagonal √(a2+ b2 + c2)is 10-60589close 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)