One of the unsolved problems in mathematics. A number theory question:

Does there exist a rectangular box all of whose edges and diagonals are integers?

A rectangular box is a solid with six rectangular faces.

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## Pythagorean Triples

For any odd number a,

[a, (a^2 - 1) / 2, (a^2 + 1) / 2] and

[(a^2 + 1) / 2, (((a^2 + 1)/2)^2 - 1) / 2, (((a^2 + 1)/2)^2 + 1) / 2]

[(a^2 + 1) / 2, (((a^2 + 1)/2)^2 - 1) / 2, (((a^2 + 1)/2)^2 + 1) / 2]

both sets represent Pythagorean triples.

Then,

the diagonal is .... (((a^2 + 1) / 2)^2 + 1) / 2

getting the equation....

a^2 + ((a^2 - 1) / 2)^2 + ((((a^2 + 1) / 2)^2 - 1) / 2)^2 = ((((a^2 + 1) / 2)^2 + 1) / 2)^2

Can "a" be an integer?

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Related topics:

Euler Brick

Perfect Cuboid

Heronian Tetrahedron

The face cuboid pentacycle solutions hints strongly that no such perfect cuboid exists. A descent proof would be adequate.

ReplyDelete- Randall Rathbun

The face cuboid pentacycle solutions hints strongly that no such perfect cuboid exists. A descent proof would be adequate.

ReplyDelete- Randall Rathbun