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