Rechenkraft.net e.V.

Zak,
the only thing you achieved, you derived that for given natural d, e, f and g the next equation must be true: Let's check for our already well-known Edge cuboid:

Edit: Where do you expect to find a perfect cuboid? And where is your upper testing limit at all? To be honest, very unlikely a perfect cuboid exists at all, so I rather expect the project probable achievement will become the only short statement: "if a perfect cuboid exists, its space diagonal exc...

Perfect Cuboid 1st Batch finish Now we are close to 1st Batch ending. For those who's interested, what's next, we inform, after complete validation of 0-2^50 range we are going to publish a new app version which will be goaled to Perfect cuboids search only. Thus it's 7 times faster than the current...

We've successfully completed the 0-10T range verification. As was expected, we found a lot of cuboids there: 54'761 edge cuboids 94'501 face cuboids 228'198 imaginary cuboids 1'132'380 twilight cuboids A total of 1'509'840 different cuboids. The most fortunate were users year02 and krzyszp who snatc...

it is quite possible that it has two sides as the legs of the smallest Pythagorean triple: a = 3, b = 4, and the third side crushes together with the spatial diagonal to infinity, and we will rather have not a brick, but a long beam, a straight line. If you know the smallest edge a , there's also k...

Zak hat geschrieben: Why not?

Because, it's a middle school task, to prove that an angle between each pair of face diagonals in your picture is acute.

Why did you decide that (d, e, f) are mutually orthogonal?



Of course, you always can build a cuboid having 3 integer lengths, but why did you decide that the space diagonal of such cuboid should be integer?

Zak, What you want to prove had already been proven many years (centuries) ago. Just open Wiki and attentively read: https://en.wikipedia.org/wiki/Euler_brick#Perfect_cuboid Some facts are known about properties that must be satisfied by a primitive perfect cuboid, if one exists, based on modular ar...

as far as a^2+b^2+c^2=g^2 and g>0, you've proved that 1 = 2 * sqrt(3/2), isn't it?

Zak hat geschrieben:the face diagonals d, e, f

Zak hat geschrieben:the radius of the sphere in both cases is equal to half the spatial diagonal: r=d/2

Excuse me, but looks like the spatial diagonal has a different designation: g.
So, the radius of the sphere is equal to half the spatial diagonal: r=g/2.