This problem looks so easy and classic. Just search a cuboid with every side, face diagonals, and inner diagonals are integers. But, I have still not getting any cause why it still unsolved when the hardest problem like Poincare Conjecture has been solved? Is this problem too easy for professional mathematicians or what?
Because you have to find some numbers with very rare properties. Take the easiest case possible: a rectangular triangle (this is by far no perfect cuboid, but it is a very simple example that shows the problem). If you try to find such an triangle, where Pythagoras holds, a² + b² = c² there are not so much numbers that solve this equation as integers.
For your perfect cuboid you also need to solve:
a² + b² = d²
a² + c² = e²
b² + c² = f²
a² + b² +c² = g²
You have to solve all of these equations within the set of integers, but by adding one of those formulars you filter out more and more elements of your integers. Hence you reach higher and higher numbers which are very hard to handle even by supercomputers.
The final problem is then to prove whether you have a very small set of solutions or if there is no solution at all.
I have tried to generalize the solution with Pythagoras Identity:
- a² + b² = d², if m and n is integer, m,n≠0,n
To be honest, I don't know how you can proof uniqueness here, because usually you proof that by assuming a second entity, say x and y, give you the same solution for you equation only if they are equal to m and n, so:
x² - y² = 2pq x = m and y = n
but doing this you will get m² - n² = x² - y²
and I am not sure if this is solvable within the set of integers, if m is not equal to x and n is not equal to y...
I think that if you can prove that $a^2+f^2=g^2=b^2+e^2$ has only one solution for $a^2+f^2=g^2$ then you have proven that there is no perfect cuboid, but I haven't even figured out how to solve the problem because I'm not an expert in number theory. But I think what I've written above is a very simple conjecture that is a reduced form of the original problem. The case of $c^2+d^2=g^2$ I think is a red herring. Its not required. It comes down to being a degree of freedom of the original problem.
I have found the proof that the internal diagonal is irrational when the edges and face diagonals are all integer numbers satisfying four equations: 1) A^2+B^2+C^2=D^2, 2) A^2+B^2=E^2, 3) A^2+C^2=F^2, and 4) B^2+C^2=G^2.
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