I can’t prove mathematically whether perfect cuboids exits or not, but I believe I have found a logical reason for them not to and would
like to know if my reasoning’s sound.
The space diagonal, S, is calculated using Pythagoras’s theorem from the height, H and the base diagonal, B, so for a perfect cuboid
the ratio H : B : S must be a Pythagorean triple.
We must also relate B to the other two face diagonals, X and Y, but X : Y : B isn’t a right angled triangle so it can only be defined using
Fermat’s last theorem. Unfortunately we don’t know which face diagonal is the longest so we can’t decide which way to define it. It could be;
X^n + Y^n = B^n, X^n + B^n = Y^n or Y^n + B^n = X^n
As it turns out it doesn’t matter, so since I’ve called the base diagonal B I’ve used the first one. Pythagoras’s theorem says B must be an
integer while Fermat’s last theorem says it can’t be.
Both facts can’t be correct therefore B^n # B^2 which can only mean these don't represent the same base diagonal and that he base must be a trapezium, or trapezoid if you prefer the term, not a rectangle, which makes it impossible to construct a perfect cuboid.
This helps us understand Fermat’s last theorem. An infinite number of integer sided acute and obtuse triangles exist but Fermat’s last
theorem applies to them all, not Pythagoras’s.
X^n + Y^n = Z^n is X^2 * X^(n-2) + Y^2 * Y^(n-2) = Z^2 * Z^(n-2), a right angled triangle distorted by the unequal multipliers.
The fact that the nth root of (Z^n)^n = Z^n means we can describe X^n + Y^n = Z^n using any root we choose. If we use square roots
we have sqrt(X^n)^2 + sqrt(Y^n)^2 = sqrt(Z^n)^2 which, when all three square roots are integers becomes a Pythagorean triple so we
know we can obtain all integer square roots for X^n + Y^n = Z^2.
Any Pythagorean triple because Z^n is Z^2
4^3 + 8^3 = 24^2
154^4 + 11,858^4 = 140,612,166^2
22^5 + 33^5 = 6,655^2
Here’s what we have now.
X^n + Y^n = Z^n
Sqrt (X^n + Y^n) = sqrt (Z^n)
Let k = sqrt(Z^n) - sqrt(X^n)
sqrt(X^n + k) = sqrt(Z^n)
But k doesn’t equal sqrt(Y^n)
Therefore when all these terms are integers X^n + Y^n # Z^n, it equals sqrt(Z^n)^2, not any other nth root(Z^n)^n
We can’t have a situation where these square roots have integer square roots of their own because then we have Fermat’s last theorem in the form A^4 + B^4 = C^4.