Why does one side of a perfect cuboid have to be odd?

Question

With the equations of:
a^2 + b^2 = d^2
b^2 + c^2 = e^2
a^2 + c^2 = f^2 and
a^2 + b^2 + c^2 = g^2
with all variables being intergers,
wouldn't it seem more logical to have all even sides?

Answer 1

All diagonals of a perfect cuboid are integers

Assume the cuboid has height h, width w and depth d and h>w>d.

The lengths of the face diagonals are:

* fd12 = h2 + w2
* fd22 = h2 + d2
* fd32 = w2 + d2

The length of the space diagonal is:

* sd2 = h2 + w2 + d2

If h,w and d are all even, then every fd and the sd are divisible by 2, and so we can divide h,w and d by 2.

The sum of 2 odd integers squared is never a square, and so the only possiblity is for exactly one of h,w,d to be odd.

In turn, this implies that two of the fdi are odd, the other even. Consider fd1 to be even.

The aim of this proof is to demonstrate that any such fd1 must contain an infinite power of 2 as a factor.

If we expand the fd equations, using a,b,c as parameters (h=2a, w=2b, d=2c+1), then we arrive at:

4[a2 + b2 + c2 + c] + 1 = sd2

a and b must be of the same parity, as 4k+1 is only square for even k.

However a2 + b2 = fd12/4

And so fd12/4 must be even, i.e. h=4a, w=4b.

If we repeat the process with the new parameterization, then we arrive at:

4[8a2 + 8b2 + c2 + c] + 1 = sd2

But now fd12/16 must be even, i.e. h=8a, w=8b.

This is obviously absurd.