Let , , , , , and be positive integers. We suppose , and have no common prime factors and , and are the edges, , and are the diagonals and is the space diagonal. If a perfect cuboid exists, the following equations hold.
is odd and and are even because , and are not all even and if two edges are odd, then the diagonal is not a square number. And and are odd and is even. Let , , and be positive inetgers and hods. We supopse and are relatively prime and the odd/even number of and are different. By the form of Pythagorean triple, the following equations hold.
is even and the others are all odd. We suppose , and have no common prime factors since the greatest common diviosor of , and is one. Let be an odd positive integer. Since has factors , and ,
holds. Let and be odd positive integers and be an even positive integer. By this equation, the following equations hold.
By these equations,
holds. Let and be odd positive integers and be an even positive integer. By this equation, the following equations hold.
By these equations,
holds. If has a common prime factor of , then have the common prime factor and and have it also. In this case, , and have the common prime factor and this contradicts that , and have no common prime factors. Therefore, and are coprime. In the same way, and , and , and , and and and are relatively prime. Comparing the equation (2) with the equation (1), we find that the following equations hold since there exists a perfect cuboid whose three sides of the equation (1) are twice as long.
It becomes a contradiction since and are odd. From the above, there are no perfect cuboids. (Q.E.D.)