Perfect cuboid cube Is there any proof that there is no cubic perfect cuboid? Here is a description of the problem: . I'm currently using trying to get an empty set to solve it...
[ A "perfect cuboid" is one whose edges, face, and body diagonals are all integers. ]
 A: A proof that a perfect cuboid does not exist was posted to arxiv.org/abs/1506.02215 by Walter Wyss.  Randall Rathbun announced to the Number Theory List that he has checked the proof carefully and found no errors.  The first version of Wyss's paper was posted in June 2015 and the current version was posted June 2016.  Perhaps it is still being checked.
A: let $ ℕ=\{1,2,3,4,5,6,...\}$ 

Find solutions to the following equations or prove that there exist no solutions:
\begin{align}
d^2 &= a^2+b^2 \\
e^2 &= a^2+c^2 \\
f^2 &= b^2+c^2 \\
g^2 &= a^2+b^2+c^2
\end{align}
$$\text{Such that } (a,b,c,d,e,f,g) ∈ ℕ$$
First lets consider the space diagonal $g$ to be a diameter of circle $R$, and assume that the equation $g^2 = a^2+b^2+c^2$ is satisfied such that $(a,b,c,g)∈ ℕ$. as illustrated in fig.1;  $$Fig.1$$
Second We observe the following in fig.1;


*

*Fig.1 X, Y, and Z demonstrates that if the equation $g^2 = a^2+b^2+c^2$ is satisfied such that $(a,b,c,g)∈ ℕ$, by implication only a maximum of two from the three face diagonals $(d,e,f)∈ ℕ$ can have solutions.

*The implication in fig.1 is that if the equation $g^2=a^2+b^2+c^2$ is satisfied then the three axis $(a,b,c)$ and the space diagonal $g$ will form a trapezoid which can only be arranged to produce one or a maximum of two from the three  $(d,e,f)$  face diagonals because we note that the trapezoid formed by $(a,b,c,g)$ can be arranged to only have a maximum of two right angle triangles while the sides $(a,b,c,g)$ remain unequal once you arrange a trapezoid $(a,b,c,g)$ to accommodate three right angle triangles it becomes a square or rectangle implying that two opposite or symmetrical sides of this trapezoid must be equal. Thus only a maximum of two right angle triangles can be accommodate in a trapezoid as shown in fig.1 the trapezoid will be ether fig.1 X, or Y, or Z. that might have solutions. 

*Fig.1 can be used to explain why near perfect cuboids can exist, example known near-perfect cuboid $a=672,\ b=104,\ c=153,\ d=680,\ f=185,\ g=697,$ which satisfies $d^2=a^2+b^2,\ f^2=b^2+c^2,\ and\ g^2=a^2+b^2+c^2,$ but fails to satisfy $e^2 = a^2+c^2$.
Third lets consider the space diagonal $g$ to be a diameter of a circle $S$, and assume that the equations $g^2 = a^2+f^2$, $g^2 = b^2+e^2$, and $g^2 = c^2+d^2$ are satisfied such that $(a,b,c,d,e,f)∈ ℕ$ as illustrated in fig.2;  $$Fig.2$$
Fourth We observe the following in fig.2;


*

*Fig.2 demonstrates that if the equations $g^2 = a^2+f^2$, $g^2 = b^2+e^2$, and $g^2 = c^2+d^2$  are satisfied such that $(a,b,c,d,e,f)∈ ℕ$, by implication the equation $g^2 = a^2+b^2+c^2$ cannot have a solution.

*The implication in fig.2 is that if the three face diagonal can be satisfied such that $g^2 = a^2+f^2$, $g^2 = b^2+e^2$, and $g^2 = c^2+d^2$ then by implication there exist no trapezoid that can be arranged to accommodate the three face diagonals because we note that the trapezoid formed by $(a,b,c,g)$ can be arranged to only have a maximum of two right angle triangles while the sides $(a,b,c,g)$ remain unequal once you arrange a trapezoid $(a,b,c,g)$ to accommodate three right angle triangles it becomes a square or rectangle implying that two opposite or symmetrical sides of this trapezoid must be equal.  Thus the equation $g^2=a^2+b^2+c^2$ will not have a solution.

*Fig.2 can be used to explain why all Euler bricks are not perfect cuboids as they all fail to satisfy $g^2=a^2+b^2+c^2$;  
Conclusion as observed in fig.1 and .2 only a maximum of three equations from the given four equations of the perfect cuboid problem can have solutions such that $(a,b,c,d,e,f,g)∈ ℕ$, by implication; 
$$∴∄\{perfect\ cuboids\}\ Such\ that\ (a,b,c,d,e,f,g)∈ℕ$$
A: From Wikipedia "Euclid's formula is a fundamental formula for generating Pythagorean triples given an arbitrary pair of integers m and n with m > n > 0. Since every Pythagorean triple can be divided through by some integer k to obtain a primitive triple, every triple can be generated uniquely by using the formula with m and n to generate its primitive counterpart and then multiplying through by k."
$${\displaystyle a=k*(m^{2}-n^{2}),\ \, b=k*(2mn),\ \, c=k*(m^{2}+n^{2}})$$
Also according to Wikipedia:
"A perfect cuboid (also called a perfect box) is an Euler brick whose space diagonal also has integer length. In other words, the following equation is added to the system of Diophantine equations defining an Euler brick:
$${\displaystyle a^{2}+b^{2}+c^{2}=g^{2}}$$
where g is the space diagonal. As of May 2015, no example of a perfect cuboid had been found and no one has proven that none exist."
If you plug these two equations into one another you inevitably end up with $$√2(m^2+n^2)*k=g$$
And since the square root of 2 is irrational it stands to reason that you will never have a rational diagonal g that is an integer.
