Proof of non-existence of perfect cuboids
Let $a$, $b$, $c$, $d$, $e$, $f$ and $g $be positive integers. If a perfect cuboid exists, the following equations hold where $a$, $b$, $c$ are the edges, $d$, $e$, $f$ are the diagonals and $g$ is the space diagonal.
$$a^2+b^2=d^2$$
$$b^2+c^2=e^2$$
$$c^2+a^2=f^2$$
$$a^2+b^2+c^2=g^2$$
If $a$ and $b $are odd, $a^2+b^2$ is not a square number. Therefore, we consider $a$ is odd and $b$ and $c$ are even. In this way, it does not lose its generality. In this case, $d$, $f$ and $g$ are odd and $e$ is even. Let $i$, $m_i$, $n_i$ and $k_i$ be positive integers and $1≦i≦6$ holds. We suppose $GCD(a, b, c)=1$ holds and $m_i$ and $n_i$ are coprime. According to the primitive Pythagorean numbers, the following equations hold.
$$a=(m_1^2-n_1^2)k_1$$
$$b=2m_1n_1k_1$$
$$d=(m_1^2+n_1^2)k_1$$
$$(m_1^2-n_1^2)^2k_1^2+(2m_1n_1k_1)^2=(m_1^2+n_1^2)^2k_1^2$$
$$a=(m_1^2-n_1^2)k_1=(m_2^2-n_2^2)k_2$$
$$c=2m_2n_2k_2$$
$$f=(m_2^2+n_2^2)k_2$$
$$(m_1^2-n_1^2)^2k_1^2+(2m_2n_2k_2)^2=(m_2^2+n_2^2)^2k_2^2$$
$$b=2m_1n_1k_1=(m_3^2-n_3^2)k_3$$
$$c=2m_3n_3k_3$$
$$e=(m_3^2+n_3^2)k_3$$
$$(2m_1n_1k_1)^2+(2m_3n_3k_3)^2=(m_3^2+n_3^2)^2k_3^2$$
$$g=(m_4^2+n_4^2)k_4=(m_5^2+n_5^2)k_5=(m_6^2+n_6^2)k_6$$
$$(m_1^2-n_1^2)^2k_1^2+(m_3^2+n_3^2)^2k_3^2=(m_4^2+n_4^2)^2k_4^2 ...(1)$$
$$(2m_1n_1k_1)^2+(m_2^2+n_2^2)^2k_2^2=(m_5^2+n_5^2)^2k_5^2 ...(2)$$
$$(2m_3n_3k_3)^2+(m_1^2+n_1^2)^2k_1^2=(m_6^2+n_6^2)^2k_6^2 ...(3)$$
It is assumed that $b=(m_3^2-n_3^2)k_3$ but does not lose generality. $k_1$, $k_2$, $k_4$, $k_5$ and $k_6$ are odd and $k_3$ is even. By the equation (1),
$$(m_1^2-n_1^2)^2k_1^2≡0\ (mod\ k_4)$$
holds. Let $k_4'$ be a positve integer and a prime factor of $k_4$. If $k_1≡0\ (mod\ k_4')$ holds, then it becomes a contradiction since $a$, $b$ and $c$ have a common prime factor. Therefore,
$$m_1^2-n_1^2≡0\ (mod\ k_4)$$
holds. In the same way, the following congruent expressions hold by the equations (1), (2) and (3).
$$m_2^2-n_2^2≡0\ (mod\ k_4)$$
$$m_3^2+n_3^2≡0\ (mod\ k_4)$$
$$m_1n_1≡0\ (mod\ k_5)$$
$$m_2^2+n_2^2≡0\ (mod\ k_5)$$
$$m_3^2-n_3^2≡0\ (mod\ k_5)$$
$$m_1^2+n_1^2≡0\ (mod\ k_6)$$
$$m_2n_2≡0\ (mod\ k_6)$$
$$m_3n_3≡0\ (mod\ k_6)$$
Let $p$, $q$, $r$, $s$, $t$, $u$, $v$, $w$ and $x$ be positive integers.
$$m_1^2-n_1^2=pk_4$$
$$m_2^2-n_2^2=qk_4$$
$$m_3^2+n_3^2=rk_4$$
$$m_1n_1=sk_5$$
$$m_2^2+n_2^2=tk_5$$
$$m_3^2-n_3^2=uk_5$$
$$m_1^2+n_1^2=vk_6$$
$$m_2n_2=wk_6$$
$$m_3n_3=xk_6$$
By these equations,
$$(pk_4)^2+(2sk_5)^2=(vk_6)^2 ...(4)$$
$$(qk_4)^2+(2wk_6)^2=(tk_5)^2 ...(5)$$
$$(uk_5)^2+(2xk_6)^2=(rk_4)^2 ...(6)$$
hold. Let $k_6'$ be a positve integer and a prime factor of $k_6$. By the equation (4), $p≡0\ (mod\ k_6)$ and $s≡0\ (mod\ k_6)$ hold. At this time, $m_1^2-n_1^2≡0\ (mod\ k_6')$ and $m_1n_1≡0\ (mod\ k_6')$ hold. It becomes a contradiction since $m_1≡0\ (mod\ k_6')$ and $n_1≡0\ (mod\ k_6')$ hold and $m_1$ and $n_1$ are not relatively prime. Therefore, $k_6=1$ holds. In the same way, $k_4=1$ and $k_5=1$ hold by the equations (5) and (6). When $k_4=1$, $k_5=1$ and $k_6=1$ hold, the following equatioins hold by the form of Pythagorean triple and the equations (1), (2) and (3).
$$(m_1^2-n_1^2)k_1=m_4^2-n_4^2$$
$$m_1n_1k_1=m_5n_5$$
$$(m_1^2+n_1^2)k_1=m_6^2-n_6^2$$
By these equations,
$$(m_4^2-n_4^2)^2+(2m_5n_5)^2=(m_6^2-n_6^2)^2$$
holds. In this case, the following equation must hold slimilarly as above.
$$m_6^2-n_6^2=m_4^2+n_4^2$$
At this time, $(m_1^2+n_1^2)k_1=m_4^2+n_4^2$ holds and this is a contradiction since $c=0$ holds. From the above, it is proved that there are no perfect cuboids. (Q.E.D.)
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