Let the edge lengths of a rectangular prism be $x, y, z$. According to the problem, we have $x^{2}+y^{2}+z^{2}=x y z$.
The problem is to prove that the equation $x^{2}+y^{2}+z^{2}=x y z$ has infinitely many positive integer solutions $\left(a_{n}, b_{n}, c_{n}\right)$, and that among $a_{n}, b_{n}, c_{n}$, the product of any two numbers can be expressed as the sum of the squares of two positive integers.
First, define the sequence $\left\{F_{n}\right\}:$
$$F_{0}=0, F_{1}=1, F_{n+2}=F_{n+1}+F_{n} \text {. }$$

Lemma (1) $F_{n+1} F_{n-1}-F_{n}^{2}=(-1)^{n}(n \geqslant 1)$, particularly $F_{2 n+1} F_{2 n-1}=F_{2 n}^{2}+1$;
(2) $F_{n+m}=F_{n} F_{m-1}+F_{m} F_{n+1}(m \geqslant 1$, $n \geqslant 0)$, particularly $F_{2 n+1}=F_{n}^{2}+F_{n+1}^{2}$;
$$\text { (3) } 1+F_{2 n-1}^{2}+F_{2 n+1}^{2}=3 F_{2 n-1} F_{2 n+1}(n \geqslant 1) \text {. }$$

Proof of the lemma: (1) Let $x_{n}=F_{n+1} F_{n-1}-F_{n}^{2}$. Then $x_{1}=-1$.
$$\begin{array}{l} \text { Since } x_{n+1}=F_{n+2} F_{n}-F_{n+1}^{2} \\ =\left(F_{n+1}+F_{n}\right) F_{n}-F_{n+1}^{2} \\ =F_{n}^{2}-F_{n+1}\left(F_{n+1}-F_{n}\right) \\ =F_{n}^{2}-F_{n+1} F_{n-1}=-x_{n}, \end{array}$$

Therefore, $x_{n}=(-1)^{n}$, i.e.,
$$F_{n+1} F_{n-1}-F_{n}^{2}=(-1)^{n} \text {. }$$
(2) Induction on $n$: For $n=0$, it is clearly true. Assume for $n=k$, $F_{k+m}=F_{k} F_{m-1}+F_{m} F_{k+1}$. For $n=k+1$,
$$\begin{array}{l} F_{k+m+1}=F_{k} F_{m}+F_{m+1} F_{k+1} \\ =F_{k} F_{m}+\left(F_{m}+F_{m-1}\right) F_{k+1} \\ =F_{k+1} F_{m-1}+F_{m}\left(F_{k}+F_{k+1}\right) \\ =F_{k+1} F_{m-1}+F_{m} F_{k+2}, \end{array}$$

Thus, it holds for $n=k+1$.
Therefore, $F_{n+m}=F_{n} F_{m-1}+F_{m} F_{n+1}$.
Taking $m=n+1$ as a special case.
(3) For $n=1$, $1+F_{1}^{2}+F_{3}^{2}=6=3 F_{1} F_{3}$, which is true.
Assume for $n=k$,
$$1+F_{2 k-1}^{2}+F_{2 k+1}^{2}=3 F_{2 k-1} F_{2 k+1} \text {. }$$

For $n=k+1$, since $F_{2 k-1}$ is a root of the equation
$$x^{2}-3 F_{2 k+1} x+1+F_{2 k+1}^{2}=0$$

the other root is
$$\begin{array}{c} 3 F_{2 k+1}-F_{2 k-1}=2 F_{2 k+1}+F_{2 k} \\ =F_{2 k+1}+F_{2 k+2}=F_{2 k+3}, \end{array}$$

Thus, $1+F_{2 k+1}^{2}+F_{2 k+3}^{2}=3 F_{2 k+1} F_{2 k+3}$.
Hence, $1+F_{2 n-1}^{2}+F_{2 n+1}^{2}=3 F_{2 n-1} F_{2 n+1}$.
Returning to the original problem.
By Lemma (3),
$$(x, y, z)=\left(3,3 F_{2 n-1}, 3 F_{2 n+1}\right)(n \geqslant 2)$$

is a solution to $x^{2}+y^{2}+z^{2}=x y z$, and by Lemmas (2) and (1),
$$\begin{array}{l} 3 \times 3 F_{2 n-1}=\left(3 F_{n-1}\right)^{2}+\left(3 F_{n}\right)^{2}, \\ 3 \times 3 F_{2 n+1}=\left(3 F_{n}\right)^{2}+\left(3 F_{n+1}\right)^{2}, \\ 3 F_{2 n-1} \times 3 F_{2 n+1}=\left(3 F_{2 n}\right)^{2}+3^{2} . \end{array}$$

Therefore, the original equation has infinitely many positive integer solutions $\left(a_{n}, b_{n}, c_{n}\right)$, such that among $a_{n}, b_{n}, c_{n}$, the product of any two numbers can be expressed as the sum of the squares of two positive integers.
Thus, the conclusion of the original problem is true.