<s>[INST] Cuboid with inner and outer Cubes A cuboid with integer side lengths consists of cubes with side length 1. Such a cube is called an inner cube if none of its sides are visible from the outside, otherwise, i.e. if at least one side is visible from the outside, the cube is called an outer cube. The $3 \times 4 \times 7$ cuboid shown consists of 10 inner and 74 outer cubes.
For which a is there an $a \times b \times c$ cuboid with $a \le b$ and $a \le c$, with the same number of inner and outer cubes?
I have:
Number inner cubes: $(a-2)(b-2)(c-2)$
Number of outer cubes: $abc - (a-2)(b-2)(c-2)$
$(a-2)(b-2)(c-2) = abc - (a-2)(b-2)(c-2)$
$a=4+(8c+8b-16)/(bc-4c-4b+8)$
How can I determinate $a, b$ and $c$???

 [/INST] We will show that the set of all solutions are finite (solutions with length greater than $2$), and we give some bounds to find all of the solutions.

Without lose of generality we can assume $2 < a \leq b \leq c$ (Note that if $a=1, 2$ then every cube is a visible cube). We will show that there are only finitely many natural numbers that are satisfying $2(a-2)(b-2)(c-2)=abc$, with $2 < a \leq b \leq c$.

Lemma (1): If $10 \leq d$, then $d < \sqrt[3]2(d-2)$.
(Hint: Prove that the statement holds for $d=10$, and then, by induction, prove that if the statement holds for $d=r$, then it holds for $d=r+1$.)
Lemma (2): Let $2 < a \leq b \leq c$ be integers that are satisfying $2(a-2)(b-2)(c-2)=abc$, then $a\leq 9$.
Proof: Suppose on contrary that $10 \leq a \leq b \leq c$, then Lemma (1) implies that
$$a.b.c < \sqrt[3]2(a-2).\sqrt[3]2(b-2).\sqrt[3]2(c-2)=2(a-2)(b-2)(c-2)=abc,$$
which is an obvious contradiction.
So, we'e proved that $a\in \{3, 4, \cdots, 9 \}$.

In the following we will consider two separated cases. The argument for the First case is due to the answer by "Jean Marie":
First case ($a=3 , 4$): In this case we have $2(a-2)\leq a$, and also clearly we have $b-2 < b$ and $c-2 < c$. Since all of them are positive, we can conclude that
$$2(a-2)(b-2)(c-2) < (a)(b)(c),$$
So we can conclude that there are no solution for $a=3, 4$.

Second case ($a=5, 6, 7, 8, 9$): In this case we will give some bounds for $b$, to show that there are only finitly many solutions.
Remark: Let's (ignore $t=3, 4$ and) fix some $t\in \{5, 6, \cdots, 9 \}$, and suppose that $a=t$, and let $\lambda_t=\frac{2(t-2)}{t}$ (You can easily verify that $1 < \lambda_t$.) Now consider these simple manipulation (we just set $a=t$!):
$$2(a-2)(b-2)(c-2)=abc \Leftrightarrow 2(t-2)(b-2)(c-2)=tbc \Leftrightarrow \lambda_t(b-2)(c-2)=bc,$$
so we are going to deal with equation $\lambda_t(b-2)(c-2)=bc$. [end of Remark]
Lemma (3): If $\frac{4\lambda_t}{\lambda_t -1} \leq d$, then $d < \sqrt[2]\lambda_t(d-2)$.
Proof: First verify that $\frac{4\lambda_t}{\lambda_t -1} \leq d$
implies $4 <\frac{4\lambda_t}{\lambda_t -1} \leq d$, so both of $d$ and $d-2$ are positive numbers (to obtain this we used $1 < \lambda_t$). Now consider the following:
$$\frac{4\lambda_t}{\lambda_t -1} \leq d \Rightarrow 
4\lambda_t d \leq (\lambda_t -1) d^2  \Rightarrow 
4\lambda_t d < (\lambda_t -1) d^2 +4\lambda_t \Rightarrow 
$$
$$ 
\Rightarrow 
d^2 < \lambda_t d^2 - 4\lambda_t d +4\lambda_t  \Rightarrow 
d^2 < \lambda_t (d-2)^2  \Rightarrow d < \sqrt[2]\lambda_t(d-2), 
$$
and we are done.
Lemma (4): Let $4 < t=a \leq b \leq c$ be integers that are satisfying $2(a-2)(b-2)(c-2)=abc$, then $t \leq b< \frac{4\lambda_t}{\lambda_t -1}$.
Proof: To prove it, first we will consider the above remark, so we can consider the equation $\lambda_t(b-2)(c-2)=bc$, and we will switch into this equation; i.e we are dealing with solutions of this last equation satisfying $4 < t \leq b \leq c$, and we will prove that $ b < \frac{4\lambda_t}{\lambda_t -1}$.
Suppose on contrary that $\frac{4\lambda_t}{\lambda_t -1}  \leq b \leq c$, then Lemma (3) implies that
$$b.c < \sqrt[2]\lambda_t(b-2) . \sqrt[2]\lambda_t(c-2)=\lambda_t (b-2)(c-2)=bc,$$
which is an obvious contradiction.
So, we'e proved that $ b < \frac{4\lambda_t}{\lambda_t -1}$, and we are done.

So we are dealing with finite sets of paired integers $(a, b)$ (with $2<a$), and if we suppose that $a$ and $b$ are given constants, then the equation turns into a linear equation containing just one variable $c$. Also, the reader can check this method works for higher dimentional cubes.
Also, note that the bounds in my solution are not sharp, but they are so small, and perhaps one can find all solution just by hand-based-calculation in less than 1 hour.

Algorithm:
(I): Fix some $t \in \{ 5, 6, \cdots, 9\}$, and set $a=t$.
(II): Now fix some $t \leq s < 4 + \frac{4t}{t-4}=8+\frac{16}{t-4}$, and set $b=s$.
(III): Now consider the equation $2(a-2)(b-2)(c-2)=abc$, and replace $a=t$ and $b=s$, and solve for the variable $c$, which would be equal $c=\frac{4(t-2)(s-2)}{2(t-2)(s-2)-ts}=2+\frac{2ts}{2(t-2)(s-2)-ts}$. If $c=2+\frac{2ts}{2(t-2)(s-2)-ts}$ is a natural number, then $(a, b, c)$ is a solution, otherwise it is not a solution.
Remark: Considering the bounds given in parts (I) and (II) in the algorithm, we have just to consider the following pairs:

*

*$t=5$: In this case $5 \leq s < 24$, and there are at most $19$ potential answers with $(a, b)=(t, s)$.

*$t=6$: In this case $6 \leq s < 16$, and there are at most $11$ potential answers with $(a, b)=(t, s)$.

*$t=7$: In this case $7 \leq s < 13+\frac{1}{3}$, and there are at most $7$ potential answers with $(a, b)=(t, s)$.

*$t=8$: In this case $8 \leq s < 12$, and there are at most $4$ potential answers with $(a, b)=(t, s)$.

*$t=9$: In this case $9 \leq s < 11+\frac{1}{5}$, and there are at most $3$ potential answers with $(a, b)=(t, s)$ (There are not any pairs in this case).

I think all the answers are listed in the answer by "Michael Hoppe" (I didn't verify all of the possibilites and also his answer).
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