Cuboid with inner and outer Cubes

Question

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 $c$???

Answer 1

(see Edit Below)

In fact, your equation is

$2(a-2)(b-2)(c-2)=abc \tag{1}$

I have found a solution which is

$$\color{red}{a=5, b=14, c=72}$$

(may be there are others).

Indeed one can verify that

$$2 \times 3 \times 12 \times 70 = 5 \times 14 \times 72$$

because they have the same factorisation into power of primes $2^4*3^2*5*7$.

How did I reached such a result ?

I have first applied Gauss theorem to (1): necessarily, due to the presence of $2$ in the LHS, one of the terms on the RHS must be even. Assume it is $c$ (WLOG, due to the symmetry in (1)); therefore we have $c=2k$.

$$2(a-2)(b-2)(2k-2)=ab2k$$

$$4(a-2)(b-2)(k-1)=ab2k$$

$$2(a-2)(b-2)(k-1)=abk \tag{2}$$

• $a=3$ cannot be a solution for (2) because this would mean that $2(b-2)(k-1)=3bk$ which is clearly impossible.

• nor $a=4$ can be a solution because it would mean that $(b-2)(k-1)=bk$ which is impossible as well.

Therefore the smallest value possible for $a$ is $a=5$.

At this step, I have been asking to Wolfram Alpha:

"solve $2(k-1)*(b-2)*(c-2)=k*b*c$ over the integers"

with answer (please note that I don't say "solution"):

$$\begin{cases}a \ge 5&(i)\\ b > \dfrac{4(a-2)}{a-4}&(ii)\\ k=\dfrac{2(a-2)(b-2)}{a(b-4)-4b+8}&(iii)\end{cases}\tag{3}$$

Only the second information is new (indeed, the equality (3)(iii) is nothing more than (2), but we will need it anyway).

Let us try the first value possible for $a$, i.e. $a=5$. Plugging it into (3)(ii), I had $b>12$. I chose the first even number following $12$, i.e., $14$ taking into account the fact that, in this way, one could factor out a $2$ in the denominator in (3)(iii), making possible a simplification with the $2$ present in the numerator.

And, happily, taking these values $a=5$, $b=14$, I obtained an integer value for $k=36 \implies c=2k=72$ as said in (1).

Edit: seing the list of all possible values for $(a,b,c)$ obtained by @Michael Hoppe, I had the idea to plot them as points in a 3D cartesian system. In fact, I chose to suppress the condition $a \le \le b \le c$ by taking all possible permutations $(a,c,b), (b,a,c)...$ etc. The initial points are plotted in red, all others are plotted in blue.

In this way, one sees that the different points belong to certain hyperbolas according to the value of the coordinate with the smallest value. For example, if the smallest value is $c=5$, substituting this value is (1) gives:

$$2(a-2)(b-2)(3)=ab3 \implies b=12 \frac{a-2}{a-12}=12\left( 1+\frac{10}{a-12}\right)\tag{4}$$

which is indeed the equation of a (branch of) hyperbola. In particular, (4) has a finite number of integer solutions ; this is the consequence of an immediate reasoning on the limit of this expression which is $12_{-}$ when $b \to \infty$.

Remark: in (4), taking minimum value $a=13$ give the maximum value $b=132$

Let us consider as well case $c=6$. Plugged into (1), we get:

$$8(a-2)(b-2)=6ab \ \implies \ b=8\dfrac{c-2}{c-8}$$

which is again the equation of a branch of hyperbola.

Same operation for $a=7$, $a=8$.

Answer 2

(79 chars too long for a comment ...)

All solution satisfying $2<a\le b\le c$ are

$$\left( \begin{array}{ccc} a\to 5 & b\to 13 & c\to 132 \\ a\to 5 & b\to 14 & c\to 72 \\ a\to 5 & b\to 15 & c\to 52 \\ a\to 5 & b\to 16 & c\to 42 \\ a\to 5 & b\to 17 & c\to 36 \\ a\to 5 & b\to 18 & c\to 32 \\ a\to 5 & b\to 20 & c\to 27 \\ a\to 5 & b\to 22 & c\to 24 \\ a\to 6 & b\to 9 & c\to 56 \\ a\to 6 & b\to 10 & c\to 32 \\ a\to 6 & b\to 11 & c\to 24 \\ a\to 6 & b\to 12 & c\to 20 \\ a\to 6 & b\to 14 & c\to 16 \\ a\to 7 & b\to 7 & c\to 100 \\ a\to 7 & b\to 8 & c\to 30 \\ a\to 7 & b\to 9 & c\to 20 \\ a\to 7 & b\to 10 & c\to 16 \\ a\to 8 & b\to 8 & c\to 18 \\ a\to 8 & b\to 9 & c\to 14 \\ a\to 8 & b\to 10 & c\to 12 \\ \end{array} \right)$$

Seems not so easy to show that the shortest side must not exceed $8$.

Answer 3

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.

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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 $2 < a \leq b \leq c$.

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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 $a\in \{3, 4, \cdots, 9 \}$.

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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 $a=3, 4$.

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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.

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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.

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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).