There Can't Be Too Many?

Consider a rectangular prism with dimensions $x \times y \times z$ and WLOG $x \leq y \leq z$ . The total number of cubes is $xyz$ . We also have that the total number of interior cubes is $(x-2)(y-2)(z-2)$ . Thus, we have the following:

$\begin{aligned} xyz &= 2(x-2)(y-2)(z-2) \quad \quad \quad (1)\\ \dfrac{1}{2} &= \left (1-\dfrac {2}{x} \right )\left (1-\dfrac {2}{y} \right )\left (1-\dfrac {2}{z} \right ) \end{aligned}$

Since $x \leq y \leq z$ , we have $1-\dfrac {2}{x} \leq 1-\dfrac {2}{y} \leq 1-\dfrac {2}{z}$ . From this, we get

$\begin{aligned} 1-\dfrac {2}{x} &\leq \dfrac {1}{\sqrt[3]{2}}\\ x &\leq \dfrac {2}{1-\sqrt[3]{\frac{1}{2}}}\\ x &\leq 9 \end{aligned}$

Thus, we have a bound for how large the smallest dimension of the rectangular prism is. Geometrically, we also know that $x > 3$ . Going back to statement $(1)$ , expanding, we get

$\begin{aligned} 2(xyz - 2xy - 2yz - 2zx + 4x + 4y + 4z - 8) &= xyz\\ xyz-4(xy+yz+zx)+8(x+y+z)-16 &= 0 \end{aligned}$

We will now break this down into the cases for when the value of $x$ tends from 4 to 9.

Case 1: $x=4$

Substituting this, we get

$\begin{aligned} 4yz - 4(4y + yz + 4z) + 8(4+y+z)-16 &= 0\\ 16 - 8(y + z) &= 0\\ y+z &= 2 \end{aligned}$

This has no solutions since we must have $x \leq y \leq z$ .

Case 2: $x=5$ :

Substituting this, we get

$\begin{aligned} 5yz - 4(5y + yz + 5z) + 8(5+y+z)-16 &= 0\\ yz -12y-12z+24 &= 0\\ (y-12)(z-12) &= 120 \end{aligned}$

Factorising 120, we get

$(y-12) \in \{1, 2, 3, 4, 5, 6, 8, 10\} \text{ and } (z-12) \in \{120, 60, 40, 30, 24, 20, 15, 12\}$

$(y,z) = (13, 132), (14, 72), (15, 52), (16, 42), (17, 36), (18, 32), (20, 27), (22, 24)$

Here, we have 8 solutions which satisfy.

Case 3: $x=6$

Substituting this, we get

$\begin{aligned} 6yz - 4(6y + yz + 6z) + 8(6+y+z)-16 &= 0\\ 2yz -16y - 16z + 32 &= 0\\ (y-8)(z-8) &= 48 \end{aligned}$

Factorising, we get

$(y-8) \in \{1, 2, 3, 4, 6\} \text{ and } (z-8) \in \{48, 24, 16, 12, 8\}$

$(y, z) = (9, 56), (10, 32), (11, 24), (12, 20), (14, 16)$

Here, we have 5 solutions.

Case 4: $x=7$

Substituting this, we get

$\begin{aligned} 7yz - 4(7y + yz + 7z) + 8(7+y+z)-16 &= 0\\ 3yz - 20y - 20z + 40 &= 0\\ 9yz - 60y - 60z + 120 &= 0\\ (3y - 20)(3z-20) &= 280 \end{aligned}$

Factorising, we get

$(3y-20) \in \{1,2,4,5,7,8,10,14\} \text{ and } (3z - 20) \in \{280, 140, 70, 56, 40, 35, 28, 20\}$

$(y, z) = (7,100), (8, 30), (9, 20), (10, 16)$

(All fractional solutions were removed). Here, we have 4 solutions.

Case 5: $x = 8$

Substituting this, we get

$\begin{aligned} 8yz - 4(8y + yz + 8z) + 8(8+y+z)-16 &= 0\\ 4yz-24y-24z+48 &= 0\\ (y-6)(z-6) &= 24 \end{aligned}$

Factorising, we get

$(y-6) \in \{1, 2, 3, 4\} \text{ and } (z-6) \in \{24, 12, 8, 6\}$

$(y, z) = (8, 18), (9, 14), (10, 12)$

(The solution which contradicts $x \leq y \leq z$ has been removed). Here, we have 3 solutions.

Case 6: $x = 9$

Substituting this, we get

$\begin{aligned} 9yz - 4(9y + yz + 9z) + 8(9+y+z)-16 &= 0\\ 5yz - 28y - 28z + 56 &= 0\\ 25yz - 140y - 140z + 280 &= 0\\ (5y-28)(5z-28) &= 504 \end{aligned}$

Factorising, we have

$(5y-28) \in \{1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 18, 21\} \text{ and } (5z-28) \in \{504, 252, 168, 126, 84, 72, 63, 56, 42, 36, 28, 24\}$

This has no solutions which satisfy $x \leq y \leq z$ and $x, y, z$ being positive integers.

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Over all the cases, we get a total of $8 + 5 + 4 + 3 = \boxed{20}$ different rectangular prisms which satisfy.

Preparation

Let $3\leq x,\:y ,\: z\in\mathbb{N}$ be the lengths of the prism's sides. They need to be greater 2 for the prism to have an interior. Calculate the number of units in the interior and the exterior of the prism:

If the interior equals the exterior, we have $\begin{aligned} xyz&=2(x-2)(y-2)(z-2)&\Rightarrow&&\frac{1}{2}&=\left(1-\frac{2}{x}\right) \left(1-\frac{2}{y}\right) \left(1-\frac{2}{z}\right),&x,\:y,\:z\geq 3&&&&(*) \end{aligned}$

Notice that every factor is an increasing function in its variable, e.g. $\begin{aligned} \frac{1}{3}\leq 1-\frac{2}{x}<1&&\text{for any}&&x\geq 3 \end{aligned}$

We learned that we cannot obtain a solution if all factors are greater than $\frac{1}{\sqrt[3]{2}}$ (then the LHS in $(*)$ would be smaller than the RHS) or if any factor is smaller or equal to $\frac{1}{2}$ (then the RHS would be smaller than the LHS)! For one factor, let it contain $x$ , we need $\begin{aligned} 1-\frac{2}{x}&\in\left(\frac{1}{2};\:\frac{1}{\sqrt[3]{2}}\right]&\Rightarrow&& x&\in\{5;\:\ldots;\:9\} \end{aligned}$

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The cases in $x,\:y,\:z$

Insert each value of $x$ into the main equation $(*)$ , the results are given in the second column of the table below. Now we reuse the trick to obtain bounderies for $y$ . We cannot obtain a solution if both remaining factors are greater than the square root of the new LHS or if any factor is smaller than the new LHS. For one factor, let it contain $y$ , we need: $\begin{aligned}\begin{array}{r | c | c | c} & \quad\ldots=\left(1-\frac{2}{y}\right) \left(1-\frac{2}{z}\right) & 1-\frac{2}{y}\in\ldots\quad &y\in\ldots\quad\\ \hline x=5& \frac{5}{6}&\left(\frac{5}{6};\:\sqrt{\frac{5}{6}}\right]&\{13;\:\ldots;\:22\}\\ x=6& \frac{3}{4}&\left(\frac{3}{4};\:\sqrt{\frac{3}{4}}\right]&\{9;\:\ldots;\:14\}\\ x=7& \frac{7}{10}&\left(\frac{7}{10};\:\sqrt{\frac{7}{10}}\right]&\{7;\:\ldots;\:12\}\\ x=8& \frac{2}{3}&\left(\frac{2}{3};\:\sqrt{\frac{2}{3}}\right]&\{7;\:\ldots;\:10\}\\ x=9& \frac{9}{14}&\left(\frac{9}{14};\:\sqrt{\frac{9}{14}}\right]&\{6;\:\ldots;\:10\} \end{array}\end{aligned}$

In the first variant of the main equation $(*)$ we isolate $z$ . The result is a function in $x,\:y$ : $\begin{aligned} \green{xy}z&=2(x-2)(y-2)(z-\red{2})&\Rightarrow&&z&=\frac{\red{4}(x-2)(y-2)}{2(x-2)(y-2)-\green{xy}}=:f(x,\:y) \end{aligned}$

Calculate $z=f(x,\:y)$ for all 31 pairs of the table above - only if $z$ is an integer we have found a valid solution. After some hard work, we are done and the only valid solutions are $\begin{aligned}\begin{array}{r | l} & (y,\:z)\in\ldots\\ \hline x=5& \left\{(13;\:132),\: (14;\:72),\: (15;\:52),\: (16;\:42),\: (17;\:36),\: (18;\:32),\: (20;\:27),\: (22;\:24)\right\}\\[.25em] x=6& \left\{\green{(9;\:56)},\:(10;\:32),\:(11;\:24),\:(12;\:20),\:(14;\:16)\right\}\\[.25em] x=7& \left\{(7;\:100),\:\green{(8;\:30)},\:\green{(9;\:20)},\:(10;\:16)\right\}\\[.25em] x=8& \left\{\red{(7;\:30)},\:(8;\:18),\:\green{(9;\:14)},\:(10;\:12)\right\}\\[.25em] x=9& \left\{\red{(6;\:56),\:(7;\:20),\:(8;\:14)}\right\} \end{array}\end{aligned}$

Removing $\red{\text{four dublicates}}$ from the table, there are $\fbox{20}$ distinct prisms with equal interior end exterior volume!