Misbehaving Integer Cubes

Make one vertex the origin and the three adjacent ones $(2k,2k,k),(2k,-k,-2k),(-k,2k,-2k)$ where $k$ is any positive integer. There are infinitely many such cubes.

This is a long and boring, non-rigorous proof, so if you're feeling brave keep reading!

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What would such a misbehaving integer cube look like?

Let's imagine such a cube $C'$ that has a vertex at the origin without loss of generality.

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Let's call the three vertices that are adjacent to the one at the origin $A = ( a_1, a_2, a_3 ), B = ( b_1, b_2, b_3 )$ and $C = ( c_1, c_2, c_3 )$ . Since no face is parallel to the $xy, yz, xz$ planes we cannot have any of these vertices having an entry as zero. This is because if any of these entries are $0$ that means that one of the sides of the cube does not leave that corresponding plane.

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What can we say about the volume of $C'$ ?

Well, those familiar with determinants might know that the volume of a cube with a vertex at the origin can be given by the determinant of the vectors of the $3$ adjacent vertices. We won't go into detail here but it if the these vertices have integer co-ordinates then the cube has integer volume.

So we now know that the cube has integer volume!

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What does this mean for the length of each side of the $C'$ ?

We know that the volume of the cube is given by the cube of the lengths of it's sides.

Then the volume of the cube is given by

$Vol( C' ) = \sqrt{ a_1^2 + a_2^2 + a_3^2 }\sqrt{ b_1^2 + b_2^2 + b_3^2 }\sqrt{ c_1^2 + c_2^2 + c_3^2 }.$

Now since we know that the volume is an integer we must have this right hand side equation be an integer too. When is this true? Well since it is a cube, all sides have the same length, say

$\begin{aligned} L = \sqrt{l} &= \sqrt{ a_1^2 + a_2^2 + a_3^2 } \\ &= \sqrt{ b_1^2 + b_2^2 + b_3^2 } \\ &= \sqrt{ c_1^2 + c_2^2 + c_3^2 } \end{aligned}$

for some integer $l$ . So this simplifies our equation to $Vol( C' ) = L^3 = \sqrt{l}^3 = l\sqrt{l}$ . Now the only way we can make this an integer is if $l$ is a square number, $l = k^2$ for some non-zero integer $k$ .

This means that we must have

$\begin{aligned} k^2 &= a_1^2 + a_2^2 + a_3^2 \\ k^2 &= b_1^2 + b_2^2 + b_3^2 \\ k^2 &= c_1^2 + c_2^2 + c_3^2 \end{aligned}$ .

Now these look a lot like Pythagorean Tripples ! In fact, they're all Pythagorean Quadruples . With this we're almost done (well done for getting this far).

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Now there is just one more observation about cubes we must make - our $3$ vectors $A, B , C$ must all be at right angles to one another. This means that they're dot product must all be $0$ with one another. Explicitly we must have

$\begin{aligned} 0 &= a_1b_1 + a_2b_2 + a_3b_3 \\ 0 &= b_1c_1 + b_2c_2 + b_3c_3 \\ 0 &= c_1a_1 + c_2a_2 + c_3a_3 \end{aligned}$ .

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So now we've reduced our problem down to hunting down $3$ vectors $A = ( a_1, a_2, a_3 ), B = ( b_1, b_2, b_3 )$ and $C = ( c_1, c_2, c_3 )$ such that the entries are all non-zero integers for some non-zero integer $k$ and

$\begin{aligned} k^2 &= a_1^2 + a_2^2 + a_3^2 \\ k^2 &= b_1^2 + b_2^2 + b_3^2 \\ k^2 &= c_1^2 + c_2^2 + c_3^2 \end{aligned}$

$\begin{aligned} 0 &= a_1b_1 + a_2b_2 + a_3b_3 \\ 0 &= b_1c_1 + b_2c_2 + b_3c_3 \\ 0 &= c_1a_1 + c_2a_2 + c_3a_3 \end{aligned}$ .

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We can make a parametrisation of

$\begin{aligned} A &= (n, n+1, n(n+1)) \\ B &= (n(n+1), n, -(n+1))\\ C &= (n+1, -n(n+1), n) \end{aligned}$ .

Note that $n^2+ (n+1)^2 + n^2(n+1)^2 = (n(n+1) + 1)^2$ and all of the checks to see if they are perpendicular work too. They're all clearly non-zero for non-zero integers and distinct for different values of $n$ .

So there must be infinitely many misbehaving integer cubes!

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This is just one parametrisation and it doesn't capture all the cubes with this property. If you find any more parametrisations - let me know :)