Linear algebra - side of a cube is an integer?

[Pagan Harpoon]

Homework Statement

A cube of sides a*a*a is in 3 dimensional space. All eight of its corners have integer coordinates. Prove that a is an integer.

Homework Equations

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The Attempt at a Solution

First, I considered three corners of the cube p, q and r, with these, two vectors that run along the sides of the cube and have magnitude of a are q-p and r-p and these vectors have coordinates of (q₁-p₁, q₂-p₂ etc...). Taking the cross product of these two vectors, I arrive at a vector that has magnitude of a². I had hoped that the expression for the length of (q-p)x(r-p) would turn out to be of the form (expression in p₁, p₂ etc...)² and then this would show that a=that expression in the components of p, q and r therefore, since those are integers, a is an integer.

Unfortunately, the expression for the length of that cross product is very, very ugly. I could, perhaps have put it into a computer programme and had that simplify it for me, but I'm sure there is some more efficient way of doing this. It is also possible to shift the cube such that one vertex is on the origin, this simplifies the calculation a lot, but it still isn't pretty and I still think that there is a more efficient method. There is a note beneath the question saying "This question is not assessed, so your only reward is aesthetical pleasure from solving it" which leads me to believe that the solution is clever somehow.

Ideas?

Thank you.

 

[txy]

hmm how about you consider different cases?

The simplest case is when the edges of the cube are parallel to the axes. Then it is simple to show that the length of one side is an integer.

The second case is when 2 pairs of opposing edges (a total of 4 edges) are parallel to one of the axes, for example the z-axis. Then the length of each of these edges is an integer, so case closed.

The third case is when none of the edges is parallel to any of the axes. Now this is a bit difficult. We can consider rotating it about one of the corners so that it becomes the second case above. If we can show that all the corners still have integer coordinates after the rotation, then we're done. But then this is not an easy rotation.

 

[Pagan Harpoon]

The solution has been released, it is quite nice -

Refer to the area of each face as S. S is equal to the magnitude of the cross product of two vectors than run along the edges of the cube. Since all of the corners have integer coordinates, it should be clear that this cross product's magnitude (and S) is an integer.

Similarly, refer to the volume of the cube as V. V is equal to the absolute value of the dot product of the third vector on the cube's edge with the cross product previously mentioned. Since all of the corners have integer coordinates, it should again be clear that V is an integer.

S=a², V=a³ therefore a=V/S.

This proves that a is rational as V and S are integers. We also know that a can be simplified to an integer as any primes in its denominator that wouldn't cancel wouldn't cancel in a² either and we know that a² is an integer.

 

[Dick]

That is very nice. I was scratching my head over that for a while.