As requested by Alex, here is an old fashioned proof that packing spheres into a cube N on a side uses the same number of spheres as arranging them into an N-high hexagonal pyramid.

We already know that a 1^{3} cube uses just a single sphere — the same as a “pyramid” of height 1 — which is just a single sphere.

We just need to show that adding an Nth layer to a hexagonal pyramid uses the same number of spheres as incrementing from a cube of width N-1 to a cube of width N.

To get from an (N-1)^{3} cube to an N^{3} cube requires N^{3} – (N-1)^{3} more spheres. That’s N^{3} – (N^{3} – 3N^{2} + 3N – 1) spheres, which simplifies to 3N(N-1)+1.

If we can show that a packed hexagon with N spheres on a side uses the same number of spheres, then we’re done. We do that as follows:

To go from our single sphere to a packed hexagon with 2 spheres on each side, we need to add 1 sphere for each of the hexagon’s six sides.

Similarly, to go from that to a packed hexagon with 3 spheres on each side, we need to add 2 spheres for each side of the hexagon.

So to build up to a packed hexagon with N spheres on a side, we need to add 1 + 2 + …. (N-1) spheres for each side of the hexagon.

The sum of the numbers from 1 through N-1 is N(N-1)/2 (here’s the proof on Wikipedia).

Putting it all together, we have six of those sums (because the hexagon has six sides), plus the one sphere in the center.

Which gives us 6×( N(N-1)/2 ) + 1, or 3N(N-1)+1.

As they say in latin class, QED. But I think the visual proof is a lot more elegant — and it didn’t require linking to the Wikipedia to explain picky details. 🙂

Thank you, but please allow me to ask one silly question. I think in many aspects of life, what scientist does is defining a problem, then finding out a solution. It seems like there is”solution” now. are there any problems we can solve with this fact, or it’s just a pure observation?