this post was submitted on 01 Jul 2025
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What? No. The divisibility of the side lengths have nothing to do with this.
The problem is what's the smallest square that can contain 17 identical squares. If there were 16 squares it would be simply 4x4.
And the next perfect divisor one that would hold all the ones in the OP pic would be 5x5. 25 > 17, last I checked.
He's saying the same thing. Because it's not an integer power of 2 you can't have a integer square solution. Thus the densest packing puts some boxes diagonally.