The 4 packing takes up 100% of its square; it's trivially optimal. The 6 packing only takes up 2/3 of it, so it's not necessarily obvious that you can't do better.
for me it is obvious. If I am reading s=3 as multiplier of side of smaller square to side of bigger square, which means that bigger square side is 3 times the side of smaller one, than it is obvious that it should be poossible to squeeze 9 small squares into bigger square. It is children puzzle after all. What is not obvious here?
Because it could be possible, as we just saw with 5, that by some rotation of some number of cubes, you could fit six unit cubes in to box smaller than area 9. Since we just saw the unusual result in 5, it is worth verifying.
i believe 4, 9, 16, 25 etc are just subdivisions of the unit square (they're perfect squares)
but the text also says "For the $n ≤ 324$ not pictured, the trivial packing (with no tilted squares) is the best known packing." applying 'trivial' to numbers that aren't perfect squares so iunno
It's two different trivial things. For each, it's just the case which doesn't require doing anything special.
One is trivial proofs, which are where 100% is covered. This doesn't really leave much to prove in terms of whether or not more area can be covered by a different layout.
The other is trivial packings, the very simple type without any tilting or need of gaps between squares. Trivial packings are only sometimes optimal. Of optimal trivial packings, only some can be shown optimal with an aforementioned trivial proof.