Post #2672388
2026-01-22 17:59 UTC
The same is true for the n=6 solution, which adds four new rectangles to the n=2 solution (by, for example, wrapping four larger rectangles around the original two).
And you can keep doing that for any square: do the "wrap-around" technique to add four more rectangles. n=10 is just n=6 with another layer!
So for any solution n, we also have n+4, n+8, ... ad inf.
Replies (1)
-
@joshmillard@mastodon.social 2026-01-22 18:04
And here's one of the big flourishes: if we know we have solutions for n=5 and n=6, and we know we can make solutions for n=5+3k and n=6+4j, we can make solutions for ALL values of n>=8. Proof? n=5, do the +3 rectangle split trick and we have an n=8 solution. n=6, do the +3 split trick and we have n=9. n=6, do the +4 rectangle wraparound trick, we have n=10. Now we know we have solutions for 8, 9, and 10. And we can get +3 solutions for each: 11, 12, 13, and then 14, 15, 16, and then...