I am trying to do an exercise on this topic. I have realized that base cases should be $n = 10$ and $n = 12$. Also I realized that I would need to use $(n+4)\times 3$ during my inductive step. But that is pretty much it. I really don't understand the structure which I am meant to prove this. Or the steps after Any help given would be much appreciated. Thanks!
2026-05-06 01:22:54.1778030574
How to prove that chessboard of size $n \times 3$, with even $n$ and $n \geq 10$, has a closed knight's tour with induction?
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The key inductive part after setting the base cases for $3\times 10$ and $3\times 12$ boards is to show how to extend a tour on the board that is four squares shorter. This is very straightforward for extending a closed tour, because we know the shape of the path at the end of the board through the corner squares (shown on the left):
and we can then break into that path to cover a four-column-longer board as follows:
(where the two solid red links and the whole of the blue path replace the removed dotted red link).