I'm currently stuck on a problem from my computer theory class homework. The question asks me to prove, using induction, that any chessboard of size $n \times 3$, where $n$ is even and $\geq 10$, will have a closed knight's tour. I've worked on the problem for a couple hours but can't seem to find any insight into the solution. I can't even figure out the tour for what I think is the base case, $n = 10$. Any tips or hints as to how I might go about proving this?
The question itself comes with a hint, saying that there will be two base cases. At first I thought this meant that I would have to show a closed knight's tour for one value of $n < 10$, which might give me a helpful property during the induction step, but having checked the Wiki article on closed knight's tours I understand now that there is no value of $n < 10$ for which a closed knight's tour is possible. In that case, maybe one base case chessboard can be broken down into subchessboards in one way that reveals a closed knight's tour, that extends to some values of $n$, while another can be broken down in a different way that extends to some other values of $n$? I'm not really sure how to interpret the hint and since I can't even figure out a single closed knight's tour I'm effectively stumped.
Some definitions that come with the question:
In chess, a knight’s move consists of travelling two squares in one direction (horizontally or vertically, but not diagonally) and then one square at 90 degrees to these.
A closed knight’s tour is a sequence of knight moves that touch upon every square on the board exactly once and end on a square from which one is a knight’s move away from the beginning square.
I don't want the full solution to the problem, but would appreciate some help as to how I should go about approaching it, since I'm currently completely stuck.