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I am having trouble understanding the solution to the question, how does when it starts on a black square must end on another black square, and if the king's moves change the colors of the squares how does that constitute making the number of traversed squares odd?

Question: A king Visited all squares of the usual 8 by 8 chessboard exactly once starting from the lower left square (a1) and finishing at the upper right square (h8). Prove that the king made at least one diagonal move.

Answer: Assume on the contrary that the king made no diagonal moves. then on each move it changed the color of it's square. Consider a list of all squares in the order the king visited them. It starts with a white square and ends with a white square, and the colors of the squares alternate. Therefore it has an odd number of squares. but the king visited all 64 squares exactly once. a contradiction. therefore the king made at least 1 diagonal move.

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    $\begingroup$ There are $64$ squares in total, which means that the King must make $63$ moves. If each move changes the color, then the King will end in a square of a different color, as $63$ is odd. To visualize it, I suggest working with a smaller board. For $2\times 2$, there isn't much to it. Look at $4\times 4$. $\endgroup$
    – lulu
    Mar 1 at 12:00
  • $\begingroup$ @lulu That's an answer to the question. Why are you putting it in the comment section and not as an answer? $\endgroup$
    – Arthur
    Mar 1 at 12:00
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    $\begingroup$ @Arthur, because I expect the question to be closed soon. But maybe I'm wrong about that, so I'll post it. $\endgroup$
    – lulu
    Mar 1 at 12:02

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There are $64$ squares in total, which means that the King must make $63$ moves. If each move changes the color, then the King will end in a square of a different color, as $63$ is odd. To visualize it, I suggest working with a smaller board. For $2\times 2$, there isn't much to it, so look at $4\times 4$.

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