First move: select two adjacent holes, fill them with pegs.

Second move: select two diagonally opposite holes, fill them with pegs.

Now there are at least 3 holes with pegs in them, as you could not have selected the same two holes twice there. At this point you have either won or the final hole is empty. If you have not won yet,

Third move: select two diagonally opposite holes, if one is empty, fill it and you win. If both are pegged, empty one of them.

Now you know that the board has two pegged holes adjacent to eachother, and two empty holes adjacent to eachother.

Fourth move: select two adjacent holes, if both are filled, empty them. If both are empty, fill them. If one is empty and one is full, switch them

Now you have either won by filling the empty holes or emptying the filled ones, or the board is in a configuration with two pegged holes that are diagonally opposite, and two empty holes that are diagonally opposite. Now you are set to win.

Fifth move: select two diagonally opposite holes, if they are empty fill them, if they are pegged empty them.

Now you win.

This strategy will solve the puzzle for N=5. I would be greatly surprised if there are solutions with less moves, but it is not clear at all the me how to prove it. If anybody out there has a better solution or a proof that there isn't one, feel free to post it in the comments.

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