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So, say you have 16 points (0-indexed):

You want to walk across the vertices, skipping over a specific number of them, landing on the end vertex. You can always 1-connect them:

But 2-connecting 3-connecting, 4-connecting, $X$-connecting is only possible if ${N-1}\mod{X}=0$, $N$ is the number of vertices.

Why isn't it $N \mod x=0$?

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Well, when there are $N$ vertices, $N-1$ is the number of steps you need, in order to advance from the first vertex to the last. (If you number them $0$ to $N-1$, then $(N-1) - (0) = N-1$. If you number them $1$ to $N$, then $(N) - (1) = N-1$. However you number them, the "distance" from the first to the last is $N-1$. It makes sense that this distance is what matters.)

Specifically, if this number $N-1$ of steps has a factorisation $N-1 = xy$, then you can cover the distance $N-1$ by taking $y$ steps of length $x$ each. (With "1-connecting" ($x=1$, that corresponds to taking $N-1$ of length $1$ each.)