Lemma 9

Assume that X is a chain with vertices 1, 2, ..., l and edges {1, 2}, {2, 3}, ..., {l - 1, l}.

(i) If one of the edges {2, 3}, {3, 4}, ..., {l - 2, l - 1} is of order $\geq 4$, this edge is of order 4 and the graph is the following:

!Graph with 4 nodes connected in a chain

(ii) If the edge {1, 2} is of order 5, the graph is one of the following:

!Graphs with 5 nodes connected in chains

We can assume that $l > 2$ (Lemma 4). Assume that $\{i, i + 1\}$ is of order $\geq 4$, with $1 \leq i \leq l - 1$. Put

By Lemma 8, $\| x \| ^2 = \frac{1}{2} i(i + 1)$, $\| y \| ^2 = \frac{1}{2} j(j + 1)$. On the other hand, $(x|y) = ij(e_i|e_{i+1}) = -ij \cos \frac{\pi}{m}$ with $m = 4$ or 5 (Lemma 4). Now

which gives

so

This gives, first of all, $ij - i - j - 1 < 0$, or $(i - 1)(j - 1) < 2$. If

then $1 < j < l - 1$, so $i = j = 2$, and so $^1$

and hence $m = 4$. This proves (i). If $i = 1$ and $m = 5$, then $2j + 2 > 4j \frac{3+\sqrt{5}}{8}$, or $j \frac{\sqrt{5}-1}{2} < 2, j < \sqrt{5} + 1 < 4$, and so $l = j + 1 \leq 4$. This proves (ii).

$^1$ The 5th roots of 1 distinct from 1 are the solutions of $z^4 + z^3 + z^2 + z + 1 = 0$. Putting $x = \frac{1}{2}(z + \frac{1}{z})$, this equation becomes $(2x)^2 - 2 + 2x + 1 = 0$, or $4x^2 + 2x - 1 = 0$, so $x = \frac{-1 \pm \sqrt{5}}{4}$. Hence,