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\begin{remark*}[Silly remark]
  If $G$ is a \gls{complete_graph}, then $b$ is
  not well-defined -- $\complete_t$ is $\strp(t -
  1, t - 2, b)$ is \gls{streg} for any $b$. Also,
  if $b = 0$, then $G$ need not be
  \gls{connected} -- for example two copies of
  $\complete_t$ is $\strp(t - 1, t - 2, 0)$.
\end{remark*}
\vspace{-1em}

Being \gls{streg} is actually \emph{extremely}
restrictive:

\begin{flashcard}[rationality-criterion-for-streg-graphs]
\begin{theorem}[Rationality criterion for
  \gls{streg} \glspl{graph}]
  % Theorem 3
  \label{streg_rationality_crit}
  Let $G$ be a \gls{streg} \gls{graph} on $n$
  \glspl{vertex}, with parameters $\strp(k, a,
  b)$. ($G$ not \glsref[complete_graph]{complete},
  $b \ge 1$). Then the numbers $\half \left( k - 1
  \pm \frac{(n - 1)(b - a) - 2k}{\sqrt{(a - b)^2 +
  4(k - b)}} \right)$ are integers.
\end{theorem}

\fcscrap{
\begin{note*}
  Denominator $\sqrt{(a - b)^2 + 4(k - b)} \neq 0$
  -- else $a = b$, $b = k$, contradicting $a \le k
  - 1$.
\end{note*}
}

\begin{proof}
  \cloze{
  Have $G$ \gls{connected} (since $b \ge 1$). We
  also have that $G$ is \kreg{k}, so $k$ is an
  \gls{eval} with multiplicity $1$ and \gls{evec}
  $(1, 1, \ldots, 1)$.

  Consider matrix $A^2$. We know
  \[ (A^2)_{ij} = \#\text{\glspl{walk} of length $2$ from
  $i$ to $j$} \]
  so
  \[ (A^2)_{ij} = \begin{cases}
    k & \text{if $i = j$} \\
    a & \text{if $i, j$ adjacent} \\
    b & \text{if $i, j$ non-adjacent}
  \end{cases} \]
  and thus $A^2 = kI + aA + b(J - I - A)$, where
  $J$ is the matrix with all entries being $1$. So
  \[ A^2 + (b - a)A + (bI - bJ - kI) = 0 ,\]
  which is not quite quadratic, as there exists a
  $J$-term. So for \gls{evec} $x$, \gls{eval}
  $\lambda \neq k$: have $\langle x, (1, \ldots,
  1)\rangle = 0$, so $Jx = 0$. Hence
  \begin{align*}
    0x
    &= A^2 x + (b - a) Ax + (b - k) Ix \\
    &= \lambda^2 x + (b - a)\lambda x + (b - k)x
    \\
    &= (\lambda^2 + (b - a)\lambda + (b - k)) x
  \end{align*}
  Thus $\lambda^2 + (b - a)\lambda + (b - k) = 0$.
  Hence the \glspl{eval} $\neq k$ are $\lambda, \mu$ given
  by $\half (a - b \pm \sqrt{(b - a)^2 + 4(k -
  b)})$. Let their multiplicities be $r, s$
  respectively. Then $r + s =  - 1$ (as $A$ is
  diagonalisable), and $r \lambda + s\mu + k = 0$
  (as $\Trace A = 0$). Solving for $r, s$ gives
  the numbers in the theorem.
  }
\end{proof}
\end{flashcard}

\subsubsection*{Back to Moore Graphs}

\begin{flashcard}[moore-graph-values]
\begin{theorem}
  % Theorem 4
  Let $G$ be a \gls{moore_graph} of degree $k$.
  Then $k \in \{2, 3, 7, 57\}$.
\end{theorem}

\begin{proof}
  \cloze{
  Have $G$ \gls{streg}, parameters $\strp(k, a,
  b)$. Hence by \cref{streg_rationality_crit},
  either the number $(n - 1)(b - a) - 2k = k^2 -
  2k$ is $0$ or $\sqrt{(a - b)^2 + 4(k - b)}
  = \sqrt{4k - 3}$ is an integer.

  If $k^2 - 2k = 0$: then we have $k = 2$.

  If not: write $t = \sqrt{4k - 3}$. $t$ divides
  $k^2 - 2k = \left( \frac{t^2 + 3}{4} \right)^2 -
  2 \left( \frac{t^2 + 3}{4} \right)$. So $t$
  divides
  \[ (t^2 + 3)^2 - 8(t^2 + 3) = t^4 - 2t^2 - 15 .\]
  Hence $t$ divides $15$, so $t = 1, 3, 5$ or
  $15$. These give $k = \frac{t^2 + 3}{4}$ gives
  $1$ (not allowed) or $3$ or $7$ or $57$.
  }
\end{proof}
\end{flashcard}

Which of these can occur?

We saw earlier that for $k = 2$, $C_5$ works, and
for $k = 3$ the Petersen graph works.
It helps to think of the drawing of the Petersen
graph drawn like this:
\begin{center}
  \begin{tsqx}
    ! size(3cm);
     A .= dir 126
     B .= dir 54
     C .= dir -18
     D .= dir -90
     E .= dir -162
     A--B--C--D--E--A
     label 0 @ 1.1*A
     label 1 @ 1.1*B
     label 2 @ 1.1*C
     label 3 @ 1.1*D
     label 4 @ 1.1*E
  \end{tsqx}
  \begin{tsqx}
    ! size(3cm);
     A .= dir 126
     B .= dir 54
     C .= dir -18
     D .= dir -90
     E .= dir -162
     A--C--E--B--D--A
     label 0 @ 1.1*A
     label 1 @ 1.1*B
     label 2 @ 1.1*C
     label 3 @ 1.1*D
     label 4 @ 1.1*E
  \end{tsqx}
\end{center}
where we join $t$ in the pentagon to $t$ in the
pentagram.

Now for $k = 7$: The \emph{Hoffman-Singleton
graph}: \kreg{7}, $50$ \glspl{vertex}, \gls{diam}
$2$.
Take $5$ pentagons $P_0, \ldots, P_4$ and $5$
pentagrams $Q_0, \ldots, Q_4$, and join
\gls{vertex} $t$ in $P_i$ to vertex $t + ij$ in
$Q_j$. This works!

$k = 57$: \gls{graph}, \kreg{57}, on $3250$
\glspl{vertex}, \gls{diam} $2$. It is unknown
whether this graph exists. Sometimes called the
``missing \gls{moore_graph}''. If it exists, turns
out that it cannot be transitive (meaning the
automorphism group is transitive, or more
intuitively ``all vertices look the same'').
