I am asked to show that if $G$ is an incomplete, strongly $(k,0,3)$ regular graph, then $|G| = n \in \{6,162\}$
Now I know the rationality condition here tells us that: $\frac{1}{2}(n-1 \pm \frac{3(n-1)-2k}{\sqrt{4k-3}}) \in \mathbb Z$
Given this we have a few cases to check:
Case 1: $3(n-1) = 2k \Rightarrow k = 3l, \; l \in \mathbb N$
$\Rightarrow n = 2l+ 1 \Rightarrow \frac{1}{2}(n-1) \notin \mathbb Z$
Case 2: $4k-3 = u^2, \; u \in \mathbb N$ and $u \mid 2k \Rightarrow u \mid 3(n-1)$
$4k-3 = u^2 \Rightarrow u \mid u^2 + 3 \Rightarrow u \in \{1,3\}$
$u = 1 \Rightarrow k = 1 \Rightarrow $ Rationality condition holds trivially
$u = 3 \Rightarrow k = 3 \Rightarrow $ Rationality condition holds trivially
Case 3: $u \nmid 2k \Rightarrow u \notin \{1,3\}$ but $u \mid 3(n-1) - 2k \Rightarrow u \mid 3(n-1) - \frac{u^2 + 3}{2} \Rightarrow u \mid 6(n-1) -3 $
But $u \neq 3 \Rightarrow u \mid 2(n-1) - 1$
But now at this point I can't really show that $n \in \{1,3\}$ and additionally I can't make sense of my result in case 2. It seems that for those values of $k$, then the rationality condition is just always satisfied. Did I make a mistake there?
Any help would be greatly appreciated, thank you!