Only five feasible degrees: $k \in \{2, 4, 14, 22, 112, 994\}$

Let $g$ be a strongly regular graph with parameters $(n, k, 1, 2)$ on a finite vertex set with at least two vertices. Then

Together with the counting identity $2n = k^2 + 2$ these degrees correspond to the parameter tuples $(3,2,1,2)$ (the triangle $K_3$, where the condition on non-adjacent pairs is vacuous), $(9,4,1,2)$, $(99,14,1,2)$, $(243,22,1,2)$, $(6273,112,1,2)$ and $(494019,994,1,2)$. Beyond the complete case, the restriction comes from the integrality conditions: the adjacency matrix has eigenvalues $k$ and the two roots of $x^2 + x - (k-2) = 0$, so $4k - 7$ must be a perfect square $t^2$, and the multiplicities of the two non-principal eigenvalues are integers only when $t$ divides $63$. This is the classical argument showing that the family $\lambda = 1$, $\mu = 2$ contains only five non-degenerate feasible parameter tuples, two of which are realised and three of which — including $(99,14,1,2)$ — are open.