> k(1+(\mu+1)/s)$. It therefore suffices to show that $1+(\mu+1)/s \geqslant 0$.

Since $\lambda = k + r + s + rs$, the inequality $\lambda+1 < -k/s$ becomes $$k + (r+1)(s+1) < -k/s.$$ Since $s < -1$, it follows that $r+1 > -k/s$. Multiplying this inequality by $-s$ gives $-s(r+1) > k$. Since both $s$ and $r$ are integers, we have $-s(r+1) \geqslant k + 1$ Now by rearranging and substituting $\mu = k+rs$, we obtain the inequality $1+(\mu+1)/s \geqslant 0$ as required.

\[lem:NoOtherZeros\] Let $\Gamma$ be an edge-regular graph with parameters $(v,k,\lambda)$ such that $C_\Gamma(b,\lambda + 2)\geqslant 0$ for all integers $b$. Then $C_\Gamma(b,c)\geqslant 0$ for all $c\in \{2,\ldots,\lambda+2\}$ and all integers $b$.

Let $c\in \{2,\ldots,\lambda+2\}$ and let $b$ be an integer. If $b\leqslant 0$, then from the definition of the clique adjacency polynomial $C_\Gamma(x,y)$, we see that $C_\Gamma(b,c)\geqslant 0$, so we now assume that $b$ is positive.

A calculation (see Appendix \[sec:comp\]) shows that $$C_\Gamma(b,c)-C_\Gamma(b,\lambda+2) = (\lambda+2-c)(b-c)(b-c+1) + 2b(\lambda+2-c)(k-\lambda-1).$$ This quantity is nonnegative since $b$ and $\lambda+2-c$ are nonnegative integers, the product of two consecutive integers is nonnegative, and $k-\lambda-1$ is also nonnegative by Proposition \[pro:SRGbound\]. Hence $$C_\Gamma(b,c)\geqslant C_\Gamma(b,\lambda+2)\geqslant 0,$$ as required.

Now we prove Theorem \[thm:limitation\].

Firstly, if 