Let $\mathcal D$ be a symmetric $2$-$(v,k,\lambda)$ design, and put $n=k-\lambda$. If $v$ is even, then $n$ is a square. If $v$ is odd, then the Diophantine equation
\begin{align*}
z^2=n x^2+(-1)^{(v-1)/2}\lambda y^2
\end{align*}
has a nonzero integer solution $(x,y,z)\in\mathbb Z^3$.