Using quadratic residue we have that $$\alpha=\frac{1\pm \sqrt{-11}}{2}$$$$\beta=\frac{1\pm \sqrt{-99}}{2}=\frac{1\pm 3\sqrt{-11}}{2}$$Since $p$ must be an odd prime, $\frac{1}{2}$ has inverse mudulo $p$. Supose that there exists $\alpha$ then $\sqrt{-11}$ exists modulo $p$. Hence $\beta$ also exists, the case $p=3$ can be hand checked. The reciprocal is the same.

ESCHen99 wrote: Given a positive prime number $p$. Prove that there exist a positive integer $\alpha$ such that $p|\alpha(\alpha-1)+3$, if and only if there exist a positive integer $\beta$ such that $p|\beta(\beta-1)+25$. It's followed from $(\frac{-11}{p}) =(\frac{-99}{p})$