If $ p=2 $ then $ p+2=4 $ and $ p^{2}+2p-8=0 $. So $ p=2 $ is not a solution.

If $ p=3 $ then $ p+2=2 $ and $ p^{2}+2p-8=0 $. So $ p=3 $ is a solution.

Now, assume $ p> 3 $, so $ p $ is odd.

We consider two cases. Now you've to solve these two cases by using modular arithmetic