$ 5^{2p}\equiv 1\mod q$ and $ 5^{2q}\equiv 1\mod p$ Now, $ ord_{q}(5)|2p$, so $ ord_{q}(5)=1$, $ 2$, $ p$, or $ 2p$. If $ ord_{q}(5)=1$ or $ 2$ then either $ 5\equiv 1\mod q$ or $ 5^2\equiv 1\mod q$. Thus $ q=2$ or $ q=3$. If $ q=2$, then $ p|5^2+1=26$, so $ p=2$ or $ p=13$. We can check that $ (2,2)$ and $ (13,2)$ both work. If $ q=3$, then $ p|5^3+1=126$, so $ p=2$, $ p=3$, or $ p=7$. Again, we see that $ (3,3)$, and $ (7,3)$ are all solutions, but $ (2,3)$ is not. We now have solutions $ (2,2),(13,2),(3,3),(7,3)$. By considering $ ord_{p}(5)=1$ or $ 2$ then we get the reversed solutions $ (2,13)$ and $ (3,7)$. Now suppose that $ ord_{q}(5)=p$ or $ 2p$ and $ ord_{p}(5)=q$ or $ 2q$. Then $ p|q-1$ and $ q|p-1$. But $ q>p$ follows from $ p|q-1$ and $ p>q$ follows from $ q|p-1$, a contradiction. Thus the only solutions are $ (2,2),(3,3),(13,2),(7,3),(2,13),(3,7)$.