Fall Mathematical Competition

It is obvious that $p=2$ and then $r-2=q$. Hence $q(q-2)-54=u^2$ which can be written as $(q-1{)}^2-u^2=55⟺(q-1-u)(q-1+u)=55=1\cdot 55=5\cdot 11$. Hence we have the following two cases:

Case 1. $q-1-u=1$ and $q-1+u=55$. Then $q=u+2$, whence $q=29$ and $r=29+2=31$.

Case 2. $q-1-u=5$ and $q-1+u=11$. Then $q=u+6$, whence $q=9$, a contradiction.