The given equation modulo 3 becomes: $1,2,0 - 1,2,0 = 1,2 + 2pq$ where $x,y$ means $x$ or $y$. Thus $2pq = 1,2,0$ $mod$ $3$ Case 1: $2pq = 0$ $mod$ $3$ either $p = 0$ $mod$ $3$ hence $p=3$ or $q = 0$ $mod$ $3$ hence $q=3$ if $p=3$ then $18 = q(q^4 + q + 6)$ $q=3,2$ and both cases lead to contradictions. otherwise if $q=3$ then $p(p^2 - p -6) = 252 = 2^2 * 7 * 3^2 $ so $p=2,3,7$ which gives only one pair as a solution $(p,q) = (7,3)$ Case 2: $2pq = 1$ $mod$ $3$ $\iff$ $pq = 2$ $mod$ $3$ either $p = 1$ $mod$ $3$ and $q = 2$ $mod$ $3$ or $q = 1$ $mod$ $3$ and $p = 2$ $mod$ $3$ Trying both of these pairs in the original equation gives contradictions. Case 3: $2pq = 2$ $mod$ $3$ $\iff$ $pq = 1$ $mod$ $3$ either $p = q =1$ $mod$ $3$ or $p = q = 2$ $mod$ $3$ both cases tried in the original equation also lead to contradictions. Hence, $(p,q) = (7,3)$ is the only solution.