$(n(x!-xy-x-y+2)+2,n(x!-xy-x-y+3)+3)>1$ $\Leftrightarrow$ $n(x!-xy-x-y+2)+2,n+1)>1$. Choosing $n=p-1$ where $p \in \mathbb{P}$ imposes that $p|(p-1)(x!-xy-x-y+2)+2$ $\Leftrightarrow$ $p|x!-xy-x-y+2-2$ $\Leftrightarrow$ $p|x!-xy-x-y$. Hence, choosing $p>|x!-xy-x-y|$ imposes $x!-xy-x-y=0$ $\Leftrightarrow$ $x!-x=y(x+1)$. This implies that $x+1|x!+1$ which yields $x+1=q>2$ where $q \in \mathbb{P}$ and that $y=\dfrac{(q-1)!-(q-1)}{q}$. This way the set $(x,y)= (q-1,\dfrac{(q-1)!-(q-1)}{q})$, $q>3 \in \mathbb{P}$ is obtained and it remains to check that the initial condition is satisfied. It is though, since $(2n+2,3n+3)=n+1>1$. $\blacksquare$

puzld wrote: $(n(x!-xy-x-y+2)+2,n(x!-xy-x-y+3)+3)>1$ $\Leftrightarrow$ $n(x!-xy-x-y+2)+2,n+1)>1$. Choosing $n=p-1$ where $p \in \mathbb{P}$ imposes that $p|(p-1)(x!-xy-x-y+2)+2$ $\Leftrightarrow$ $p|x!-xy-x-y+2-2$ $\Leftrightarrow$ $p|x!-xy-x-y$. Hence, choosing $p>|x!-xy-x-y|$ imposes $x!-xy-x-y=0$ $\Leftrightarrow$ $x!-x=y(x+1)$. This implies that $x+1|x!+1$ which yields $x+1=q>2$ where $q \in \mathbb{P}$ and that $y=\dfrac{(q-1)!-(q-1)}{q}$. This way the set $(x,y)= (q-1,\dfrac{(q-1)!-(q-1)}{q})$, $q>3 \in \mathbb{P}$ is obtained and it remains to check that the initial condition is satisfied. It is though, since $(2n+2,3n+3)=n+1>1$. $\blacksquare$ It wrong my friend.