Find all pairs $(x,y)$, $x,y\in \mathbb{N}$ for which 
$gcd(n(x!-xy-x-y+2)+2,n(x!-xy-x-y+3)+3)>1$
for $\forall$ $n\in \mathbb{N}$.
1. We start with the given condition:
   \[
   \gcd(n(x! - xy - x - y + 2) + 2, n(x! - xy - x - y + 3) + 3) > 1 \quad \forall \, n \in \mathbb{N}
   \]

2. Let us denote:
   \[
   a = n(x! - xy - x - y + 2) + 2
   \]
   \[
   b = n(x! - xy - x - y + 3) + 3
   \]

3. We need to find the greatest common divisor (gcd) of \(a\) and \(b\):
   \[
   \gcd(a, b) = \gcd(n(x! - xy - x - y + 2) + 2, n(x! - xy - x - y + 3) + 3)
   \]

4. Notice that:
   \[
   b - a = (n(x! - xy - x - y + 3) + 3) - (n(x! - xy - x - y + 2) + 2) = 1
   \]

5. Therefore, we have:
   \[
   \gcd(a, b) = \gcd(a, a + 1) = \gcd(a, 1)
   \]

6. For the gcd to be greater than 1, \(a\) must be such that it shares a common factor with 1, which is impossible unless \(a\) itself is 1. Hence, we need to find conditions under which \(a\) and \(b\) are not coprime.

7. Choose \(n = p - 1\) where \(p \in \mathbb{P}\) (prime numbers). This imposes that:
   \[
   p \mid (p-1)(x! - xy - x - y + 2) + 2
   \]

8. Simplifying, we get:
   \[
   p \mid x! - xy - x - y
   \]

9. Choosing \(p > |x! - xy - x - y|\) implies:
   \[
   x! - xy - x - y = 0 \quad \Rightarrow \quad x! - x = y(x + 1)
   \]

10. This implies that \(x + 1 \mid x! + 1\). Let \(x + 1 = q\) where \(q \in \mathbb{P}\) and \(q > 2\).

11. Therefore, we have:
    \[
    y = \frac{(q-1)! - (q-1)}{q}
    \]

12. Thus, the pairs \((x, y)\) are given by:
    \[
    (x, y) = (q-1, \frac{(q-1)! - (q-1)}{q})
    \]
    where \(q > 3\) and \(q \in \mathbb{P}\).

13. Finally, we need to check that the initial condition is satisfied. We have:
    \[
    \gcd(2n + 2, 3n + 3) = n + 1 > 1
    \]

Thus, the solution is verified.

The final answer is \( \boxed{ (x, y) = (q-1, \frac{(q-1)! - (q-1)}{q}) } \) where \(q > 3\) and \(q \in \mathbb{P}\).