(a) Determine a natural number $n$ such that $n(n+2022)+2$ is a perfect square.
[hide=Spoiler]In case you didn't realize, $n=1$ works lol[/hide]
(b) Determine all natural numbers $a$ such that for every natural number $n$, the number $n(n+a)+2$ is never a perfect square.
### Part (a)
1. We need to find a natural number \( n \) such that \( n(n+2022) + 2 \) is a perfect square.
2. Let \( n(n+2022) + 2 = k^2 \) for some integer \( k \).
3. Rearrange the equation:
   \[
   n^2 + 2022n + 2 = k^2
   \]
4. To find a solution, we can try \( n = 1 \):
   \[
   1(1 + 2022) + 2 = 2023 + 2 = 2025
   \]
5. Check if 2025 is a perfect square:
   \[
   2025 = 45^2
   \]
6. Since 2025 is indeed a perfect square, \( n = 1 \) works.

### Part (b)
1. We need to determine all natural numbers \( a \) such that for every natural number \( n \), the number \( n(n+a) + 2 \) is never a perfect square.
2. Assume \( n(n+a) + 2 = k^2 \) for some integer \( k \).
3. Rearrange the equation:
   \[
   n^2 + an + 2 = k^2
   \]
4. Consider the discriminant of the quadratic equation \( n^2 + an + (2 - k^2) = 0 \):
   \[
   \Delta = a^2 - 4(2 - k^2) = a^2 + 4k^2 - 8
   \]
5. For \( n \) to be an integer, \( \Delta \) must be a perfect square. Let \( \Delta = m^2 \) for some integer \( m \):
   \[
   a^2 + 4k^2 - 8 = m^2
   \]
6. Rearrange to find:
   \[
   a^2 - m^2 = 8 - 4k^2
   \]
7. Factorize the left-hand side:
   \[
   (a - m)(a + m) = 8 - 4k^2
   \]
8. Since \( 8 - 4k^2 \) must be a product of two integers, we need to check if there are any \( k \) such that \( 8 - 4k^2 \) is a product of two integers. Notice that \( 8 - 4k^2 \) must be non-negative:
   \[
   8 \geq 4k^2 \implies 2 \geq k^2 \implies k = 0 \text{ or } k = 1
   \]
9. For \( k = 0 \):
   \[
   8 - 4(0)^2 = 8
   \]
   \[
   (a - m)(a + m) = 8
   \]
   Possible pairs \((a - m, a + m)\) are \((1, 8)\), \((2, 4)\), \((4, 2)\), and \((8, 1)\). Solving these pairs, we get:
   \[
   a = \frac{1 + 8}{2} = 4.5 \quad (\text{not an integer})
   \]
   \[
   a = \frac{2 + 4}{2} = 3
   \]
   \[
   a = \frac{4 + 2}{2} = 3
   \]
   \[
   a = \frac{8 + 1}{2} = 4.5 \quad (\text{not an integer})
   \]
10. For \( k = 1 \):
    \[
    8 - 4(1)^2 = 4
    \]
    \[
    (a - m)(a + m) = 4
    \]
    Possible pairs \((a - m, a + m)\) are \((1, 4)\) and \((2, 2)\). Solving these pairs, we get:
    \[
    a = \frac{1 + 4}{2} = 2.5 \quad (\text{not an integer})
    \]
    \[
    a = \frac{2 + 2}{2} = 2
    \]

11. Therefore, the only natural number \( a \) such that \( n(n+a) + 2 \) is never a perfect square for any natural number \( n \) is \( a = 2 \).

The final answer is \( \boxed{ a = 2 } \)