Let $p$ be some prime number. 
          a) Prove that there exist positive integers $a$ and $b$ such that $a^2 + b^2 + 2018$ is multiple of $p$. 

          b) Find all $p$ for which the $a$ and $b$ from a) can be chosen in such way that both these numbers aren’t multiples of $p$. 
### Part (a)

1. **Restate the problem in general terms:**
   We need to prove that for any prime number \( p \), there exist positive integers \( a \) and \( b \) such that \( a^2 + b^2 + 2018 \) is a multiple of \( p \).

2. **Define sets \( S_1 \) and \( S_2 \):**
   Let \( S_1 = \{ a^2 \mid 0 \leq a \leq p-1 \} \) and \( S_2 = \{ -2018 - b^2 \mid 0 \leq b \leq p-1 \} \) as subsets of \( \mathbb{Z}_p \).

3. **Cardinality of sets:**
   Since \( a^2 \) and \( b^2 \) can take on \(\frac{p+1}{2}\) distinct values in \( \mathbb{Z}_p \) (as each quadratic residue has exactly two square roots modulo \( p \)), we have \( |S_1| = |S_2| = \frac{p+1}{2} \).

4. **Intersection of sets:**
   If \( S_1 \) and \( S_2 \) were disjoint, then \( p \geq |S_1 \cup S_2| = |S_1| + |S_2| = p + 1 \), which is a contradiction. Therefore, \( S_1 \) and \( S_2 \) must have at least one common element.

5. **Existence of \( a \) and \( b \):**
   This common element implies there exist \( a \) and \( b \) such that \( a^2 \equiv -2018 - b^2 \pmod{p} \). Hence, \( a^2 + b^2 + 2018 \equiv 0 \pmod{p} \).

Thus, we have proven that there exist positive integers \( a \) and \( b \) such that \( a^2 + b^2 + 2018 \) is a multiple of \( p \).

### Part (b)

1. **Special cases:**
   - For \( p = 2 \), we can choose \( a = 1 \) and \( b = 1 \), since \( 1^2 + 1^2 + 2018 = 2020 \equiv 0 \pmod{2} \).
   - For \( p = 1009 \), since \( 1009 \equiv 1 \pmod{4} \), \( -1 \) is a quadratic residue modulo \( 1009 \). Thus, we can find \( b \) such that \( b^2 \equiv -1 \pmod{1009} \) and choose \( a = 1 \).

2. **General case analysis:**
   - We need to check when \( -2018 \) is a quadratic residue modulo \( p \). If \( -2018 \) is not a quadratic residue modulo \( p \), then there exist \( a \) and \( b \) such that \( p \nmid a \) and \( p \nmid b \) and \( p \mid a^2 + b^2 + 2018 \).

3. **Case when \( -2018 \) is a quadratic residue modulo \( p \):**
   - If \( -2018 \) is a quadratic residue modulo \( p \), then there exists \( n_0 \) such that \( n_0^2 \equiv -2018 \pmod{p} \). We can use the Pythagorean triple \( (3, 4, 5) \) to construct \( a \) and \( b \) such that \( a \equiv 3 \cdot 5^{-1} \cdot n_0 \pmod{p} \) and \( b \equiv 4 \cdot 5^{-1} \cdot n_0 \pmod{p} \).

4. **Exclusion of specific primes:**
   - For \( p = 3 \), we must have either \( a \) or \( b \) divisible by \( 3 \). For \( p = 5 \), we can choose \( a \equiv 1 \pmod{5} \) and \( b \equiv 1 \pmod{5} \).

5. **Conclusion:**
   - Except for \( p = 3 \), the job can be done for every prime \( p \).

The final answer is \( \boxed{ p \neq 3 } \).