Determine all integers $a$ and $b$ such that
\[(19a + b)^{18} + (a + b)^{18} + (a + 19b)^{18}\]
is a perfect square.
1. **Reduction to Modulo 19:**
   We need to determine all integers \(a\) and \(b\) such that 
   \[
   (19a + b)^{18} + (a + b)^{18} + (a + 19b)^{18}
   \]
   is a perfect square. To simplify the problem, we will consider the expression modulo 19.

2. **Using Fermat's Little Theorem:**
   Fermat's Little Theorem states that for any integer \(x\) and a prime \(p\), \(x^{p-1} \equiv 1 \pmod{p}\) if \(x\) is not divisible by \(p\). Here, \(p = 19\), so for any integer \(x\) not divisible by 19, we have:
   \[
   x^{18} \equiv 1 \pmod{19}
   \]

3. **Applying Fermat's Little Theorem:**
   We apply this theorem to each term in the expression:
   \[
   (19a + b)^{18} \equiv b^{18} \equiv 1 \pmod{19} \quad \text{(since \(19a \equiv 0 \pmod{19}\))}
   \]
   \[
   (a + b)^{18} \equiv (a + b)^{18} \equiv 1 \pmod{19}
   \]
   \[
   (a + 19b)^{18} \equiv a^{18} \equiv 1 \pmod{19} \quad \text{(since \(19b \equiv 0 \pmod{19}\))}
   \]

4. **Summing the Terms:**
   Summing these congruences, we get:
   \[
   (19a + b)^{18} + (a + b)^{18} + (a + 19b)^{18} \equiv 1 + 1 + 1 \equiv 3 \pmod{19}
   \]

5. **Checking for Perfect Square:**
   We need to check if 3 modulo 19 can be a perfect square. We need to find an integer \(x\) such that:
   \[
   x^2 \equiv 3 \pmod{19}
   \]
   By checking all possible values of \(x\) from 0 to 18, we find that there is no such \(x\) that satisfies this congruence. Therefore, 3 is not a quadratic residue modulo 19.

6. **Special Cases:**
   We need to check the cases where either \(a = 0\) or \(b = 0\):
   - If \(a = 0\), then the expression becomes:
     \[
     (19 \cdot 0 + b)^{18} + (0 + b)^{18} + (0 + 19b)^{18} = b^{18} + b^{18} + (19b)^{18} = 2b^{18} + (19b)^{18}
     \]
     Since \(19b \equiv 0 \pmod{19}\), this simplifies to:
     \[
     2b^{18} \equiv 2 \pmod{19}
     \]
     This is not a perfect square modulo 19.

   - If \(b = 0\), then the expression becomes:
     \[
     (19a + 0)^{18} + (a + 0)^{18} + (a + 19 \cdot 0)^{18} = (19a)^{18} + a^{18} + a^{18} = (19a)^{18} + 2a^{18}
     \]
     Since \(19a \equiv 0 \pmod{19}\), this simplifies to:
     \[
     2a^{18} \equiv 2 \pmod{19}
     \]
     This is not a perfect square modulo 19.

7. **Conclusion:**
   Since neither \(a = 0\) nor \(b = 0\) yields a perfect square, and the general case modulo 19 does not yield a perfect square, the only solution is \(a = 0\) and \(b = 0\).

\(\blacksquare\)

The final answer is \( \boxed{ (a, b) = (0, 0) } \)