Determine all pairs $(a, b)$ of integers having the following property: 
there is an integer $d \ge  2$ such that $a^n + b^n + 1$ is divisible by $d$ for all positive integers $n$.
To determine all pairs \((a, b)\) of integers such that there exists an integer \(d \ge 2\) for which \(a^n + b^n + 1\) is divisible by \(d\) for all positive integers \(n\), we proceed as follows:

1. **Initial Consideration:**
   Let \(d\) be a divisor of \(a^n + b^n + 1\) for all positive integers \(n\). This implies that \(a^n + b^n + 1 \equiv 0 \pmod{d}\) for all \(n\).

2. **Using Euler's Totient Function:**
   Consider \(n = \phi(d)\), where \(\phi\) is Euler's totient function. Since \(\phi(d)\) is the number of integers less than \(d\) that are coprime to \(d\), we have:
   \[
   a^{\phi(d)} \equiv 1 \pmod{d} \quad \text{and} \quad b^{\phi(d)} \equiv 1 \pmod{d}
   \]
   if \(\gcd(a, d) = 1\) and \(\gcd(b, d) = 1\). Therefore,
   \[
   a^{\phi(d)} + b^{\phi(d)} + 1 \equiv 1 + 1 + 1 \equiv 3 \pmod{d}
   \]
   This implies \(d\) must divide 3, so \(d\) can be 1 or 3. However, since \(d \ge 2\), we have \(d = 3\).

3. **Considering \(d = 3\):**
   If \(d = 3\), then:
   \[
   a^n + b^n + 1 \equiv 0 \pmod{3}
   \]
   This implies:
   \[
   a^n + b^n \equiv -1 \pmod{3}
   \]
   For this to hold for all \(n\), \(a\) and \(b\) must satisfy:
   \[
   a \equiv 1 \pmod{3} \quad \text{and} \quad b \equiv 1 \pmod{3}
   \]
   or
   \[
   a \equiv 2 \pmod{3} \quad \text{and} \quad b \equiv 2 \pmod{3}
   \]
   because \(1^n \equiv 1 \pmod{3}\) and \(2^n \equiv (-1)^n \pmod{3}\).

4. **Considering \(d = 2\):**
   If \(d = 2\), then:
   \[
   a^n + b^n + 1 \equiv 0 \pmod{2}
   \]
   This implies:
   \[
   a^n + b^n \equiv -1 \equiv 1 \pmod{2}
   \]
   For this to hold for all \(n\), \(a\) and \(b\) must have opposite parities (one even and one odd).

5. **Considering \(d = 6\):**
   If \(d = 6\), then \(d\) must be divisible by both 2 and 3. Therefore, \(a\) and \(b\) must satisfy the conditions for both \(d = 2\) and \(d = 3\):
   \[
   a \equiv 1 \pmod{6} \quad \text{and} \quad b \equiv 4 \pmod{6}
   \]
   or
   \[
   a \equiv 4 \pmod{6} \quad \text{and} \quad b \equiv 1 \pmod{6}
   \]

Conclusion:
The pairs \((a, b)\) of integers having the property that there is an integer \(d \ge 2\) such that \(a^n + b^n + 1\) is divisible by \(d\) for all positive integers \(n\) are:
- For \(d = 2\): \(a\) is even and \(b\) is odd, or vice versa.
- For \(d = 3\): \(a \equiv b \equiv 1 \pmod{3}\) or \(a \equiv b \equiv 2 \pmod{3}\).
- For \(d = 6\): \(a \equiv 1 \pmod{6}\) and \(b \equiv 4 \pmod{6}\), or \(a \equiv 4 \pmod{6}\) and \(b \equiv 1 \pmod{6}\).

The final answer is \( \boxed{ (a, b) } \) pairs as described above.