Find all pairs $(a,b)$ of integers satisfying: there exists an integer $d \ge 2$ such that $a^n + b^n +1$ is divisible by $d$ for all positive integers $n$.
We need to find all pairs \((a, b)\) of integers such that there exists an integer \(d \ge 2\) for which \(d\) divides \(a^n + b^n + 1\) for all positive integers \(n\).

We will consider three cases based on the divisibility properties of \(a\) and \(b\) with respect to \(d\).

### Case 1: Both \(a\) and \(b\) are divisible by \(d\)
Let \(a = da_1\) and \(b = db_1\), where \(a_1\) and \(b_1\) are integers. Then:
\[
a^n + b^n + 1 = d^n a_1^n + d^n b_1^n + 1 = d^n (a_1^n + b_1^n) + 1
\]
For \(d\) to divide \(a^n + b^n + 1\), we must have:
\[
d^n (a_1^n + b_1^n) + 1 \equiv 0 \pmod{d} \implies 1 \equiv 0 \pmod{d}
\]
This is a contradiction since \(d \ge 2\). Therefore, this case is not possible.

### Case 2: One of \(a\) or \(b\) is coprime with \(d\)
Without loss of generality, assume \((b, d) = 1\) and \(a = da_1\). Then:
\[
a^n + b^n + 1 = d^n a_1^n + b^n + 1
\]
For \(d\) to divide \(a^n + b^n + 1\), we must have:
\[
d^n a_1^n + b^n + 1 \equiv 0 \pmod{d} \implies b^n + 1 \equiv 0 \pmod{d} \implies b^n \equiv -1 \pmod{d}
\]
Since this must hold for all \(n\), we have:
\[
b \equiv -1 \pmod{d} \implies b^2 \equiv 1 \pmod{d}
\]
Thus:
\[
b(b - 1) \equiv 0 \pmod{d} \implies b \equiv 1 \pmod{d} \text{ or } b \equiv -1 \pmod{d}
\]
Since \(b \equiv -1 \pmod{d}\) and \(b \equiv 1 \pmod{d}\), we get:
\[
d \mid 2 \implies d = 2
\]

### Case 3: Both \(a\) and \(b\) are coprime with \(d\)
Since \(a\) and \(b\) are coprime with \(d\), we have:
\[
a^{3n} + b^{3n} \equiv -1 \pmod{d}
\]
This implies:
\[
(a^n + b^n)(a^{2n} - a^n b^n + b^{2n}) \equiv -1 \pmod{d}
\]
Since \(a^n + b^n \equiv -1 \pmod{d}\), we have:
\[
a^{2n} - a^n b^n + b^{2n} \equiv 1 \pmod{d}
\]
Thus:
\[
a^n b^n \equiv -2 \pmod{d}
\]
So:
\[
ab \equiv -2 \pmod{d} \implies (ab)^2 \equiv 4 \pmod{d}
\]
Since \(ab \equiv -2 \pmod{d}\) and \(ab \equiv 1 \pmod{d}\), we get:
\[
d \mid 3 \implies d = 3
\]

### Verification:
For \(d = 2\):
- Let \(a \equiv 0 \pmod{2}\) and \(b \equiv 1 \pmod{2}\). Then:
\[
a^n + b^n + 1 \equiv 0 + 1 + 1 \equiv 0 \pmod{2}
\]
This satisfies the condition.

For \(d = 3\):
- Consider the cases:
  1. \((a, b) \equiv (-1, -1) \pmod{3}\): This is a contradiction for \(n = 1\).
  2. \((a, b) \equiv (1, -1) \pmod{3}\): This is a contradiction for \(n = 1\).
  3. \((a, b) \equiv (1, 1) \pmod{3}\): Then:
\[
a^n \equiv 1 \pmod{3} \text{ and } b^n \equiv 1 \pmod{3}
\]
Thus:
\[
a^n + b^n + 1 \equiv 1 + 1 + 1 \equiv 0 \pmod{3}
\]
This satisfies the condition.

Therefore, the possible pairs \((a, b)\) are:
- For \(d = 2\): \((a, b) \equiv (0, 1) \pmod{2}\)
- For \(d = 3\): \((a, b) \equiv (1, 1) \pmod{3}\)

The final answer is \( \boxed{ (a, b) \equiv (0, 1) \pmod{2} } \) and \((a, b) \equiv (1, 1) \pmod{3}\)