3. Find all integer pairs $(a, b)$ such that there exists an integer $d > 1$ satisfying that for any positive integer $n$, $a^{n} + b^{n} + 1$ is a multiple of $d$.
(Chen Yonggao \quad supplied the problem)
3. When $a, b$ are one odd and one even, $a^{n}+b^{n}+1$ is always a multiple of 2, satisfying the condition.

When $a, b$ are both odd or both even, by $d \mid\left(a^{n}+b^{n}+1\right)$, we know $d$ is an odd number.
$$
\begin{array}{l}
\text { Also, } a^{2}+b^{2}+1=(a+b)^{2}-2 a b+1, \\
d \mid (a+b+1), d \mid \left(a^{2}+b^{2}+1\right),
\end{array}
$$

Thus, $d \mid \left[(-1)^{2}-2 a b+1\right]$, i.e.,
$d \mid 2(1-a b)$.
Therefore, $d \mid (a b-1)$.
By $a^{3}+b^{3}+1=(a+b)\left(a^{2}+b^{2}-a b\right)+1$ $\equiv(-1)(-1-1)+1=3(\bmod d)$, $d \mid \left(a^{3}+b^{3}+1\right)$,
Thus, $d \mid 3$.
Since $d>1$, then $d=3$.
Because $(a-b)^{2}=a^{2}+b^{2}-2 a b$
$\equiv-1-2 \equiv 0(\bmod 3)$,
So, $a \equiv b(\bmod 3)$.
By $0 \equiv a+b+1 \equiv 2 a+1(\bmod 3)$, we know $a \equiv 1(\bmod 3)$.
Thus, $a \equiv b \equiv 1(\bmod 3)$.
On the other hand, when $a \equiv b \equiv 1(\bmod 3)$, for any positive integer $n$ we have
$$
a^{n}+b^{n}+1 \equiv 1+1+1 \equiv 0(\bmod 3) .
$$

Therefore, $a^{n}+b^{n}+1$ is a multiple of 3, satisfying the condition.
In summary, the integer pairs $(a, b)$ are $(2 k, 2 l+1)$,
$$
(2 k+1,2 l),(3 k+1,3 l+1)(k, l \in \mathbf{Z}) \text {. }
$$