3. Given $P(n)=n^{2}+n+1$. For any positive integers $a, b$, if the set
$$
\{P(a), P(a+1), P(a+2), \cdots, P(a+b)\}
$$

has each element not coprime with the product of the other elements, then this set is called "fragrant". Find the minimum number of elements in a fragrant set.
$$
\begin{array}{l}
=\left(n^{2}+n+1, n^{2}+3 n+3\right) \\
=\left(n^{2}+n+1,2 n+2\right) \\
=\left(n^{2}+n+1, n+1\right) \\
=(1, n+1)=1 .
\end{array}
$$

Lemma 2 For $n \neq 2(\bmod 7)$, we have
$$
(P(n), P(n+2))=1 \text {; }
$$

For $n \equiv 2(\bmod 7)$, we have
$$
(P(n), P(n+2))=7 \text {. }
$$

Proof of Lemma 2 Since
$$
(2 n+7) P(n)-(2 n-1) P(n+2)=14 \text {, }
$$

and $P(n)$ is odd, then $(P(n), P(n+2))$ must be a divisor of 7.

For $n \equiv 0,1, \cdots, 6(\bmod 7)$, direct verification yields the conclusion.

Lemma 3 For $n \neq 1(\bmod 3)$, we have
$(P(n), P(n+3))=1$;
For $n \equiv 1(\bmod 3)$, we have
$$
3 \text { divides }(P(n), P(n+3)) \text {. }
$$

Proof of Lemma 3 Since
$$
(n+5) P(n)-(n-1) P(n+3)=18 \text {, }
$$

and $P(n)$ is odd, then $(P(n), P(n+3))$ must be a divisor of 9.

For $n \equiv 0,1,2(\bmod 3)$, direct verification yields the conclusion.

Lemmas 1 to 3 are proved.
Assume there exists an aromatic set with at most five elements. We can assume it exactly contains the elements $P(a), P(a+1)$, $\cdots, P(a+4)$. This is because the following proof still holds for fewer than five elements.

By Lemma 1, we know $P(a+2)$ is coprime with both $P(a+1)$ and $P(a+3)$.
Without loss of generality, assume $(P(a), P(a+2))>1$.
By Lemma 2, we know $a \equiv 2(\bmod 7)$.
Since $a+1 \equiv 3(\bmod 7)$, by Lemma 2, we know $(P(a+1), P(a+3))=1$.
Since the set is aromatic, $(P(a), P(a+3))$ and $(P(a+1), P(a+4))$ are both greater than 1.
By Lemma 3, we know
$a \equiv 1(\bmod 3)$ and $a+1 \equiv 1(\bmod 3)$.
Contradiction.
Below, we construct a set with six elements that is aromatic.
By the Chinese Remainder Theorem, there exists a positive integer $a$ such that $a \equiv 7(\bmod 19)$, $a+1 \equiv 2(\bmod 7)$, and $a+2 \equiv 1(\bmod 3)$.
For example, $a=197$.
By Lemma 2, we know $P(a+1)$ and $P(a+3)$ are both divisible by 7. By Lemma 3, we know $P(a+2)$ and $P(a+5)$ are both divisible by 3.

Since $19|P(7)=57$ and $19| P(11)=133$, $P(a)$ and $P(a+4)$ are both divisible by 19.

This shows that $\{P(a), P(a+1), \cdots, P(a+5)\}$ is an aromatic set.
Therefore, the minimum number of elements in an aromatic set is 6.