, 2016

We have the following observations.
(i) $(P(n), P(n+1)) = 1$ for any $n$.
We have $(P(n), P(n+1)) = \left(n^{2} + n + 1, n^{2} + 3n + 3\right) = \left(n^{2} + n + 1, 2n + 2\right)$. Noting that $n^{2} + n + 1$ is odd and $\left(n^{2} + n + 1, n + 1\right) = (1, n + 1) = 1$, the claim follows.

(ii) $(P(n), P(n+2)) = 1$ for $n \not\equiv 2 \pmod{7}$ and $(P(n), P(n+2)) = 7$ for $n \equiv 2 \pmod{7}$.
From $(2n + 7) P(n) - (2n - 1) P(n+2) = 14$ and the fact that $P(n)$ is odd, $(P(n), P(n+2))$ must be a divisor of $7$. The claim follows by checking $n \equiv 0, 1, \ldots, 6 \pmod{7}$ directly.

(iii) $(P(n), P(n+3)) = 1$ for $n \not\equiv 1 \pmod{3}$ and $3 \mid (P(n), P(n+3))$ for $n \equiv 1 \pmod{3}$.
From $(n + 5) P(n) - (n - 1) P(n+3) = 18$ and the fact that $P(n)$ is odd, $(P(n), P(n+3))$ must be a divisor of $9$. The claim follows by checking $n \equiv 0, 1, 2 \pmod{3}$ directly.

Suppose there exists a fragrant set with at most $5$ elements. We may assume it contains exactly $5$ elements $P(a), P(a+1), \ldots, P(a+4)$ since the following argument also works with fewer elements. Consider $P(a+2)$. From (i), it is relatively prime to $P(a+1)$ and $P(a+3)$. Without loss of generality, assume $(P(a), P(a+2)) > 1$. From (ii), we have $a \equiv 2 \pmod{7}$. The same observation implies $(P(a+1), P(a+3)) = 1$. In order that the set is fragrant, $(P(a), P(a+3))$ and $(P(a+1), P(a+4))$ must both be greater than $1$. From (iii), this holds only when both $a$ and $a+1$ are congruent to $1 \pmod{3}$, which is a contradiction.

It now suffices to construct a fragrant set of size $6$. By the Chinese Remainder Theorem, we can take a positive integer $a$ such that
$$a \equiv 7 \pmod{19}, \quad a+1 \equiv 2 \pmod{7}, \quad a+2 \equiv 1 \pmod{3} .$$
For example, we may take $a = 197$. From (ii), both $P(a+1)$ and $P(a+3)$ are divisible by $7$. From (iii), both $P(a+2)$ and $P(a+5)$ are divisible by $3$. One also checks from $19 \mid P(7) = 57$ and $19 \mid P(11) = 133$ that $P(a)$ and $P(a+4)$ are divisible by $19$. Therefore, the set $\{P(a), P(a+1), \ldots, P(a+5)\}$ is fragrant.

Therefore, the smallest size of a fragrant set is $6$.