We call a set of positive integers fragrant if it has at least 2 elements, and for every element, there is at least one other element such that the two elements have a common prime divisor. Let $P(n)=n^{2}+n+1$. What is the smallest positive integer $b$ for which there exists a non-negative integer $a$ such that the set

$$
\{P(a+1), P(a+2), \ldots, P(a+b)\}
$$

is fragrant?
Baran Zsuzsanna's solution. We will see that the smallest such $b$ positive integer is $b=6$.

The task is to find as few consecutive positive integers as possible such that each of their $P$ values has a common prime divisor with the $P$ value of another element. For this, we will examine what common prime divisor can exist for the $P$ values of close numbers, that is, for what prime $p$ can $p \mid P(n)$ and $p \mid P(n+x)$ (where $n$ is a positive integer). We will examine $x=1,2,3$ and 4.

First, since $n^{2}+n=n(n+1)$ is even, $P(n)$ is always odd, so $p$ cannot be 2.

For $x=1$:

$$
\begin{aligned}
\left(n^{2}+n+1 ;(n+1)^{2}+(n+1)+1\right) & =\left(n^{2}+n+1 ; 2 n+2\right)= \\
& =\left(n^{2}+n+1 ; n+1\right)=\left(n^{2} ; n+1\right)=1
\end{aligned}
$$

Thus, $P(n)$ and $P(n+1)$ cannot have a common prime divisor.

For $x=2$:

$$
\begin{aligned}
0 & \equiv n^{2}+n+1 \equiv(n+2)^{2}+(n+2)+1=n^{2}+5 n+7(\bmod p) \\
0 & \equiv 4 n+6(\bmod p) \\
2 n & \equiv-3(\bmod p) \\
0 & \equiv 4 n^{2}+4 n+4 \equiv(-3)^{2}+2(-3)+4=7(\bmod p)
\end{aligned}
$$

Thus, $p \mid P(n)$ and $p \mid P(n+2)$ is only possible if $p=7$ and $2 n \equiv-3 \equiv 4(\bmod 7)$, so $n \equiv 2(\bmod 7)$. In this case, the divisibility indeed holds: $2^{2}+2+1 \equiv 4^{2}+4+1 \equiv 0(\bmod 7)$.

For $x=3$:

$$
\begin{aligned}
0 & \equiv n^{2}+n+1 \equiv(n+3)^{2}+(n+3)+1=n^{2}+7 n+13(\bmod p) \\
0 & \equiv 6 n+12(\bmod p) \\
3 n & \equiv-6(\bmod p) \\
0 & \equiv 9 n^{2}+9 n+9 \equiv(-6)^{2}+3 \cdot(-6)+9=27(\bmod p)
\end{aligned}
$$

Thus, only $p=3$ can be considered.

Then $1^{2}+1+1 \equiv 0(\bmod 3)$, but $0^{3}+0+1 \equiv 2^{2}+2+1 \not \equiv 0(\bmod 3)$, so $n \equiv 1(\bmod 3)$ is good, but not others. Therefore, the only common prime divisor of $P(n)$ and $P(n+3)$ can be 3, and only if $n \equiv 1(\bmod 3)$.

For $x=4$:

$$
\begin{aligned}
0 & \equiv n^{2}+n+1 \equiv(n+4)^{2}+(n+4)+1=n^{2}+9 n+21(\bmod p) \\
0 & \equiv 8 n+20(\bmod p) \\
2 n & \equiv-5(\bmod p) \\
0 & \equiv 4 n^{2}+4 n+4 \equiv(-5)^{2}+2 \cdot(-5)+4=19(\bmod p)
\end{aligned}
$$

Thus, only $p=19, 2 n \equiv-5 \equiv 14(\bmod 19)$, so $n \equiv 7(\bmod 19)$ can be considered. This is also correct: $7^{2}+7+1 \equiv 9^{2}+9+1 \equiv 0(\bmod 19)$. Therefore, the only common prime divisor of $P(n)$ and $P(n+4)$ can be 19, and only if $n \equiv 7(\bmod 19)$.

Now, based on the above, let's try to create a suitable 6-element set. We want to find an $a$ such that $P(a+1)$ and $P(a+5)$, $P(a+2)$ and $P(a+4)$, and $P(a+3)$ and $P(a+6)$ have a common prime divisor.

Let $b=6$ and $a \equiv 1(\bmod 3)$, $a \equiv 6(\bmod 19)$, and $a \equiv 0(\bmod 7)$ (this has a positive integer solution according to the Chinese Remainder Theorem).

Then $a+1 \equiv 7(\bmod 19)$, so $19 \mid P(a+1), P(a+1+4)$, while $a+2 \equiv 2(\bmod 7)$, so $7 \mid P(a+2), P(a+2+2)$, and finally $a+3 \equiv a+6 \equiv 1(\bmod 3)$, so $3 \mid P(a+3), P(a+6)$.

Thus, with the above $a$, each element of the set $\{P(a+1), P(a+2), \ldots, P(a+6)\}$ has a common prime divisor with another element, making the set suitable.

Suppose there is a smaller suitable $b$ as well, that is, there exists a positive integer $b \leq 5$ for which there is a positive integer $a$ such that $\{P(a+1), \ldots, P(a+b)\}$ is suitable.

It cannot be $b=2$ or $b=3$, because $P(a+2)$ is relatively prime to both $P(a+1)$ and $P(a+3)$.

It cannot be $b=4$, because then $P(a+2)$ is relatively prime to both $P(a+1)$ and $P(a+3)$, so it must have a common prime divisor with $P(a+4)$, which can only be 7 ($P(n)$ and $P(n+2)$ can only have 7 as a common prime divisor). Similarly, $P(a+3)$ must have a common prime divisor with $P(a+1)$, but this can also only be 7. This would mean that both $P(a+1)$ and $P(a+2)$ are divisible by 7, which is a contradiction.

It cannot be $b=5$ either, because then $P(a+3)$ is relatively prime to its neighbors, so it must have a common prime divisor with $P(a+1)$ or $P(a+5)$, which can only be 7. If $7 \mid P(a+3)$, then $7 \nmid P(a+2), P(a+4)$. Meanwhile, $P(a+2)$ is relatively prime to its neighbors and, since it is not divisible by 7, it is also relatively prime to $P(a+4)$. Then $P(a+5)$ must have a common prime divisor with $P(a+2)$, which can only be 3 ($P(n)$ and $P(n+3)$ can only have 3 as a common prime divisor).

Similarly, $P(a+4)$ must have a common prime divisor with $P(a+1)$, which can only be 7. However, this would mean that $7 \mid P(a+1)$ and $7 \mid P(a+2)$, which is a contradiction.

Therefore, it cannot be $b \leq 5$, the smallest suitable $b$ number is $b=6$.