Problem 1859

To show it less than $3$, we let $b_n = n^6 - 2017$, $n = 0, 1, 2, \dots$, and prove that
$$\text{gcd}(b_{n-1}, b_n, b_{n+1}) = \begin{cases} 1 & \text{if } n \equiv 0, \pm 1 \pmod 7, \\ 7 & \text{if } n \equiv \pm 2, \pm 3 \pmod 7. \end{cases}$$
Fix an index $n$, and let $d$ be a positive integer dividing $b_{n-1}, b_n$ and $b_{n+1}$. Since at least one of $b_{n-1}, b_n, b_{n+1}$ is odd, so is $d$. The numbers $b_{n+1}-2b_n+b_{n-1} = 2(15n^4+15n^2+1)$ and $b_{n+1}-b_{n-1} = 4n(3n^4+10n^2+3)$ are both divisible by $d$. Since $d$ is odd, it divides $15n^4+15n^2+1$, so $d$ is coprime to $n$, hence it divides $3n^4+10n^2+3$ as well. Write $7 \cdot (5n^2+2) = 5 \cdot (3n^4+10n^2+3) - (15n^4+15n^2+1)$ to infer that $7 \cdot (5n^2+2)$ is divisible by $d$. Since $3 \cdot (5n^2+2)(5n^2+3) = 5 \cdot (15n^4+15n^2+1)+13$, it follows that $d$ is one of $1, 7, 13, 7 \cdot 13$. To conclude the proof, notice that no $b_k$ is divisible by $13$.