## Problem 3.

Let $a_{n}=1+n^{3}$ be the sequence $\{2,9,28,65, \ldots\}$ and $\delta_{n}=\operatorname{gcd}\left(a_{n+1}, a_{n}\right)$. Find the maximum value that $\delta_{n}$ can take.
Solution: $\delta_{n}$ divides $a_{n+1}$ and $a_{n}$, and therefore their difference $b_{n}=a_{n+1}-a_{n}=3 n^{2}+3 n+1$.

It also divides $c_{n}=3 a_{n}-n b_{n}=3-n-3 n^{2}$ and the sum $d_{n}=b_{n}+c_{n}=4+2 n$. But then $\delta_{n}$ also divides $e_{n}=2 b_{n}-3 n d_{n}=2-6 n$. Finally, it divides $3 d_{n}+e_{n}=14$.

But $b_{n}=3 n^{2}+3 n+1=3 n(n+1)+1$ is an odd number, so $\delta_{n}$ can only be 1 or 7. The maximum is 7 since $\operatorname{mcd}\left(5^{3}+1,6^{3}+1\right)=7$.