<s>[INST] Find the GCD of $10$ integers 
Find the greatest common divisor of the following ten integers: 
  \begin{align}
& 2000^3+3\cdot2000^2+2\cdot2000,\  2001^3+3\cdot2001^2+2\cdot2001,\  \dots, \\[5pt]
& \dots, \  2008^3+3\cdot2008^2+2\cdot2008,\  2009^3+3\cdot2009^2+2\cdot2009.
\end{align}

How are these numbers related to each other? Is there a formula for this?
 [/INST] Let $n=2000$; the first number is $n^3+3n^2+2n$ and the second number is $(n+1)^3+3(n+1)^2+2(n+1)$, so their difference is
$$
(n+1)^3-n^3+3(n+1)^2-3n^2+2(n+1)-2n=3n^2+9n+6
$$
The third number is $(n+2)^3+3(n+2)^2+2(n+2)$ and the difference with the first number is
$$
(n+2)^3-n^3+3(n+2)^2-3n^2+2(n+2)-2n=6n^2+24n+24
$$
The gcd should be a divisor of
$$
(6n^2+24n+24)-2(3n^2+9n+6)=6n+12=6(n+2)
$$
The same computation from $n=2001$ shows the gcd must be a divisor of both $6\cdot2002$ and $6\cdot2003$, but $2002$ and $2003$ are coprime.
Hence the gcd is a divisor of $6$; now, $n^3+3n^2+2n=n(n+1)(n+2)$, so…
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