Question about G.C.D.

Question

Let,
$$a_{n}=n^2+20$$ $$d_{n}=\gcd(a_{n},a_{n+1})$$
where $n$ is a positive integer. Find the set of all values attained by $d_{n}$

I tried,
$d_{n}=\gcd(n^2+2n+21,n^2+20)$
$=\gcd(n^2+2n+21,2n+1)$
$=\gcd(n^2+20+2n+1,2n+1)$
$=\gcd(n^2+20,2n+1)$
However, after this I'm stuck.
Please Help!
Thanks!

Answer 1

$(n^2+20,2n+1)=(4n^2+80,2n+1)=(81,2n+1)$. It's easy to show that any divisor of 81 is possible. And from the information above we can see that it's the all possible numbers. So the answer is $1,3,27,9,81$

Answer 2

If $d$ divides both $n^2+20,(n+1)^2+20$

$d$ must divide $(n+1)^2+20-(n^2+20)=2n+1$

$d$ must divide $2(n^2+20)-n(2n+1)=40-n$

$d$ must divide $2(40-n)+(2n+1)=81$