Question about G.C.D. 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!


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 $2(40-n)+(2n+1)=81$
