2010/038) For a natural number n, let An = n^2 + 20

If Dn denotes the greatest common divisor of An and An+1,?
then show that Dn divides 81.

proof:
GCD(An , An+1)as we need to get a constant
= GCD(n^2+20, (n+1)^2 + 20) (one of 1st or 2nd term need to be constant by proper transform)
= GCD(n^2+20, n^2+2n + 21)
= GCD(n^2+20, 2n+1) as GCD(a,b) = GCD(a,b-a)
[now we need to eliminate n^2 from 1st term so we double the 1st term as 2nd term is odd so GCD(a,b) = GCD(2a,b) when b is odd]
= GCD(2n^2+40, 2n+1)
= GCD(2n^2+40-n(2n+1),2n+21) as GCD(a,b) = GCD(a-kb,b) for any k
= GCD(40-n, 2n+1)
= GCD(80-2n, 2n+1) [ same argument above as 2n+1 is odd double the 1st term
= GCD(81, 2n+1) as GCD(a,b) = GCD(a+b.b)

now 81 is constant and GCD must devide 81