The sequence {S(n)} is defined by the relationship: S(n) = 100 + n², whenever n is a positive integer.

If G(n) = gcd(S(n), S(n+1)), then find the maximum value of G(n).

(In reply to computer exploration by Charlie)

Fair disclosure: the following was written after seeing Charlie’s results.

 

Consider the function f =a+x^2.

It is easily shown that f  is  a multiple  of  4a+1  both for  x=2a and x=2a+1.

Indeed  :

a+(2a)^2=a*(4a+1)
a+(2a+1)^2=4a^2+5a+1= (a+1)*(4a+1)

Since 4a+1 is a common divisor and a dominant factor (in the only valid general factorization)
 it is a g.c.d.

The 4a+1 is a basic solution and   any member of the sequence 2a+401*k will do.

Maybe that a formal proof requires showing the impossibility of getting a g.c.d. greater than 4a+1,
so I did not post it as a solution, it is rather an observation.  

 

Posted by Ady TZIDON on 2014-10-12 21:33:00