回複：過程。。。。

Transformation (1)
161^n-1=(161-1)(161^(n-1)+161^(n-2)+.....+1)

so n=even number. Let n=2a

Transformation (2)
161^2a-1=(161^a1+1)(161^a1-1)

(161^a1+1) and (161^a1-1) are 2 consecutive even number. Only one of them can be divided by 4, which must be (161^-1). Because the above transformation (1).

Repeat transformation (1) and (2), we'll get a2 follow the same pattern. Repeat until it just give (161^ai-1) is in a form of (161-1), which gives the smallest n.

Each round give 1 more factor 2, and (161-1) gives 5 factor 2s.

(161^a1+1)(161^a1-1) must be in a form of (161^2000+1)(161^2000-1)

so n=4000

所有跟帖： 

• 回複：回複：過程。。。。 -亂彈- ♂ (28 bytes) () 03/03/2009 postreply 11:48:01

• 回複：回複：回複：過程。。。。 -passenger101- ♂ (57 bytes) () 03/04/2009 postreply 06:34:03

• 佩服！似有筆誤？ -endofsuburbia- ♂ (0 bytes) () 03/03/2009 postreply 17:33:41