>  gcd(a^m - 1, a^n - 1) (m > n)
> 
> if m = p*n + r
> then 
>    a^m - 1 = a^(p*n+r) - a^r + a^r - 1
>            = a^r(a^(p*n)-1) + (a^r-1)
> 
>    since (a^n-1) | (a^(p*n)-1)
> 
> gcd(a^m - 1, a^n - 1) = gcd(a^r - 1, a^n-1)
> 
> so  gcd(a^m - 1, a^n - 1) = a^gcd(m,n) - 1
> 
> so ......
> 
> 
> 
> 

• Re:prove it like this (691) gemenhao 2006-04-07 14:48:19
• good prove (0) hello__world 2009-02-11 15:18:17