> 1/a = (1/b + 1/c)/ (1 - 1/(b*c))
> => bc-1 = a(b+c)
> assume b=a+m and c=a+n (b and c is always bigger than a)
> (a+m)(a+n)-1=a(a+m+a+n)
> => a*a+a*n+a*m+m*n-1=2*a*a+m*a+n*a
> => m*n=a*a+1
> and then
> for(m=a;m>=1;m--)
>    if((a*a+1)%m==0)
>       break;
> n=(a*a+1)/m

• Re:我的推导过程 (29) 19871026 2009-05-07 01:14:24