回复：A General Solution

Thanks to fzy, I'v worked a more general solution to share with you.

x*(10^k+1)=y^2=(10^k+1)^2*p^2/q^2;
x=(10^k+1)^2*p^2/q^2 (0.1

Since mod(10^k+1, 11)=0, when k is odd, we pick q=11.
Also notice (10^k+1)/11=9090...9091, with (k-1)/2 9's.

We only need to solve for k such that (10^k+1)^2 can be devided by 11^2=121.

This is easy, solong as 9*(k-1)/2-1 can be divided by 11.

The first k=11, the second k=11*13...

We need 0.1

With k, p, q=11, solutions for x are:
x=(10^k+1)*p^2/q^2.

所有跟帖： 

• 回复：A General Solution -fzy- ♀ (287 bytes) () 02/07/2005 postreply 07:48:07

• 回复：回复：A General Solution -jinjing- ♀ (498 bytes) () 02/07/2005 postreply 16:48:20