回复：有没有好办法？

let a^2+4*a*b+b^2=（a+2*b)^2-3b^2=t^2-3*b^2;

if x=t1^2-3*b1^2, y==t2^2-3*b2^2;

so x*y=(t1^2-3*b1^2)*(t2^2-3*b2^2)=(t1*t2+3*b1*b2)^2-3*(t1*b2+t2*b1)^2;

so x*y also belongs to A

所有跟帖： 

• 回复：回复：有没有好办法？ -cma- ♂ (48 bytes) () 03/11/2012 postreply 15:49:44

• ＳＯＲＲＹ，ＬＡＳＴ　ＩＳ　ＮＯＴ　ＲＩＧＨＴ。 -jinjing- ♀ (0 bytes) () 03/11/2012 postreply 16:05:20

• Great! This is related to Pell's equation etc. -乱弹- ♂ (0 bytes) () 03/11/2012 postreply 19:33:45