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
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
• 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