p1=A+B
p2=A^2+(A+1)B
p3=A^2+(A^2+A+1)B

p1p2p3=A^6+...<=2000000000

1.  if A>0  && B>0
==>A<=36=2000000000^(1/6)

let A=1 p1p2p3=(1+B)(1+2B)(1+3B)=6B^3+...<2000000000

==>B<=694

2.  if A>0  ， B为负   

p1=d=A-B>=3   
p2= (B+d)d-B=d^2+(d-1)B>d^2+2B
p3=(B+d)(d^2+(d-1)B)-B=d^3+.....

p1p2p3=d^6+.....

==>d<=36

let d=3
p1p2p3=3*((B+3)3-B) *(B+3)(9+2B)-B=12B^3+     

B<=551

A<=587


3.  if A<0    无解

• 精彩 这样推才是硬道理 (0) ecjtunh 2006-10-01 17:37:36