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Posted by hansechang at 2008-05-22 22:30:02 on Problem 1320 and last updated at 2008-05-23 10:03:30
In Reply To:推导: Posted by:WuChengWei at 2008-05-03 15:13:01
>  
> 1+2+...+(a-1) = (a+1)+(a+2)+...+b
> ==> a(a-1)    = (a+1+b)(b-a)  //等差数列求和
> ==> a^2 - a   = b^2-a^2 +b-a
> ==> 2*a^2     = b^2+b
> ==> 8*a^2     = 4*b^2 +4*b +1-1
> ==> 2*(2*a)^2 = (2*b+1) -1
> ==> (2*b+1)^2 - 2*(2*a)^2 = 1
> 
> 令  x = 2*b+1, y= 2*a
> 则  x^2 - 2*y^2 = 1 *(1)
> 
> 设 (x1,y1) (x2,y2) 是方程的解
> 则有 x1^2 - 2*y1^2 = 1   (2)
>      x2^2 - 2*y2^2 = 1   (3)
> 
> (2)*(3) ==>  (x1^2 - 2*y1^2)*(x2^2 - 2*y2^2) =1
>         ==>  x1^2*x2^2 - (2*x1^2*y2^2+2*y1^2*x2^2) + 4*y1^2*y2^2 = 1
>         ==>  (x1*x2)^2+(2*y1*y2)^2+4*x1*x2*y1*y2
>             - (2*x1^2*y2^2+2*y1^2*x2^2 + 4*x1*x2*y1*y2) = 1
>        ==> (x1*x2+2*y1*y2)^2 - 2*( x1*y2+x2*y1)^2 = 1  (4)
> 令 p = x1*x2+2*y1*y2
>    q = x1*y2+x2*y1
> 则 (4) ==> p^2 -2*q^2 = 1 所以 (p,q)也是方程 1的解
> 
> ps: 在(2)*(3)的推导中,辅助项“+4*x1*x2*y1*y2” 可以是减号
>      则 p = abs( x1*x2 - 2*y1*y2 )
>        q = abs( x1*y2 - x2*y1)
> 
> 则由题目中的 a1=6, b1=8 ==> x1 = 17, y1=12
>            a2=35, b2=49 ==> x2 = 99, y2=70  
>         ==> p = abs(17*99-2*12*70) = abs(1683-1680) = 3 = 2*b+1 ==> b=1
>             q = abs(17*70-12*99)   = abs(1190-1188) = 2 = 2*a ====> a=1
> 得到的 (a,b) = (1,1)是最小对
> 
> 应用辅助项“+4*x1*x2*y1*y2”(加号)可以
> 由 (1,1)和(a1,b1)=(6,8)       推出(a2,b2) = (35, 49)
> 由 (1,1)和(a2,b2)=(35, 49)    推出(a3,b3) = (204, 288)
> 由 (1,1)和(a3,b3)=(204,288)   推出(a4,b4) = ...
> .......................

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