國際數學奧林匹克!!!!!! [ 1 ]

chengclement, it seems you are writing your solution from the answer...
See how I solve this question

Let x, y be the two numbers. Condition gives
x+y = xy - 2006
Factorization gives
(x-1)(y-1) = 2007
Note 2007 = 32 x 223
The number 2007 has 6 positive factors, namely 1, 3, 9, 223, 669, 2007
As one of x and y (say x) is a perfect square, x-1 must be one less than a perfect square
Among the 6 factors, only the factor 3 has this property (3 = 22- 1)
Thus we take x-1 = 3 and y-1 = 669
This follows x=4 and y=670
Finally, difference = 670-4 = 666

Hope the above information helps =) By 小儒

let 兩個未知數為A, B
AB-2006 = A+B
AB-A-B =2006
AB-A-B+1 = 2007
(A-1) (B-1) = 2007 = 3 X 3 X 667
SO, (A,B)可能性為(4,670), (10,667)
只有4 為平方數, 所以(A,B) 只可以是(4,670)
兩數之差為666。

(外國數是容易解過亞洲數學的, 所以國際數學程度並不如亞洲數學, 如中國, 日本, 香港等)

2007-04-06 12:18:33 補充：
因(A-1) (B-1) = 3 X 2001, 或(A-1) (B-1) = 9 X 667所以A-1 = 3, B-1 = 2001, 或A-1 = 9, B-1 = 667所以, A=4, B=2002, 或A=10, B=668 (沒有平方數, REJECTED)所以, 相差應是2002 -4 = 1998(之前計錯)