Question

如果直角三角形三条边长是正整数.且有一条直角边是2008.那么另一条直角边长得取值个数最多为几个.

Answer 1

即直角三角形直角边A=2008、B,斜边C,求直角边B.

由勾股定理 A² + B² = C²

C² - B² = A² = 2008*2008

(C + B) (C - B) = 2008*2008 = 2^6×251^2

因C + B、C - B这两个数必同奇偶,且两数的积是偶数,则两数必都是偶数,

C + B > C - B

根据2008*2008 = 2^6×251^2的因数分配,有如下不同的7组方程组：

①

C + B = 251*251*2

C - B = 2^5

解得

C = 63017

B = 62985

②

C + B = 251*251*2*2

C - B = 2^4

解得

C = 126010

B = 125994

③

C + B = 251*251*2*2*2

C - B = 2^3

解得

C = 252008

B = 252000

④

C + B = 251*251*2*2*2*2

C - B = 2^2

解得

C = 504010

B = 504006

⑤

C + B = 251*251*2*2*2*2*2

C - B = 2

解得

C = 1008017

B = 1008015

⑥

C + B = 251*2*2*2*2*2

C - B = 251*2

解得

C = 4267

B = 3765

⑦

C + B = 251*2*2*2*2

C - B = 251*2*2

解得

C = 2510

B = 1506