Question

如果A是一个直角三角形的面积,那么存在u,v,w有理数,u^2-v^2=v^2-w^2吗?

Answer 1

若A是一个有理边长直角三角形的面积,

则存在有理数a,b,c > 0,满足a²+b² = c²,ab/2 = A.

取u = (a+b)/2,v = c/2,w = (a-b)/2,可知u,v,w都是有理数.

且u²-v² = (a²+2ab+b²)/4-(a²+b²)/4 = ab/2 = A,

v²-w² = (a²+b²)/4-(a²-2ab+b²)/4 = ab/2 = A.

即存在满足要求的u,v,w.

反过来,若存在有理数u,v,w使u²-v² = v²-w² = A.

由A > 0,有|u| > |v| > |w|.

取a = |u|+|w|,b = |u|-|w|,c = 2|v|.

可知a,b,c > 0且都为有理数.

并有a²+b² = 2u²+2w² = 2(v²+A)+2(v²-A) = 4v² = c²,

且ab/2 = (u²-w²)/2 = (v²+A)/2-(v²-A)/2 = A.

因此A是以a,b,c为三边的直角三角形的面积.