Question

求方程x+y/x2-xy+y2=3/7的整数解

Answer 1

7(x+y)=3(x^2-xy+y^2)=3[(x+y)^2-3xy]

必有x+y=3k,k为整数

因此7k=9k^2-3xy

故也有k=3n--> x+y=9n,n为整数

7n=27n^2-xy--> xy=27n^2-7n=n(27n-7)

x,y为方程Z^2-9nZ+(27n^2-7n)=0的根

delta1=81n^2-108n^2+28n=28n-27n^2=p^2

27n^2-28n+p^2=0-->delta2=28^2-4*27p^2=4(196-27p^2)=4q^2,p