设x=根号（n+1）+根号n分之根号（n+1）—根号n,y=根号（n+1）—根号n分之根号（n+1）+根号n（n为不为0的自然数）,如果x^2+1108xy+y^2=2006,求n的值

Question

设x=根号（n+1）+根号n分之根号（n+1）—根号n,y=根号（n+1）—根号n分之根号（n+1）+根号n（n为不为0的自然数）,如果 x^2+1108xy+y^2=2006,求n的值

Answer 1

x=根号（n+1）+根号n分之根号（n+1）—根号n,y=根号（n+1）—根号n分之根号（n+1）+根号n
x^2+y^2=4n+2+2(n+1)/n-4根号(n+1)
xy=2n+1-(n+1)/n+2根号(n+1)
x^2+1108xy+y^2=4n+2+2(n+1)/n-4根号(n+1)+2216n+1108-1108(n+1)/n+2216根号(n+1)=2220n+1110-1106(n+1)/n+2212根号(n+1)=2006
n=