Question

已知x^2=x+1,y^2=y+1,且x≠y.（1）求证：x+y=1.（2）求x^4+y^4的值

Answer 1

两式相减得x^2-y^2=(x+1)-(y+1),即(x+y)(x-y)=(x-y),因为x不等于y,所以x-y不等于0,故x+y=1

x^4+y^4=(x+1)^2+(y+1)^2=x^2+2x+1+y^2+2y+1=(x+1)+(y+1)+2(x+y)+2=3(x+y)+4=3+4=7