由m^2=m+1,得，m^2-m-1=0,
又n^2-n-1=0，知，m,n是方程x^2-x-1=0的两根，
由根与系数关系，得，m+n=1,mn=-1,
所以m^2+n^2
=(m+n)^2-2mn
=1+2
=3,
m^4+n^4
=(m^2+n^2)^2-2m^2n^2
=9-2
=7
又(m^2+n^2)(n^4+n^4)=m^6+m^2n^4+m^4n^2+n^6
即21=m^6+n^6+m^2n^2(m^2+n^2)
解得m^6+n^6=21-3=18
所以m^7+n^7
=(m+n)(m^6-m^5n+m^4n^2+m^3n^3-m^2n^4-mn^5+n^6)
=m^6-m^5n+m^4n^2+m^3n^3-m^2n^4-mn^5+n^6
=m^6+n^6+m^4-m^3n+m^2n^2-mn^3+n^4(mn=-1代入)
=m^6+n^6+m^4+n^4+m^2+n^2+1
=18+7+3+1
=29
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