a,b是一元二次方程x^2-x-1=0
即a^2-a-1=0
a^2-a=1
a^2=a+1
由韦达定理得
a+b=1
a^4+3b
=(a+1)^2+3b
=a^2+2a+1+3b
=a^2-a+3a+1+3b
=a^2-a+3a+3b+1
=a^2-a+3(a+b)+1
=1+3*1+1
=5