已知a2+4a+1=0，且a4+ma2+13a3+ma2+3a＝5，则m=_．

∵a2+4a+1=0，∴a2=-4a-1，a4+ma2+13a3+ma2+3a=(−4a−1)2+ma2+13a(−4a−1)+ma2+3a=(16+m)a2+8a+2(m−12)a2=(16+m)a2+8a+2(m−12)(−4a−1)=(16+m)(−4a−1)+8a+2(m−12)(−4a−1)=5，∴（16+m）（-4a-1）+8a+2=5（...