已知数列｛an｝满足a1=1，a2=2，a3=3，a4=4，a5=5，当n≥5时，a(n+1)=a1a2……an-1，若数列{bn}（n∈N*）

b1=a1-a1^2=0,b2=-3,b3=-8,b4=-6,b5=65
n≥5时，a(n+1)=a1a2……an-1 ,a6=a1a2a3a4a5-1,a6-a1a2a3a4a5=1
b6=(a1a2...a5a6-a1^2-a2^2-a3^2-...-a^5^2-a6^2=(a1a2a3a4a5)^2-2a1a2a3a4a5+(a1a2a3a4a5-a1^2-a2^2-a3^2-a4^2-a5^2)-(a1a2a3a4a5-1)^2=(a1a2a3a4a5-a1^2-a2^2-a3^2-a4^2-a5^2)-1=b5-1
b7=b5-(7-5)
bn=b5-(n-5)
n≥5时，a(n+1)=a1a2……an-1,
n≥5，(a1a2a3...an+1)-an+1^2=(a1a2a3...an)(a1a2a3...an-1)-(a1a2a3..an-1)^2
=(a1a2a3...an)-1
b6=b5-1
b7=b6-1=b5-(7-5)
..
bn=b5-(n-5)
n=70时,b70=0
n=1,n=70时,bn=0, 10,n>70时,bn<0
所以仅存在两个整数m,m=1,m=70使得bm=0

当m=1时,b1=0.
显然an>0,m≠n时,am≠an.
当m=1时,b1=a1-a1^2=1-1=0.
当m>1时,
bm=a1a2……am-(a1^2+a2^2+……+am^2)
=-a1a2……am<0
所以仅存在m=1时,使bm=0.
供参考.