解:由题意:设两根为x1,x2 x1+x2=a2-3a-1 x1x2=a2+aa-1 ∵两根均为整数∴a2-3a-1=a+1-2a-1 a2+aa-1=a+2+2a-1∴a-1=±2±1∴a=3-120∵△=[-(a2-3)]2-4(a-1)(a2+a)≥0 ∴a=-1或a=0