∵an+2= a +2 an ,∴an+2an=an+12+2,∵若存在常数p,q,使得对一切n∈N*都有an+2=pan+1+qan,∴panan+1+qan2=an+12+2,①又a1=a2=1,令n=1,代入①,得:p+q=3,a3=pa2+qa1=p+q=3,令n=2,代入①得:3p+q=9+2=11,联立②③得:p=4,q=-1,∴存在p=4,q=-1,使得an+2=pan+1+qan.