令x=1, 代入,得:f(f(1))=1, 所以 f(f(f(1)))=f(1), 即 f(1)^2-f(1)+1=f(1), 解得 f(1)=1
再令x=0, 代入,得:f(f(0))=1, 所以 f(f(f(0)))=f(1), 即 f(0)^2-f(0)+1=1, 得 f(0)=0 或 f(0)=1
又,如果f(0)=0, 得 f(f(0))=f(0)=0, 與上述f(f(0))=1矛盾,故應舍去,所以 f(0)=1是唯一解
令x=1, 代入,得:f(f(1))=1, 所以 f(f(f(1)))=f(1), 即 f(1)^2-f(1)+1=f(1), 解得 f(1)=1
再令x=0, 代入,得:f(f(0))=1, 所以 f(f(f(0)))=f(1), 即 f(0)^2-f(0)+1=1, 得 f(0)=0 或 f(0)=1
又,如果f(0)=0, 得 f(f(0))=f(0)=0, 與上述f(f(0))=1矛盾,故應舍去,所以 f(0)=1是唯一解
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