\(填空:a^{16}-a+1是(正数，0，负数)\)

\[ \begin{align*} 原式：a^{16}-a+1 \\ & = a^{16}-a^8+\frac{1}{4}+a^8+\frac{3}{4} \\ & = (a^8-\frac{1}{2})^2 + a^8 -a^4+\frac{1}{4} +a^4-a+\frac{2}{4} \\ & = (a^8-\frac{1}{2})^2 + (a^4-\frac{1}{2})^2 +a^4-a+\frac{2}{4} \\ & = (a^8-\frac{1}{2})^2 + (a^4-\frac{1}{2})^2 +a^4 -a^2 +\frac{1}{4} +a^2 -a +\frac{1}{4} \\ & = (a^8-\frac{1}{2})^2 + (a^4-\frac{1}{2})^2 +(a^2-\frac{1}{2})^2 +(a-\frac{1}{2})^2 \\ 由于：(a+b)^2 \geq 0： \\ (a^8-\frac{1}{2})^2 + (a^4-\frac{1}{2})^2 +(a^2-\frac{1}{2})^2 +(a-\frac{1}{2})^2 \geq 0 \\ 当：a^8 = \frac{1}{2},a^4=\frac{1}{2},a^2=0,a=\frac{1}{2}时， \\ (a^8-\frac{1}{2})^2 + (a^4-\frac{1}{2})^2 +(a^2-\frac{1}{2})^2 +(a-\frac{1}{2})^2 均不能同时为0， \\ 所以：a^{16}-a+1是：正数。 \end{align*} \]