Three. (25 points) Given the function
$$
y=(a+2) x^{2}-2\left(a^{2}-1\right) x+1 \text {, }
$$

where the independent variable $x$ is a positive integer, and $a$ is also a positive integer. Question: For what value of $x$ is the function value minimized?
Three, the function is organized as
$$
y=(a+2)\left(x-\frac{a^{2}-1}{a+2}\right)^{2}+1-\frac{\left(a^{2}-1\right)^{2}}{a+2} \text {. }
$$

Then its axis of symmetry is
$$
x=\frac{a^{2}-1}{a+2}=(a-2)+\frac{3}{a+2} \text {. }
$$

Since $a$ is a positive integer, we have
$$
\begin{array}{l}
0<1$ when, $a-2<0$, i.e., $a=1$; \\
0<1$ when, $a-2>0$, i.e., $a=2$ or 3, $x$ takes $a-1$, making $y_{2}$ the minimum value; \\
(2) When $4-a=0$, i.e., $a=4$, we have $y_{1}=y_{2}$, at this time, $x$ takes 2 or 3; \\
(3) When $4-a<0$, i.e., $a>4$ and is an integer, $x$ takes $a-2$, making $y_{1}$ the minimum value.
\end{array}
$$
In summary, 
$$ 
x=\left\{\begin{array}{ll}
1, & a=1 ; \\
a-1, & a=2 \text { or } 3 ; \\
2 \text { or } 3, & a=4 ; \\
a-2, & a>4 \text { and } a \text { is an integer }
\end{array}\right.
$$

when the function value is the smallest.
（Sun Yan, Anhui Provincial Anqing City Teaching and Research Office, 246004 Huang Quanfu, Anhui Provincial Huaining County Jiangzhen Middle School, 246142）