One, (20 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. Find the value of $x$ for which the function value is minimized.
One, the function is organized as
$$
y=(a+2)\left(x-\frac{a^{2}-1}{a+2}\right)^{2}+1-\frac{\left(a^{2}-1\right)^{2}}{a+2},
$$

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, hence
$$
0 < \frac{3}{a+2} \leq 1,
$$
we have
$$
a-2 < x \leq a-1.
$$
Let $x_{1}=a-2$ and $x_{2}=a-1$, then
$$
y_{1}=(a+2)\left(\frac{3}{a+2}\right)^{2}+1-\frac{\left(a^{2}-1\right)^{2}}{a+2},
$$
and
$$
y_{2}=(a+2)\left(-\frac{1}{a+2}\right)^{2}+1-\frac{\left(a^{2}-1\right)^{2}}{a+2}.
$$
Thus,
$$
y_{1}-y_{2}=\frac{10-4a}{a+2}.
$$
Therefore,
(i) When $4-a>0$, i.e., $a=2$ or 3, $x$ takes $a-1$ to make $y_{2}$ the minimum value;
(ii) When $4-a=0$, i.e., $a=4$, we have $y_{1}=y_{2}$, at this time, $x$ takes 2 or 3;
(iii) When $4-a<0$, i.e., $a>4$, $x$ takes $a-2$ to make $y_{1}$ the minimum value.
$$
\text { In summary, when } x=\left\{\begin{array}{ll}
1, & a=1 ； \\
a-1, & a=2 \text { or } 3 \text { ； } \\
2 \text { or } 3, & a=4 ； \\
a-2, & a>4
\end{array}\right.
$$

the function value is minimized.