One. (20 points) Find all ordered integer pairs $(a, b)$ such that the roots of the equation
$$
x^{4}+\left(2 b-a^{2}\right) x^{2}-2 a x+b^{2}-1=0
$$

are all integers.
$$
\begin{array}{l}
\Rightarrow\left\{\begin{array}{l}
x_{1}=\frac{-a-1}{2}, \\
x_{2}=\frac{-a+1}{2}, \\
x_{3}=\frac{a+3}{2}, \\
x_{4}=\frac{a-3}{2} ;
\end{array}\right. \\
\left\{\begin{array}{l}
x_{1}=\frac{-a+1}{2}, \\
x_{2}=\frac{-a-1}{2}, \\
x_{3}=\frac{a-3}{2}, \\
x_{4}=\frac{a+3}{2}
\end{array}\right. \\
\left\{\begin{array} { l } 
{ x _ { 1 } = \frac { - a + 1 } { 2 } , } \\
{ x _ { 2 } = \frac { - a - 1 } { 2 } , } \\
{ x _ { 3 } = \frac { a + 3 } { 2 } , } \\
{ x _ { 4 } = \frac { a - 3 } { 2 } ; }
\end{array} \quad \left\{\begin{array}{l}
x_{1}=\frac{-a-1}{2}, \\
x_{2}=\frac{-a+1}{2}, \\
x_{3}=\frac{a-3}{2}, \\
x_{4}=\frac{a+3}{2} .
\end{array}\right.\right. \\
\end{array}
$$

Since $a$ is an odd number, take $a=2 n-1$ and we get
$$
\begin{array}{l}
\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \\
=(-n,-n+1, n+1, n-2), \\
(-n+1,-n, n-2, n+1), \\
(-n+1,-n, n+1, n-2), \\
(-n,-n+1, n-2, n+1) .
\end{array}
$$

Therefore, $a=2 n-1, b=n^{2}-n-1(n \in \mathbf{Z})$.
$$