One. (20 points) Find all integers $a$ such that the quadratic equation in $x$
$$
x^{2}-\sqrt{5 a^{2}-26 a-8} x-\left(a^{2}-4 a+9\right)=0
$$
has both roots as integers.
Let the two roots of the equation be $x_{1}$ and $x_{2}$, then,
$\sqrt{5 a^{2}-26 a-8}=x_{1}+x_{2}=$ integer,
which means equation (1) is a quadratic equation with integer coefficients, and its roots are integers, so its discriminant $\Delta$ must be a perfect square. Let
$\left(5 a^{2}-26 a-8\right)+4\left(a^{2}-4 a+9\right)=b^{2}, b \in \mathbf{N}$, that is, $(3 a-7)^{2}-b^{2}=21$.
Thus, $(3 a-7-b)(3 a-7+b)=21$.
And $21=3 \times 7=1 \times 21=(-7) \times(-3)$
$=(-21) \times(-1)$,
then $\left\{\begin{array}{l}3 a-7-b=3, \\ 3 a-7+b=7\end{array}\right.$ or $\left\{\begin{array}{l}3 a-7-b=1, \\ 3 a-7+b=21\end{array}\right.$
or $\left\{\begin{array}{l}3 a-7-b=-7, \\ 3 a-7+b=-3\end{array}\right.$ or $\left\{\begin{array}{l}3 a-7-b=-21, \\ 3 a-7+b=-1 .\end{array}\right.$
Solving these respectively gives $a=4,6, \frac{2}{3},-\frac{4}{3}$.
Since $a$ is an integer, and when $a=4$, $\sqrt{5 a^{2}-26 a-8}$ is undefined, so only $a=6$ is valid. In this case, equation (1) becomes $x^{2}-4 x-$ $21=0$, which has two integer roots 7 and -3.
Therefore, the required integer is $a=6$.