One. (20 points) Determine for which positive integers $a$, the equation
$$
5 x^{2}-4(a+3) x+a^{2}-29=0
$$

has positive integer solutions? And find all positive integer solutions of the equation.
Rewrite the equation as
$$
(x-6)^{2}+(a-2 x)^{2}=65 \text {. }
$$

Since 65 can only be expressed as the sum of squares of two positive integers in two different ways: $65=1^{2}+8^{2}=4^{2}+7^{2}$, then
$$
\left\{\begin{array} { l } 
{ | x - 6 | = 8 , } \\
{ | a - 2 x | = 1 }
\end{array} \text { or } \left\{\begin{array}{l}
|x-6|=7, \\
|a-2 x|=4
\end{array}\right.\right.
$$

or $\left\{\begin{array}{l}|x-6|=1, \\ |a-2 x|=8\end{array}\right.$ or $\left\{\begin{array}{l}|x-6|=4, \\ |a-2 x|=7 \text {. }\end{array}\right.$
From the first set of equations, we get
$$
x=14, a=29 \text { or } 27 \text {; }
$$

From the second set of equations, we get
$$
x=13, a=22 \text { or } 30 \text {; }
$$

From the third set of equations, we get $x=5, a=2$ or $18 ; x=7, a=6$ or 22 ; from the fourth set of equations, we get $x=2, a=11 ; x=10, a=13$ or 27.