Three. (25 points) Let $a$ be a positive integer, the quadratic function $y=x^{2}+(a+17) x+38-a$, and the reciprocal function $y=\frac{56}{x}$. If the intersection points of the two functions are all integer points (points with both coordinates as integers), find the value of $a$.

---

The above text has been translated into English, preserving the original text's line breaks and format.
Three, by eliminating $y$ from the two equations, we get
$$
x^{2}+(a+17) x+38-a=\frac{56}{x} \text {, }
$$

which simplifies to $x^{3}+(a+17) x^{2}+(38-a) x-56=0$.
Factoring, we get
$$
(x-1)\left[x^{2}+(a+18) x+56\right]=0 \text {. }
$$

Clearly, $x_{1}=1$ is a root of equation (1), and $(1,56)$ is one of the intersection points of the two function graphs.
Since $a$ is a positive integer, the discriminant of the equation
$$
x^{2}+(a+18) x+56=0
$$

is $\Delta=(a+18)^{2}-224>0$, so it must have two distinct real roots.

Since the intersection points of the two function graphs are all integer points, the roots of equation (2) must all be integers. Therefore, its discriminant $\Delta=(a+18)^{2}-224$ should be a perfect square.
Let $(a+18)^{2}-224=k^{2}$ (where $k$ is a non-negative integer), then $(a+18)^{2}-k^{2}=224$,

which simplifies to $(a+18+k)(a+18-k)=224$.
Following the solution method of the third problem in the A paper, we get
$$
\left\{\begin{array} { l } 
{ a = 3 9 , } \\
{ k = 5 5 }
\end{array} \text { or } \left\{\begin{array}{l}
a=12, \\
k=26 .
\end{array}\right.\right.
$$

When $a=39$, equation (2) becomes $x^{2}+57 x+56=0$, whose roots are -1 and -56. In this case, the two function graphs have two additional intersection points $(-1,-56)$ and $(-56,-1)$.

When $a=12$, equation (2) becomes $x^{2}+30 x+56=0$, whose roots are -2 and -28. In this case, the two function graphs have two additional intersection points $(-2,-28)$ and $(-28,-2)$.