4. If the parabola $y=\frac{x^{2}}{2}-m x+m-1$ intersects the $x$-axis at integer points, then the axis of symmetry of the parabola is ( ).
(A) $x=1$
(B) $x=-1$
(C) $y=1$
(D) Cannot be determined
4. A.

Let the parabola intersect the $x$-axis at integer points $\left(x_{1}, 0\right)$ and $\left(x_{2}, 0\right)$ $\left(x_{1}, x_{2} \in \mathbf{Z}, x_{1}<x_{2}\right)$. Then the equation $\frac{x^{2}}{2}-m x+m-1=0$ has integer roots $x_{1}$ and $x_{2}$, yielding the identity
$$
\frac{1}{2} x^{2}-m x+m-1=\frac{1}{2}\left(x-x_{1}\right)\left(x-x_{2}\right) \text {. }
$$

Substituting $x=1$ to eliminate $m$ gives
$$
\left(1-x_{1}\right)\left(1-x_{2}\right)=-1 \text {. }
$$

Therefore, $\left\{\begin{array}{l}1-x_{1}=1, \\ 1-x_{2}=-1 .\end{array}\right.$
Solving yields $x_{1}=0, x_{2}=2$.
Substituting these into the equation both yield $m=1$.
Thus, the axis of symmetry of the parabola is $x=1$.