It is known that ILP with a fixed number of variables is in $\mathsf{P}$[Les83]. (For convenience, the decision version of the problem is given in the post.)
I have the following two questions regarding it.
Q1. Where does it lie within $\mathsf{P}$?
Q2. Is it $\mathsf{P}$-hard, or is it in $\mathsf{NC}$?
[The decision problem]
Input: Let $m,n\in \mathbb{Z^+}$, $A$ is ${m\times n}$ matrix with intgeral coefficients, and $b\in \mathbb{Z}^m$.
Promise: The value of $n$ will be kept fix. It won't change from one instance of the problem to another.
Goal: Does there exist a vector $x\in \mathbb{Z}^n$ that satisfies the set of inequalities $Ax\leq b$?
Additionally, the size of an instance is defined as $m\cdot n\cdot \log(a+2)$, where $a$ is the maximum magnitude of the entries of the matrix $A$ and the vector $b$.
My attempt:
Remark 1: I don't see any (obvious) way to reduce Linear programming to it. The key hurdle is that the restriction on $n$ is removed for LP. (This is my failed attempt to show $\mathsf{P}$ hardness.)
Remark 2: In [Les83], there is a use of the LLL algorithm (a lattice basis reduction algorithm) as a subroutine, which is a sequential algorithm [See page 3 of this]. (This is my failed attempt to show containment in $\mathsf{NC}$.)