To solve linear equations over $\mathbb{Z}$ we have a system of linear equations represented by some integer matrix $A$ of $n \times m$ dimension and $b \in \mathbb{Z}^n$. Such that solving $Ax = b$ gives our solutions.
We can solve $Ax = b$ by using the smith normal form of $A$ given by $A = UDV$ with $U \in GL(m, \mathbb{Z})$, $V \in GL(n,\mathbb{Z})$ and $D$ is some integer diagonal matrix following the properties of Smith normal form (the diagonals $d_i | d_{i+1}$).
Then we can solve $Ax = b$ in the following way, let $y = Vx$ and $c = U^{-1}b$, so we have $Dy = c$, solving $Dy = c$ is simple as $D$ is diagonal, then for the elements of $y$ we have \begin{equation*} \begin{pmatrix} y_1 \\ y_2 \\ \vdots \\ y_l \\ \vdots \\ y_n \end{pmatrix} = \begin{pmatrix} c_1/d_1 \\ c_2/d_2 \\ \vdots \\ c_l / d_l \\ \vdots \\ 0 \end{pmatrix} \end{equation*} where rank$(D) = l$, we then have $x = V^{-1}y$ to acquire our solutions.
My question is as we are looking for integer solutions, what if $c_i/d_i \in \mathbb{Q}$ is not an integer? Certainly just because $c_i/d_i$ is not an integer it doesn't meant for $x = V^{-1}y$, $x_i$ is not an integer, but if $x$ attained through this method also does not have all integer solutions does that mean no integer solutions exist - afterall SNF is unique for some integer matrix $A$.