Inverse of an Integer Matrix I found a problem on the 
Open Problem Garden which asks about the conditions on a rectangular, full-rank, integer matrix such that its right inverse (given by:  $A^T (AA^T)^{-1}$ ) is also an integer matrix. The rectangular matrix is constructed in the following way :

Let D be a square diagonal matrix (size $m \times  m$) with integer elements $\geq$ 2 along the main diagonal (in order to ensure full rank and thus existence of a right inverse)

Let X be an integer matrix (size $m \times  n$) with $n\geq m$.

Now, concatenate the matrices to make a new rectangular matrix M = [D X], giving it dimension $m \times  (m+n)$. I am interested in the right inverse of this matrix M.

I have written code to test some matrices, and I have yet to find even one integral element, let alone an entire matrix. I've done some algebraic analysis on a general 2x4 matrix, and intuitively it looks as though some elements will never be a non-zero integer, but it is difficult to prove. If anyone has any advice on how to proceed or any insight, that would be great.

Edits: Clarified the characterization of the matrices in question. Renamed matrices for consistency.
Here's an example for a square 2x2 matrix with each element larger than or equal to 2:

$$\begin{bmatrix}3 &amp; 2\\4 &amp; 3\end{bmatrix}^{-1} = \begin{bmatrix}3 &amp; -2\\-4 &amp; 3\end{bmatrix}$$
You can find a necessary condition for the inverse of a square integer matrix to also be an integer matrix as follows:

If $A$ is a square integer matrix then its determinant $\det (A)$ is an integer. Similarly if $A^{-1}$ exists and is an integer matrix then $\det (A^{-1})$ is an integer too. But $\det(A)\det(A^{-1}) = \det(AA^{-1}) = \det(I) = 1$. So you are looking for two integers whose product is 1 ... there are not many choices ...