Check my proof: What determinants are possible for integer matrices whose inverses are also integer matrices? I claim this is an adequate proof, but would appreciate criticism.
Trivially, we can see that this would be true for the special linear group ($|det(A)|=1$).
For the inverse condition to be true, the determinant must divide each element of the cofactor matrix, where $c_{ij} = (-1)^{i+j}M_{ij}$. Thus $det(A)|M_{ij} \forall i,j $ .
As the $(n-1)$st minor is a diagonal element, the "multiplicative trace" of $A$ must divide each element in the cofactor matrix.
Thus any integer determinant for A is attainable.
 A: I appear to have made an incorrect move regarding the minor.
I made the incorrect assumption that the minor preserves the diagonal of $A$, removing diagonal elements only one at a time.
We may recursively take minors of any matrix until arriving at a 1x1 matrix of any arbitrary element in $A$ (not only diagonal elements, as I had earlier assumed).
As $det(A)|M_{ij}$, and $M_{ij}|M_{ik,jl}, k\ne i, l\ne j$, and so on up to each single element matrix, $det(A)|a_{ij}$, that is, the determinant divides every element of $A$.
Suppose $A:n\times n;$ $|det A | \ne 1$. Let $|c|\ne1$ be the largest such element dividing each element in $A$. We may multiply each row in A by $c^{-1}$ to get $A'$ with $\det A'=c^{-n}\det A=1$,with the latter given by the condition that $c$ is the largest possible element.
Possible determinants for $A(\mathbb F^{n\times n})$ are therefore $c^n$ $\forall c\in\mathbb F$. In the case of integer matrices, $c\in\mathbb Z$, but then $c$ does not divide the elements of  $A^{-1}$. Therefore $|c|=1$ and A belongs to the special linear group.
