$M_n(\mathbb{Z}) = SL_n(\mathbb{Z}) + SL_n(\mathbb{Z})$ I came across the following problem and I'm stuck.

Let $n>1$ be an even integer, and $A\in \mathcal{M}_n(\mathbb{Z})$.
Show that there exist $B,C \in \mathrm{SL}_n(\mathbb{Z})$, such that  $A=B+C$.

Any hint is appreciated.
 A: Ideas :
Use Smith normal form. Then $M=PDQ$ where $D$ is a diagonal matrix and $P,Q$ are invertible matrices in $\mathcal{M}_n(\mathbb{Z})$ with $det(P)=det(Q)=\pm{1}$.
A: You may prove the following conjecture first. If it's true, the statement in question is almost trivial.

Conjecture. When $m\ge2$, every $m\times m$ integer matrix $B$ is the sum of two integer matrices $C$ and $D$ with $\det(CD)=(-1)^m$.

Proof of the main statement when the conjecture is true.
Let $n=2m\ge4$. Partition the $n\times n$ integer matrix $A$ into four subblocks of equal sizes, i.e. $A=\pmatrix{X&Y\\ Z&W}$. Write $X=C_1+D_1$ and $W=C_2+D_2$ according to the conjecture. Interchange the roles of $C_1$ and $D_1$ if necessary, so that $\det(C_1)=\det(C_2)$. Then $A=\pmatrix{X&Y\\ Z&W}=\pmatrix{C_1&Y\\ 0&C_2}+\pmatrix{D_1&0\\ Z&D_2}$ is the required sum. $\square$

Outline proof of the conjecture. While I believe that the following ideas work, I may be mistaken and you should verify each step carefully.
We shall prove the above conjecture by mathematical induction. Note that $B=C+D$ with $\det(CD)=(-1)^m$ if and only if $PBQ=PCQ+PDQ$ with $\det\left((PCQ)(PDQ)\right)=(-1)^m$ for any integer matrices $P$ and $Q$ with determinants $\pm1$. In other words, the conjecture is invariant under left or right multiplication of matrices with unit determinants to $B$.
There are three kinds of integer matrices that have unit determinants and are pertinent to our conjecture. The first one is a permutation matrix. The second one, which I shall call a "GCD matrix", is used to turn one of two nonzero matrix entries $a$ and $b$ on the same row or the same column into $d=\operatorname{gcd}(a,b)$. To illustrate, suppose $ax+by=d$ for some integers $x$ and $y$. Then
$$
\underbrace{\pmatrix{x&y\\ -\frac bd &\frac ad}}_{\large\det=1}
\pmatrix{a\\ b}
=\pmatrix{d\\ kd}
$$
for some integer $k$. Consequently, one can use a third kind of integer matrix -- an elementary matrix for Gaussian elimination -- to turn an entry into zero:
$$
\pmatrix{1&0\\ -k&1}
\pmatrix{x&y\\ -\frac bd &\frac ad}\pmatrix{a\\ b}
=\pmatrix{d\\ 0}.
$$
Now, suppose the base case $m=2$ is true. In the inductive step, we can use some appropriate permutation matrices, GCD matrices and Gaussian eliminations to turn the last entry of $B$ into zero. Hence one can decompose $B$ into the desired sum in a manner akin to how we decompose $A$ (with $0=-1+1$ when decomposing the last entry).
For the base case $m=2$, modulo permutations of rows or columns, it suffices to consider the following three cases.


*

*$B$ is diagonal. We have $\pmatrix{a&0\\ 0&d}=\pmatrix{a&1\\ 1&0}+\pmatrix{0&-1\\ -1&d}$.

*$B=\pmatrix{a&b\\ 0&0}$ with $a,b\ne0$. By right multiplications of a GCD matrix and a Gaussian elimination matrix, one can kill the $(1,2)$-th element and turn $B$ into a diagonal matrix.

*$B=\pmatrix{a&b\\ c&d}$ with $a,b,c\ne0$. By left multiplications of GCD matrix and Gaussian elimination matrix, we can kill $c$ and assume that $B$ is of the form $\pmatrix{a&b\\ 0&d}$ with $a,b\ne0$. In view of case 2, we may further assume that $d\ne0$. By a left multiplication of GCD matrix followed by a right multiplication of another GCD matrix, one can turn the $(1,2)$-th entry of $B$ into $\operatorname{gcd}(a,b,d)$. At this point, the new $B$ is no longer upper triangular, but all its entries are integer multiples of the new $b$ (which is the GCD of the original $a,b,d$). So, one can use Gaussian eliminations the two diagonal elements, and the resultant $B$ becomes an antidiagonal matrix, which reduces to case 1 by permutation.

