Coprime Symmetric Matrices We define two $ n\times n $ matrices $ c,d \in M_{n}(\mathbb{Z}) $ to be co-prime if
$$\{\gamma \in M_{n}(\mathbb{Q})|\gamma c,\gamma d \in M_{n}(\mathbb{Z})\} = M_{n}(\mathbb{Z}) .$$
We also define a pair of matrices $ (c,d) $ to be a symmetric pair if
$$cd^{T} = dc^{T} .$$
We define an equivalence relation $ \sim $ on the set of all coprime pairs of symmetric matrices by
$$ (c,d) \sim (c',d') \textit{ if there exists } u \in \textit{SL}_{n}(\mathbb{Z}) \textit{ such that } (c,d) = (uc',ud') ,$$
and set $ \Delta $ equal to the set of all equivalence classes. Now the paper I'm reading makes the following claim:
The set
$$\{(c,d) \in \Delta:|c| \neq 0\} $$
is in bijective correspondence with the set
$$ \{a\in M_{n}(\mathbb{Q}|a^{T} = a\} $$
via the prescription $ (c,d) \mapsto c^{-1}d $.
The paper claims this is a well known result however I am am not aware of this result. Does anybody know the name of this result, or where I might find a proof. I get this impression that this is from a paper of Siegel's, however I am neither able to read German, nor get a copy quickly.
As usual any help is much appreciated.
 A: The original question as is stated is FALSE. 
If you replace $SL_n(\mathbb Z)$ with $GL_n(\mathbb Z)$ in the equivalence relation, then it is true. Alternatively you may use the set $\{(c,d)\in\Delta:\  \det(c)>0\}$ instead of $\{(c,d)\in\Delta:\ \det(c)\neq 0\}$.
Here a proof.
First, let me note that if $d=I$ is the identity matrix, then any matrix $c$ is coprime with $I$ and for any symmetric $c$, the pair $(c,d)$ is coprime and symmetric. 
Since the OP uses $c^{-1}$ I suppose that $c$ is invertible.
Let call $F$ the map $F(c,d)=c^{-1}d$. We need to check that
1) $F$ is well defined on equivalence classes
2) The image of $F$ is in the set of symmetric matrices
3) $F$ is injective
4) $F$ is surjective on the set of symmetric matrices.
We see 1). Clearly if $a=uc$ and $b=ud$ then $$F(a,b)=a^{-1}b=c^{-1}u^{-1}ud=c^{-1}d=F(c,d).$$
Wee see 2). If $dc^T=cd^T$ then 
$$c^{-1}dc^{T}=d^T$$ whence, since $(c^T)^{-1}=(c^{-1})^T$,
$$c^{-1}d=d^T(c^{T})^{-1}=d^T(c^{-1})^T=(c^{-1}d)^T$$
So $F(c,d)$ is symmetric.
We see $3)$. Suppose that $F(c,d)=F(a,b)$. Then 
$$c^{-1}d=a^{-1}b\qquad b=ac^{-1}d\qquad a=ac^{-1}c$$
thus, setting $u=ac^{-1}$ we see that $$(a,b)=(uc,ud)\qquad (u^{-1}a,u^{-1}b)=(c,d)$$ 
Since $(a,b)$ and $(c,d)$ are both coprime pairs, this forces both $u$ and $u^{-1}$ to belong to $M_n(\mathbb Z)$. Therefore $u\in GL_n(\mathbb Z)$ and $\det(u)=\pm1$.
Thus $(a,b)$ and $(c,d)$ are in the same $GL_n(\mathbb Z)$-equivalence class. Note that if we restrict to the set of comprime symmetric pairs $(x,y)$ where $\det x>0$ then we have $\det u=\det a\det c^{-1}>0$, hence $(a,b)$ and $(c,d)$ are in the same $SL_n(\mathbb Z)$-equivalente class.
We see $4)$. If $a$ is symmetric, then we can write 
$$a=\dfrac{1}{n}d$$ where $d$ has integers coefficients and $n$ is the minimum common denominator of its coefficents (set $n=1$ if $d=0$). Thus, setting $c=nId$ we have $a=c^{-1}d=F(c,d)$. Clearly $(c,d)$ is coprime by the choice of $n$, and symmetric beacuse $a$ is symmetric, so $d$ is, $c$ is a multiple of $Id$ so it is symmetric and commutes with any matrix.
Counterexample to the original question.
If $n=2k+1$ then the $-Id\in GL_n(\mathbb Z)$ but not in $SL_n(\mathbb Z)$.
Thus, given any symmetric coprime pairs $(c,d)$ we have that $(-c,-d)$ is not equivivalent to $(c,d)$, however $F(c,d)=F(-c,-d)$. This shows that the correspondence is not a bijection.
If $n$ is $2k$ than consider the block diagonal matrix
$u=\begin{pmatrix}U& & & \\& I & & \\ & & \ddots & \\ & & & I\end{pmatrix}$ where 
$U=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ and $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$
Then, given any symmetric coprime pair $(c,d)$ the pair $(uc,ud)$ is symmetric and coprime (note that $u^{-1}=u=u^{T}$). 
However, $u\in GL_n(\mathbb Z)$ but $\det u=-1$ so $u$ is not in $SL_n(\mathbb Z)$. It there is $M\in GL_n(\mathbb Z)$ such that $(Muc,Mud)=(c,d)$ then, since $c$ is invertible we would have $Mu=Id$ and thus $1=\det M\det u=-\det M$. This shows that $(uc,ud)$ is not equivalent to $(c,d)$. However, $F(uc,ud)=F(c,d)$.
In conclusion, the set $\{(c,d)\in\Delta \det c\neq 0\}$ is NOT in bijection with the set $\{a\in M_n(\mathbb Q): a^T=a\}$.
