Let $n$ be a positive integer and let $Q$ be an integral quadratic form in $n$ variables. Let $M$ be the symmetric "two's in" matrix associated with $Q$ so that $Q$ can be expressed as the $1 \times 1$ matrix product \begin{align*} [x_1 \: x_2 \: \ldots \: x_n] \frac{M}2 [x_1 \: x_2 \: \ldots \: x_n]^T \end{align*}
For $i=1, \ldots, n$ we let $m_i$ denote the $i^{th}$ leading principal minor of $M$, that is, the determinant of the $i^{th}$ leading principal submatrix of $M$. (To be clear, the $i^{th}$ leading principal submatrix of a square matrix is the $i \times i$ submatrix containing the first $i$ rows and columns.)
Let $\displaystyle \Delta = \prod_{i=1}^{n-1} m_i$ denote the product of the first $n-1$ minors of $M$. We call this the associated minor product of $M$. It is well known that if $\Delta \neq 0$, then $M$ admits an $LDL^T$ decomposition, where $L$ is unit lower triangular and $D$ is diagonal.
Assume that $\Delta \neq 0$.
As $\Delta \neq 0$, it is well known that the quadratic form $Q$ can be diagonalized over the field $\mathcal{Q}$. In fact, from the above we see that \begin{align*} Q &= [x_1 \: x_2 \: \ldots \: x_n] L \frac{D}2 L^T [x_1 \: x_2 \: \ldots \: x_n]^T\\ &= [X_1 \: X_2 \: \ldots \: X_n] \frac{D}2 [X_1 \: X_2 \: \ldots \: X_n]^T, \end{align*}
where the variables $X_i$ are rational functions of $x_i, x_{i+1}, \ldots, x_n$. Thus, we have expressed our integral quadratic form $Q$ as a diagonal form, where our variables are linear combinations of variables with rational coefficients.
What's more, it is straight forward to determine an expression for the diagonal elements. Suppose we let $D = (d_i)$. Then it can be recursively determined that $\displaystyle d_i = \frac{m_i}{m_{i-1}}$, where for convenience we let $m_0 = 1$.
This diagonalization was first asserted by V. A. Lebesgue (without proof) in 1856. It was likely proven earlier but I cannot find such a proof. Any links to a proof anywhere would be very helpful. Current diagonalizations of quadratic forms focus on spectral decomposition, which isn't helpful in my case.
$\textbf{If no proof can be found}$
I would like to determine an expression for the lower triangular matrix entries.
If we let $L = (\ell_{ji})$ for $1 \leq i \leq j \leq n$ where $\ell_{ii}=1$, then I believe that \begin{align*} \ell_{ji} = \frac{\det(M_i[i,j])}{m_i}, \end{align*} where $\displaystyle M_i[i,j]$ is the $i \times i$ leading principal submatrix, of the matrix obtained by interchanging columns $i$ and $j$ in $M$. Currently, I can show that \begin{align*} \ell_{ji} = \frac{\det(LD_i[i,j])}{m_i} \end{align*} where $\displaystyle LD_i[i,j]$ is the $i \times i$ leading principal submatrix obtained by replacing the $i^{th}$ column in $LD$ with the $j^{th}$ column of $M$.
It is not obvious that $\displaystyle \det(M_i[i,j]) = \det(LD_i[i,j])$, although it is almost certainly true; i.e., I have tested it for small $n$ and the formula establishes itself, but not in a recursive way or a way which identifies a pattern.