Let $M \in \mathbb{R}^{n \times n}$ be a symmetric positive-definite matrix whose entries happen to be integers. We can write $$M = R^T R$$ with some invertible matrix $R \in \mathbb{R}^{n \times n}$ which is not unique by any means.
Usually we can't choose $R$ to be integral. Even multiplying $M$ by a scalar (integer) and trying to solve $$a M = R^T R$$ with $R \in \mathbb{Z}^{n \times n}$ is impossible in general when $n$ is even because $$a^n \mathrm{det}(M) = \mathrm{det}(a M) = \mathrm{det}(R^T R) = \mathrm{det}(R)^2$$ would force $\mathrm{det}(M)$ to be square.
This restriction does not happen when $n$ is odd. Question: given $M$ as above, if $n$ is odd, can we always find a scalar $a$ such that $aM$ has the form $R^T R$ with $R \in \mathbb{Z}^{n \times n}$?
No. Let $$ M = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array} \right) $$ One of your forms, is going to be anisotropic in the 2-adic numbers but isotropic in all other $p$-adic numbers. This one, the reverse situation, is isotropic in the $2$-adics, but anisotropic in the $3$-adic numbers.
Give it a try, by your methods.
Analogous, even though the determinant is already a square: $$ N = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 10 \end{array} \right) $$