This is inspired by a recent question about the existence of *orthogonal Cauchy-like matrices*. It is proved that there are indeed such matrices, i.e. there are vectors $x,y,r,s\in\mathbb R^n$ such that for the matrix $C$ defined by
$$
C_{ij} = \frac{r_i s_j}{ x_i - y_j},
$$ we have $C C^T = I$.

In the pdf with the solution, the following fact is used: If $C$ is a $n \times n$ real *Cauchy matrix* $C$, i.e. there are vectors $x,y\in\mathbb R^n$ such that $$C_{ij} = \frac{1}{ x_i - y_j},$$ then its inverse admits the factorization
$$C^{-1} = D_aC^TD_b,$$
where $D_a,D_b$ denote the diagonal matrices with the entries of $a, b\in\mathbb R^n$, which can be computed by explicit formulae obtained by Lagrangian interpolation. BTW, it is not clear to me why the RHS features $C^T=(( \frac{1}{ x_j - y_i}))_{i,j=1}^n$, as the given formula $(7)$ in Theorem 1 of the original article (accessible here with different notation, as it calls $a,b$ what we call $x,y$ here) seems to contain rather $(( { x_j - y_i}))_{i,j=1}^n$ than the reciprocals.

In any case, my question is

whether the vectors $x,y,r,s\in\mathbb R^n$ defining a Cauchy-like matrix can all have

entries, i.e. whether there is a Cauchy matrix $C$ such that each entry of the corresponding vectors $a,b$ in the factorization of $C^{-1}$ quoted above is a (rational) square?rational

Note that the given construction shows that there are Cauchy matrices where $a,b$ have only positive entries and uses their square roots at some point. But for *rational* square roots, I wouldn't know how to go about that even for $n=2$.