Does Anyone Know Anything about the Determinant and/or Inverse of this Matrix? The matrix I am inquiring about here is the $n \times n$ matrix where the entry $A_{ij}$ is $\frac{1}{(i+j-1)^2}$. The $2 \times 2$ matrix looks like 
$$
\begin{pmatrix}
1 & 1/4 \\
1/4 & 1/9
\end{pmatrix}.
$$
The $3 \times 3$ case looks like 
$$
\begin{pmatrix}
1 & 1/4 & 1/9 \\
1/4 & 1/9 & 1/16 \\
1/9 & 1/16 & 1/25 
\end{pmatrix}.
$$
The $n \times n$ matrix looks like
$$
 \begin{pmatrix}
1 & 1/4 & \cdots & 1/(n^2) \\
1/4 & 1/9 & \cdots & 1/{(n+1)^2} \\
\vdots & \vdots & \cdots & \vdots \\
1/(n^2) & 1/{(n+1)^2} & \cdots & 1/{(2n-1)^2}
\end{pmatrix}.
$$
If anyone has any information or knows of any papers talking about this matrix please let me know. Thanks!
 A: The determinant formula for $\det (1/(x_i+y_j)^2)_{i,j}$ is due to Borchardt, see Krattenthaler's article given in Steve's answer, which contains a $(q)$-deformation of it as well. I want to mention, that Emmanuel Preissmann conjectured also a formula for the generalized Cauchy determinant identity 
$$\det \left(\frac{1}{(x_i+y_j)^k}\right)_{i,j}$$  for every $k\ge 1$, see http://ipg.epfl.ch/~leveque/Conjectures/cauchy.pdf. 
Further generalizations of Cauchy's determinant identity are given in Okada's paper 
http://www.math.kobe-u.ac.jp/publications/rlm18/9.pdf, where $(1.6)$ is again Borchardt's identity.
A: See (3.9) and (3.10)/Theorem 25 in Krattenthaler's "Advanced determinant calculus",  http://arxiv.org/abs/math/9902004
A: Just a couple observations, building upon and explaining why Igor Rivin found that the denominators are often perfect squares or close to being such.
Let $A$ be the matrix with components $A_{ij} = 1/(x_i + y_j)^2$.  Expressing its determinant as a sum over permutations and putting everything on the same denominator gives
$$
\det A = \frac{P(\{x_i\}, \{y_i\})}{\prod_{i,j} (x_i + y_j)^2} \,,
$$
for some polynomial $P$ with integer coefficients.  Also, if two $x_i$ or two $y_i$ are equal, then the corresponding rows/columns are equal, and the determinant vanishes.  Hence
$$
\det A = \frac{Q(\{x_i\}, \{y_i\}) \prod_{i\neq j} (x_i - x_j) (y_i - y_j)}{\prod_{i,j} (x_i + y_j)^2} \,.
$$
Actually, as pointed out in other answers, the Cauchy formula provides $Q$ as the permanent of some matrix, but all I need here is to know $Q$ has integer coefficients.
In your particular case, the products give
$$
\frac{\prod_{i\neq j} (i - j) ((i - 1) - (j - 1))}{\prod_{i,j} (i + j - 1)^2}
=
\frac{\prod_{i < j} (j - i)^4}{\prod_{i,j} (i + j - 1)^2}
=
\prod_{j=1}^N \left[(j - 1)!^4 \frac{(j - 1)!^2}{(j + N - 1)!^2} \right]
=
\frac{\prod_{j=0}^{N-1} j!^8}{\prod_{j=0}^{2N-1} j!^2}
$$
Note that $j!^8$ divides $(2j)!(2j+1)!$, so the numerator divides the denominator.  All in all,
$$
\det A = \frac{Q}{\prod_{j=0}^{2N-1} j!^2 \bigg/ \prod_{j=0}^{N-1} j!^8}
$$
for some integer $Q$.  The denominator appearing in this formula is typically rather close to the ones given in Igor Rivin's answer, and are trivially squares: $1$, $12^2$, $2160^2$, $6048000^2 = 672000^2 \cdot 9^2$, etc.  The discrepancy is of course accounted for by cancellations between $Q$ and the product of factorials.  Again using Igor Rivin's data, I find (indices denote the size of the matrix)
$$
\begin{aligned}
  Q_1 &= 1 \\
  Q_2 &= 7 \\
  Q_3 &= 647 = 2^3 \cdot 3^4 - 1 = 8 \cdot (16 \cdot (6 - 1) + 1) - 1 \\
  Q_4 &= 878769 = 16 \cdot (12 \cdot (32 \cdot (144 - 1) + 1) - 1) + 1 \\
  Q_5 &= 18203480001 = 40000 \cdot ( 48 \cdot ( 120 \cdot (80 - 1) + 1) - 1) + 1 \\
  Q_6 &= 5850859031888599 \,.
\end{aligned}
$$
Here I've included some expressions which may or may not generalize.  In any case, it is certain that $Q$ will not be expressed as a (simple) product, as it involves some very large prime factors.
