# determinant of divisor functions

Let $$A$$ be the $$(n-1) \times (n-1)$$ matrix whose entries $$a_{ij}=d(\gcd(i+1,j+1))$$. Here, $$d(n)$$ means the number of divisors of $$n$$. It seems that the determinant of it is the number of square-free positive integers less than or equal to $$n$$. How to prove it?

For $$a_{ij}=d(\gcd(i,j))$$, the determinant is $$1$$, because it is the product $$CC^T$$,where $$c_{ij}=$${if j|i then 1 else 0} But this method seems not to work on the $$\gcd(i+1,j+1)$$ case.

Let $$X$$ be the $$n \times n$$-matrix obtained as follows: in the first row, the $$i$$-th entry equals $$\mu(i)$$ where $$\mu$$ denotes the Möbius function, and in the $$j$$-th row ($$j>1$$), the $$i$$-th entry is $$1$$ if $$i$$ divides $$j$$ and $$0$$ otherwise.
For example, for $$n=8$$ we obtain $$X=\left( \begin{array}{cccccccc} 1 & -1 & -1 & 0 & -1 & 1 & -1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \\ \end{array} \right).$$
Now $$XX^t$$ contains the matrix of interest $$A$$ as a submatrix: $$XX^t = Y = \left( \begin{array}{cccccccc} 6 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 2 & 1 & 2 & 1 & 2 \\ 0 & 1 & 2 & 1 & 1 & 2 & 1 & 1 \\ 0 & 2 & 1 & 3 & 1 & 2 & 1 & 3 \\ 0 & 1 & 1 & 1 & 2 & 1 & 1 & 1 \\ 0 & 2 & 2 & 2 & 1 & 4 & 1 & 2 \\ 0 & 1 & 1 & 1 & 1 & 1 & 2 & 1 \\ 0 & 2 & 1 & 3 & 1 & 2 & 1 & 4 \\ \end{array} \right).$$ Let $$k$$ be the number of squarefree integers in $$\{1,2,\ldots,n\}$$. Then we claim that for general $$n$$ we have $$\det X = k$$ and that $$XX^t$$ is the block matrix $$Y = \left( \begin{array}{cc} k & 0 \\ 0 & A \end{array} \right).$$ This would imply that $$k \det(A) = \det(Y) = \det (XX^t) = \det(X)^2 = k^2$$ or $$\det(A)=k$$, as desired.
Let us first show that $$XX^t = Y$$. We check the equality of the element in row $$i$$ and column $$j$$. For $$i,j > 1$$ this follows as in the $$\gcd(i,j)$$-case. For $$i=j=1$$ this follows from $$k = \sum_{i=1}^n \mu(i)^2$$. For the other cases this follows from the formula $$\sum_{d \mid n} \mu(d) = 0 \qquad \mbox{for } n > 1.$$
It remains to show that $$\det X = k$$. For each $$2 \leq i \leq n$$, we add the $$i$$-th column $$\mu(i)$$ times to the first column. Then the determinant does not change and the first column will contain a $$k$$ in the first row and $$0$$ in other rows by the formula $$\sum_{d \mid n} \mu(d) = 0$$. In our $$n=8$$ example we find the following: $$\left( \begin{array}{cccccccc} 6 & -1 & -1 & 0 & -1 & 1 & -1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \\ \end{array} \right).$$ It is now clear that (in general) the determinant equals $$k$$ by noting that the term corresponding to the diagonal is the only nonzero term.