Freshman's dream and the commutativity of the square root of the Möbius function over the divisors.

Question

Let the infinite matrix $A$ be:

$$A(n,k)=\left[ k \mid n \right] \left(\frac{\sqrt{k \, \mu(k)}}{n^s}\right)$$

where $n=1,2,3,4,5,...$ and $k=1,2,3,4,5,...$

Multiply $A$ with its transpose $A^{\mathsf{T}}$:

$$T=A.A^{\mathsf{T}}$$

which I think can be thought of as squaring it: $T=A^2$ compared to if we first take the sum total of matrix $A(n,k)$ and then square it we get the same result as long as $s$ is a real number $>1$:

$\text{Claimed Freshman's dream equality:}$ $$\Re\left(\left(\sum _{n=1}^{\infty} \left(\sum\limits_{k=1}^{\infty} A(n,k)\right)\right)^2\right)=\sum _{n=1}^{\infty}\sum _{k=1}^{\infty}\left(A.A^{\mathsf{T}}\right)(n,k)$$

which is also equal to:

$$=\left(\sum _{n=1}^{\infty} \left(\sum\limits_{\substack{k=1\\k\mid n}}^{\infty} \frac{\sqrt{k \mu (k)}}{n^s}\right)\right)^2=\sum _{n=1}^{\infty} \left(\sum _{k=1}^{\infty} a(\gcd (n,k))\right)=\frac{\zeta (s) \zeta (s)}{\zeta (2 s-1)}$$

where $a(n) = \sum\limits_{d|n} d \, \mu(d)$

Why does this happen, and is this unique to the Möbius function or is it an example of some general matrix factorization identity?

Mathematica program to demonstrate the equality numerically:

(*start*)nn = 200;
s = N[8, 20];
(Sum[Sum[If[Mod[n, k] == 0, (MoebiusMu[k]*k)^(1/2)/n^s, 0], {k, 1, 
     nn}], {n, 1, nn}])^2
(A = Table[
    Table[If[Mod[n, k] == 0, (MoebiusMu[k]*k)^(1/2)/n^s, 0], {k, 1, 
      nn}], {n, 1, nn}]);
N[Total[Total[A.Transpose[A]]], 20]
Zeta[s]*Zeta[s]/Zeta[2*s - 1]
(*end*)

(*start*)
nn = 200;
s = N[8, 20];
(Sum[Sum[If[Mod[n, k] == 0, (MoebiusMu[k])^(1/2)/n^s, 0], {k, 1, 
     nn}], {n, 1, nn}])^2
(A = Table[
    Table[If[Mod[n, k] == 0, (MoebiusMu[k])^(1/2)/n^s, 0], {k, 1, 
      nn}], {n, 1, nn}]);
N[Total[Total[A.Transpose[A]]], 20]
Zeta[s]*Zeta[s]/Zeta[2*s]
(*end*)

Freshman's dream

Answer 1

If you take the form $A(n,k)=[k | n]f(k)n^{-s}$, then the squared sum of the entries of $A$ is \begin{eqnarray*} \left(\sum_{k, n\ge 1} A(n,k)\right)^2&=&\left(\sum_{k, n\ge 1,\ k|n} f(k)n^{-s}\right)^2\\ &=& \left(\sum_{k, a\ge 1} f(k)(ka)^{-s}\right)^2\\ &=& \left(\sum_{k\ge 1} f(k) k^{-s}\right)^2 \left(\sum_{a\ge 1}a^{-s}\right)^2\\ &=& \left(\sum_{k\ge 1} f(k) k^{-s}\right)^2 \zeta(s)^2 \end{eqnarray*} and the sum of the entries of $A A^T$ is \begin{eqnarray*} \sum_{k, m, n\ge 1} A(n, k) A(m, k)&=&\sum_{k, m, n\ge 1,\ k|m, k|n} f(k)^2 m^{-s} n^{-s}\\ &=&\sum_{k, a, b\ge 1} f(k)^2 (ka)^{-s} (kb)^{-s}\\ &=& \left(\sum_{k\ge 1} f(k)^2 k^{-2s}\right) \left(\sum_{a\ge 1} a^{-s}\right) \left(\sum_{b\ge 1} b^{-s}\right)\\ &=& \left(\sum_{k\ge 1} f(k)^2 k^{-2s}\right) \zeta(s)^2. \end{eqnarray*} So, the question reduces to looking at the sum of the squares of $f(k)k^{-s}$ and the real part of the squared sum of $f(k) k^{-s}$.

Assuming that $k\mapsto f(k)^2$ is multiplicative ($f(ab)^2=f(a)^2f(b)^2$ for $a$ and $b$ relatively prime), then $k\mapsto f(k)^2 k^{-2s}$ is as well so $$ \sum_{k\ge 1} f(k)^2 k^{-2s}=\prod_{\hbox{$p$ prime}} (1+f(p)^2 p^{-2s} + f(p^2)^2 p^{-4s} + \cdots) $$ If $f(k)=\sqrt{k \mu(k)}$, then $f(k)^2=k \mu(k)$ is multiplicative and, using the definition of the Möbius function, the product equals $$ \prod_{\hbox{$p$ prime}} (1 - p \cdot p^{-2s})=\zeta(2s-1)^{-1}. $$ For the squared sum, observe that since $\mu(k)$ is always 0 or $\pm 1$, then $$ \sqrt{\mu(k)}= \frac{1+i}{2} \mu(k)^2+\frac{1-i}{2} \mu(k) $$ and so \begin{eqnarray*} \sum_{k\ge 1} f(k) k^{-s}&=&\frac{1+i}{2} \sum_{k\ge 1} \mu(k)^2 k^{(1/2)-s}+\frac{1-i}{2} \sum_{k\ge 1} \mu(k) k^{(1/2)-s}\\ &=& \frac{1+i}{2} \prod_{\hbox{$p$ prime}} (1+p^{(1/2)-s})+\frac{1-i}{2} \prod_{\hbox{$p$ prime}} (1-p^{(1/2)-s})\\ &=& A+B, \qquad \qquad \hbox{say.} \end{eqnarray*} Since $A^2$ and $B^2$ are purely imaginary and $AB$ is real the real part of the square of this will equal $2AB$, which is \begin{eqnarray*} \prod_{\hbox{$p$ prime}} (1+p^{(1/2)-s}) \prod_{\hbox{$p$ prime}} (1-p^{(1/2)-s})&=& \prod_{\hbox{$p$ prime}} (1-p^{1-2s})\\ &=& \zeta(2s-1)^{-1}. \end{eqnarray*} If you were to take the more general form $f(k):=\sqrt{\mu(k)} g(k)$, for some real multiplicative $g$, then you would get $$ \sum_{k\ge 1} f(k)^2 k^{-2s} =\hbox{Re} \left(\sum_{k\ge 1} f(k) k^{-s}\right)^2=\prod_{\hbox{$p$ prime}} (1 - g(p)^2 p^{-2s}). $$