Möbius function summation How can I evaluate ( estimate ) the sum $s(n)=\sum_{k=1}^{n-1}\mu ^{2}(k(n-k))$ ($\mu$ is the Möbius function). Trivial estimate
$s(n)<\varphi (n)$ follows from the fact that $\mu (k(n-k))$ is zero if $k$ is not prime to $n$. It seem that in the case of $n$ - primorial,  $s(n)$ is pretty close to $\varphi (n)$
 A: Let $f_n(x)=x(n-x)$. You count squarefree values of $f_n(k)$ for $k$ between $1$ and $n$.
Given a prime $p$ let $a_{p,n}$ be the number of solutions $k\in \mathbb{Z}/p^2\mathbb{Z}$ to the congruence $f_n(k)\equiv 0\bmod p^2$.
The heuristic asymptotic answer is $n$ times the following infinite product over primes
$$\prod_{p} \left(1-\frac{a_{p,n}}{p^2} \right),$$
where the $p$th term stands for the probability that $f_n(k)$ is indivisible by $p^2$. This answer can be established unconditionally for quadratic polynomials $f$; this is due to Ricci (1933). Reference and proofs of special cases are given in these lecture notes of Rudnick:
http://www.math.tau.ac.il/~rudnick/courses/sieves2015/squarefrees.pdf
The mentioned result of Ricci deals with a fixed polynomial, while your $f_n$ varies with $n$. However, the arguments can be made uniform in your choice of $f_n$.
Let us investigate the above product. The value of $a_{p,n}$ is $2$ if $n$ is indivisible by $p$ (corresponding to $k\equiv 0,n\bmod p^2$) and is equal to $p$ otherwise (corresponding to $k\equiv 0,p,2p,\ldots\bmod p^2$). Hence the product is $A_n B_n$ where
$$A_n=\prod_{p\nmid n} \left(1-\frac{2}{p^2}\right),$$
$$B_n=\prod_{p\mid n} \left(1-\frac{1}{p}\right)=\frac{\phi(n)}{n}.$$
Then we see that your sum, divided by $\phi(n)$, is asymptotic to $A_n$. The constant $A_n$ is asymptotic to $1$ if $n$ is the product of all primes up to $x$, because we can estimate it naively as follows, if we take its logarithm:
$$A_n=\exp\left(O\left(\sum_{p>x}\frac{1}{p^2}\right)\right)=\exp\left(O\left(\frac{1}{x}\right)\right)$$.
