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 $\varphi (n)$



Möbius function summation
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](http://www.math.tau.ac.il/%7Erudnick/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:
$$A\_n=\exp\left(O\left(\sum\_{p>x}\frac{1}{p^2}\right)\right)=\exp\left(O\left(\frac{1}{x}\right)\right)$$.


