How to prove an arithmetic convolution identity? Let $s(n) = n(n + 1)(2n + 1)/6$ the sum of squares.
How can we prove that $(\mu * s)(n) = \tfrac{1}{3} n \varphi(n) + \tfrac{1}{6} \prod_{p|n}(1-p)$?


*

*$\mu(1) = 1$

*$\mu(a^2b) = 0$

*$\mu(p_1 p_2 \cdots p_k) = (-1)^k$

 A: Just write $s(n)$ out as a polynomial.  $s(n)=n^3/3 + n^2/2 + n/6$.
Then determine the convolutions of $n^3$, $n^2,$ and $n$ with $\mu$.  These are easier to compute because $n^k$ is a multiplicitive function, so we only need to determine the values for prime powers.
We can figure out $m^k*\mu$ by evaluating on $p^\alpha$:
$$(m^k*\mu)(p^\alpha) = p^{k\alpha} - p^{k(\alpha-1)}=\phi(p^\alpha) p^{(k-1)(\alpha-1)}\frac{p^k-1}{p-1}$$
So if $n=p_1^{\alpha_1}...p_t^{\alpha_t}$ then:
$$(m^k * \mu)(n) = \phi(n) \, \prod_{i=1}^{t} {p_i^{(\alpha_i-1)(k-1)}\frac{p^k-1}{p-1}}$$
Let $n_0 = p_1p_2...p_t$.  Then 
$$(m^k * \mu) = \phi(n) (\frac{n}{n_0})^{k-1} \prod_i \frac{p^k-1}{p-1}$$
So $$(s*\mu)(n) = {\frac{\phi(n)}{6}}[1 + 3\frac{n}{n_0}\prod(1+p) + 2(\frac{n}{n_0})^2\prod(1+p+p^2)]$$
$$= \frac{\phi(n)}{6}[1 + 3n \prod(1+\frac{1}{p}) + 2n^2 \prod (1+\frac{1}{p}+\frac{1}{p^2})]$$
This has as a lower bound $\frac{\phi(n)n^2}{3}$, and so your formula is wildly incorrect.
A: You cant prove this because it is false.  Let $n=1.$  Then $$\mu*s(1)=\mu (1)s(1)=1$$ while $$\frac{1}{3}1\cdot 1+\frac{1}{6}1=\frac{1}{2}.$$  Similarly if $n=2$ the left hand side is $3$ while the right hand side is $1$.  Lastly when $n=3$, the left hand side is $8$ while the right hand side is $\frac{10}{3}.$  The reason why they can't be equal is that $s(n)$ is not a multiplicative function.
