find all $f$ satisfying $f * f = 1$ Let $f : \mathbb{N} \rightarrow \mathbb{C}$. Find all $f$ satisfying $$f*f = 1,$$ that is $$\sum_{d|n} f(d)f(\frac{n}{d}) = 1$$ for all $n \in \mathbb{N}.$
Sol. Clearly, $f(1) = 1$ or $f(1) = -1$
Assume first $f(1) = 1$. Let $p$ be prime. Then $$1 = 2f(p).$$ Now, for any $p, q$ distinct prime, $$1 = 2f(pq) + 2f(p)f(q)$$ which yields $f(pq) = 1/4.$
I guess that $f(p_1p_2...p_k) = 1/2^k$, but proving  by induction involves a complicated terms. Moreover, for general $n$, I am not sure for possible formula.
Any help, or suggestion for a more effective ways to deal with this question ?
 A: $$\sum_{d | n} f(d) f(n/d) = 1$$
$$ \implies (\sum_{n=1}^\infty f(n) n^{-s})^2 = \sum_{n=1}^\infty n^{-s} \sum_{d | n} f(d) f(n/d) = \sum_{n=1}^\infty n^{-s}$$
and the trick is 
$$ \implies (\sum_{k=0}^\infty f(p^k)p^{-sk})^2 = \sum_{k=0}^\infty p^{-sk} = \frac{1}{1-p^{-s}}$$
even when $f(n)$ is not multiplicative.
So we get (using the binomial series)
$$\sum_{k=0}^\infty f(p^k)p^{-sk} = \pm (1-p^{-s})^{-1/2} = \pm \sum_{k=0}^\infty {-1/2 \choose k} (-1)^k p^{-sk}$$
So that $f(p^k) = \pm {-1/2 \choose k} (-1)^k$.
But we have the right to choose $\pm$ only once : for  $f(1) = \pm 1$ ! Then we have $f(p^k) = f(1) {-1/2 \choose k} (-1)^k$ at every prime power.
Finally, if $f(d)$ is known for every $d | n, d < n$, then
$$\sum_{d | n} f(d)f(n/d) =1 \quad  \implies \quad f(n) = \frac{1}{2f(1)}(1-\sum_{d | n, 1 < d < n} f(d)f(n/d))$$
So once the prime powers are chosen, we have no other choices.
Hence there is the multiplicative solution $f(p^k) = {-1/2 \choose k} (-1)^k, f(n) = \prod_{p^k \| n} f(p^k) $, 
and the non-multiplicative solution $f_0(n) = -f(n)$.
$\qquad \scriptstyle \text{note : only formal series are involved here, no convergence problem}$
