I'm studying the basic analytic number theory material. As of now, I'm stuck with the practice problem.
The problem is following : Find all arithmetic functions f satisfying the given relation, f * f = e
I don't know where should I begin. Should I start with the definition of convolution? Except for e, which is Multiplicative identity function, is there any other arithmetic functions for this relation?
"Should I start with the definition of convolution?"
Yes.
$f*g(n)= \sum_{d|n}f(d)g(n/d)$.
So if $f*f(n) =e(n)$ then $f*f(1) = 1$ so $f*f(1) = f(1)f(1) = 1$ so $f*f(1) = \pm 1$.
If $p$ is prime $f*f(p) = f(1)f(p) + f(p)f(1) = \pm 2f(p)=0$ so $f(p) = 0$.
Strong induction. Suppose we assume $f(m*p) = 0$ for all $m \le n$. (we did the base case with $n = 1$) can we show $f((n+1)p)=0$?
$f*f((n+1)p) = f(1)f((n+1)p) + \sum_{d|n}f(d*p)f(n/d) + \sum_{d|n}f(d)f(p*n/d) + f((n+1)p)f(1) = $
$ f(1)f((n+1)p) + \sum_{d|n}0*f(n/d) + \sum_{d|n}f(d)*0 + f((n+1)p)f(1) =$
$\pm 2f((n+1)p)f(1)= 2f((n+1)p)=0$
So $f((n+1)p) = 0$.
Thus for all $n = m*p$ for some prime, which is to say, for all $n \ne 1$ we have $f(n) = 0$.
So $f(1) = \pm 1$ and $f(n) = 0$ if $n \ne 1$. So $f(n) = \pm e(n)$.