Problem:

Let f be the arithmetic function given by f(n) = n if n is square-free and f(n) = 0 if n is not square-free.

(a) Prove that f is a multiplicative function.

(b) Is f completely multiplicative?

(c) Let F be the summatory function of f: $\displaystyle F(n) = \sum_{d | n} f(n)$ . Find $\displaystyle F(p^{a})$ for any a. Calculate F(144).

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Attempt:

A square-free number is a number that its unique prime factors do not repeat. 4(which is 2*2), 8(which is 2*2*2), 9(which 3*3).. are not square-free.

So, for part (a), I would like to show for some m, n, then

f(mn) = f(m)f(n) by the definition of multiplicative.

$\displaystyle f(x)=\begin{cases}

n&\text{if } n \text{is square-free}\\

0&\text{if } n \text{is not square-free} \end{cases} $

In order to show this, I broke it into cases.

If m, n are not square free then f(mn) = 0 = f(m)(n)

If either m or n are not square free then f(mn) = 0, so either f(m) = 0 or f(n) = 0.

If both m and n are square free and $\displaystyle m \neq n $, then f(mn) = m*n=f(m)f(n)

(b) I say this is not completely multiplicative since if m = n, then f(mn) = f(mm) = f(nn) = not a square free number; thus, f(mn) = 0.

(c) $\displaystyle F(p^{a}) = F(p_{1}^{a1})...F(p_{t}^{at})$, but I don't know how to derive a formula for the summatory function F(n). I know that since small f is multiplicative, then so is big F.