# Thread: [SOLVED] Squarefree and Multiplicative Functions

1. ## [SOLVED] Squarefree and Multiplicative Functions

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: $F(n) = \sum_{d | n} f(n)$ . Find $F(p^{a})$ for any a. Calculate F(144).
================
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.

$f(x)=\begin{cases}
n&\text{if } n \text{is square-free}\\
0&\text{if } n \text{is not square-free} \end{cases}$

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 $m \neq n$, 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) $F(p^{a}) = F(p_{1}^{a1})...F(p_{t}^{at})$, 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.

2. The first two parts look good. For the last part, I think at first, consider p as a prime.

What is $\sum_{d|n} f(d)$ in this case?

3. Originally Posted by Paperwings
(c) $F(p^{a}) = F(p_{1}^{a1})...F(p_{t}^{at})$, 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.
You meant to say, if $n=p_{1}^{a1}...p_{t}^{at}$ then $F(n) = F(p_{1}^{a1})...F(p_{t}^{at})$ ( where all the $p_i$ are different primes)

Now $F(p^a)=f(p^0)+f(p^1)+...+f(p^a)$ (we sum over all the positive divisors of $p^a$).

So $
F\left( {p^a } \right) = \left\{ \begin{gathered}
1 + p{\text{ if }}a \geqslant 1 \hfill \\
1{\text{ if }}a = 0 \hfill \\
\end{gathered} \right.
$
(using the definition of f)

4. Thank you Simon and Paul.

However, I don't understand the first line (should it be just 1 + p if $a \geq 1$):

$

F\left( {p^a } \right) = \left\{ \begin{gathered}
1 + p + p^2 {\text{ if }}a \geqslant 2 \hfill \\
1 + p{\text{ if }}a = 1 \hfill \\
1{\text{ if }}a = 0 \hfill \\
\end{gathered} \right.
$

since for example: $F(8) = F(2^3) = f(1) + f(2) + f(4) + f(8)$.
By definition, since 4 and 8 are not square free, then f(4) = 0 and f(8) = 0.

Then,

$F(8) = f(1) + f(2) = 1 + 2 = 3$.

5. It should be, $1+p$, it was a typo I think

6. Yes, you are right, I put it as if It'd been cube-free instead of square-free.

7. Ah, ok. Thank you for the help.