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I have

$F(n) = \sum_{d\mid n} f(d)$ where $F$ is multiplicative. I need to prove that $f$ is multiplicative, without the Mobius inversion formula, which is introduced in a later chapter. The following is my work:

$\because F(1) = 1$, $$\sum_{d\mid 1}^{}f(d) = f(1) = 1$$ Let $p$ and $q$ be prime. $$F(p) = \sum_{d\mid p}^{} f(p) = f(1) + f(p) = 1 + f(p)$$ $$f(p) = F(p) - 1$$ Similarly, $$F(p^2) = \sum_{d\mid p^2}^{}f(d) = f(1) + f(p) + f(p^2) = 1 + F(p) - 1 + f(p^2) = F(p) + f(p^2)$$ $$f(p^2) = F(p^2) - F(p)$$ $$F(p^3) = \sum_{d\mid p^3}^{} f(d) = f(1) + f(p) + f(p^2) + f(p^3) = 1 + F(p) - 1 + F(p^2) - F(p) + f(p^3)$$ $$\therefore f(p^3) = F(p^3) - F(p^2)$$\newline We claim that $f(p^n) = F(p^n) - F(p^{n-1})$. We prove this by strong induction. Let the statement $f(p^k) = F(p^k) - F(p^{k-1})$ be true for $1\leq k \leq n.$Then $$F(p^{n+1}) = f(1) + f(p^{2}) + \dots + f(p^{n+1})$$ $$\therefore F(p^{n+1}) = {1} + {F(p)} - {1} + {F(p^2)} - {F(p)} + \dots + {F(p^{n-1})} - {F(p^{n-2})} + F(p^{n}) - {F(p^{n-1})} + f(p^{n+1})$$ $$f(p^{n+1}) = F(p^{n+1})-F(p^n)$$Thus our claim is true, by induction. $$f(p^r)f(q^s) = [F(p^r) - F(p^{r-1})][F(q^s) - F(q^{s-1})] = F(p^r q^s) - F(p^r q^{s-1}) - F(p^{r-1} q^{s}) + F(p^{r-1} q^{s-1}) $$ $\because F$ is multiplicative. $$f(p^r)f(q^s) = \Bigg[\sum_{d\mid p^r q^s}^{} f(d) - \sum_{d\mid p^r q^{s-1}}^{} f(d)\Bigg] - \Bigg[\sum_{d\mid p^{r-1} q^s}^{} f(d) - \sum_{d\mid p^{r-1} q^{s-1}}^{} f(d)\Bigg]$$ $$f(p^r)f(q^s) = \Bigg[f(q^{s}) + f(pq^{s}) + \dots + f(p^{r-1}q^{s}) + f(p^{r}q^{s})\Bigg] - \Bigg[f(q^{s}) + f(pq^{s}) + \dots + f(p^{r-1}q^{s}) \Bigg] = f(p^r q^s)$$

How do I get from here to $f$ is multiplicative? Thank you.

EDIT:

Okay, from the first comment I get:

$$\sum_{d\mid mn} f(d) = \sum_{u\mid m} \sum_{v\mid n} f(uv)$$

Where m an n are coprime. And $u$ and $v$ are coprime too. And there is a bijection between $d$ and $uv$.

Also:

$$\sum_{d\mid mn} f(d) = F(m)F(n) = \sum_{u\mid m} f(u) \sum_{v\mid n} f(v) = \sum_{u\mid m} \sum_{v\mid n} f(u)f(v) $$

$$\therefore f(uv) = f(u)f(v)$$ and thus $f$ is multiplicative.

  • 1
    Say $\gcd(m,n)=1$. From $F(mn)=F(m)F(n)$ you get an equation relating values of $f$. See where it leads. – 2017-02-08
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    Thank you so much. Can you check out my edit and see if I've got a proof? – 2017-02-08
  • 0
    How do you go from $\sum x=\sum yz$ to $x=yz$? – 2017-02-08

1 Answers 1

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This can be done by induction.

Assume $f$ is not multiplicative. Let $f(m)f(n) \ne f(mn)$ where $m$ and $n$ are relatively prime and $mn$ is chosen to be minimal.

Now

$$ F(m)F(n) = \sum_{d|m} f(d) \sum_{e|n} f(e) = \sum_{d|m, e|n} f(d)f(e), $$ and $$ F(mn) = \sum_{l|mn} f(l) = \sum_{d|m, e|n} f(de). $$

By the choice of $m$ and $n$, all the terms in the sums on the right are identical except the term $f(m)f(n)$ in the first equation and the term $f(mn)$ in the second, which are assumed to differ. It follows that $F(mn) \ne F(m)F(n)$, a contradiction.