The Divisor and Sigma Functions

We use the divisor-pairing technique for coprime moduli, a standard tool built directly on the definition of multiplicativity and the properties of $\gcd$. First, $F(1) = \sum_{e \mid 1} f(e) = f(1) = 1$, since $f$ is multiplicative and the only divisor of $1$ is $1$ itself. Now let $m,n$ be positive integers with $\gcd(m,n) = 1$. We must show $F(mn) = F(m)F(n)$.

Step 1: divisors of $mn$ correspond bijectively to pairs of divisors. Define a map

We claim $\Phi$ is a bijection. $\Phi$ is well-defined: if $d_1 \mid m$ and $d_2 \mid n$, then $d_1 d_2 \mid mn$ trivially, since $m = d_1 m'$ and $n = d_2 n'$ give $mn = (d_1d_2)(m'n')$. $\Phi$ is injective: suppose $d_1 d_2 = d_1' d_2'$ with $d_1, d_1' \mid m$ and $d_2, d_2' \mid n$. Since $\gcd(m,n)=1$, also $\gcd(d_1, d_2') = 1$ (as $d_1 \mid m$ and $d_2' \mid n$, and any common factor of $d_1, d_2'$ would be a common factor of $m,n$). From $d_1 d_2 = d_1' d_2'$ we get $d_1 \mid d_1' d_2'$; since $\gcd(d_1,d_2')=1$, Euclid's lemma gives $d_1 \mid d_1'$. By the symmetric argument $d_1' \mid d_1$, so $d_1 = d_1'$ (both positive). Then $d_2 = d_2'$ follows by cancellation. Hence $(d_1,d_2) = (d_1',d_2')$. $\Phi$ is surjective: let $e \mid mn$. Set $d_1 = \gcd(e,m)$ and $d_2 = \gcd(e,n)$. Clearly $d_1 \mid m$ and $d_2 \mid n$. We show $e = d_1 d_2$. Since $\gcd(m,n)=1$, we have $\gcd(d_1,d_2) = 1$ (a common divisor of $d_1,d_2$ divides both $m$ and $n$, hence divides $\gcd(m,n)=1$). Also $d_1 d_2 \mid e$: this requires a short argument. Write $e = \gcd(e,mn)$ since $e\mid mn$. Because $\gcd(m,n) = 1$, one can show $\gcd(e,mn) = \gcd(e,m)\gcd(e,n)$ — this is a standard consequence of unique factorization: for each prime $p$, $v_p(mn) = v_p(m) + v_p(n)$ and at most one of $v_p(m), v_p(n)$ is nonzero (as $\gcd(m,n)=1$), so $\min(v_p(e), v_p(m)+v_p(n)) = \min(v_p(e),v_p(m)) + \min(v_p(e),v_p(n))$ once one checks both sides agree in the two cases $v_p(m)=0$ or $v_p(n) = 0$. Taking the product over all primes gives $\gcd(e,mn) = \gcd(e,m)\gcd(e,n) = d_1 d_2$. Since $e \mid mn$, $\gcd(e,mn) = e$, so $e = d_1 d_2$, confirming surjectivity. Thus $\Phi$ is a bijection between divisors of $mn$ and pairs $(d_1,d_2)$ with $d_1 \mid m$, $d_2 \mid n$.

Step 2: use multiplicativity of $f$ and coprimality of the parts. For any divisor pair from Step 1, $d_1 \mid m$ and $d_2 \mid n$ with $\gcd(m,n)=1$ forces $\gcd(d_1,d_2) = 1$ (shown above). Since $f$ is multiplicative,

Step 3: factor the double sum. Using the bijection $\Phi$ to re-index the sum defining $F(mn)$:

The last step is legitimate distributivity of finite sums (factoring a product of sums into a double sum, and back), valid since all sums here are finite. This proves $F(mn) = F(m)F(n)$ for all coprime $m,n$, and $F(1)=1$, so $F$ is multiplicative. $\blacksquare$

Apply the theorem with $f(e) = 1$ for all $e$ (which is trivially completely multiplicative, since $1\cdot 1 = 1$ for all inputs, and in particular multiplicative) to get that $F(n) = \sum_{e\mid n} 1 = d(n)$ is multiplicative. Apply the theorem with $f(e) = e^k$. This $f$ is completely multiplicative: $f(ab) = (ab)^k = a^k b^k = f(a)f(b)$ for all $a,b$ (no coprimality needed), and $f(1) = 1^k = 1$. In particular $f$ is multiplicative, so the theorem applies and $F(n) = \sum_{e \mid n} e^k = \sigma_k(n)$ is multiplicative. $\blacksquare$

Prime power case. The positive divisors of $p^a$ are exactly $1, p, p^2, \dots, p^a$. This is because any divisor $e$ of $p^a$ must, by unique factorization, be of the form $e = p^b$ for some $0 \le b \le a$: indeed if a prime $q \ne p$ divided $e$ then $q$ would divide $p^a$, contradicting unique factorization (the only prime factor of $p^a$ is $p$); and the exponent of $p$ in $e$ cannot exceed $a$ since $e \mid p^a$. Conversely every $p^b$ with $0\le b\le a$ visibly divides $p^a$. This gives exactly $a+1$ divisors (namely $b = 0,1,\dots,a$), so

Summing $k$-th powers over these divisors,

a finite geometric series with ratio $p^k$ and $a+1$ terms. Since $p \ge 2$, we have $p^k \ne 1$ whenever $k \ne 0$ (as $p^k = 1 \iff k=0$ for $p > 1$ real), so the standard finite geometric series formula applies:

General $n$. Write $n = p_1^{a_1}\cdots p_r^{a_r}$ with the $p_i$ distinct primes; this factorization exists and is unique by the Fundamental Theorem of Arithmetic. We prove the product formula by induction on $r$. Base case $r=1$: $n = p_1^{a_1}$, and the prime-power formulas above give exactly $d(n) = a_1+1$ and $\sigma_k(n) = \frac{p_1^{k(a_1+1)}-1}{p_1^k-1}$, matching the claimed product of one factor. Inductive step: suppose the formula holds for integers with $r-1$ distinct prime factors ($r \ge 2$). Let $n = p_1^{a_1}\cdots p_r^{a_r} = m \cdot p_r^{a_r}$ where $m = p_1^{a_1}\cdots p_{r-1}^{a_{r-1}}$. Since the $p_i$ are distinct primes, $\gcd(m, p_r^{a_r}) = 1$ (no prime divides both, again by unique factorization). By the Corollary above (multiplicativity of $d$ and $\sigma_k$),

By the induction hypothesis applied to $m$ (which has exactly $r-1$ distinct prime factors $p_1,\dots,p_{r-1}$),

Multiplying by the prime-power formula for $p_r^{a_r}$ gives exactly

completing the induction. $\blacksquare$