Multiplicative Functions

How many divisors does $360$ have? Can we find that number without listing them all? A student factors $360 = 2^3 \cdot 3^2 \cdot 5$ . They compute $(3+1)(2+1)(1+1) = 24$ . Whether or not they realize it, they've exploited a deep structural fact. Many natural functions on the positive integers respect multiplication of coprime numbers. This property is called multiplicativity. It converts an arithmetic function, defined a priori on all of $\mathbb{N}$ , into something else. It becomes an object completely determined by its values on prime powers. Those values are usually trivial to compute. The idea matters for a key reason. The positive integers are built from primes via the Fundamental Theorem of Arithmetic. Multiplicative functions are precisely those arithmetic functions compatible with this multiplicative structure. Once we know a function is multiplicative, something remarkable happens. An infinite amount of information — its value at every integer — collapses dramatically. It reduces to a much smaller amount of information: its values at prime powers $p, p^2, p^3, \ldots$ for each prime $p$ . This is the mechanism behind formulas for the divisor count $d(n)$ , the sum of divisors $\sigma(n)$ , and Euler's totient $\varphi(n)$ . It is also the algebraic backbone underlying Dirichlet Convolution. Here is the big picture. Arithmetic functions form a vast, largely unstructured space. But multiplicativity carves out a subspace with rich internal structure. This subspace is closed under pointwise operations in limited ways. It is beautifully closed under Dirichlet convolution, though. It is also fully parametrized by an Euler product over primes. This turns hard combinatorial questions about all integers into easy local questions, one prime at a time. This article gives precise definitions of multiplicative and completely multiplicative functions. It proves the reduction theorem that pins a multiplicative function down by its prime-power values. It also proves a companion fact: a function is determined by, and can be reconstructed from, data at prime powers alone via unique factorization. Along the way we build intuition and a toolkit. This includes worked examples, a careful counterexample showing why the coprimality hypothesis cannot be dropped, and a look at where this machinery reappears throughout mathematics.

Key takeaways

A function $f:\mathbb{N}\to\mathbb{C}$ is multiplicative if $f(1)=1$ and $f(mn)=f(m)f(n)$ whenever $\gcd(m,n)=1$ .
It is completely multiplicative if that identity holds for all $m,n$ , no coprimality needed.
Multiplicativity plus unique factorization forces $f(n) = \prod_i f(p_i^{a_i})$ for $n=\prod_i p_i^{a_i}$ : prime powers determine everything.
Completely multiplicative functions are even simpler: $f(p^a) = f(p)^a$ , so a single value per prime suffices.
Classic examples: $d(n)$ , $\sigma(n)$ , $\varphi(n)$ , $\mu(n)$ are multiplicative but not completely multiplicative; $n^s$ and the constant function $1$ are completely multiplicative.
Key idea: coprimality of the arguments is exactly the hypothesis that makes prime-power data suffice to reconstruct the whole function.

Where this lives today

The reduction of multiplicative functions to prime-power data is the computational engine behind essentially every classical arithmetic function: the divisor function $d(n)$ and sum-of-divisors $\sigma(n)$ (see The Divisor and Sigma Functions), Euler's totient $\varphi(n)$ (see The Euler Totient Function), and the Möbius function $\mu(n)$ all owe their tractability to being multiplicative. The algebraic structure this induces — Dirichlet Convolution, under which multiplicative functions form a group — is the backbone of analytic number theory, where Euler products $\sum f(n) n^{-s} = \prod_p (\cdots)$ translate additive information about all integers into local information at each prime, the technique underlying the analytic proof of the Prime Number Theorem and the study of the Riemann zeta function and Dirichlet $L$ -functions. Beyond pure number theory, multiplicativity underlies fast algorithms in computer science and cryptography: computing Euler's totient for RSA key generation reduces to factoring $n$ and multiplying prime-power totients, exactly the mechanism proved above, making key generation efficient once factorization is known (and correspondingly hard to invert without it). In combinatorics, multiplicative structure governs counting problems on integer partitions and divisor lattices, where generating functions factor as Euler products over primes. In algebraic number theory, multiplicative arithmetic functions generalize to ideal-theoretic norms in number fields, and the same prime-power reduction underlies local-global principles used throughout the theory of $L$ -functions and modular forms.

Ask the tutor: How do I prove that Dirichlet convolution of two multiplicative functions is again multiplicative, using this reduction theorem?

f:\mathbb{N}\to\mathbb{C}

Intuition

Think of the positive integers as being assembled, brick by brick, from primes: unique factorization says $n = p_1^{a_1} p_2^{a_2}\cdots p_k^{a_k}$ in exactly one way (up to order). A general function on $\mathbb{N}$ need not respect this construction at all — its value at $n$ could depend on $n$ in some way totally unrelated to its prime factorization. Multiplicativity is the demand that $f$ "sees" the assembly process: if you build $n$ by gluing together two coprime pieces $m_1$ and $m_2$ (pieces sharing no prime factors), then $f$ of the whole is the product of $f$ of the pieces. The coprimality condition is essential, and it is exactly what lets us split $n$ along its distinct prime factors. Since $p_1^{a_1}, p_2^{a_2}, \ldots, p_k^{a_k}$ are pairwise coprime (they involve disjoint sets of primes), multiplicativity lets us peel them apart one at a time: $f(n) = f(p_1^{a_1})\,f(p_2^{a_2}\cdots p_k^{a_k}) = \cdots = f(p_1^{a_1})\cdots f(p_k^{a_k})$ . So the entire infinite table of values $f(1), f(2), f(3), \ldots$ collapses to a much shorter list: the values $f(p^a)$ for each prime $p$ and exponent $a \ge 1$ , together with $f(1)=1$ . Picture this as a two-dimensional grid, one row per prime, one column per exponent. A multiplicative function is fully specified by filling in this grid — freely, with no constraint linking different primes — and then multiplying across the rows corresponding to the primes dividing $n$ . Complete multiplicativity is a further collapse: within each row the entries $f(p), f(p^2), f(p^3),\ldots$ must themselves be $f(p), f(p)^2, f(p)^3, \ldots$ , a geometric progression, so one number per prime is enough.

Figure

Building f(360) from prime-power data via unique factorization

f

f(1) = 1

Worked examples

Example 1: Reconstructing $\sigma(n)$ , the sum-of-divisors function, at $n = 2^2\cdot 3 \cdot 7^2 = 588$ . The function $\sigma(n) = \sum_{d\mid n} d$ is known to be multiplicative (proved in the article on The Divisor and Sigma Functions); here we simply use the reduction theorem to compute it efficiently, illustrating the payoff of Part (2). First find $\sigma$ at each prime power appearing in $588 = 2^2\cdot 3^1\cdot 7^2$ :

$\sigma(2^2) = 1+2+4 = 7$ ,
$\sigma(3^1) = 1+3 = 4$ ,
$\sigma(7^2) = 1+7+49 = 57$ .

Since the theorem guarantees $\sigma(588) = \sigma(2^2)\,\sigma(3)\,\sigma(7^2)$ (the three prime-power factors $4, 3, 49$ are pairwise coprime, and their product is $588$ ), we get

\sigma(588) = 7\cdot 4\cdot 57 = 1596.

As a check, one can note the closed formula $\sigma(p^a) = \frac{p^{a+1}-1}{p-1}$ , giving $\sigma(4) = \frac{2^3-1}{1}=7$ , $\sigma(3)=\frac{3^2-1}{2}=4$ , $\sigma(49) = \frac{7^3-1}{6} = \frac{342}{6}=57$ , matching. Note that had we instead tried $\sigma(588) \overset{?}{=} \sigma(84)\sigma(7)$ (splitting as $84\times 7$ ), this would be invalid: $84 = 2^2\cdot3\cdot7$ and $7$ share the common factor $7$ , so $\gcd(84,7)=7\ne1$ , and multiplicativity simply does not apply to that split. This is exactly the coprimality hypothesis at work: only the split into prime-power pieces is guaranteed to be usable.

Example 2: A completely multiplicative function versus a merely multiplicative one, evaluated at $n = 2^3\cdot 5^2 = 200$ . Let $\chi(n) = (-1)^{n-1}$ if $n$ is odd and $\chi(n)=0$ if $n$ is even — no, take instead a cleaner completely multiplicative example: let

f(n) = n^2,

the squaring function, and compare it against Euler's totient $\varphi(n)$ , which is multiplicative but not completely multiplicative. Squaring is completely multiplicative. For any $m,n\ge1$ (coprime or not), $f(mn) = (mn)^2 = m^2 n^2 = f(m) f(n)$ , and $f(1) = 1$ . So by the Lemma, $f(p^a) = f(p)^a = (p^2)^a = p^{2a}$ , consistent with the direct computation $f(p^a) = (p^a)^2 = p^{2a}$ . At $n = 200 = 2^3 \cdot 5^2$ :

f(200) = f(2^3)\,f(5^2) = 2^{2\cdot3}\cdot 5^{2\cdot2} = 2^6 \cdot 5^4 = 64\cdot 625 = 40000 = 200^2,

consistent, and here we could equally have used $f(p)=p^2$ directly: $f(2)^3 f(5)^2 = 4^3\cdot 25^2 = 64\cdot 625=40000$ . Complete multiplicativity means one number per prime ( $f(2)=4$ , $f(5)=25$ ) suffices, with exponents just along for the ride. Totient is not completely multiplicative. Recall $\varphi(p^a) = p^a - p^{a-1} = p^{a-1}(p-1)$ (proved in the article on The Euler Totient Function). At the same $n=200$ :

\varphi(200) = \varphi(2^3)\,\varphi(5^2) = (2^3-2^2)(5^2-5^1) = (8-4)(25-5) = 4\cdot 20 = 80.

Now test whether $\varphi(p^a) \overset{?}{=} \varphi(p)^a$ , the complete-multiplicativity signature from the Lemma. Take $p=2,a=3$ : $\varphi(2^3) = 4$ , but $\varphi(2)^3 = 1^3 = 1 \ne 4$ . So $\varphi$ fails the Lemma's conclusion, and indeed $\varphi$ is not completely multiplicative — even though it is multiplicative, and Part (2) of the theorem still applies to it perfectly well (as just used). This contrast is the point: multiplicativity guarantees reduction to prime powers; complete multiplicativity further reduces prime powers to primes, and $\varphi$ only enjoys the weaker property.

\gcd(m,n)=1

Scope, edge cases & common traps

Situation	What holds	What can fail
$f$ multiplicative, $\gcd(m,n)=1$	$f(mn)=f(m)f(n)$ always	—
$f$ multiplicative, $\gcd(m,n)>1$	no general identity	$f(mn)=f(m)f(n)$ can be false (e.g. $\varphi$ at $m=n=2$ )
$f$ completely multiplicative	$f(p^a)=f(p)^a$ ; one value per prime suffices	—
$n=1$	$f(1)=1$ forced for any multiplicative $f$	Students sometimes forget this must be proved, not assumed
Pointwise sum $f+g$ of two multiplicative functions	usually not multiplicative	$(f+g)(1) = 2 \ne 1$ in general, already violating the definition
Extending values from primes alone	fine for completely multiplicative $f$	For merely multiplicative $f$ , values at $p$ alone do not determine $f(p^2), f(p^3),\ldots$

The mathematician behind it

Origins · 1 minLeonhard Euler and how this was discovered

Leonhard Euler

1707–1783 · Enlightenment-era Europe, working across St. Petersburg and Berlin

Euler did not use the modern vocabulary of "multiplicative function," which crystallized later, but he was the first to systematically exploit the phenomenon: his introduction of the totient function $\varphi(n)$ , his formula for $\sigma(n)$ via divisor sums, and above all his discovery of the Euler product \sum_{n=1}^\infty \frac{f(n)}{n^s} = \prod_{p \text{ prime}} \left(1 + \frac{f(p)}{p^s} + \frac{f(p^2)}{p^{2s}} + \cdots \right) for multiplicative $f$ (most famously for $f\equiv 1$ , giving the Euler product for the zeta function) is the direct analytic descendant of the reduction theorem proved here. Euler's genius was to see that unique factorization, translated into a statement about functions, becomes a statement about factoring an infinite series into an infinite product over primes — the seed of analytic number theory.

History

- 1737 — Euler proves the Euler product formula for

\sum n^{-s}

, implicitly using complete multiplicativity of the constant function

1

- 1760s — Euler studies the totient function

\varphi(n)

and its multiplicative behavior in the context of what is now Euler's theorem.

- 1801 — Gauss's Disquisitiones Arithmeticae systematizes divisor-sum and totient identities building on this multiplicative structure.

- 1830s–1850s — Dirichlet develops Dirichlet series and the convolution structure that formalizes multiplicative functions as an algebraic class.

- 1900s — the terminology "multiplicative function" and "completely multiplicative function" becomes standard in textbooks following Landau and later Hardy–Wright.

Why is

f(1)=1

required in the definition rather than left to be derived case by case?

Requiring $f$ not identically zero and using $\gcd(1,N)=1$ for some $N$ with $f(N)\ne 0$ , one can derive $f(1)=1$ from the multiplicative identity itself, as done in the proof above — so strictly the definition need only assume $f\not\equiv 0$ plus the multiplicative property, and $f(1)=1$ follows. Many textbooks instead build $f(1)=1$ directly into the definition for convenience, avoiding the (easy) derivation each time. Either convention is standard; the content is identical, since the derivation is a one-line consequence. What must never be dropped is some mechanism forcing $f(1)=1$ , since without it the zero function would vacuously satisfy $f(mn)=f(m)f(n)=0$ for coprime $m,n$ , trivializing the definition.

Is the sum

f+g

of two multiplicative functions multiplicative?

Generally no. Even setting aside the value at $1$ , take $f=g=\varphi$ : then $(f+g)(1) = \varphi(1)+\varphi(1) = 1+1=2 \ne 1$ , already violating the normalization condition in the definition. Even functions engineered to fix $(f+g)(1)=1$ typically fail the coprime-multiplicative identity, because $(f+g)(mn) = f(mn)+g(mn) = f(m)f(n)+g(m)g(n)$ generally does not factor as $\big(f(m)+g(m)\big)\big(f(n)+g(n)\big)$ . Multiplicative functions are not closed under pointwise addition; they are closed under pointwise multiplication and, more importantly, under Dirichlet Convolution, which is the operation that respects their structure.

If two multiplicative functions agree on all primes, must they agree everywhere?

Not necessarily — agreement on primes alone is not enough for merely multiplicative functions, only for completely multiplicative ones. For example, $d(n)$ (divisor count) and any completely multiplicative function $h$ with $h(p)=2$ for every prime $p$ agree at every prime ( $d(p)=2=h(p)$ ), but $d(4) = 3$ while $h(4) = h(2)^2 = 4$ , so they disagree at $n=4$ . Two multiplicative functions agree everywhere if and only if they agree at every prime power $p^a$ , by the reduction theorem; agreement at primes is strictly weaker unless both functions are additionally completely multiplicative.

Does multiplicativity say anything about

f(m+n)

or additive behavior?

No — multiplicativity is purely about the operation of multiplication of coprime arguments, and it makes no claim about sums. A function can be multiplicative and behave wildly under addition; for instance $\sigma(n)$ is multiplicative but $\sigma(m+n)$ has no simple formula in terms of $\sigma(m)$ and $\sigma(n)$ (e.g. $\sigma(2)+\sigma(3) = 3+4=7$ while $\sigma(5)=6\ne7$ ). This is a different regime from additive arithmetic functions (where $f(mn)=f(m)+f(n)$ for coprime $m,n$ , as with $\log$ restricted suitably, or the number of distinct prime factors $\omega(n)$ ), which form a separate but analogous theory.

How does this connect to Dirichlet convolution?

Dirichlet convolution defines a new arithmetic function $(f*g)(n) = \sum_{d\mid n} f(d)\,g(n/d)$ from two arithmetic functions $f,g$ . A key theorem, built directly on the reduction result proved here, states that if $f$ and $g$ are both multiplicative, so is $f*g$ ; the proof again reduces to checking the identity at prime powers, using unique factorization to separate divisors of $p^a$ from divisors of coprime cofactors. This makes the set of multiplicative functions closed under convolution and turns it into a group under $*$ with identity element $\varepsilon(n) = [n=1]$ , a structure explored fully in Dirichlet Convolution.

Can a multiplicative function take the value

0

somewhere without being identically zero?

Yes, and this is common. For instance the Möbius function $\mu(n)$ is multiplicative with $\mu(p)=-1$ for every prime but $\mu(p^a) = 0$ for every $a\ge 2$ — so $\mu(4)=\mu(9)=\mu(12)=0$ , yet $\mu$ is not the zero function since $\mu(1)=1\ne 0$ and $\mu(p)=-1\ne0$ . The definition only excludes $f$ being identically zero on all of $\mathbb{N}$ ; isolated or even infinitely many zero values are perfectly consistent with multiplicativity, and indeed $\mu$ 's zero pattern at non-squarefree prime powers is the entire content of its usefulness in the Möbius inversion formula.

What is the quickest way to check whether a formula defines a multiplicative function without testing all pairs

m,n

By the reduction theorem, it suffices to check $f(1)=1$ and then verify the coprime-multiplicative identity directly from the formula, or — more practically — to derive a closed form for $f(p^a)$ from the definition and confirm that plugging prime-power factorizations into the claimed product formula reproduces the original formula for general $n$ . In practice, one typically proves multiplicativity from a defining sum or product (e.g. a sum over divisors) using an explicit bijection between divisors of $mn$ and pairs of divisors of $m,n$ when $\gcd(m,n)=1$ , as done for $\sigma$ and $\varphi$ in their dedicated articles, rather than testing infinitely many pairs by hand.