Download CHAP08 Multiplicative Functions

8. MULTIPLICATIVE FUNCTIONS
§8.1. Multiplicative Functions
There are functions F(n) in number theory that have the property that F(mn) = F(m)F(n)
whenever m, n are coprime. Functions with this property are called multiplicative functions. Of
course there are trivial examples such as the constant functions F(n) = 1 or G(n) = n. Less trivial
examples are the number of divisors and the sum of the divisors. It is clear that if F(n) is
multiplicative then so is kF(n) for any integer k.
In Chapter 3 we defined the Euler function ϕ(n) to be the number of numbers from 1 to n
that are coprime with n. Equivalently it is the number of units (elements with a multiplicative
inverse) in the ring ℤm of integers modulo m. We proved as a corollary to Theorem 6 in Chapter 3
that ϕ is a multiplicative function. and that ϕ(pn) = pn−1(p − 1). We now define two other important
multiplicative functions.
Define δ(n) to be the number of divisors of n, including 1 and n. Here we are considering
only positive numbers and their positive divisors. We define σ(n) to be the sum of the (positive)
divisors of n.
Theorem 1: δ(n) and σ(n) are multiplicative functions.
Proof: Suppose m, n are coprime. If the divisors of m are a1, a2, ..., ar and the divisors of n are b1,
b2, ... bs then every divisor of mn can be expressed uniquely as aibj.
Hence δ(mn) = rs = δ(m)∆(n).
And σ(mn) = ∑ aibj = (a1 + ... + ar)(b1 + ... + bs) = σ(m)σ(n).
i,j
pn+1 − 1
Theorem 2: If p is prime, δ(p ) = n + 1 and σ(p ) =
.
p−1
Proof: The divisors of pn are 1, p, p2, ..., pn.
Hence the number of them is n + 1 and their sum is the sum of the GP 1 + p + p2 + ... + pn.
n
n
We can use these theorems to find δ(n), σ(n) and ϕ(n)for any n.
Example 1: Find δ(600), σ(600) and ϕ(600).
Solution: ∆ (600) = ∆ (23.3.52) = 4.2.3 = 24.
σ(600) = σ(23.3.52) = 15.4.31 = 1860.
ϕ(600) = ϕ(23.3.52) = 22.2.5.4 = 160.
§8.2. The Möbius Function
We define the Möbius function, µ(n), as follows.
n
µ(n) = (−1) if n is a product of n distinct primes. [In particular µ(1) = 1],
µ(n) = 0 otherwise.
We say that a number is square-free if it is not divisible by a prime squared in which case it
is a product of distinct primes. (We consider 1 as being the product of zero primes.) So if n is not
square-free then µ(n) = 0.
This rather strange function in that it takes just three values: −1, 0, 1. Yet it’s extremely
useful when used in combination with other multiplicative functions.
71
Example 2: µ(30) = µ(2.3.5) = (−1)3 = −1.
µ(330) = µ(2.3.5.11) = (−1)4 = 1.
µ(990) = µ(2.32.5.11) = 0.
Theorem 3: µ(n) is a multiplicative function.
Proof: Suppose that m, n are coprime.
If m or n is divisible by the square of a prime then so is mn and µ(mn) = µ(m)µ(n) = 0.
If m is a product of r distinct primes and n is divisible by s distinct primes then mn is divisible by
r + s distinct primes and so µ(mn) = (−1)r+s = (−1)r(−1)s = µ(m)µ(n).
[Remember that m, n are coprime so the primes dividing m are distinct from those dividing n.]
Example 3: µ(30) = µ(6.5) = µ(6)µ(5) = 1(−1) = −1.
Theorem 4:
1
if n = 1
∑µ(d) = 0 if n > 1 .
d|n
Proof: Suppose n =p1a1 p2a2 ... pkak > 1.
Then ∑µ(d) = 1 + ∑µ(pi) + ∑µ(pipj) +
d|n
i
i,j
∑ µ(pipjpk) + ...
i,j,k
 k   k   k
= 1 − 1 + 2 − 3 + ...
k
= (1 − 1)
= 0.
Example 4:
∑µ(d) = µ(1) + µ(2) + µ(4) + µ(5) + µ(8) + µ(10) + µ(20) + µ(40)
d|40
= 1 − 1
= 0.
+ 0 −
1 +
0 + 1
+
0
+
0
§8.3. The Monoid of Multiplicative Functions
Let M be the set of all multiplicative functions from ℕ to ℕ. Define the Möbius product
n
F * G of two multiplicative functions F, G by (F * G)(n) =
F(d)Gd
 
∑
d|n
=
∑
n
F dG(d) .
 
d|n
We can write this symmetrically as (F * G)(n) = ∑F(c)G(d) , and so F * G = G * F for all
cd=n
multiplicative functions.
A monoid is a generalisation of a group in that, although there is an identity under
multiplication, not every element need have an inverse. To show that M is a commutative monoid
under this product we need to check that it’s closed, that the operation is associative and has an
identity.
Theorem 5: M is a commutative monoid with identity under the Möbius product.
Proof: Suppose m, n are coprime. Then every divisor of mn has the form ab where a | m and b | n.
m
n
Moreover, the a, b will be coprime and a and b will also be coprime.
72
Suppose that F, G are multiplicative functions.
mn
F(d)G d 
Then (F * G)(mn) =
 
∑
d|mn
=
∑
mn
F(ab)G ab 
 
a | m,b|n
=
∑F(a)F(b)Gma Gnb
a|m,b|n
=
∑
∑
a|m
b|m
m
F(a)G a 
 
n
F(b)Gb
 
= (F * G)(m).(F * G)(n).
Thus F * G is a multiplicative function.
The associative law results from the fact that both (F * G) * H(n) and F * G * H)(n) can be
written as ∑F(a)G(b)H(c) .
abc=n
1 if n = 1
Finally, if let 1:ℕ → ℕ be defined by 1(n) = 0 otherwise . This is very clearly a

multiplicative function. Moreover, if F is any multiplicative function then
n
(F * E)(n) =
F(d).E b .
 
∑
d|n
This sum collapses to the single term, namely when d = n. This value is F(n).
Hence F * 1 = 1 * F = F
and so 1 is the identity of this monoid.
We’re faced with a notational problem here because there are three multiplicative functions
that could be considered as an identity function. In fact all three are identities under some
appropriate multiplication.
There’s the function that maps n to n. This is the identity under composition of functions.
Then there is the function that maps all n to 1. This is the identity under what is called “point-wise”
multiplication, where (FG)(n) = F(n)G(n). Finally we have the function, 1, defined above which is
the identity for Möbius multiplication.
We shall resolve this problem by denoting these three identities as follows:
Operation
Möbius multiplication
Identity Defined by
1 if n = 1
1
1(n) = 0 otherwise

Composition
ℑ(n) = n for all n
ℑ
Pointwise multiplication U
U(n) = 1 for all n
Note that all are multiplicative functions but 1 is the identity in M.
73
∑µ(d) = n for all n. The left hand side of this equation can be expressed as
By Theorem 4,
d|n
the Möbius product of µ and U. So µ * U = 1 and so U and µ are inverses of one another. The
Möbius inversion formula is now a simple consequence of the algebra of the monoid M.
Theorem 6 (MÖBIUS INVERSION FORMULA):
If F is a multiplicative function then G(n) = ∑F(d) if and only if F(n) =
d|n
∑µ(d)Gnd .
d|n
−1
Proof: The assumption is essentially G = F * U = F * µ .
Multiplying both sides by µ we get F = µ * G.
It is a simple exercise to show that δ = U * U and σ = ℑ * U.
Theorem 7: ϕ = µ * ℑ.
Proof: If p, q, ... are the distinct prime divisors of n then
n
n
ϕ(n) = n −
+
p
pq − …..
∑ ∑
p|n
=
∑ µ(d) nd
pq|n
d|n
=
∑
n
µdd by a suitable change of variable.
 
d|n
Hence ϕ = µ * ℑ.
Example 6: ϕ(100) = ϕ(22.52) = ϕ(22)ϕ(52) = 2.5.4 = 40.
∑ ϕ(d) = ϕ(1) + ϕ(2) + ϕ(4) + ϕ(5) + ϕ(10) + ϕ(20) + ϕ(25) + ϕ(50) + ϕ(100)
d|100
= 1 + 1
= 100.
+ 2
+ 4
+ 4
+
8
+ 20
+ 20 +
40
SUMMARY OF MULTIPLICATIVE FUNCTIONS
definition
1 1 if n = 1, 0 otherwise
U 1
ℑ n
µ −1 if product of odd number of distinct primes
0 if divisible by a prime square

 1 if product of even number of distinct primes
ϕ number of units in ℤn#
δ number of divisors
σ sum of divisors
74
1
1
1
1
1
2
0
1
2
−1
1 1
1 2
1 3
6
0
1
6
1
n
7
0
1
7
−1
3
0
1
3
−1
4
0
1
4
0
2
2
4
2 2 6
3 4 2
7 12 8
8
0
1
8
0
12
0
1
12
0
28
0
1
28
0
30
0
1
30
−1
4 4 12 8
4 6 6 8
15 28 56 72
SUMMARY OF SOME RELATIONS IN THE MONOID M
1
U
ℑ
µ
ϕ
δ
σ
1
1
U
ℑ
µ
ϕ
δ
σ
U
U
δ
σ
1
ℑ
ℑ
ℑ
σ
nδ
ϕ
µ ϕ δ σ
µ ϕ δ σ
1 ℑ
ϕ
U ℑ
σ
U σ
ℑ
75
in terms of U, ϕ
1
U
U*ϕ
U−1
ϕ
U*U
U*U*ϕ
76