Arithmetic Functions
Any real-valued function on the integers f:Z → R
(or complex-valued function f:Z → C) is called an
arithmetic function.
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Examples: ν (n) = number of divisors of n; ϕ (n) =
€
number of invertible congruence classes mod n.
The most important arithmetic functions in
€
€
number theory are the multiplicative
functions,
those which satisfy (m,n) = 1 ⇒ f (mn) = f (m)f (n).
Indeed, there are some very simple multiplicative
functions.
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Examples: I(n) = n; 1(n) = 1.
These functions are completely multiplicative:
they satsify f (mn) = f (m)f (n) for all m and n.
We have already proven that ν and ϕ (Euler’s
function) are multiplicative functions. But we have
€ way to prove these facts that provides a
another
method to identify other
functions.
€ multiplicative
€
Theorem If f(n) is multiplicative, then so is its
summatory function F(n) = ∑ f (d) .
d|n
€
Proof If (m,n) = 1, every divisor of mn has the
form ab where a|m and b|n, and necessarily,
(a,b) = 1. Conversely, if a|m and b|n, then ab|mn
(and necessarily, (a,b) = 1). Therefore, if f is
multiplicative,
F(n) = ∑ f (d)
d|mn
= ∑ ∑ f (ab)
a|m b|n
= ∑ ∑ f (a)f (b)
a|m b|n
= ∑ f (a) ∑ f (b)
a|m
b|n
= F (m)F(n)
so F is multiplicative as well. //
€ ν is multiplicative.
Corollary
Proof ν (n) is the summatory function ∑ 1(n) . //
d|n
€
Corollary σ (n) = (sum of the divisors of n) is
€multiplicative.
€
Proof
∑ I(n) . //
€ σ (n) is the summatory function d|n
€
€
It is a nontrivial fact that the converse of the
theorem we just proved is also true. This was done
by August Möbius, a protegé of Gauss, in the 1830s.
In the process he defined a new arithmetic function
that played an important role in the proof, the
Möbius function
 1
if n = 1

µ (n) =  0
if n is not square - free
(−1) k if n is the product of k distinct primes

Proposition µ is multiplicative. //
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Lemma
1 if n = 1
∑ µ (d) = 
d|n
0 if n > 1
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Proof The case n = 1 is clear. Suppose then that n
e
is a prime
power:
n
=
p
. Then
€
∑ µ (d) = µ (1) + µ ( p) + µ ( p 2 ) + L+ µ ( p e )
d|n
€
= 1 − 1 + 0 + L+ 0
=0
Since µ is multiplicative, so is its summatory
function, and the result follows. //
€
€
The Möbius Inversion Formula Let f be any
arithmetic function and let F(n) = ∑ f (d) be its
summatory function. Then
d|n
n 
n 
f (n) = ∑ µ €
  ⋅ F (d) = ∑ µ (d) ⋅ F  
d|n  d 
d|n
d 
Proof The second equality is immediate, for as d
runs through the set of divisors of n, so does n/d.
€ the first equality, note that
For
 n 

n 
∑ µ   ⋅ F (d) = ∑ µ   ∑ f (c)
d|n  d 
d|n   d c|d

n 
= ∑ ∑ µ  f (c)
d|n c|d  d 
Each term in this last double sum corresponds to a
choice of divisor d of n, whose complementary
€
n
divisor is d′ = , and a choice of divisor c of d. We
d
could equally well choose c to be some divisor of n
first, however, then select d′ to be a divisor of the
n
complementary
€
factor . It follows that this sum
c
can be rewritten in
€ the form
€


∑ ∑ µ (d′ )f (c) = ∑  ∑ µ (d′)  ⋅ f (c) = f (n)

c|n d |(n
′ /c )
c|n  d |(n/
′
c)
€
where the final step follows from the lemma as the
sum in parenthesis vanishes unless c = n. //
Corollary If f is an arithmetic function whose
summatory function F(n) = ∑ f (d) is
d|n
multiplicative, then f must be multiplicative also.
Proof Let (m, n) = 1. Then by Möbius inversion,
€
 mn 
f (mn) = ∑ µ 
 ⋅ F(d)
d|mn  d 
m n 
= ∑ µ  ⋅  ⋅ F(ab)
a|m ,b|n  a b 
m  n 
= ∑ µ   ⋅ µ   ⋅ F(a) ⋅ F (b)
a|m ,b|n  a 
b 



m 
n 
=  ∑ µ   ⋅ F (a)  ∑ µ   ⋅ F(b) 
 a|m  a 
b|n  b 

= f (m) ⋅ f (n)
which shows that f is multiplicative. //
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