MODULE 3. NUMBER THEORETIC FUNCTIONS
3.4
Lecture 4
Lecture 4
Preamble: In this lecture, we will discuss multiplicative functions. The multiplicative
property of an arithmetic function yields interesting consequences. As an application of
the notion of multiplicative function, we will characterize all even perfect numbers.
Keywords: Arithmetic function, multiplicative function, perfect numbers
3.4.1
Arithmetic Functions
A function f : N −→ C is called an arithmetic function. An arithmetic function is called
multiplicative if
f (mn) = f (m)f (n) when gcd(m, n) = 1.
Examples:
1. Nk (n) = nk is multiplicative for any integer k.
2. Let n denote a natural number. Let φ(n) denote the number of positive integers
less than n and prime to it. Recall that this function is known as Euler’s φ-function
and φ(mn) = φ(m)φ(n) if gcd(m, n) = 1.
3. ω : N −→ Z ≥ 0 given by ω(n) =
P
p|n 1
is also multiplicative:
m = pe11 . · · · .perr
=⇒ ω(m) = r
n = q1f1 . · · · .qses
=⇒ ω(n) = s
mn = pe11 . · · · .perr q1f1 . · · · .qses =⇒ ω(mn) = (−1)r+s ,
where pi ’s and qj ’s are pairwise distinct primes.
The following proposition is useful in constructing new multiplicative functions from
known ones.
PROPOSITION 3.12. Let f be a multiplicative arithmetic function and
F (n) =
X
f (d).
d|n
Then F is also a multiplicative arithmetic function.
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MODULE 3. NUMBER THEORETIC FUNCTIONS
Lecture 4
Proof: Let m and n be two coprime integers. If d|mn, we can write d uniquely as
d = d1 d2 such that d1 |m and d2 |n. Conversely, if d1 |m and d2 |n, then d1 d2 |mn. Clearly,
if m and n are coprime, so are their divisors d1 and d2 . Now,
X
f (d)
F (mn) =
d|n
X
=
f (d1 d2 )
d1 |m, d2 |n
X
=
f (d1 )f (d2 )
d1 |m, d2 |n
=
hX
f (d1 )
ih X
i
f (d2 )
d2 |n
d1 |m
= F (m)F (n)
The above proposition gives us the following multiplicative functions:
1. Consider the arithmetic function τ given by
X
τ (n) =
1.
d|n
For any natural number n, τ (n) denotes the number of divisors of n. Now
X
X
τ (n) =
1=
N0 (d),
d|n
d|n
where N0 is the multiplicative functions that we saw previously. Hence τ (n) is
multiplicative by the previous proposition.
2. Consider the function σk given by
σk (n) =
X
dk =
d|n
X
Nk (d).
d|n
As Nk is multiplicative, so is σk (d). We usually denote σ1 (n) by σ(n), thus
X
σ(n) =
d
d|n
is the sum of all the (positive) divisors of a natural number n and also a multiplicative function. Observe that τ = σ0 .
If f is a multiplicative function, we need to know its values only for the prime powers.
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MODULE 3. NUMBER THEORETIC FUNCTIONS
Lecture 4
PROPOSITION 3.13. Let n = pe11 · · · .perr be the unique factorization of n. If f is a
multiplicative function, then
f (n) = f (pe11 ). · · · .f (perr ).
Proof: It follows from the multiplicative property.
3.4.2
Perfect Numbers
We will now introduce perfect numbers. The multiplicative property of the function σ
defined above is crucial in characterizing the perfect numbers.
DEFINITION 3.14. A natural number n is called perfect if it is the sum of all its proper
divisors, in other words,
X
n=
d.
d|n, 1≤d<n
In terms of the arithmetic function σ(n), we can say that n is perfect if
2n = σ(n).
For example, 6 and 28 are perfect numbers, as 6 = 1 + 2 + 3, and 28 = 1 + 2 + 4 + 7 + 14.
Now we first give a sufficient condition for a number to be perfect.
PROPOSITION 3.15. Let p be a prime number such that 2p − 1 is also a prime. Then,
m = 2p−1 (2p − 1) is a perfect number.
Proof: It is enough to show that σ(m) = 2m. Observe that
σ(2p − 1) = 1 + 2p − 1
as 2p − 1 is a prime. Now,
σ(m) = σ(2p−1 (2p − 1))
= σ(2p−1 )σ(2p − 1)
= (1 + 2 + · · · + 2p−1 )(1 + 2p − 1)
= (2p − 1)2p
= 2m.
The examples that we had mentioned earlier are in deed of the form 22−1 (22 − 1) = 6,
and 23−1 (23 − 1) = 28. In fact any even perfect number that we find will necessarily of
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MODULE 3. NUMBER THEORETIC FUNCTIONS
Lecture 4
this form, which we are going to prove in the next proposition.
PROPOSITION 3.16. An even number is perfect only if it is of the form 2p (2p−1 − 1),
where both p and 2p − 1 are primes.
Proof: Let m = 2k−1 l, where k ≥ 2 and l is odd. We want to show that l is a prime
and that l = 2k − 1. It will then follow that k is also prime, otherwise k = rs with
r, s > 1 would give two non-factors 2r − 1 and 2s − 1 of l. Note that l must be prime if
σ(l) = l + 1. Now,
2m = σ(m)
=⇒
2k l = σ(2k−1 )σ(l)
=⇒
2k l = (1 + 2 + · · · + 2k−1 )σ(l)
=⇒
2k l = (2k − 1)σ(l)
=⇒
(2k − 1)|l, say l = (2k − 1)t
=⇒
2k t = σ(l).
But t and l = (2k − 1)t are two divisors of l, which add up to
t + l = t + (2k − 1)t = 2k t = σ(l).
Hence t and l = (2k − 1)t are the only divisors of l, and l is a prime. As 2k − 1 ≥ 22 − 1,
we must have t = 1. Thus, l = 2k − 1 is a prime, which implies k is prime too.
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