Number Theory: An Introduction to Mathematics

3 Multiplicative Functions 157

Proof Letf:N→Cbe given and putfˆ=f∗i.Then

fˆ∗μ=f∗i∗μ=f∗δ=f.

Conversely, letfˆ:N→Cbe given and putf=fˆ∗μ.Thenf∗i=fˆ∗δ=fˆ.
Moreover, by the first part of the proof, this is the only possible choice forf.

If we apply Proposition 29 to Euler’sφ-function then, by Proposition II.24, we
obtain the formula

φ(n)=n

∑

d|n

μ(d)/d.

In particular, ifn=pα,wherepis prime andα∈N,then

φ(pα)=μ( 1 )pα+μ(p)pα−^1 =pα( 1 − 1 /p).

Sinceφis multiplicative, we recover in this way the formula

φ(n)=n

∏

p|n

( 1 − 1 /p) for everyn∈N.

Theσ-function arises in the study of perfect numbers, to which the Pythagoreans
attached much significance. A positive integernis said to beperfectif it is the sum of
its (positive) divisors other than itself, i.e. ifσ(n)= 2 n.
For example, 6 and 28 are perfect, since

6 = 1 + 2 + 3 , 28 = 1 + 2 + 4 + 7 + 14.

It is an age-old conjecture that there are no odd perfect numbers. However, the even
perfect numbers are characterized by the following result:

Proposition 30An even positive integer is perfect if and only if it has the form
2 t( 2 t+^1 − 1 ),wheret∈Nand 2 t+^1 − 1 is prime.

Proof Letnbe any even positive integer and writen= 2 tm,wheret≥1andmis
odd. Then, sinceσis multiplicative,σ(n)=dσ(m),where

d:=σ( 2 t)= 2 t+^1 − 1.

Ifm = danddis prime, thenσ(m) = 1 +d = 2 t+^1 and consequently
σ(n)= 2 t+^1 m= 2 n.
On the other hand, ifσ(n)= 2 n,thendσ(m)= 2 t+^1 m.Sincedis odd, it follows
thatm=dqfor someq∈N. Hence

σ(m)= 2 t+^1 q=( 1 +d)q=q+m.

Thusqis the only proper divisor ofm. Henceq=1andm=dis prime.