Number Theory: An Introduction to Mathematics

3 Multiplicative Functions 155

h(n)=

∑

d′|n′,d′′|n′′

f(d′d′′)g(n′n′′/d′d′′)

=

∑

d′|n′,d′′|n′′

f(d′)f(d′′)g(n′/d′)g(n′′/d′′)

=

∑

d′|n′

f(d′)g(n′/d′)

∑

d′′|n′′

f(d′′)g(n′′/d′′)=h(n′)h(n′′). 

Proposition 27If f :N→Cis a multiplicative function, then its Dirichlet inverse
f−^1 :N→Cis also multiplicative.

Proof Assume on the contrary thatg:= f−^1 is not multiplicative and letn′,n′′be
relatively prime positive integers such thatg(n′n′′)=g(n′)g(n′′). We supposen′,n′′
chosen so that the productn=n′n′′is minimal. Since fis multiplicative,f( 1 )= 1
and henceg( 1 )=1. Consequentlyn′>1,n′′>1and

0 =

∑

d′|n′

g(d′)f(n′/d′)=

∑

d′′|n′′

g(d′′)f(n′′/d′′)=

∑

d|n

g(d)f(n/d).

But
∑

d|n

g(d)f(n/d)=g(n)f( 1 )+

∑

d′|n′,d′′|n′′,d′d′′<n

g(d′d′′)f(n′n′′/d′d′′)

=g(n)+

∑

d′|n′,d′′|n′′,d′d′′<n

g(d′)g(d′′)f(n′/d′)f(n′′/d′′)

=g(n)−g(n′)g(n′′)+

∑

d′|n′

g(d′)f(n′/d′)·

∑

d′′|n′′

g(d′′)f(n′′/d′′)

=g(n)−g(n′)g(n′′).

Thus we have a contradiction.

It follows from Propositions 26 and 27 thatunder Dirichlet multiplication the mul-
tiplicative functions form a subgroup of the group of all functionsf :N→Cwith
f( 1 )=0. The further subgroup generated byiandjcontains some interesting func-
tions. Letτ(n)denote the number of positive divisors ofn,andletσ(n)denote the
sum of the positive divisors ofn:

τ(n)=

∑

d|n

1 ,σ(n)=

∑

d|n

d.

In other words,

τ=i∗i,σ=i∗j,

and hence, by Proposition 26,τandσ are multiplicative functions. Thus they are
uniquely determined by their values at the prime powers. But ifpis prime andα∈N,
the divisors ofpαare 1,p,...,pαand hence

τ(pα)=α+ 1 ,σ(pα)=(pα+^1 − 1 )/(p− 1 ).