UMASS AMHERST MATH 471 FALL 2006, F. HAJIR
HOMEWORK 9: ARITHMETIC FUNCTIONS
Let us make a couple of definitions first. An arithmetic function is a function f:Z+→C
where Z+is the set of positive integers. For example, the Euler phi function ϕ:Z+→Z+
sending mto ϕ(m) = |(Z/mZ)×|is an arithmetic function. We say that an arithmetic
function fis multiplicative if
f(mn) = f(m)f(n) for all coprime integers m, n ∈Z+.
Please note the condition that the multiplicativity condition f(mn) = f(m)f(n) does not
need to hold for all pairs m, n, only for those that satisfy gcd(m, n) = 1. One way to think of
this is that a multiplicative function fis “free” to hold whatever values it wants on the prime
powers m=pt, but once those are chosen, then all the remaining values are determined by
the fact that each integer is (uniquely) a product of primes.
Given an arithmetic function f:Z+→C, we construct the divisor-sum function of f,
called F, by
F(m) = X
d|m
f(d).
Here, by convention, the notation “d|m” indicates that the sum is over the set of positive
divisors dof m.
Some other standard arithmetic functions are the nu-function ν(m) = |Div+(m)|which
counts the number of divisors of mand the sigma-function σ(m) = Pd|mdwhich gives the
sum of the divisors of m.
A very useful (but at first sight really weird) function is the M¨obius mu (pronounced
“mew”) function µ. It can be defined as follows. We have µ(1) = 1 (see problem 1 below
for why this is a popular choice) and put µ(2) = µ(3) = µ(5) = µ(7) = µ(11) = · · · =−1,
in fact let’s put µ(p) = −1 for every prime p. What about prime powers? Here you might
be surprised by the choice I am about to make: If t > 1, and pis a prime, we define
µ(pt) = 0. So, µ“kills” higher prime powers. Finally, we define µon the rest of the integers
by multiplicativity. To put this all together, the definition of µis as follows. First, µ(m) = 1
if m= 1, and µ(m) = (−1)rif m=p1· · · pris the product of rdistinct primes p1, . . . , pr.
But if mis divisible by the square of any prime, then µ(m) = 0. You’ll see below why this
is a useful function.
1. Show that if fis a multiplicative arithmetic function, then f(1) = 1.
2. Suppose Div+(m) = {1≤d≤m|dis a divisor of m}is the set of positive divisors of
an integer m. Show that if m, n are coprime, then
Div+(mn) = {dd0|d∈Div+(m) and d0∈Div+(n)}.
Hint: Show that the LHS is contained in the RHS and versa vice.
1
