66
6.8.
Exercises
6.8
Exercises
1. Show that
d
(
n
) is odd if and only if
n
is a square.
2. Show that for a multiplicative function
f
, we have
f
(1) = 0 or
f
(1) = 1. Also, show that if
f
(1) = 0 then
f
is identically the
zero function.
3. Show that
d
(
n
) is divisible by 3 if and only if in the prime
factorization
n
=
p
α
1
1
· · ·
p
α
r
r
of
n
, at least one of the
α
i
is
≡
2
(mod 3).
4. Show that a multiplicative function
f
is completely multiplica
tive if and only if for all primes
p
and all
α
≥
1, we have
f
(
p
α
) =
f
(
p
)
α
.
5. (Shoup, Exercise 2.23) Show that
φ
(
mn
) =
gcd
(
m, n
)
φ
(
lcm
(
m, n
)).
6. Show that if
n
is divisible by
r
distinct odd primes, then
2
r

φ
(
n
).
7. Prove the identities
X
r

n
μ
(
r
)
d
(
n/r
) = 1
and
X
r

n
μ
(
r
)
d
(
r
) = (

1)
ω
(
n
)
where
ω
(
n
) is the number of distinct prime divisors of
n
.
8. Prove that
X
r

n
μ
(
r
)
σ
(
n/r
) =
n
and
X
r

n
μ
(
r
)
σ
(
r
) = (

1)
ω
(
n
)
p
1
· · ·
p
r
where
p
1
,
· · ·
, p
r
are the distinct prime divisors of
n
.
9. (Shoup, Exercise 2.51) Show that
n
is squarefree if and only if
X
d

n
μ
(
d
)
2
φ
(
d
) =
n.
67
6.
Arithmetic Functions
10. Show that if
f
is an arithmetic function with
f
(1)
6
= 0, there
is a unique arithmetic function
g
so that
f
◦
g
=
δ
.
6.9
Solutions to Exercises
1. If
n
=
p
α
1
1
· · ·
p
α
r
r
then
d
(
n
) = (
α
1
+ 1)
· · ·
(
α
r
+ 1) and for this
expression to be odd, all the
α
i
must be even. Let
β
i
=
α
)
i/
2
and
m
=
p
β
1
1
· · ·
p
β
r
r
. Then
n
=
m
2
. Conversely, if
n
is a square,
all the
α
i
are even and so the expression for
d
(
n
) shows that
it is odd.
2. Again, using the formula for
d
(
n
), we see that 3

d
(
n
) if and
only if 3

(
α
i
+ 1) for some
i
.
(This is because 3 is prime.)
Hence, for this
i
, we have
α
i
≡
2
(mod 3).
3. Since
f
is multiplicative,
f
(1) =
f
(1
·
1) =
f
(1)
2
and the result
follows.
4. If
f
is the zero function, there is nothing to prove, so sup
pose that
f
is not identically zero.
If
f
is completely mul
tiplicative, then this can be proved inductively.
It holds for
α
= 1.
If it holds for smaller values of
α
, then
f
(
p
α
) =
f
(
p
·
p
α

1
) =
f
(
p
)
f
(
p
α

1
) (since
f
is completely multiplica
tive).
This last quantity is equal to
f
(
p
)
f
(
p
)
α

1
(using the
induction hypothesis) and this is equal to
f
(
p
)
α
. Conversely,
suppose
f
(
p
α
) =
f
(
p
)
α
for all primes
p
and all
α
≥
1. Notice
that this holds for
α
= 0 as well, since by the previous exercise
f
(1) = 1.
Write
n
=
p
α
1
1
· · ·
p
α
r
r
and
m
=
p
β
1
1
· · ·
p
β
r
r
where
some of the
α
i
or
β
j
could be 0. Then
mn
=
p
α
1
+
β
1
1
· · ·
p
α
r
+
β
r
r
.
Since we are given that
f
is multiplicative,
f
(
mn
) =
f
(
p
α
1
+
β
1
1
)
· · ·
f
(
p
α
r
+
β
r
r
)
.
By the supposition, the right hand side is
f
(
p
1
)
α
1
+
β
1
· · ·
f
(
p
r
)
α
r
+
β
r
and this is
f
(
p
1
)
α
1
f
(
p
1
)
β
1
· · ·
f
(
p
r
)
α
r
f
(
p
r
)
β
r
=
f
(
p
α
1
1
)
· · ·
f
(
p
α
r
r
)
·
f
(
p
β
1
1
)
· · ·
f
(
p
β
r
r
)
and again using the fact that
f
is multiplicative, this is equal
to
f
(
m
)
f
(
n
) as required.
68
6.9.
Solutions to Exercises
5. The left hand side is equal to
mn
Y
p

mn
1

1
p
.
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 Spring '15
 HenryKim
 Congruence, Integers, Prime number, Divisor, congruences