HOMEWORK SHEET 7 SOLUTIONS - UIUC MATH 453
NICOLAS ROBLES
Exercise 0.1. As follows.
a) Suppose that m and n are relatively prime positive integers. If m = 1 or n = 1 (or both),
then the proof of multiplicative is straightforward. So, let us assume that m > 1 and n > 1.
Let a and b be the number of distinct prime numbers in the prime factorizations of m and
n, respectively. Then the number of distinct prime numbers in the prime factorization of
mn is a + b and
ρ(mn) = 2a+b = 2a 2b = ρ(m)ρ(n),
as we wanted. If m = 2 and n = 4, then we have ρ(mn) = 2 and ρ(m)ρ(n) = 4 and so ρ is
in fact not completely multiplicative.
b) Since f is multiplicative by a) and the theorem that says that divisor sums of multiplicative
function is itself a multiplicative function, it is completely determined by its values at powers
of primes. Accordingly, let p ∈ P and let a ∈ N. Then
X
f (pa ) =
ρ(pa )
d|n
= ρ(1) + ρ(p) + ρ(p2 ) + · · · ρ(pa )
= 1 + 2 + 2 + ···2
= 1 + 2a.
(a times)
Hence, if pa11 pa22 · · · pamm is the prime factorization of n, then
f (n) = (1 + 2a1 )(1 + 2a2 ) · · · (1 + 2am ).
We note that this formula also holds for n = 1 if all ai are taken to be 0.
Exercise 0.2. As follows.
a) As it is often the case, the result is true for n = 1, so assume that n > 1. Then n is divisible
by some prime p and, since
1
1 − < 1,
p
(0.1)
we have φ(n) < n by the product formula for φ(n), namely
Y 1
φ(n) =
1−
,
p
p|n, p∈P
and
√ the second inequality is proved. For the first inequality, it is enough to show that
n ≤ 2φ(n). For this, we use the Theorem of the second midterm, i.e.
m Y
1
a1 a2
am
φ(n) = p1 p2 · · · pm
1−
,
pi
i=1
1
2
NICOLAS ROBLES
where pa11 pa22 · · · pamm (I think we are all getting bored of repeating this factorization...). In
other words, it is enough to show that
m
Y
a1 /2 a2 /2
a1 −1 a2 −1
am /2
am −1
p1 p2 · · · pm ≤ 2p1 p2
· · · pm
(pi − 1).
i=1
a /2
pi i
a /2
pi i
piai −1 .
If ai > 1, then
≤
If ai = 1, then
< pi −1 unless pi = 2; if ai = 1 and pi = 2,
ai /2
then pi
< 2. So, in all cases, the terms on the left-hand side of the desired inequality are
bounded above by terms on the right-hand side of the desired inequality. This means that
the desired inequality does hold.
b) Let p be the least prime divisor of n. Then
√
p ≤ n,
(0.2)
see Chapter 1 and early primality testing. Now, we have
√
1
1
= n − n,
φ(n) ≤ n 1 −
≤n 1− √
p
n
where the first inequality follows from (0.1) and the second inequality follows from (0.2).
This ends the proof.
√
√
Exercise 0.3. Let d be the number of positive divisors of√n less than or equal to n. Then d ≤ n.
Now the number of positive divisors of n greater than n is d − 1 if n is a perfect square and d
otherwise. Hence
√
d(n) ≤ 2d ≤ 2 n,
as claimed.
Exercise 0.4. Use the formulas of the lectures to get the following.
a) 8892,
b) 204288,
101
101 −1)
.
c) (2 −1)(5
4
Exercise 0.5. The first inequality is straightforward since n is a positive divisor of itself. For the
second inequality, we need a clever trick, which is as follows
σ(n) ≤ 1 + 2 + · · · + n =
n2 + n
n2 + n2
n(n + 1)
=
≤
= n2 .
2
2
2
This ends the exercise.
Exercise 0.6. This is the same procedure as with σ(n) and d(n).
a) We have σ3 (12) = 2044 and σ4 (8) = 4369.
b) Since the arithmetic function f (n) = nk is completely multiplicative then by our previous
homework (no. 6), the multiplicativity of σk (n) follows from the well known theorem of
divisor sums of multiplicative functions.
c) One has
p(a+1)k − 1
σk (pa ) = 1k + pk + p2k + · · · + pak =
.
pk − 1
d) Since the formula of c) above is valid if a = 0, then part b) yields
σk (n) =
r
(a +1)k+1
Y
p i
i
i=1
pki − 1
.