MATH 356
Homework 15
Throughout, all variables represent integers unless otherwise specified.
For some of the problems on this assignment, it is helpful to observe that d|n if and only
if (n/d)|n.
1. (a) Verify that τ (n) = τ (n + 1) = τ (n + 2) = τ (n + 3) for n = 3655 and n = 4503.
(b) When n = 14, n = 206, and n = 957, show that σ(n) = σ(n + 1).
Solution: Compute.
2. (a) Find the form of all positive integers satisfying τ (n) = 10. What is the smallest
positive integer for which this is true?
(b) Show that there are no positive integers satisfying σ(n) = 10.
Solution:
(a) If n = pe11 · · · perr , then τ (n) = (e1 + 1) · · · (er + 1) = 10. Thus e1 + 1 = 2 and
e2 + 1 = 5 (or vice-versa), or e1 + 1 = 10. Thus n has the form p9 or pq 4 for some
primes p and q. The smallest such is 3 · 24 = 48.
(b) Since σ(n) ≥ n + 1 > n for n > 1, it suffices to compute σ(n) for n < 10. None
equal 10.
3. Let f and g be multiplicative functions that are not identically zero and such that
f (pk ) = g(pk ) for each prime p and k ≥ 1. Prove that f = g.
r
r
Solution: Write n = pe11 · · · perr . Then f (n) = f (p1 )e1 · · · f (per ) = g(p1 )e1 · · · g(per ) =
g(n) since f and g are multiplicative. Thus f = g.
4. Let ω(n) denote the number of distinct prime divisors of n > 1, with ω(1) = 0.
(Example: ω(6) = ω(12) = ω(24) = 2.) Show that 2ω(n) is a multiplicative function.
Solution: Let m and n be relatively prime integers. Then ω(mn) = ω(m) + ω(n), so
2ω(mn) = 2ω(m)+ω(n) = 2ω(m) 2ω(n) , as desired.
5. Given n ≥ 1, let σs (n) =
X
ds (the sum of the sth powers of all divisors of n).
d|n
(a) Show that σ0 = τ and σ1 = σ.
(b) Show that σs is multiplicative.
Solution:
(a) Definition.
(b) Notice that f (n) = ns is multiplicative, so σs =
X
f (d) is multiplicative by a
d|n
theorem from class.
6. (a) Show that for each n ∈ Z+ , µ(n)µ(n + 1)µ(n + 2)µ(n + 3) = 0.
(b) Show that if n ≥ 3, then
n
X
k=1
Solution:
µ(k!) = 1.
(a) One of n, n + 1, n + 2, and n + 3 will contain a factor of 4. Whichever it is, µ
applied to that factor will yield a 0.
(b) For n = 3, we have µ(1!) + µ(2!) + µ(3!) = 1 − 1 + 1 = 1. If n ≥ 4, then for each
k ≥ 4, k! contains a factor of 4, making µ(k!) = 0. Thus only the first three terms
are nonzero.
7. Show that if n = pe11 · · · pekk is the prime factorization of n > 1, then
‚
Œ
‚
X
d|n
Œ
µ(d)
=
d
1
1
1−
··· 1 −
.
p1
pk
µ(d)
1
, the function f in question is f (n) = .
d
n
Applying Exam Problem 4 (below) to this gives the desired result.
Solution: Since the summands are
8. Find formulas for
X
d|n
µ2 (d)
and
τ (d)
X
d|n
µ2 (d)
in terms of the prime factorization of n.
σ(d)
Solution:
(a) Let f (n) =
µ(n)
. Then f is multiplicative, so by Exam Problem 4,
τ (n)
X
µ(d)f (d) = (1 − f (p1 )) · · · (1 − f (pr ))
d|n
= 1−
r
=
(b) Now let f (n) =
3
2
−1
−1
··· 1 −
1+1
1+1
.
µ(n)
. Then f is multiplicative, so by Exam Problem 4,
σ(n)
X
d|n
µ(d)f (d) = (1 − f (p1 )) · · · (1 − f (pr ))
‚
Œ
‚
−1
−1
··· 1 −
= 1−
p1 + 1
pr + 1
p1 + 2
pr + 2
=
···
.
p1 + 1
pr + 1
Œ
√
9. TURN IN: If n ≥ 1, then τ (n) ≤ 2 n.
10. TURN
IN: Show that if n = pe11 · · · pekk is the prime factorization of n > 1, then
X
dµ(d) = (1 − p1 ) · · · (1 − pk ).
d|n