THEORY OF NUMBERS
HOMEWORK 5 SOLUTIONS
Computations
(1) Compute the following which are related to the Möbius function:
(a) Compute the values of the MöbiusP
function µ(n) for n P
= 1, · · · , 20
(b) In class, we showed that σ(n) = d|n d and τ (n) = d|n 1. Use the Möbius
inversion formula to recover the functions f (n) = n and g(n) = 1 for n = 60.
Solution:
(a) The following table lists the values of the Möbius function:
µ(1) = 1
µ(8) = 0
µ(15) = 1
µ(2) = −1
µ(9) = 0
µ(16) = 0
µ(3) = −1
µ(10) = 1
µ(17) = −1
µ(4) = 0
µ(11) = −1
µ(18) = 0
µ(5) = −1
µ(12) = 0
µ(19) = −1
µ(6) = 1
µ(13) = −1
µ(20) = 0
µ(7) = −1
µ(14) = 1
(b) The following table lists the values of the Möbius function, τ , and σ for the
divisors of 60
µ(1) = 1
τ (1) = 1
σ(1) = 1
µ(2) = −1
τ (2) = 2
σ(2) = 3
µ(3) = −1
τ (3) = 2
σ(3) = 4
µ(4) = 0
τ (4) = 3
σ(4) = 7
µ(5) = −1
τ (5) = 2
σ(5) = 6
µ(6) = 1
τ (6) = 4
σ(6) = 12
µ(10) = 1
τ (10) = 4
σ(10) = 18
µ(12) = 0
τ (12) = 6
σ(12) = 28
µ(15) = 1
τ (15) = 4
σ(15) = 24
µ(20) = 0
τ (20) = 6
σ(20) = 42
µ(30) = −1
τ (30) = 8
σ(30) = 72
µ(60) = 0
τ (60) = 12
σ(60) = 168
Then, the Möbius inversion formula gives the following:
µ(1)τ (60) + µ(2)τ (30) + µ(3)τ (20) + µ(4)τ (15) + µ(5)τ (12) + µ(6)τ (10) + µ(10)τ (6)
+ µ(12)τ (5) + µ(15)τ (4) + µ(20)τ (3) + µ(30)τ (2) + µ(60)τ (1)
1
= 12 − 8 − 6 + 0 − 6 + 4 + 4 + 0 + 3 + 0 − 2 = 1
and
µ(1)σ(60) + µ(2)σ(30) + µ(3)σ(20) + µ(4)σ(15) + µ(5)σ(12) + µ(6)σ(10) + µ(10)σ(6)
+ µ(12)σ(5) + µ(15)σ(4) + µ(20)σ(3) + µ(30)σ(2) + µ(60)σ(1)
= 168 − 72 − 42 + 0 − 28 + 8 − 12 + 0 + 7 + 0 − 3 = 60
Therefore, the values of the Möbius inversion formula are f (60) = 60 and g(60) =
1.
(2) For all primes
p between 1 and 20 and for all 1 ≤ n < p, compute the Legendre
symbol np .
Solution: The primes between 1 and 20 are 2, 3, 5, 7, 11, 13, 17, 19. Then the Legendre symbols are as follows:
1
1
1
2:
=1
3:
=1
5:
=1
2
3
5
2
2
= −1
= −1
3
5
3
= −1
5
4
=1
5
1
1
1
=1
11 :
=1
13 :
=1
7:
7
11
13
2
2
2
=1
= −1
= −1
7
11
13
3
3
3
= −1
=1
=1
7
11
13
4
4
4
=1
=1
=1
7
11
13
5
5
5
= −1
=1
= −1
7
11
13
6
6
6
= −1
= −1
= −1
7
11
13
7
7
= −1
= −1
11
13
8
8
= −1
= −1
11
13
9
9
=1
=1
11
13
2
17 :
1
=1
17
2
=1
17
3
= −1
17
4
=1
17
5
= −1
17
6
= −1
17
7
= −1
17
8
=1
17
9
=1
17
10
= −1
17
11
= −1
17
12
= −1
17
13
=1
17
14
= −1
17
15
=1
17
16
=1
17
19 :
10
11
= −1
1
=1
19
2
= −1
19
3
= −1
19
4
=1
19
5
=1
19
6
=1
19
7
=1
19
8
= −1
19
9
=1
19
10
= −1
19
11
=1
19
12
= −1
19
13
= −1
19
14
= −1
19
15
= −1
19
16
=1
19
3
10
=1
13
11
= −1
13
12
=1
13
17
=1
19
18
= −1
19
(3) If a and n are relatively prime, then the order of a mod n is the smallest integer x
such that ax ≡ 1 mod n. For n = 1, · · · , 15 and 1 ≤ a < n, compute the order of a
mod n.
Hint: For example, the order of 4 mod 9 is 3 since 4 6≡ 1 mod 9, 42 ≡ 16 ≡ 7 6≡ 1
mod 9, and 43 ≡ 64 ≡ 1 mod 9.
Solution:
1:
o(1) = 1
2:
o(1) = 1
3:
o(1) = 1, o(2) = 2
4:
o(1) = 1, o(3) = 2
5:
o(1) = 1, o(2) = 4, o(3) = 4, o(4) = 2
6:
o(1) = 1, o(5) = 2
7:
o(1) = 1, o(2) = 3, o(3) = 6, o(4) = 3, o(5) = 6, o(6) = 2
8:
o(1) = 1, o(3) = 2, o(5) = 2, o(7) = 2
9:
o(1) = 1, o(2) = 6, o(4) = 3, o(5) = 6, o(7) = 3, o(8) = 2
10 :
o(1) = 1, o(3) = 4, o(7) = 4, o(9) = 2
11 :
o(1) = 1, o(2) = 10, o(3) = 5, o(4) = 5, o(5) = 5,
o(6) = 10, o(7) = 10, o(8) = 10, o(9) = 5, o(10) = 2
12 :
o(1) = 1, o(5) = 2, o(7) = 2, o(11) = 2
13 :
o(1) = 1, o(2) = 12, o(3) = 3, o(4) = 6, o(5) = 4, o(6) = 12,
o(7) = 12, o(8) = 4, o(9) = 3, o(10) = 6, o(11) = 12, o(12) = 2
14 :
o(1) = 1, o(3) = 6, o(5) = 6, o(9) = 3, o(11) = 3, o(13) = 2
15 :
o(1) = 1, o(2) = 4, o(4) = 2, o(7) = 4, o(8) = 4, o(11) = 2, o(13) = 4, o(14) = 2
Proofs
(4) (a) Suppose that f is a multiplicative function with f (1) = 1. Show that
X
µ(d)f (d) = (1 − f (p1 ))(1 − f (p2 )) · · · (1 − f (pk ))
d|n
where n = pa11 · · · pa22 · · · pakk is the prime factorization for n.
(b) Use the formula above to find a simple formula for
X
µ(d)τ (d)
d|n
Solution:
4
P
(a) Let G(n) = d|n µ(d)f (d), then we first show that G is multiplicative. Let m
and n be relatively prime. Then
X
G(mn) =
µ(d)f (d)
d|mn
=
X
µ(d1 d2 )f (d1d2 )
X
µ(d1 )f (d1)µ(d2 )f (d2 )
d1 |m,d2 |n
=
d1 |m,d2 |n
=
X
µ(d1 )f (d1) ·
d1 |m
X
µ(d2 )f (d2)
d2 |n
= G(m)G(n)
Where the second inequality follows from the fact that m, n are relatively prime
and is a proof we did in class. The second equality is because µ and fQare multiplicative. Since G is multiplicative, if n = pa11 · · · pakk , then G(n) = G(pai i ).
Therefore, if we can show that for any prime p and a ≥ 1, G(pa ) = 1 − f (p),
then the result follows.
P
Consider G(pa ) = d|pa µ(d)f (d). Now, if d|pa , then d must also be a power
of p since all prime divisors of d would then divide p. In addition, the power
of d must be less than a by the divisiblity relation. Therefore d is of the form
pb where P
0 ≤ b ≤ a. Substituting this into the expression above, we have
a
G(p ) = 0≤b≤a µ(pb )f (pb ). For all b ≥ 2, µ(pb) = 0 since pb is not a product
of discinct primes. For b = 1, µ(p1 ) = −1 since p is a product of one distinct
prime, and for b = 0, µ(p0 ) = 1 since µ(1) = 1. Therefore, this sum simplifies to
f (1) − f (p). In the statement of the problem, f (1) = 1, so this equals 1 − f (p),
completing the proof.
(b) Since τ (1) = 1 as 1 has only one divisor and τ is multiplicative, the result
above applies. Moreover, for any prime p, τ (p) = 2 since there
P are only two
ak
a1
positive divisors for p, 1, p. Therefore, if n = p1 · · · pk , the sum d|n µ(d)τ (d) =
Q
Q
Q
(1 − τ (pi )) = (1 − 2) = (−1) = (−1)k .
(5) (a) Show that if b is a positive integer not divisible by the prime p, then
b
2b
(p − 1)b
+
+·+
=0
p
p
p
(b) Let a, b be integers not divisible by p. Show that either one or all three of the
integers a, b, ab are residues mod p.
Solution:
(a) This
question is only true for p an odd prime because when p = 2, the sum is
1
= 1.
Assume now that
2
p is an odd prime and that b = 1. In this case, the
which is the sum over 1, · · · , p−1. We showed in class
sum is p1 +· · ·+ (p−1)
p
that half of 1, · · · , p − 1 are residues and half of 1, · · · , p − 1 are nonresidues.
Therefore, half of the symbols above are 1 and half are −1, so that the sum is 0.
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Now, consider an arbitrary b. We showed in the proof of Fermat’s little theorem that {b, 2b, · · · , (p − 1)b} consists of elements congruent to {1, · · · , (p − 1)}
mod p. In particular, there is a bijection between the two sets via congruence. Therefore, mod p, the set {b, 2b, · · · , (p − 1)b} is the same as the set
{1, · · · , (p − 1)} just possibly in a differentorder.
Then, since Legendre symbols
b
do not change under congruency the sum p + · · · + (p−1)b
is the same sum
b
up to reodering and we know the second sum to be 0.
as p1 + · · · + (p−1)
p
(b) We do this by a case-by-case analysis. If a, b are nonresidues, then ab
=
p
b
a
= (−1)2 and so ab is a residue and there is one residue in {a, b, ab}.
p
p
If a is a residue, butb is
a nonresidue
(or equivalently a is a nonresidue, but
ab
b
b is a residue), then p = ap
= (1)(−1) = (−1), so ab is a nonresidue
p
and there is one residue in {a, b, ab}. If a, b are residues, then let x, y be square
roots, so x2 ≡ a mod p and y 2 ≡ b mod p. Then xy is a square root of ab since
(xy)2 ≡ x2 y 2 ≡ ab mod p. Therefore all three of {a, b, ab} are residues.
Note: We could also do the last case by using the same formula as above.
Challenge
(6) Show that if the equation φ(n) = k, where k is a positive integer, has exactly one
solution n then 36|n.
Hint: Prove this by contradiction, i.e., assume that 2 ∤ n or 4 ∤ n and construct a
number m different from n such that φ(n) = φ(m) contradicting the assumption that
there is only one solution.
Solution: We prove this using a case-by-case analysis and contradiction: Let n be
the solution to φ(n) = k.
Assume first that 2 ∤ n, then φ(2n) = φ(2)φ(n) = 1 · φ(n) = φ(n) = k, which
contradicts the assumption that φ(n) = k has only one solution.
Assume now that 2|n. If 2|n, but 4 ∤ n, then n = 2m and 2 ∤ m. Then k =
φ(n) = φ(2m) = φ(2)φ(m) = 1 · φ(m) = φ(m), which contradicts the assumption
that φ(n) = k has only one solution.
Assume now that 4|n. If 3 ∤ n, then n = 2l m where l ≥ 2, 3 ∤ m and 2 ∤ m. Then
φ(n) = φ(2l m) = φ(2l )φ(m) = (2l − 2l−1 )φ(m). On the other hand, φ(3 · 2l−1 m) =
φ(3)φ(2l−1)φ(m) = 2(2l−1 − 2l−2 )φ(m) = (2l − 2l−1 )φ(m), which contradicts the
assumption that φ(n) = k has only one solution.
Assume now that 4|n and 3|n. If 3|n, but 9 ∤ n, then n = 2l · 3m where l ≥ 2, 2 ∤ m,
and 3 ∤ m. Then φ(n) = φ(2l )φ(3)φ(m) = (2l − 2l−1 )(3 − 1)φ(m) = (2l+1 − 2l )φ(m).
On the other hand, φ(2l+1 m) = φ(2l+1)φ(m) = (2l−1 − 2l )φ(m), which contradicts
the assumption that φ(n) = k has only one solution.
Therefore, 4|n and 9|n which implies that 36|n.
6