Download Homework #3

Due Friday, 15 July
1.
Homework #3
Math 4400
Consider the two integers n = 210 and n = 225. For each n, find all of the divisors of
n, compute ϕ(d) for each divisor d of n, and then compute the sum of all of the ϕ(d),
i.e., compute
X
ϕ(d).
d|n
2.
Repeat #1 for the integer n = pqr, where p, q, r are three distinct prime numbers.
[Hint: there are 8 divisors of n.]
3.
Suppose d, n are positive integers such that d | n. Prove that ϕ(d) | ϕ(n).
4.
Compute the last three digits of 31215 .
[Hint: compute ϕ(1000) and use Euler’s theorem.]
5.
Compute the last three digits of 61215 .
[Hint: compute it modulo 8 and modulo 125, then use Chinese Remainder Theorem.]
6.
Let F be a field and n a natural number. Define µn to be the subset of F× given by
µn = a ∈ F× : an = 1 .
Prove that µn is a group under the multiplication inherited from the field.
[Hint: recall the group axioms about closure, associativity, identity, and inverses.]
Answers
1.
X
ϕ(d) = 210:
d|210
d
ϕ(d)
X
1
1
2
1
3
2
5
4
6
2
7
6
10 14 15
4 6 8
21 30 35
12 8 24
42 70 105
12 24 48
210
48
ϕ(d) = 225:
d|225
d
ϕ(d)
2.
1
1
3
2
5
4
9
6
15 25 45
8 20 24
75 225
40 120
The divisors of n = pqr are 1, p, q, pq, r, pr, qr, pqr, so
X
ϕ(d) = ϕ(1) + ϕ(p) + ϕ(q) + ϕ(pq) + ϕ(r) + ϕ(pr) + ϕ(qr) + ϕ(pqr)
d|n
= 1 + (p − 1) + (q − 1) + (p − 1)(q − 1)
+ (r − 1) 1 + (p − 1) + (q − 1) + (q − 1)
= 1 + (r − 1) 1 + (p − 1) + (q − 1) + (pq − p − q + 1)
= r pq
= pqr.
3.
Suppose that d | n and that the prime factorization of d is
d = pe1 1 pe2 2 · · · per r ,
where the pi are distinct primes and the exponents satisfy ei > 1. Since n is a multiple
of d, the prime factorization of n must include all of the primes dividing n, with possibly
larger exponents, as well as possibly additional primes. In other words, the factorization
of n is of the form
n = m pf11 pf22 · · · pfrr ,
where fi > ei and m is relatively prime to each of the pi . Now we compute:
ϕ(d) = (pe1 1 − pe1 1 −1 ) · · · (per r − per r −1 ),
and
ϕ(n) = ϕ(m)(pf11 − p1f1 −1 ) · · · (pfrr − prfr −1 ).
Since fi > ei > 1, each pei i − piei −1 term divides the corresponding pfi i − pfi i −1 term.
This proves that ϕ(d) divides ϕ(n).
Alternatively, consider the set P of primes dividing n. We can write this as the
disjoint union P = Q ∪ R, where Q is the set of primes dividing d and R is the set of
primes which divide n but not d. Then
Y
n
(1 − p1 )
ϕ(n)
nY
p∈P
= Y
=
(1 − p1 ).
ϕ(d)
d
d
(1 − 1 )
p
p∈R
p∈Q
We know that n/d is an integer, but the product over p ∈ R of (1 − 1/p) is a rational
number. The denominator of this rational number is the product of all the primes in R.
Luckily,
Qthe integer n/d contains at least one copy of each prime in R, so the number
(n/d) p∈R (1 − 1/p) is indeed an integer, proving that ϕ(d) divides ϕ(n).
4.
We compute ϕ(1000) = ϕ(8)ϕ(125) = 4 · 100 = 400. Since 3 is relatively prime to 1000,
Euler’s theorem says that 3400 ≡ 1 (mod 1000). Therefore,
31215 ≡ (3400 )3 · 315 ≡ 13 · 315 ≡ 315
2
4
(mod 1000).
8
We have 3 ≡ 9, 3 ≡ 81, and 3 = 6561 ≡ 561, so
315 ≡ 38 34 32 3 ≡ 561 · 81 · 9 · 3 ≡ 907
1215
So, the last 3 digits of 3
5.
(mod 1000).
are 907.
This is similar to the last problem, except that 6 is not relatively prime to 1000. Instead,
we will separately compute 61215 modulo 125 and modulo 8. Once we know these two
values, the Chinese Remainder Theorem will guarantee a unique solution modulo 1000.
We start modulo 8 because it’s so simple:
63 ≡ 0 (mod 8)
→
61215 ≡ 63 61212 ≡ 0
(mod 8).
1215
Now we compute 6
modulo 125. Since ϕ(125) = 100 and 6 is relatively prime to
125, we know that 6100 ≡ 1 (mod 125), so
61215 ≡ (6100 )12 · 615 ≡ 11 2 · 615 ≡ 615
2
4
(mod 125).
8
We have 6 ≡ 36, 6 ≡ 46, and 6 ≡ 116, so
615 ≡ 68 64 62 6 ≡ 116 · 46 · 36 · 6 ≡ 76 (mod 125).
So, we know that 61215 is congruent to 0 modulo 8 and congruent to 76 modulo 125.
This means it must be of the form 8u and of the form 76 + 125v. Setting these equal, we
get 8u − 125v = 76. A first solution is u0 = −53 and v0 = −4, so all solutions are of the
form u = −53−125t and v = −4−8t. In particular, 8u = 8(−53−125t) = −424−1000t.
When t = −1, we get 8u = 576, so 576 are the last 3 digits of 61215 .
6.
We need to prove that the set µn is closed under multiplication, that multiplication
is associative, that there is an identity element in µn , and that each element of µn is
invertible.
Closure: suppose that a, b are two elements of µn . We want to prove that the
product ab is an element of µn ; that is, we want to prove that (ab)n = 1. We raise ab
to the nth power and use the fact that multiplication in a field is commutative to get
(ab)n = an bn . Since a ∈ µn , we know that an = 1; and since b ∈ µn , we know that
bn = 1. Therefore, (ab)n = an bn = 1 · 1 = 1, which means that ab ∈ µn .
Associativity: for a, b, c ∈ µn , we want to prove that a(bc) = (ab)c. Since a, b, c
are also elements of the field, and multiplication in a field is associative by definition,
multiplication in a subset of the field is also associative. In other words, associativity is
inherited from the multiplication in F.
Identity: we already know that F× has an identity element which we call 1; that is,
1a = a for all a ∈ F. In particular, 1a = a for all a ∈ µn ⊂ F. And, since 1 is the
identity, we know that 1n = 1, which shows that 1 ∈ µn . Therefore, µn contains the
multiplicative identity.
Inverses: we want to prove that each element a ∈ µn has an inverse in µn . Since
µn ⊂ F× , we know that a is invertible in F× ; that is, there exists a−1 ∈ F× such that
aa−1 = 1. Since (a−1 )n = a−n = (an )−1 = 1−1 = 1, we can say that the inverse of a
is actually in µn , which is exactly what we needed to prove.