Download MTH 440/540 FALL 2016 HOMEWORK 4 SOLUTIONS (1) (exercise

MTH 440/540 FALL 2016 HOMEWORK 4 SOLUTIONS
(1) (exercise 2.19) Let φ : N → N be the Euler φ function.
(a) Find all natural numbers such that φ(n) = 1.
Solution. Since φ(n) is multiplicative and φ(pk ) = pk − pk−1 for prime powers, we see that
φ(n) = 1 just for n = 1 and n = 2.
(b) Do there exist natural numbers n and m such that φ(nm) 6= φ(n) · φ(m)?
Solution. We note φ(4) = 2, but φ(2) = 1. Thus m = n = 2 gives on example of such natural
numbers.
(2) (exercise 2.30) Compute the last two digits of 345 . (Do this by hand, do not use a computer).
Solution. We are computing 345 (mod 100). We note that 45 = 32 + 8 + 4 + 1 and so we first
compute the table of values
31 ≡ 3
(mod 100),
32 ≡ 9
(mod 100),
4
3 ≡ 81 ≡ −19
8
3 ≡ 361 ≡ 61
3
16
≡ 3721 ≡ 21
32
≡ 441 ≡ 41
3
(mod 100),
(mod 100),
(mod 100),
(mod 100).
We then find that
345 ≡ 3 · (−19) · 61 · 41
≡ (−57) · 2501
≡ 43
(mod 100)
(mod 100)
(mod 100).
Thus the last two digits of 345 are 43.
(3) (exercise 2.24) Prove that for any positive integer n the fraction
12n+1
30n+2
is in reduced form.
Solution. We show that 12n + 1 and 30n + 2 are relatively prime. For this, we have that
gcd(12n + 1, 30n + 2) = gcd(12n + 1, 30n + 2 − 2(12n + 1)) = gcd(12n + 1, 6n)
= gcd(12n + 1 − 2(6n), 6n) = gcd(1, 6n) = 1.
(4) (not from the text) Prove that if p is prime then
2p
p
≡ 2 (mod p).
Solution. We note that when ab is an integer and b is relatively prime to n, we can obtain ab (mod n)
by multiplying a by the inverse of b modulo n. We note (p − 1)! is invertible modulo p by Wilson’s
Theorem, thus
2p
(2p)!
=
p
p!p!
(2p)(2p − 1)(2p − 2) · · · (2p − (p − 1))
=
p!
2 · (2p − 1)(2p − 2) · · · (2p − (p − 1))
=
(p − 1)!
1
≡ 2 · (−1)(−2) · · · (−(p − 1)) · ((p − 1)!)−1
≡ 2 · (−1)
≡2
p−1
−1
· (p − 1)! · ((p − 1)!)
(mod p)
(mod p)
(mod p).
(5) (exercise 2.21) If G and H are both groups and a function ψ : G → H satisfies ψ(a · b) = ψ(a) · ψ(b)
for all a, b ∈ G, then we call ψ a group homomorphism. The kernel of ψ is the subset of G given
by ker(ψ) = {a ∈ G : ψ(a) = 1}. Here · denotes the binary operation of the appropriate group and
1 denotes the identity element of the appropriate group.
(a) Prove that if ψ : G → H is a group homomorphism, then ker(ψ) is a subgroup of G. (That is
to say, prove that ker(ψ) is also a group under the same operation as G.)
Solution. Since ker(ψ) is a subset of G, we do not need to check the associative properties of
the operation. We need only check that 1 ∈ ker(ψ), if a, b ∈ ker(ψ) then a · b ∈ ker(ψ), and if
a ∈ ker(ψ) then a−1 ∈ ker(ψ).
We note that ψ(1) = 1 for any group homomorphism, due to cancelling a ψ(1) in both sides of
the identity
ψ(1) = ψ(1 · 1) = ψ(1) · ψ(1).
Thus 1 ∈ ker ψ.
If a, b ∈ ker(ψ) then ψ(a) = ψ(b) = 1. Then ψ(a·b) = ψ(a)·ψ(b) = 1·1 = 1, and so a·b ∈ ker(ψ).
We first verify the more general fact that ψ(a−1 ) = ψ(a)−1 . This follows from the identity
1 = ψ(1) = ψ(a · a−1 ) = ψ(a) · ψ(a−1 ).
Now if a ∈ ker(ψ), then ψ(a) = 1. Thus ψ(a−1 ) = 1−1 = 1 and so a−1 ∈ ker(ψ).
(b) Prove that ker(ψ) is normal, i.e., if a ∈ G and b ∈ ker(ψ), then a−1 · b · a ∈ ker(ψ).
Solution. Suppose a ∈ G and b ∈ ker(ψ). Using the fact proved above that ψ(a−1 ) = ψ(a)−1 ,
we have that
ψ(a−1 · b · a) = ψ(a−1 ) · ψ(b · a) = ψ(a−1 ) · ψ(b) · ψ(a)
= ψ(a−1 ) · 1 · ψ(a) = ψ(a−1 ) · ψ(a) = 1.
Thus a−1 · b · a ∈ ker(ψ).
(6) (not from the text) Show that if p is an odd prime, then 2(p − 3)! ≡ −1 (mod p).
Solution. We note that (p − 1) is invertible modulo p and by Wilson’s Theorem (p − 1)! ≡ −1
(mod p). Thus
2(p − 3)! ≡ −(p − 2) · (p − 3)! ≡ −(p − 1)−1 · (p − 1) · (p − 2) · (p − 1)! ≡ −(p − 1)−1 · (p − 1)!
≡ −(p − 1)−1 · (−1) ≡ −(−1) · (−1) ≡ −1
(mod p).
The following exercises marked with a ? are required for students taking the course as MTH 540. These
exercises can be completed for extra credit for students taking the course as MTH 440.
(7) (?, exercise 2.25) Suppose a and b are positive integers.
(a) Prove that gcd(2a − 1, 2b − 1) = 2gcd(a,b) − 1.
Solution. See solution in part (c).
(b) Does it matter if 2 is replaced by an arbitrary prime p?
Solution. See solution in part (c).
(c) What if 2 is replaced by an arbitrary positive integer n?
2
Solution. The proof follows by induction on the larger of a and b, and uses that fact that if
gcd(x, y) = 1 then gcd(xz, y) = gcd(z, y).
We note that base case is when a = b = 1. This case is true because
gcd(n − 1, n − 1) = n − 1 = ngcd(1,1) − 1.
For the inductive step, we assume that gcd(na − 1, 2b − 1) = 2gcd(a,b) − 1 for all positive integers
a, b ≤ M , where M is a fixed positive integer. When then must prove that gcd(na − 1, 2b − 1) =
2gcd(a,b) − 1, where at least one of a or b is M + 1.
We note that in the case of a = b = M + 1, there is little to show as
gcd(nM +1 − 1, nM +1 − 1) = nM +1 − 1 = ngcd(M +1,M +1) − 1.
So what remains is the case when a = M + 1 and b < a. Here we have that
gcd(nM +1 − 1, nb − 1) = gcd(nM +1 − 1 − nb + 1, nb − 1) = gcd(nM +1 − nb , nb − 1)
= gcd(nb (nM +1−b − 1), nb − 1).
But gcd(nb , nb − 1) = 1 and so
gcd(nb (nM +1−b − 1), nb − 1) = gcd(nM +1−b − 1, nb − 1).
By the inductive hypothesis,
gcd(nM +1−b − 1, nb − 1) = ngcd(M +1−b,b) − 1.
Thus
gcd(nM +1 − 1, nb − 1) = ngcd(M +1−b,b) − 1 = ngcd(M +1,b) − 1,
which is what we were to show for the inductive step.
(8) (?, not from the text) For which positive integers n, with gcd(n, 5) = 1, is n4 + 4n prime?
Solution. We note if n is even then n4 + 4n is an even number that is at least 32, so n4 + 4n is not
prime.
When n is odd, we then have by Euler’s Theorem that
n4 + 4n ≡ 1 + (−1)n ≡ 1 − 1 ≡ 0
(mod 5).
Thus n4 + 4n is divisible by 5, and so n4 + 4n is not prime unless n4 + 4n = 5, which happens only
in the case of n = 1.
One can also use the factorization
a4 + 4b4 = (a2 + 2b2 + 2ab)(a2 + 2b2 − 2ab).
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