Solution Sheet 6

Solution Sheet 6
1. (i) (7, 11, 13) , (13, 17, 19) , (37, 41, 43) and (67, 71, 73) .
(ii) We will show that if p ≡ 1 or 2 mod 3 then at leat one of p, p + 3
and p + 4 is a multiple of 3 and thus not prime.
Consider three cases.
a) If p ≡ 1 mod 3, i.e. p = 1 + 3k, then p + 2 = 3 (1 + k) is not prime.
b) If p ≡ 2 mod 3, i.e. p = 2 + 3k, then p + 4 = 3 (2 + k) is not prime.
c) If p ≡ 0 mod 3, then the only prime divisible by 3 is 3. This leads
to the given triplet of (3, 5, 7).
2. (i) φ (1) = 1, φ (2) = 1, φ (3) = 2, φ (4) = 2, φ (5) = 4, φ (6) = 2,
φ (7) = 6, φ (8) = 4, φ (9) = 6, φ (10) = 4, φ (11) = 10, φ (12) = 4.
(ii) For p prime all the integers 1 ≤ r ≤ p − 1 are coprime with p and
so all are counted by Euler’s phi function. Hence φ (p) = p − 1.
(iii) For a power of a prime, pk , count the number of integers 1 ≤ r ≤
pk − 1 that are not coprime with pk . Such integers are of the form mp
where
1 ≤ m ≤ pk − 1
p = pk−1 − 1
p .
But m is an integer so, in fact, 1 ≤ m ≤ p k−1 − 1. Thus there are
p k−1 − 1 integers not coprime to p k out of a possible total of p k − 1
integers. Hence
φ p k = p k − 1 − p k−1 − 1 = p k − p k−1 = p k−1 (p − 1) .
3. ⋆ (i) We have 5555 ≡ 4 mod 7 and (4, 7) = 1 and so, by Fermat’s Little
Theorem, 4 6 ≡ 1 mod 7. But 2222 = 370 × 6 + 2. Thus
5555 2222 ≡ 4 6 370 4 2 ≡ 1 370 16 ≡ 2 mod 7.
We also have 2222 ≡ 3 mod 7 and (3, 7) = 1 and so, by Fermat’s Little
Theorem, 3 6 ≡ 1 mod 7. But 5555 = 925 × 6 + 5. So
2222 5555 ≡ 3 6 925 3 5 ≡ 1 925 × 9 × 9 × 3
≡ 2 × 2 × 3
≡ 5 mod 7.
1

Hence
(ii)
5555 2222 + 2222 5555 ≡ 2 + 5 ≡ 0 mod 7.
5555 2222 + 2222 5555 ≡ 2 2222 + 8 5555
≡ 2 2222 + (−1) 5555
≡ 2 2222 − 1 mod 9,
using 5555 is odd. Then Fermat’s Little Theorem states that m 6 ≡
1 mod 9 when (m, 9) = 1 since φ (9) = 6 from the solution to Question
2. Since 2222 = 370 × 6 + 2 we get
Hence
2 2222 − 1 ≡ 2 6 370 2 2 − 1
≡ 2 2 − 1 = 3 mod 9.
5555 2222 + 2222 5555 ≡ 3 mod 9.
Thus 5555 2222 + 2222 5555 is of the form 3 + 9ℓ for some ℓ ∈ Z. Here
3 + 9ℓ = 3 (1 + 3ℓ) is a multiple of 3, i.e. divisible by 3, but not by 9.
(iii) First
3333 7777 + 7777 3333 ≡ 33 7777 + 77 3333 mod 100.
But also m 40 ≡ 1 mod 100 if (m, 100) = 1 since φ (100) = 40m. Note
that 7777 = 194 × 40 + 17 while 3333 = 83 × 40 + 13, so
33 7777 + 77 3333 = 33 40 194 33 17 + 77 40 83 77 13
since (33, 100) = (77, 100) = 1.
≡ 33 17 + 77 13 mod 100,
Finally, using the method of successive squaring we get 33 17 ≡ 73 mod 100
and 77 13 ≡ 17 mod 100. Hence the last two digits of 3333 7777 + 7777 3333
are 90.
4. No. It is not well defined on Q. For instance,
1 1
∗
2 1
2
= 2
3 .

But in Q we have 1
2
= 2
4 yet
2 1
∗
4 1
5. (i) No. 1 + 1 = 2 which is not odd.
= 3
5
= 2
3 .
(ii) Yes. If a and b are even integers then a = 2k and b = 2ℓ for some
integers k, ℓ. But then ab = (2k) (2ℓ) = 2 (2kℓ) is even.
(iii) Yes. If a and b are odd integers then a = 2k + 1 and b = 2ℓ + 1 for
some integers k, ℓ. Thus
which is odd.
a ◦ b = a + b − ab
= 2k + 1 + 2ℓ + 1 − (2k + 1) (2ℓ + 1)
= 2 (−2kℓ) + 1
6. ⋆ (i) Is commutative. Proof: addition on R is commutative,
Is not associative. Counterexample: 1 ∗ (2 ∗ 3) = 1 ∗ 10 = 22 while
(1 ∗ 2) ∗ 3 = 18.
(ii) Is not commutative. Counterexample: 1∗−1 = 1 while −1∗1 = −1.
Is associative. Proof: a ∗ (b ∗ c) = a ∗ (b |c|) = a |b |c|| = a |b| |c| and
(a ∗ b) ∗ c = (a |b|) ∗ c = a |b| |c| .
(iii) Is commutative. Proof: both addition and multiplication are commutative
on R.
Is not associative. Counterexample:
1 ∗ (2 ∗ 3) =
(1 ∗ 2) ∗ 3 =
1 + (2 ∗ 3)
1 × (2 ∗ 3)
(1 ∗ 2) + 3
(1 ∗ 2) × 3 =
= 1 + 2+3
2×3
2+3
2×3
1+2
2
1+2
2
+ 3
= 11
5 .
= 9
3 .
(iv) Is commutative Proof: addition and multiplication are commutative
on Z.
3

Is associative. Proof:
a ∗ (b ∗ c) = a ∗ (b + c − bc)
= a + (b + c − bc) − a (b + c − bc)
= a + b + c − bc − ab − ac + abc
= a + b − ab + c − ac − bc + abc
= (a + b − ab) + c − (a + b − ab) c
= (a ∗ b) + c − (a ∗ b) c
= (a ∗ b) ∗ c.
(v) Is commutative, Proof: x ∗ y = max (x, y) = max (y, x) = y ∗ x.
Is associative. Proof:
x ∗ (y ∗ z) = max (x, y ∗ z) = max (x, max (y, z))
= max (x, y, z) = max (max (x, y) , z)
= max (x ∗ y, z) = (x ∗ y) ∗ z.
(vi) Is not commutative. Counterexample: (1, 2) ∗ (3, 4) = (2, 6) but
(3, 4) ∗ (1, 2) = (3, 10) .
Is associative. Proof:
((a, b) ∗ (c, d)) ∗ (e, f) = (ac, ad + b) ∗ (e, f)
= ((ac) e, (ac) f + (ad + b))
= (ace, acf + ad + b) ,
(a, b) ∗ ((c, d) ∗ (e, f)) = (a, b) ∗ (ce, cf + d)
(vii) Is commutative, Proof:
= (a (ce) , a (cf + d) + b)
= (ace, acf + ad + b) .
(a, b) ∗ (c, d) = (ad + bc, db)
4
= (cb + da, bd)
= (c, d) ∗ (a, b) .

Is associative. Proof:
((a, b) ∗ (c, d)) ∗ (e, f) = (ad + bc, db) ∗ (e, f)
= ((ad + bc) f + (db) e, f (db))
= (adf + bcf + bde, bdf) .
(a, b) ∗ ((c, d) ∗ (e, f)) = (a, b) ∗ (cf + de, fd)
= (a (fd) + b (cf + de) , (fd) b)
= (adf + bcf + bde, bdf) .
7. (i) Yes. max (1, n) = max (n, 1) = n for any n ≥ 1 and so 1 is the
identity.
(ii) No. If e ∈ Z is an identity, then we can choose an integer x < e
and for this integer we find that x ∗ e = max (x, e) = e = x.
(iii) Yes. 0. We have seen that this x ∗ y is commutative so we need
only examine x ∗ 0 = x + 0 − x × 0 = x.
(iv) Yes.
1
2
1
2
1
2
1
2
8. Many reasons. Perhaps because subtraction is not associative.
9. ⋆ a)
1 − (2 − 3) = 2 but (1 − 2) − 3 = −4.
×15 1 4 7 13
1 1 4 7 13
4 4 1 13 7
7 7 13 4 1
13 13 7 1 4
× A B C D
A A B C D
B B A D C
C C D A B
D D C B A
5
×15 3 6 9 12
3 9 3 12 6
6 3 6 9 12
9 12 9 6 3
12 6 12 3 9

where
and
10. a)
A =
1 0
0 1
1 0
, B =
0 −1
D =
−1 0
0 −1
−1 0
, C =
0 1
(b) In ({1, 4, 7, 13} , ×15), the identity is 1 and the inverses are 4 −1 = 4,
7 −1 = 13 and 13 −1 = 7.
In ({3, 6, 9, 12} , ×15) the identity is 6 and the inverses are 9 −1 = 9,
3 −1 = 12 and 12 −1 = 3.
In the matrix group the identity is A and all matrices are self-inverse.
The identity element is 8.
.
×28 4 8 12 16 20 24
4 16 4 20 8 24 12
8 4 8 12 16 20 24
12 20 12 4 24 16 8
16 8 16 24 4 12 20
20 24 20 16 12 8 4
24 12 24 8 20 4 16
4 −1 = 16, 8 −1 = 8, 12 −1 = 24, 16 −1 = 4, 20 −1 = 20 and 24 −1 = 12.
b) ({4, 8, 16} , ×28) is a closed subset.
×28 4 8 16
4 16 4 8
8 4 8 16
16 8 16 4
6

({8, 20} , ×28) is a closed subset:
×28 8 20
8 8 20
20 20 8
7