112
Chapter 7: Integers
Exercises for Section 7.1. Divisibility Properties
1. (a) 7 | − 70, c = −10.
(b) 16 ! −40.
(c) 1 | 10, c = 10.
(d) 8 | − 8, c = −1. (e) 14 | 0, c = 0. (f) 0 ! 14 (not defined).
2. (a) 3, 6, 9
(b) 5, 10, 15
(c) 30 (d) 12
(e) 12
(f) 12
3. Proof. Assume that a | b. Then b = ax for some integer x. Thus
−b = −(ax) = a(−x) and b = (−a)(−x).
Since −x is an integer, a | (−b) and (−a) | b.
4. Proof. Assume that a | b and a | (b + c). Then b = ax and b + c = ay for integers x and y.
Thus b + c = ax + c = ay. Therefore, c = ay − ax = a(y − x). Since y − x is an integer, a | c.
5. Proof. First, assume that ac | bc. Then bc = (ac)x = c(ax) for some integer x. Since c ̸= 0,
we can divide by c, obtaining b = ax. So a | b.
For the converse, assume that a | b. Then b = ax for some integer x. Multiplying by c, we
have
bc = (ax)c = (ac)x.
Since x is an integer, ac | bc.
6. Proof. Assume that a | b. Then b = ax for some integer x. Since b ̸= 0, it follows that x ̸= 0.
Thus |b| = |ax| = |a||x| ≥ |a| since |x| ≥ 1.
7. Let a = 1 and b = −1. Since 1 | (−1) and (−1) | 1 but 1 ̸= −1, it follows that a = 1 and
b = −1 is a counterexample.
8. Proof. Assume that 3 | b. Since 3 | a and 3 | b, it follows that a = 3x and b = 3y for integers
x and y. Thus a + b = 3x + 3y = 3(x + y). Since x + y is an integer, 3 | (a + b).
9. Proof. Assume that a | b or a | c, say a | b. Then b = ax for some integer x. Thus
bc = (ax)c = a(xc). Since xc is an integer, a | bc.
10. (a) Proof. Assume that a is a multiple of 36. Then a = 36x for some integer x. Thus
a = 12(3x). Since 3x is an integer, 12 | a.
(b) The result in (a) is a consequence of Theorem 7.5.
11. Proof. We proceed
Since 3 | (4 · 03 + 5 · 0), the statement is true for n = 0.
! 3by induction.
"
Assume that 3 | 4k + 5k , where k is a nonnegative integer. Then 4k 3 + 5k = 3ℓ for some
integer ℓ. We show that 3 divides 4(k + 1)3 + 5(k + 1). Observe that
4(k + 1)3 + 5(k + 1) = 4(k 3 + 3k 2 + 3k + 1) + 5k + 5
= 4k 3 + 12k 2 + 12k + 4 + 5k + 5
= (4k 3 + 5k) + 12k 2 + 12k + 9
= 3ℓ + 3(4k 2 + 4k + 3) = 3(ℓ + 4k 2 + 4k + 3).
Since ℓ + 4k 2 + 4k + 3 is an integer, 3 divides 4(k + 1)3 + 5(k + 1). It follows by the Principle
of Mathematical Induction that 3 divides 4n3 + 5n for every nonnegative integer n.
113
0
12. Proof. !We proceed
is true for n = 0. Assume
" by induction. Since 3 | (2 − 1), the statement
2k
that 3 | 2 − 1 , where k is a nonnegative integer. Then 22k − 1 = 3ℓ for some integer ℓ and
so 22k = 3ℓ + 1. We show that 3 divides 22k+2 − 1. Observe that
22k+2 − 1
=
=
=
4 · 22k − 1
4(3ℓ + 1) − 1 = 12ℓ + 3
3(4ℓ + 1).
Since 4ℓ + 1 is an !integer, "3 divides 22k+2 − 1. It follows by the Principle of Mathematical
Induction that 3 | 22n − 1 for every nonnegative integer n.
3·0
13. Proof. We
proceed
is true for n = 0. Assume
! 3k
" by induction. Since 7 | (2 − 1), the statement
that 7 | 2 − 1 , where k is a nonnegative integer. Then 23k − 1 = 7ℓ for some integer ℓ.
Thus 23k = 7ℓ + 1. We show that 7 divides 23(k+1) − 1. Observe that
23(k+1) − 1 =
=
=
23k+3 − 1 = 23 23k − 1
8 · 23k − 1 = 8(7ℓ + 1) − 1
56ℓ + 7 = 7(8ℓ + 1).
Since 8ℓ + 1 is an integer,
7" divides 23(k+1) − 1. It follows by the Principle of Mathematical
! 3n
Induction that 7 | 2 − 1 for every nonnegative integer n.
0
0
14. Proof. We
is true for n = 0. Assume
! k proceed
" by induction. Since 4 | (7 − 3 ), the statement
k
k
k
that 4 | 7 − 3 , where k is a nonnegative
integer.
Then
7
−
3
=
4ℓ for some integer ℓ and
!
"
so 7k = 4ℓ + 3k . We show that 4 | 7k+1 − 3k+1 . Observe that
7k+1 − 3k+1
=
=
7 · 7k − 3 · 3k
7(4ℓ + 3k ) − 3 · 3k
28ℓ + 4 · 3k = 4(7ℓ + 3k ).
!
"
Since 7ℓ + 3k is an integer, 4 | 7k+1 − 3k+1 . By the Principle of Mathematical Induction,
4 | (7n − 3n ) for every nonnegative integer n.
=
15. If n < 0, then n = −k for some positive integer k. Then apply Result 7.6 and Exercise 3.
16. Proof. We proceed by induction. For n = 1, 32n−1 + 1 = 4. Since 4 | 4, the statement is true
for n = 1. Assume that 4 | (32k−1 + 1) for a positive integer k. We show that 4 | (32k+1 + 1).
Since 4 | (32k−1 + 1), it follows that 32k−1 + 1 = 4x for some integer x. Thus 32k−1 = 4x − 1.
Therefore,
32k+1 + 1 = 9 · 32k−1 + 1 = 9(4x − 1) + 1 = 36x − 8 = 4(9x − 2).
Since 9x − 2 is an integer, 4 | (32k+1 + 1). By the Principle of Mathematical Induction,
4 | (32n−1 + 1) for every positive integer n.
17. Proof. We proceed by induction. For n = 1, 11n+1 +122n−1 = 112 +12 = 121+12 = 133. Since
133 | 133, the statement is true for n = 1. Assume that 133 | (11k+1 + 122k−1 ) for a positive
integer k. Then 11k+1 + 122k−1 = 133x for some integer x and so 122k−1 = 133x − 11k+1 . We
show that 133 | (11k+2 + 122k+1 ). Thus
11k+2 + 122k+1
=
11k+2 + (144)122k−1
=
11k+2 + 144(133x − 11k+1 )
=
=
=
144(133x) + (11 − 144)11k−1
133(144x) − (133)11k−1
133(144x − 11k−1 ).
114
Since 144x − 11k−1 is an integer, 133 | (11k+2 + 122k+1 ). By the Principle of Mathematical
Induction, 133 | (11n+1 + 122n−1 ) for every positive integer n.
18. (a) Proof. Let m be an integer such that 1 ≤ m ≤ 2n. Let ℓ be the greatest nonnegative
integer such that 2ℓ | m. Then m = 2ℓ k for some integer k. Necessarily k is an odd
integer with 1 ≤ k < 2n, for otherwise this would contradict the definition of ℓ.
(b) Proof. Let S be a subset of {1, 2, . . . , 2n} having cardinality n + 1. By (a), every element
of S can be expressed as 2ℓ k, where ℓ ≥ 0 and k is an odd integer with 1 ≤ k < 2n. Since
there are exactly n odd integers in the set {1, 2, . . . , 2n}, there must exist two elements
a and b in S such that a = 2i k and b = 2j k for the same odd integer k. Since a ̸= b, it
follows that i ̸= j, say 0 ≤ i < j. Then
b = 2j k = 2j−i 2i k = 2j−i a.
Since 2j−i is an integer, a | b.
Exercises for Section 7.2. Primes
1. (a) 250 = 2 · 53
(d) 1225 = 52 · 72
2. (a) 127 is a prime
(b) 297 = 33 · 11
(e) 891 = 34 · 11.
(c) 2662 = 2 · 113
(b) 129 = 3 · 43
(d) 133 = 7 · 19.
√
3. 1009 is a prime. Since 1009 < 32, we need only show that none of the primes less than 32
divide 1009. Since none of 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 divide 1009, it follows that 1009
is a prime.
(c) 131 is a prime
4. Let n = 2000!
5. Proof. Assume, to the contrary, that there is a finite number of primes greater than 1, 000, 000.
Since there is certainly a finite number of primes less than 1, 000, 000, it follows that there is
a finite number of primes. This is a contradiction.
6. Proof. For each prime p, the integer p2 is composite. Since there infinitely many primes,
{p2 : p is a prime} is an infinite set of composite numbers.
7. (a) 111 = 3 · 37
(b) 1111 = 11 · 101
(c) 111, 111 = 111 · 1001
(d) No, 11, 111 = 41 · 271.
8. Suppose that p is an odd prime. Then p = 2k + 1 for some positive integer k. Then p =
(k + 1)2 − k 2 . (Note: This also shows that every odd integer can be expressed in this way.) We
show that 2 cannot be expressed as a2 − b2 for any positive integers a and b. If 2 = a2 − b2 ,
then a > b. So a = b + ℓ, where ℓ ≥ 1. Then a2 − b2 = (b + ℓ)2 − b2 = 2bℓ + ℓ2 ≥ 3, which is a
contradiction.
9. First, ab may be prime, such as 31. However, no other integer in the list can be prime. If
abab · · · ab contains 2k digits, where k ≥ 2, then abab · · · ab = ab (1010 · · · 01).
10. Proof. Suppose that p is a prime such that p = n3 − 1 for some positive integer n. Thus n ≥ 2.
Then p = n3 − 1 = (n − 1)(n2 + n + 1). Since either n − 1 or n2 + n + 1 is 1, either n − 1 = 1
or n2 + n + 1 = 1. In the first case, n = 2, while in the second case, n = 0 or n = −1. Since
n ∈ N, it follows that n = 2 and so p = 7.