1. Section 1.5 - Problem 8 (Problem 16 in the 6th edition)
Yes there are. For example, 6 | 2 · 3 = 6, but 66 | 2 and 66 | 3.
2. Section 1.5 - Problem 18 (Problem 26 in the 6th edition)
First we prove existence. By the division algorithm there exist integers q and r such that
a = qb + r with 0 ≤ r < b. If r ≤ b/2, then we have a = bq + r with −b/2 < r ≤ b/2
as required. Otherwise, we have b/2 < r < b. This implies −b/2 < r − b < 0 and then
a = b(q + 1) + (r − b) gives us the required form.
Now we prove uniqueness. Suppose we have two solutions, a = bq1 +r1 = bq2 +r2 , satisfying
−b/2 < r1 ≤ b/2 −b/2 < r2 ≤ b/2. Then bq1 +r1 −bq2 −r2 = 0 and so b(q1 −q2 ) = r1 −r2 and
thus b | r1 −r2 . We have −b/2 < r1 ≤ b/2 and −b/2 ≤ −r2 ≤ b/2 so we get −b < r1 −r2 < b.
There are no nonzero multiples of b in this range and therefore r1 − r2 = 0 from which it follow
r1 = r2 and hence q1 = q2 , and our solution must be unique.
3. Section 1.5 - Problem 28 (Problem 36 in the 6th edition)
Solution 1: For any a we have a3 − a = a(a − 1)(a + 1). By the previous problem we know
that a = 3q + r for some integer q and for some r ∈ {−1, 0, 1}. In each case of r we have one
of a + 1, a or a − 1 is divisible by 3. Thus 3 divides a3 − a.
Solution 2: We may asume without loss of generality that a ≥ 0. Now use induction on a.
Base case: a = 0. Here a3 − a = 03 − 0 = 0 is divisible by 3.
Induction step. Suppose a3 − a is divisible by 3. Then
(a + 1)3 − (a + 1) = a3 + 3a2 + 3a + 1 − (a + 1) = (a3 − a) + 3a3 + 3a .
Each term is divisible by 3, so (a + 1)3 − (a + 1) is divisible by 3, and the induction step is
complete.
4. Section 3.3 - Problem 6
We have (a + 2, a) = (a, 2) ≤ 2. If a is even then 2 | a and so (a, 2) = 2. Otherwise if a is
odd then 26 | a and hence (a, 2) = 1.
5. Section 3.3 - Problem 8 (Problem 10 in the 6th edition)
As (a, b) = 1, there exist m, n such that am + bn = 1. Then
(a + b)(m + n) + (a − b)(m − n) = am + an + bm + bn + am − an − bm + bn = 2am + 2bn = 2
Hence (a + b, a − b) ≤ 2 and is therefore 1 or 2.
6. Section 3.3 - Problem 10 (Problem 12 in the 6th edition)
As a and b are even and not both 0, (a, b) and (a/2, b/2) are well defined. We will show
that (a, b) = 2(a/2, b/2) in two steps: first we will show that (a, b) ≥ 2(a/2, b/2), then that
(a, b) ≤ 2(a/2, b/2).
Step 1: There exist integers m, n such that am + bn = (a, b). As a, b are even, we have
2 | (a, b) and so (a, b)/2 is an integer. Then (a/2)m + (b/2)n = (a, b)/2. Since (a/2, b/2) is
the minimal positive integer linear combination of a/2 and b/2, we conclude that (a, b)/2 ≥
(a/2, b/2), that is, (a, b) ≥ 2(a/2, b/2).
Step 2: There exist integers x, y such that (a/2)x + (b/2)y = (a/2, b/2). Then ax + by =
2(a/2, b/2). Since (a, b) is the minimal positive integer linear combination of a and b, we
conclude that (a, b) ≤ 2(a/2, b/2).
Therefore (a, b) ≥ 2(a/2, b/2) and (a, b) ≤ 2(a/2, b/2). We conclude that (a, b) = 2(a/2, b/2).
7. Section 3.3 - Problem 12 (Problem 14 in the 6th edition)
Since c divides a + b, gcd(a, c) divides gcd(a, a + b) = gcd(a, b) = 1. Thus gcd(a, c) = 1.
Analogously we also get (b, c) = 1.
8. Section 3.3 - Problem 22 (Problem 24 in the 6th edition)
We see that 5(3k + 2) − 3(5k + 3) = 1 and therefore 3k + 2 and 5k + 3 are relatively prime.
9. Section 3.4 - Problem 2
a)
87 = 51 + 36
51 = 36 + 15
36 = 2 · 15 + 6
15 = 2 · 6 + 3
6=2·3
Therefore (51, 87) = 3.
b)
300 = 2 · 105 + 90
105 = 90 + 15
906 · 15
Therefore (105, 300) = 15.
c)
1234 = 981 + 253
981 = 3 · 253 + 222
253 = 222 + 31
222 = 7 · 31 + 5
31 = 6 · 5 + 1
5=5·1
Therefore (981, 1234) = 1.
10. Section 3.4 - Problem 6
a) (15, 35, 90) = ((15, 35), 90) = (5, 90) = 5
b) First we use the Euclidean algorithm and calculate:
2160 = 7 · 300 + 60
300 = 6 · 60
So (300, 2160) = 60. Then as 60 | 5040 we have (300, 2160, 5040) = ((300, 2160), 5040) =
(60, 5040) = 60.