Homework 6
Exercise 22 Section 4 Let p ∈ N such that n > 1. Suppose for all x, y ∈ Z, if p | xy,
then p | x or p | y. Then p is a prime.
Proof. Let p ∈ N such that n > 1. Suppose p is not a prime. Since p 6= 1, there is x ∈ N
such that x 6= 1, x 6= p and x | p. Therefore, there exists y ∈ Z such that p = xy. Then
y | p and p | xy. Note that either y ≤ 0 or y > 0. If y ≤ 0, then 0 < p = xy ≤ 0, which is
absurd. Hence, y > 0. Since x, y ∈ N, x | p and y | p, x ≤ p and y ≤ p. Since x 6= p, x < p.
Therefore, p - x. Since x ≥ 2, p = xy ≥ 2y = y + y ≥ y + 1 > y. Hence, p - y.
Exercise 3(a) Section 5. For each n ∈ N,
n
X
i=1
i3 =
n2 (n + 1)2
.
4
Proof. We proceed by induction on n ∈ N. Let n ∈ N. If n = 1, then
1
X
i3 = 13 =
i=1
Assume
n
X
i=1
i3 =
4
12 22
12 (1 + 1)2
=
=
.
4
4
4
n2 (n + 1)2
. Then
4
n+1
X
3
i =
i=1
n
X
i3 + (n + 1)3
i=1
=
n2 (n + 1)2
+ (n + 1)3
4
= (n + 1)2 (
n2
+ n + 1)
4
n2 + 4n + 4
= (n + 1) (
)
4
2
=
(n + 1)2 (n + 2)2
.
4
1
Exercise 6 Section 5. For each x ∈ Z, 6 | x3 − x.
Proof. First we will prove that for each n ∈ N, 6 | n3 − n. We proceed by induction on
n ∈ N. Let n ∈ N. If n = 1, 6 | 03 − 0 since 03 − 0 = 0 and 6 | 0. Assume 6 | n3 − n. Note
that (n + 1)3 − (n + 1) = n3 + 3n2 + 3n + 1 − n − 1 = (n3 − n) + 3n(n + 1) and n is either
even or odd.
Case1. Assume n is even. Then there is m ∈ Z such that n = 2m. Since m(n + 1) ∈ Z and
3n(n + 1) = 3(2m)(n + 1) = 6m(n + 1), 6 | 3n(n + 1). Since 6 | n3 − n, 6 | 3n(n + 1) and
(n + 1)3 − (n + 1) = n3 − n + 3n(n + 1), 6 | (n + 1)3 − (n + 1).
Case2. Assume n is odd. Then there is m ∈ Z such that n = 2m + 1. Hence n + 1 =
2m + 1 + 1 = 2(m + 1). Since m(n + 1) ∈ Z and 3n(n + 1) = 3n(2(m + 1)) = 6n(m + 1),
6 | 3n(n + 1). Since 6 | n3 − n, 6 | 3n(n + 1) and (n + 1)3 − (n + 1) = n3 − n + 3n(n + 1),
6 | (n + 1)3 − (n + 1).
Hence 6 | (n + 1)3 − (n + 1).
To finish this proof, let x ∈ Z. Then x ≥ 0 or x < 0. If x ≥ 0, then x ∈ N and so
6 | x3 − x. Suppose x < 0. Then −x ∈ N. Hence 6 | (−x)3 − (−x). Since −1 ∈ Z and
(−1)((−x)3 − (−x)) = −(−x)3 − x = x3 − x, 6 | x3 − x.
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