HW 3 SOLUTIONS (PROBLEMS 2.2 (P. 19))
] 1. Suppose that a, b ∈ Z with b > 0. Then, by Division algorithm,
∃ ! q 0 , r0 ∈ Z s.t. a = bq 0 + r0 , 0 ≤ r0 < b. Let q = q 0 − 2, r = r0 + 2b. Then
we have
a = q 0 b + r0 = (q + 2)b + r0 = qb + (r0 + 2b) = qb + r
and
2b ≤ r = r0 + 2b < 3b.
The q and r exist uniquely due to the uniqueness of q 0 and r0 .
] 3 (c). Any integer n is of the form n = 5q + r, 0 ≤ r ≤ 4. Consider the
five cases.
n4 = (5q)4 = 5(125q 4 ) = 5k1
n4 = (5q + 1)4 = (5q(5q + 2) + 1)2 = 5k2 + 1
(check!!)
n4 = (5q + 2)4 = (5q(5q + 4) + 4)2 = 5k3 + 1
(check!!)
..
.
] 4. Suppose 3a2 − 1 = b2 for some b ∈ Z. Note that the remainder in the
division of 3a2 − 1 by 3 is 2. But
b = 3k ⇒ b2 = 9k 2
b = 3k + 1 ⇒ b2 = 9k 2 + 6k + 1
b = 3k + 2 ⇒ b2 = 9k 2 + 12k + 4,
and so the remainder in the division of b2 by 3 is either 0 or 1. This is a
contradiction.
] 9. Suppose a = b2 and a = c3 for some b, c ∈ Z. Then
√
b = c c,
b √
= c.
c
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2
HW 3 SOLUTIONS (PROBLEMS 2.2 (P. 19))
√
√
Since cb is rational, so is c. But c ∈ Z. So c ∈ Z, i.e., c is a perfect
square number. Thus a = d6 , where d ∈ Z. It is enough to show that 7 | a
or 7 | a − 1. Now
d = 7k ⇒ a = d6 = 7(75 k 6 )
d = 7k + 1 ⇒ a = (7k + 1)6 = 7(75 k 6 + · · · + 6 · k) + 16
d = 7k + 2 ⇒ a = (7k + 2)6 = 7(75 k 6 + · · · + 6 · k · 25 ) + 7 · 9 + 1
d = 7k + 3 ⇒ a = (7k + 3)6 = 7(75 k 6 + · · · + 6 · k · 35 ) + 7 · 104 + 1
d = 7k + 4 ⇒ a = (7k + 4)6 = 7(75 k 6 + · · · + 6 · k · 45 ) + 7 · 585 + 1
d = 7k + 5 ⇒ a = (7k + 5)6 = 7(75 k 6 + · · · + 6 · k · 55 ) + 7 · 2232 + 1
d = 7k + 6 ⇒ a = (7k + 6)6 = 7(75 k 6 + · · · + 6 · k · 65 ) + 7 · 6665 + 1.
Hence 7 | a or 7 | a − 1.