Download Math 2112 Solutions Assignment 1

Math 2112 Solutions
Assignment 1
3.1.32 For all integers n and m, if n − m is even then n3 − m3 is even.
Proof: Let n and m be integers such that n − m is even. Therefore, n − m = 2k
for some k ∈ Z. Note that
n3 − m3 = (n − m)(n2 + nm + m2 ) = 2k(n2 + nm + m2 ) = 2(k(n2 + nm + m2 )).
Let r = k(n2 + nm + m2 ). Since k(n2 + nm + m2 ) is an integer, n3 − m3 = 2r,
with r ∈ Z. Therefore n3 − m3 is even.
3.1.33 For all integers n, if n is prime the (−1)n = (−1).
Counterexample: Let n = 2. Then n is prime, but (−1)n = 1.
3.1.38 Any product of four consecutive integers is one less than a
perfect square.
Proof: Let n, n + 1, n + 2, n + 3 be any four consecutive integers. Consider
n(n + 1)(n + 2)(n + 3) + 1
=
=
=
n(n + 1)(n2 + 5n + 6) + 1
n(n3 + 6n2 + 11n + 6) + 1
n4 + 6n3 + 11n2 + 6n + 1
=
(n2 + 3n + 1)2 .
Therefore, any four consecutive integers is one less than a perfect square.
√
3.1.42 If m and n are perfect squares, then m + n + 2 mn is also a
perfect square. Why?
Proof: Let m and n be perfect squares. Then there exist integers a and b such
that m = a2 and n = b2 . But now
√
√
m + n + 2 mn = a2 + b2 + 2 a2 b2
=
=
=
a2 + b2 + 2ab
a2 + 2ab + b2
(a + b)2
√
Therefore, m + n + 2 mn is also a perfect square.
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3.2.24 Suppose that a, b, c and d are integers and a 6= c. Suppose also
that x is a real number that satisfies the equation
ax + b
= 1.
cx + d
Must x be rational? If so, express x as a ratio of two integers.
Proof: Let a, b, c and d be integers with a 6= c. Suppose that x is a real number
such that
ax + b
= 1.
cx + d
But then
ax + b
cx + d
ax + b
=
1
=
cx + d
ax − cx =
(a − c)x =
d−b
d−b
d−b
.
a−c
x =
Since d and b are integers, so is d − b. Likewise, since a and c are integers, so is
a − c. Note also that since a 6= c, a − c 6= 0. Therefore, x is a rational number.
3.3.12 If n = 4k + 3, does 8 divide n2 − 1?
Proof: Let n = 4k + 3. Then
n2 − 1
= (4k + 3)2 − 1
= 16k 2 + 24k + 9 − 1
=
=
16k 2 + 24k + 8
8(2k 2 + 3k + 1).
Therefore, 8 divides n2 − 1.
3.3.24 For all integers a,b and c, if a|bc then a|b or a|c.
Counterexample: Let a = 4, b = 2 and c = 2. Then a|bc but a 6 |b and a 6 |c.
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