Solutions to Assignment 1
Question 1. [Exercises 1.1, # 6]
Use the division algorithm to prove that every odd integer is either of the form 4
k
+ 1 or of the form 4
k
+ 3
for some integer
k.
Solution:
For each positive integer
n,
the only possible remainders when
n
is divided by 4 are
r
= 0
,
1
,
2
,
3
,
so that
n
has exactly one of the forms
4
k,
4
k
+ 1
,
4
k
+ 2
,
4
k
+ 3
for some integer
k.
Since 4
k
and 4
k
+ 2 are even, then every odd integer has the form 4
k
+ 1 or 4
k
+ 3 for
some integer
k.
Question 2. [Exercises 1.1, # 8]
(a) Divide 5
2
,
7
2
,
11
2
,
15
2
,
and 27
2
by 8 and note the remainder in each case.
(b) Make a conjecture about the remainder when the square of an odd integer is divided by 8
.
(c) Prove your conjecture.
Solution:
(a) We have
5
2
= 25
= 24 + 1
= 3
·
8 + 1
7
2
= 49
= 48 + 1
= 6
·
8 + 1
11
2
= 121 = 120 + 1 = 15
·
8 + 1
15
2
= 225 = 224 + 1 = 28
·
8 + 1
27
2
= 729 = 728 + 1 = 91
·
8 + 1
,
and the remainder is 1 in each case.
(b) The conjecture is that for any odd integer
n,
we have
n
2
= 8
k
+ 1 for some integer
k.
(c) If
n
is an odd integer. then
n
= 4
m
+ 1 or
n
= 4
m
+ 3 for some integer
m,
so that either
n
2
= (4
m
+ 1)
2
= 16
m
2
+ 8
m
+ 1 = 8(2
m
+ 1) + 1
or
n
2
= (4
m
+ 3)
2
= 16
m
2
+ 24
m
+ 1 = 8(2
m
+ 3) + 1
and in both cases
n
2
= 8
k
+ 1
.
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Question 3. [Exercises 1.2, # 8].
If
r
∈
Z
and
r
is a nonzero solution of
x
2
+
ax
+
b
= 0 (where
a, b
∈
Z
), prove that
r

b.
Solution:
If
r
is a nonzero solution of
x
2
+
ax
+
b
= 0
,
then
r
2
+
ar
+
b
= 0
,
so that
b
=

r
2

ar
=

r
(
r
+
a
)
,
and
r

b.
Question 4. [Exercises 1.2, # 10].
Prove that (
n, n
+ 1) = 1 for every positive integer
n.
Solution:
Note that
1 = 1
·
(
n
+ 1) + (

1)
·
n,
so that 1 is a linear combination of
n
+ 1 and
n,
and is the smallest such positive integer.
Therefore
(
n, n
+ 1) = 1
.
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 Fall '10
 GIESEKER,D.
 Algebra, Division, Remainder, Natural number, positive integer

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