Solutions to Homework Set 3
(Solutions to Homework Problems from Chapter 2)
Problems from
§
2.1
2.1.1. Prove that
a
≡
b
(mod
n
) if and only if
a
and
b
leave the same remainder when divided by
n
.
Proof.
⇒
Suppose
a
≡
b
(mod
n
). Then, by deﬁnition, we have
a

b
=
nk
for some
k
∈
Z
. Now by the Division Algorithm,
a
and
b
can be written uniquely in form
(1)
a
=
nq
+
r
b
=
nq
0
+
r
0
with 0
≤
r, r
0
< n
. But then
(2)
a
=
b
+
nk
= (
nq
0
+
r
0
) +
nk
=
n
(
q
0
+
k
) +
r
0
Comparing (1) and (2) we have
a
=
nq
+
r
,
0
≤
r < n
a
=
n
(
q
0
+
k
) +
r
0
,
0
≤
r
0
< n
.
By the uniqueness property of the division algorithm, we must therefore have
r
=
r
0
.
⇐
If
a
and
b
leave the same remainder when divided by
n
then we have
a
=
nq
+
r
b
=
nq
0
+
r
.
Subtracting these two equations yields
a

b
=
n
(
q

q
0
)
,
so
a
≡
b
(mod
n
)
.
2.1.2. If
a
∈
Z
, prove that
a
2
is not congruent to 2 modulo 4 or to 3 modulo 4.
•
Proof.
By the Division Algorithm any
a
∈
Z
must have one of the following forms
a
=
4
k
4
k
+ 1
4
k
+ 2
4
k
+ 3
This implies
a
2
=
16
k
2
= 4(4
k
2
) = 4
q
16
k
2
+ 8
k
+ 1 = 4(4
k
2
+ 2
k
) + 1 = 4
r
+ 1
4(4
k
2
+ 16
k
+ 4 = 4(4
k
2
+ 8
k
+ 1) = 4
s
16
k
2
+ 24
k
+ 9 = 4(4
k
2
+ 6
k
+ 2) + 1 = 4
t
+ 1
1
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So
a
2
≡
±
0 (mod 4)
1 (mod 4)
.
2.1.3. If
a, b
are integers such that
a
≡
b
(mod
p
) for every positive prime
p
, prove that
a
=
b
.
•
Proof.
Since the set of prime numbers in
Z
is inﬁnite, we can always ﬁnd a prime number
p
larger
than any given number. In particular we can ﬁnd a prime number
p
such that
0
≤ 
a

b

< p
.
Now by hypothesis, we have, for this prime
p
,
a

b
=
kp
for some
k
∈
Z
(by the deﬁnition of congruence modulo
p
). Thus,
p
divides

a

b

. But 0 is the
only nonnegative number less than
p
that is also divisible by
p
. Thus,

a

b

= 0 or
a
=
b
.
2.1.4. Which of the following congruences have solutions:
(a)
x
2
≡
1 (mod 3)
•
We need
x
2

1 = 3
k
By the Division Algorithm,
x
must have one of three forms
x
=
3
t
3
t
+ 1
3
t
+ 2
⇒
x
2

1 =
9
t
2

1
9
t
2
+ 6
t
9
t
2
+ 12
t
+ 3
Thus, if
x
has the form
x
= 3
t
+ 1, then
x
2

1 = 3(3
t
2
+ 2
t
) and so
x
2
≡
1 (mod 3).
(b)
x
2
≡
2 (mod 7)
•
We need
x
2

2 = 3
k
By the Division Algorithm,
x
must have one of the seven forms
x
=
7
k
7
k
+ 1
7
k
+ 2
7
k
+ 3
7
k
+ 4
7
k
+ 5
7
k
+ 6
⇒
x
2

1 =
49
k
2

2
=
7
(
7
k
2
)
+ 2
49
k
2
+ 14
k

1
=
7
(
7
k
2
+ 2
k

1
)
+ 6
49
k
2
+ 28
k
+ 2
=
7
(
7
k
2
+ 4
k
)
+ 2
49
k
2
+ 42
k
+ 7
=
7
(
7
k
2
+ 6
k
+ 1
)
49
k
2
+ 70
k
+ 14
=
7
(
7
k
2
+ 8
k
+ 2
)
49
k
2
+ 70
k
+ 23
=
7
(
7
k
2
+ 10
k
+ 3
)
+ 2
49
k
2
+ 84 + 34
=
7
(
7
k
2
+ 12
k
+ 4
)
+ 6
Thus, if
x
has the form
x
= 7
k
+ 3 or the form
x
= 7
k
+ 4, then
x
2

2 is an integer multiple of 7
and so
x
2
≡
2 (mod 7).
(c)
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 Fall '10
 GIESEKER,D.
 Algebra, Remainder, Zn, zp

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