Math 110A homework 3 solutions
October 21, 2010
§
2.2
1
(c)
+
0
1
2
3
4
5
6
0
0
1
2
3
4
5
6
1
1
2
3
4
5
6
0
2
2
3
4
5
6
0
1
3
3
4
5
6
0
1
2
4
4
5
6
0
1
2
3
5
5
6
0
1
2
3
4
6
6
0
1
2
3
4
5
·
0
1
2
3
4
5
6
0
0
0
0
0
0
0
0
1
0
1
2
3
4
5
6
2
0
2
4
6
1
3
5
3
0
3
6
2
5
1
4
4
0
4
1
5
2
6
3
5
0
5
3
1
6
4
2
6
0
6
5
4
3
2
1
2
(a)
In
Z
8
,
0
2
= 0
1
2
= 1
2
2
= 4
3
2
= 9 = 1
4
2
= 16 = 0
5
2
= 25 = 1
6
2
= 36 = 4
7
2
= 49 = 1
So the solutions in
Z
8
to
x
2
= 1 are 1, 3, 5, and 7.
1
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(b)
In
Z
5
,
0
4
= 0
1
4
= 1
2
4
= 16 = 1
3
4
= (

2)
4
= 16 = 1
4
4
= (

1)
4
= 1
So the solutions in
Z
5
to
x
4
= 1 are 1, 2, 3, and 4.
(c)
Note that in
Z
6
,
x
2
+ 3
x
+ 2 = (
x
+ 1)(
x
+ 2).
In
Z
6
,
(0 + 1)(0 + 2) = 1
·
2 = 2
(1 + 1)(1 + 2) = 2
·
3 = 6 = 0
(2 + 1)(2 + 2) = 3
·
4 = 12 = 0
(3 + 1)(3 + 2) = 4
·
5 = 4
·
(

1) =

4 = 2
(4 + 1)(4 + 2) = 5
·
6 = 5
·
0 = 0
(5 + 1)(5 + 2) = 6
·
7 = 0
·
1 = 0
So the solutions in
Z
6
to (
x
+ 1)(
x
+ 2) = 0 are 1, 2, 4, and 5.
(d)
If there is a solution to
x
2
+ 1 = 0 in
Z
12
, then there is an integer
x
for which
x
2
+ 1
≡
0 (mod 12).
But then
x
2
+ 1
≡
0 (mod 3). That is, there is a solution to
x
2
+ 1 = 0 in
Z
3
. There is no such
thing, since in
Z
3
, 0
2
+ 1 = 1, 1
2
+ 1 = 2, and 2
2
+ 1 = 2.
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 Fall '10
 GIESEKER,D.
 Math, Algebra, Konrad Zuse, Initialisms, Z6, Z12

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