Math 541
Solutions to HW #4
1. Use the Euclidean algorithm to compute the greatest common divisor of 320353 and 257642.
320353 = 257642(1) + 62711
257642 = 62711(4) + 6798
62711 = 6798(9) + 1529
6798 = 1529(4) + 682
1529 = 682(2) + 165
682 = 165(4) + 22
165 = 22(7) + 11
22 = 11(2) + 0
We conclude that g.c.d.(320353, 257642) = 11.
2. Prove that the equation
320353
x
+ 257642
y
= 1
has no solutions with
x, y
in
Z
. (Please do this question directly without referring to theorems from
class.)
•
From (1), we know that g.c.d(320353, 257642) = 11. We can use this to determine that 320353 =
11
·
29123, and 257642 = 11
·
23422. Then 320353
x
+ 257642
y
= (11
·
29123)
x
+ (11
·
23422)
y
=
11
·
(29123
x
+23422
y
). That is, 11 divides any integral solution of the equation 320353
x
+257642
y
.
Since 11 does not divide 1, we conclude that the equation has no integer solutions.
3. Compute the following values:
φ
(100),
φ
(40),
φ
(101).
[Recall that we have formulas
φ
(
p
n
) =
p
n

p
n

1
and
φ
(
mn
) =
φ
(
m
)
φ
(
n
) if
m
and
n
are relatively
prime.]
•
φ
(100) =
φ
(2
2
·
5
2
) = (2

1)(2
1
)(5

1)(5
1
) = (1)(2)(4)(5) = 20
•
φ
(40) =
φ
(2
3
·
5
1
) = (2

1)(2
2
)(5

1)(5
0
) = (1)(4)(4)(1) = 16
•
φ
(101) = (101

1)(101
0
) = 100 (101 is a prime!)
4. Give 5 examples of groups with 8 elements. Do these groups have distinct multiplication tables up to
reordering?
•
(
Z
8
,
+), i.e.
Z
8
under addition
•
U
(15), since
φ
(15) =
φ
(3
·
5) = (3

1)(5

1) = (2)(4) = 8.
•
U
(16), since
φ
(16) =
φ
(2
4
) = (2

1)(2
3
) = (1)(8) = 8.
•
U
(20), since
φ
(20) =
φ
(2
2
·
5) = (2

1)(2
1
)(5

1) = (1)(2)(4) = 8.
•
U
(24), since
φ
(24) =
φ
(2
3
·
3) = (2

1)(2
2
)(3

1) = (1)(4)(2) = 8.
•
The multiplication tables for
Z
8
,
U
(15)
,
U
(16)
,
U
(20)
,
U
(24) follow:
1
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–
(
Z
8
,
+) =
+
0
1
2
3
4
5
6
7
0
0
1
2
3
4
5
6
7
1
1
2
3
4
5
6
7
0
2
2
3
4
5
6
7
0
1
3
3
4
5
6
7
0
1
2
4
4
5
6
7
0
1
2
3
5
5
6
7
0
1
2
3
4
6
6
7
0
1
2
3
4
5
7
7
0
1
2
3
4
5
6
–
U
(15) =
·
1
2
4
7
8
11
13
14
1
1
2
4
7
8
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 Fall '09
 Pollack
 Math, Algebra, Prime number

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