MATH 74 HOMEWORK 11
1.
Since gcd(
a
1
,b
1
) = 1, a theorem from class gives us integers
M
and
N
with
Ma
1
+
Nb
1
= 1
.
Multiplying through by
b
2
we obtain
Ma
1
b
2
+
Nb
1
b
2
=
b
2
, and using the fact that
a
1
b
2
=
a
2
b
1
we get
Ma
2
b
1
+
Nb
1
b
2
=
b
2
.
so that
b
1
(
Ma
2
+
Nb
2
) =
b
2
. We conclude that
b
1

b
2
and thus
b
1
≤
b
2
by a
theorem from class.
Since gcd(
a
2
,b
2
) = 1, a theorem from class gives us integers
A
and
B
with
Aa
2
+
Bb
2
= 1
.
Multiplying through by
b
1
we get
Aa
2
b
1
+
Bb
2
b
1
=
b
1
, and using the fact that
a
2
b
1
=
a
1
b
2
we obtain
Aa
1
b
2
+
Bb
2
b
1
=
b
1
,
so that
b
2
(
Aa
1
+
Bb
2
) =
b
1
. We conclude that
b
2

b
1
and thus
b
2
≤
b
1
by a theorem
from class. We conclude that
b
1
=
b
2
. Cancelling it from both sides of
a
1
b
2
=
a
2
b
1
we conclude that
a
1
=
a
2
.
2(a).
The elements of
D
(216) are 1, 2, 4, 8, 3, 6, 12, 24, 9, 18, 36, 72, 27, 54, 108,
216, and each of these multiplied by

1. That’s 32 divisors in all.
2(b).