Math
465 Solution Set 2
Ae Ja Yee
1. If (
a, c
) = 1 and (
b, c
) = 1, then (
ab, c
) = 1.
Solution.
Method 1: Suppose (
ab, c
)
>
1, namely
x

ab
and
x

c
for some
x >
1. Since
(
a, c
) = 1,
x
cannot divide
a
, that is (
x, a
) = 1. However,
x

ab
, so
x

b
, which is a
contradiction.
Method 2: We can write 1 =
ax
1
+
cy
1
=
bx
2
+
cx
2
. Thus
ax
1
bx
2
= (1

cy
1
)(1

cy
2
) = 1

c
(
y
1
+
y
2

cy
1
y
2
)
,
which is equivalent to
1 =
abx
+
cy,
where
x
=
x
1
x
2
and
y
=
y
1
+
y
2

cy
1
y
2
. Therefore, (
ab, c
) = 1.
2
2. Show that if (
a, b
) = 1, then (
a

b, a
+
b
) = 1 or 2. When is the value 2?
Solution.
Method 1: Let
n
=
a

b
,
m
=
a
+
b
, and (
n, m
) =
d
. Then
2
a
=
m
+
n,
2
b
=
m

n
Since (
n, m
) =
d
,
d

2
a
and
d

2
b
, that is
d
is a common divisor of 2
a
and 2
b
. We now
compute (2
a,
2
b
):
(2
a,
2
b
) = 2(
a, b
) = 2
since (
a, b
) = 1. Therefore,
d

2, that is
d
= 1 or 2.
Method 2: Since gcd is the least positive linear combination, (
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 Spring '08
 YEE
 Math, Number Theory, Greatest common divisor, Euclidean algorithm, Binary GCD algorithm, Ae Ja Yee, ab  cbx

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