25
Therefore, 1 is a linear combination of
rr
0
and
m
; as 1 is obviously the
smallest positive linear combination, it must be their gcd.
1.59
Let
a
,
b
and
d
be integers. If
d
=
sa
+
tb
, where
s
and
t
are integers,
f
nd
in
f
nitely many pairs of integers
(
s
k
,
t
k
)
with
d
=
s
k
a
+
t
k
b
.
Solution.
If
d
=
sa
+
tb
, and if we de
f
ne
s
k
=
s
+
kb
and
t
k
=
t
−
ka
,
then
d
=
s
k
a
+
t
k
b
for all
k
.
1.60
If
a
and
b
are relatively prime and if each divides an integer
n
, prove that
their product
ab
also divides
n
.
Solution.
Assume that
(
a
,
b
)
=
1 and
n
=
ak
=
b
`
. By Corollary 1.40,
b

ak
implies
b

k
. Thus,
k
=
bk
0
and so
n
=
ak
=
abk
0
.
1.61
Prove, for any (possibly negative) integers
a
and
b
, that
(
b
,
a
)
=
(
b
−
a
,
a
)
.
Solution.
If
c
is a common divisor of
a
and
b
, then
c

a
and
c

b
; hence,
c

b
−
a
, and
c
is a common divisor of
a
and
b
−
a
. This does not yet show
that the two gcd
’
s are equal. However, if
c
0
is a common divisor of
a
and
b
−
a
, then
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 Fall '11
 KeithCornell
 Greatest common divisor, Euclidean algorithm, common divisor

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