Partial Solutions to Homework 3 for MATH 336
5E # 8
By definition, there is a solution to
ax
≡
b
(mod
m
) if and only if there is an integer solution (
x, y
)
to the equation
ax
+
my
=
b
. By prop 4. (3C), this solution exists if and only if (
a, m
)

b
. Therefore
if (
a, m
)

b
there are 0 solutions.
Otherwise, there is at least one solution.
Let
x
0
be the least
solution such that 0
≤
x
0
< m
. If
x
0
≥
m
(
a,m
)
, then we claim
x
00
=
x
0

m
(
a,m
)
is also a solution. Since
0
≤
x
00
< m
, this would contradict the fact that
x
0
is the least solution in that range. We now show
x
00
is a solution (remember that
m
(
a,m
)
is an integer):
ax
00
≡
a
(
x
0

m
(
a, m
)
)
(mod
m
)
≡
ax
0

am
(
a, m
)
(mod
m
)
≡
b

0
(mod
m
)
≡
b
(mod
m
)
Therefore, we have a contradiction, and hence,
x
0
<
m
(
a,m
)
. By prop. 5 (3C) the other solutions are
x
=
x
0
+
n
m
(
a,m
)
for all integers
n
. As argued above, the least solution in our range is
x
0
when
n
= 0.
If
n
= (
a, m
), then
x
=
x
0
+
(
a,m
)
m
(
a,m
)
=
x
0
+
m
≥
0 +
m
=
m
, which is not in the desired range. If
0
< n <
(
a, m
), then
x
=
x
0
+
n
m
(
a,m
)
<
m
(
a,m
)
+ ((
a, m
)

1)
m
(
a,m
)
= (
a, m
)
m
(
a,m
)
=
m
and hence is
in the desired range. Therefore,
n
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 Spring '08
 BILLERA
 Math, Algebra, #, complete set, Z\mZ

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