Proof.
The integer
x
=
ba
0
, where
a
0
is a multiplicative inverse of
a
modulo
n
, is clearly a solution.
For any integer
x
0
, we have
ax
0
≡
b
(mod
n
) if and only if
ax
0
≡
ax
(mod
n
), which by Theorem 2.3
holds if and only if
x
≡
x
0
(mod
n
).
2
In particular, this theorem implies that multiplicative inverses are uniquely determined modulo
n
.
More generally, we have:
Theorem 2.5
Let
n
be a positive integer and let
a, b
∈
Z
. Let
d
= gcd(
a, n
)
. If
d

b
, then the
congruence
ax
≡
b
(mod
n
)
has a solution
x
, and any integer
x
0
is also a solution if and only if
x
≡
x
0
(mod
n/d
)
. If
d

b
, then the congruence
ax
≡
b
(mod
n
)
has no solution
x
.
Proof.
Let
n, a, b, d
be as defined above.
For the first statement, suppose that
d

b
.
In this case, by Theorem 2.3, we have
ax
≡
b
(mod
n
) if and only if (
a/d
)
x
≡
(
b/d
) (mod
n/d
), and so the statement follows immediately from
Theorem 2.4.
For the second statement, assume that
ax
≡
b
(mod
n
) for some integer
x
. Then since
d

n
,
we have
ax
≡
b
(mod
d
). However,
ax
≡
0 (mod
d
), since
d

a
, and hence
b
≡
0 (mod
d
), i.e.,
d

b
.
2
Next, we consider systems of congruences with respect to moduli that that are relatively prime
in pairs. The result we state here is known as the Chinese Remainder Theorem, and is extremely
useful in a number of contexts.
6
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Theorem 2.6 (Chinese Remainder Theorem)
Let
k >
0
, and let
a
1
, . . . , a
k
∈
Z
, and let
n
1
, . . . , n
k
be positive integers such that
gcd(
n
i
, n
j
) = 1
for all
1
≤
i < j
≤
k
.
Then there ex
ists an integer
x
such that
x
≡
a
i
(mod
n
i
)
(
i
= 1
, . . . , k
)
.
Moreover, any other integer
x
0
is also a solution of these congruences if and only if
x
≡
x
0
(mod
n
)
,
where
n
:=
Q
k
i
=1
n
i
.
Proof.
Let
n
:=
Q
k
i
=1
n
i
, as in the statement of the theorem. Let us also define
n
0
i
:=
n/n
i
(
i
= 1
, . . . , k
)
.
It is clear that gcd(
n
i
, n
0
i
) = 1 for 1
≤
i
≤
k
, and so let
m
i
be a multiplicative inverse of
n
0
i
modulo
n
i
for 1
≤
i
≤
k
, and define
z
i
:=
n
0
i
m
i
(
i
= 1
, . . . , k
)
.
By construction, one sees that for 1
≤
i
≤
k
, we have
z
i
≡
1 (mod
n
i
)
and
z
i
≡
0 (mod
n
j
)
for 1
≤
j
≤
k
with
j
6
=
i
.
That is to say, for 1
≤
i, j
≤
k
,
z
i
≡
δ
ij
(mod
n
j
), where
δ
ij
:= 1 for
i
=
j
and
δ
ij
:= 0 for
i
6
=
j
.
Now define
x
:=
k
X
i
=1
z
i
a
i
.
One then sees that for 1
≤
j
≤
k
,
x
≡
k
X
i
=1
z
i
a
i
≡
k
X
i
=1
δ
ij
a
i
≡
a
j
(mod
n
j
)
.
Therefore, this
x
solves the given system of congruences.
Moreover, if
x
0
≡
x
(mod
n
), then since
n
i

n
for 1
≤
i
≤
k
, we see that
x
0
≡
x
≡
a
i
(mod
n
i
)
for 1
≤
i
≤
k
, and so
x
0
also solves the system of congruences.
Finally, if
x
0
solves the system of congruences, then
x
0
≡
x
(mod
n
i
) for 1
≤
i
≤
k
. That is,
n
i

(
x
0

x
) for 1
≤
i
≤
k
.
Since gcd(
n
i
, n
j
) = 1 for
i
6
=
j
, this implies that
n

(
x
0

x
), i.e.,
x
0
≡
x
(mod
n
).
2
2.3
Residue Classes
It is easy to see that for a fixed value of
n
, the relation
· ≡ ·
(mod
n
) is an
equivalence relation
on
the set
Z
; that is, for all
a, b, c
∈
Z
, we have
•
a
≡
a
(mod
n
)
,
•
a
≡
b
(mod
n
)
implies
b
≡
a
(mod
n
)
,
and
•
a
≡
b
(mod
n
)
and
b
≡
c
(mod
n
)
implies
a
≡
c
(mod
n
)
.
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 Spring '13
 MRR
 Math, Algebra, Number Theory

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