1.11
Solutions of Polynomial Congruences
Let
f
be a polynomial with integer coefficients, and let
m
be a positive
integer.
If
x
and
x
0
are integers with
x
≡
x
0
(mod
m
) then
f
(
x
)
≡
f
(
x
0
)
(mod
m
).
It follows that the set of integers
x
satisfying the congruence
f
(
x
)
≡
0 (mod
m
) is a union of congruence classes modulo
m
. The
number
of solutions modulo
m
of the congruence
f
(
x
)
≡
0 (mod
m
) is defined to
be the number of congruence classes of integers modulo
m
such that an
integer
x
satisfies the congruence
f
(
x
)
≡
0 (mod
m
) if and only if it belongs
to one of those congruence classes. Thus a congruence
f
(
x
)
≡
0 (mod
m
)
has
n
solutions modulo
m
if and only if there exist
n
integers
a
1
, a
2
, . . . , a
n
satisfying the congruence such that every solution of the congruence
f
(
x
)
≡
0
(mod
m
) is congruent modulo
m
to exactly one of the integers
a
1
, a
2
, . . . , a
n
.
Note that the number of solutions of the congruence
f
(
x
)
≡
0 (mod
m
)
is equal to the number of integers
x
satisfying 0
≤
x < m
for which
f
(
x
)
≡
0
(mod
m
). This follows immediately from the fact that each congruence class
of integers modulo
m
contains exactly one integer
x
satisfying 0
≤
x < m
.
Theorem 1.24
Let
f
be a polynomial with integer coefficients, and let
p
be
a prime number. Suppose that the coefficients of
f
are not all divisible by
p
.
Then the number of solutions modulo
p
of the congruence
f
(
x
)
≡
0
(mod
p
)
is at most the degree of the polynomial
f
.
13
Proof
The result is clearly true when
f
is a constant polynomial. We can
prove the result for nonconstant polynomials by induction on the degree of
the polynomial.
First we observe that, given any integer
a
, there exists a polynomial
g
with
integer coefficients such that
f
(
x
) =
f
(
a
) + (
x

a
)
g
(
x
). Indeed
f
(
y
+
a
) is a
polynomial in
y
with integer coefficients, and therefore
f
(
y
+
a
) =
f
(
a
)+
yh
(
y
)
for some polynomial
h
with integer coefficients. Thus if
g
(
x
) =
h
(
x

a
) then
g
is a polynomial with integer coefficients and
f
(
x
) =
f
(
a
) + (
x

a
)
g
(
x
).
Suppose that
f
(
a
)
≡
0 (mod
p
) and
f
(
b
)
≡
0 (mod
p
).
Let
f
(
x
) =
f
(
a
) + (
x

a
)
g
(
x
), where
g
is a polynomial with integer coefficients. The
coefficients of
f
are not all divisible by
p
, but
f
(
a
) is divisible by
p
, and
therefore the coefficients of
g
cannot all be divisible by
p
.
Now
f
(
a
) and
f
(
b
) are both divisible by the prime number
p
, and therefore
(
b

a
)
g
(
b
) is divisible by
p
. But a prime number divides a product of integers
if and only if it divides one of the factors. Therefore either
b

a
is divisible
by
p
or else
g
(
b
) is divisible by
p
. Thus either
b
≡
a
(mod
p
) or else
g
(
b
)
≡
0
(mod
p
). The required result now follows easily by induction on the degree
of the polynomial
f
.
1.12
Primitive Roots
Lemma 1.25
Let
m
be a positive integer, and let
x
be an integer coprime
to
m
. Then there exists a positive integer
n
such that
x
n
≡
1
(mod
m
).
Proof
There are only finitely many congruence classes modulo
m
. Therefore
there exist positive integers
j
and
k
with
j < k
such that
x
j
≡
x
k
(mod
m
).
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 Winter '13
 DavidMackinnon
 Number Theory, Integers, Prime number, Coprime