quadratic congruence

quadratic congruence

Let

a

,

b

,

c

be known

integers

and

p

an

odd prime

number not dividing

a

.  The number of non-congruent roots of the

quadratic congruence

a

⁢

x

2

+

b

⁢

x

+

c

≡

0

(

mod

p

)

(1)

is

•

two, if

b

2

-

4

⁢

a

⁢

c

is a

quadratic residue

modulo

p

;

•

one, if

b

2

-

4

⁢

a

⁢

c

≡

0

(

mod

p

)

;

•

zero, if

b

2

-

4

⁢

a

⁢

c

is a

quadratic nonresidue

modulo

p

.

Proof.

Since

gcd

⁡

(

p

,

4

⁢

a

)

=

1

,  multiplying (1) by

4

⁢

a

gives an

equivalent

(

http://planetmath.org/Equivalent3

)

congruence

4

⁢

a

2

⁢

x

2

+

4

⁢

a

⁢

b

⁢

x

+

4

⁢

a

⁢

c

≡

0

(

mod

p

)

which may furthermore be written as

(

2

⁢

a

⁢

x

+

b

)

2

≡

b

2

-

4

⁢

a

⁢

c

(

mod

p

)

.

Accordingly, one can obtain the the solution of the given congruence from the solution of the pair of congruences

{

y

2

≡

b

2

-

4

⁢

a

⁢

c

(

mod

p

)

(

2

)

2

a

x

+

b

≡

y

(

mod

p

)

.

(

3

)

Case 1:

b

2

-

4

⁢

a

⁢

c

is a quadratic residue

(

mod

p

)

.  Then (2) has a root

y

=

y

0

≠

0

,  and therefore also the second root

y

=

-

y

0

.  The roots

y

=

±

y

0

are incongruent, because otherwise one had

p

∣

2

y

0

and thus

p

⁢

∣

y

0

∣

⁢

y

0

2

≡

b

2

-

4

⁢

a

⁢

c

which is not possible in this case.

Case 2:

b

2

-

4

⁢

a

⁢

c

≡

0

(

mod

p

)

.  Now (2) implies that

y

≡

0

(

mod

p

)

,  whence the corresponding root

x

0

of the

linear congruence

(3) does not allow other incongruent roots for (1).

Case 3:

b

2

-

4

⁢

a

⁢

c

is a quadratic nonresidue

(

mod

p

)

.  The congruence (2) cannot have solutions; the same concerns thus also (1).

Example.

Solve the congruence

4

⁢

x

2

+

6

⁢

x

-

3

≡

0

(

mod

43

)

.

We have

b

2

-

4

⁢

a

⁢

c

=

36

+

4

⋅

4

⋅

3

=

84

≡

-

2

(

mod

43

)

and the

Legendre symbol

(

-

2

43

)

=

(

-

1

43

)

⁢

(

2

43

)

=

-

1

⋅

(

-

1

)

=

1

(see values of the Legendre symbol) says that

-

2

is a quadratic residue modulo 43.  The congruence corresponding (2) is

y

2

≡

-

2

(

mod

43

)

,

which is satisfied by

y

≡

±

16

(

mod

43

)

as one finds after a little experimenting.  Then we have the two linear congruences

2

⋅

4

⁢

x

+

6

≡

±

16

(

mod

43

)

,  i.e.

4

⁢

x

≡

±

8

-

3

(

mod

43

)

corresponding (3).  The first of them,

4

⁢

x

≡

5

(

mod

43

)

,  is satisfied by

x

=

12

and the second,

4

⁢

x

≡

-

11

(

mod

43

)

,  by

x

=

8

.  Thus the solution of the given congruence is

x

≡

8

(

mod

43

)

or

x

≡

12

(

mod

43

)

.

Title

quadratic congruence

Canonical name

QuadraticCongruence

Date of creation

2013-03-22 17:45:30

Last modified on

2013-03-22 17:45:30

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

11

Author

pahio (2872)

Entry type

Theorem

Classification

msc 11A15

Classification

msc 11A07

Related topic

LinearCongruence

Related topic

LegendreSymbol

Related topic

QuadraticFormula

Related topic

ConditionalCongruences

Defines

quadratic congruence

Generated on Fri Feb 9 15:12:00 2018 by

LaTeXML