example of gcd

example of gcd

If one calculates values of the

polynomial

f

⁢

(

x

)

:=

x

4

+

x

2

+

1

on the successive

positive integers

n

, one may observe an interesting thing when

factoring

the values:

f

⁢

(

1

)

=

3

f

⁢

(

2

)

=

21

=

3

⋅

7

f

⁢

(

3

)

=

91

=

7

⋅

13

f

⁢

(

4

)

=

273

=

3

⋅

7

⋅

13

f

⁢

(

5

)

=

651

=

3

⋅

7

⋅

31

f

⁢

(

6

)

=

1333

=

31

⋅

43

f

⁢

(

7

)

=

2451

=

3

⋅

19

⋅

43

f

⁢

(

8

)

=

4161

=

3

⋅

19

⋅

73

f

⁢

(

9

)

=

6643

=

7

⋅

13

⋅

73

f

⁢

(

10

)

=

10101

=

3

⋅

7

⋅

13

⋅

37

f

⁢

(

11

)

=

14763

=

3

⋅

7

⋅

19

⋅

37

. . .

It seems as if two

consecutive

values always have at least one common

odd prime

factor, i.e. they have the

greatest common divisor

>

2

.

It is indeed true.  The reason of this fact is not particularly
deep.  It is easily understood if we can
factorize this polynomial (

http://planetmath.org/GroupingMethodForFactoringPolynomials

):

f

⁢

(

x

)

=

(

x

2

-

x

+

1

)

⁢

(

x

2

+

x

+

1

)

Then

f

⁢

(

n

)

=

(

n

2

-

n

+

1

)

⁢

(

n

2

+

n

+

1

)

,

(1)

and the next value is

f

⁢

(

n

+

1

)

=

[

(

n

+

1

)

2

-

(

n

+

1

)

+

1

]

⁢

[

(

n

+

1

)

2

+

(

n

+

1

)

+

1

]

=

(

n

2

+

3

⁢

n

+

3

)

⁢

(

n

2

+

n

+

1

)

.

(2)

Thus

f

⁢

(

n

)

and

f

⁢

(

n

+

1

)

have as their common factor at least the number

n

2

+

n

+

1

, which is

≧

3

.

Moreover, we may show that the greatest common divisor of

f

⁢

(

n

)

and

f

⁢

(

n

+

1

)

is

n

2

+

n

+

1

except in the case

n

≡

3

(

mod

7

)

where it is

7

⁢

(

n

2

+

n

+

1

)

.

Let

d

be a common

divisor

, greater than 1, of the first factors

n

2

-

n

+

1

and

n

2

+

3

⁢

n

+

3

of (1) and (2).  It’s clear that

d

is odd (

http://planetmath.org/OddNumber

).  Then

d

must divide the

difference

(

n

2

+

3

⁢

n

+

3

)

-

(

n

2

-

n

+

1

)

=

2

⁢

(

2

⁢

n

+

1

)

and the sum

(

n

2

+

3

⁢

n

+

3

)

+

3

⁢

(

n

2

-

n

+

1

)

=

4

⁢

n

2

+

6

and hence also the difference

2

⁢

(

2

⁢

n

+

1

)

+

(

4

⁢

n

2

+

6

)

-

(

2

⁢

n

+

1

)

2

=

7

.  It means that

d

=

7

.  Thus the only possible common

prime factor

of

n

2

-

n

+

1

and

n

2

+

3

⁢

n

+

3

is 7.  If we denote

n

=

7

⁢

k

+

r

where

r

∈

{

0

,

1

,

…

,

6

}

,  we see that

n

2

-

n

+

1

=

49

⁢

k

2

+

14

⁢

k

⁢

r

-

7

⁢

k

+

r

2

-

r

+

1

⁢

≡

¯

⁢

r

2

-

r

+

1

(

mod

7

)

,

n

2

+

3

⁢

n

+

3

=

49

⁢

k

2

+

14

⁢

k

⁢

r

+

21

⁢

k

+

r

2

+

3

⁢

r

+

3

⁢

≡

¯

⁢

r

2

+

3

⁢

r

+

3

(

mod

7

)

.

It’s easy to check that the right hand of these

polynomial congruences

are

divisible

by 7 only if

r

=

3

,  i.e. if

n

≡

3

(

mod

7

)

.

Note.

The

sequence

formed by the successive values
of

gcd

⁡

(

f

⁢

(

n

)

,

f

⁢

(

n

+

1

)

)

is number A111002 in Sloane’s
register (

https://oeis.org/A111002

http://www.research.att.com/ njas/sequences/).

Title

example of gcd

Canonical name

ExampleOfGcd

Date of creation

2014-10-25 7:09:30

Last modified on

2014-10-25 7:09:30

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

19

Author

pahio (2872)

Entry type

Example

Classification

msc 11A51

Related topic

Divisibility

Related topic

GreatestCommonDivisor

Related topic

Congruences

Related topic

PolynomialCongruence

Related topic

GroupingMethodForFactorizingPolynomials

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