On continued fractions of the square root of prime numbers

On continued fractions of the square root of prime numbers

Alexandra Ioana Gliga March 17, 2006

Nota Bene: Conjecture 5.2 of the numerical results at the end of this paper was not correctly derived from the Mathematica code written for this investigation. Thus, if you wish to check the validity of the numerical conjectures please run your own Mathematica code.

1 Introduction

This paper presents numerical testings concerning the following conjecture exhibited by Chowla and Chowla in [1]. For any positive integer k there exist infinitely many primes P with the continued fraction expansion of P having period k . The conjecture improves the similar already proved results for positive integers and, as a special case, for square free numbers. The validation of this conjecture would prove in the case k = 1 , for example, that there are infinitely many primes of the form m2 + 1 , m Z .

2 Basic Definitions and Notations

Definition 2.1 An expression of the form

1+ 1

a2+

a3

+

1 a4

1 +...

is called a simple continued fraction.We shall denote it more convenien-

tely by the symbol [a1, a2, a3, ..., an, ...]. The terms a1, a2, a3, ... are called the partial quotients of the continued fraction. We will discuss only the cases

when the partial quotients are positive integers.

1

Definition

2.2

We

denote

by

pn qn

= [a1, a2, ..., an]

the

n

th

convergent

of

the

simple continued fraction from (1) . Thus, pn and qn are the positive integer

numerator and denominator of the n th convergent.

Definition 2.3 A continued fraction which is periodic from the first partial quotient is called purely periodic. If the period starts with the second partial quotient, the continued fraction is called simply periodic . We shall denote a simply periodic continued fraction by [a0, a1, a2, ? ? ? , an]

Definition 2.4 A quadratic irrational is said to be reduced if > 1 is

the root of a quadratic equation with integral coefficients whose conjugate root

~ lies between -1and 0. A reduced quadratic irrational associated to D can

be

written

as

P+ Q

D,

where

P,

D,

Q

are

integers,

D, Q

>

0.

Definition 2.5 For a given k Z+ and a set of positive integers {an}n=0,1,???,k-1 we define P-1 = 1, Q-1 = 0 P0 = a0, Q0 = 1 Pn = anPn-1 + Pn-2, Qn = anQn-1 + Qn-2 for n = 1, 2, ? ? ? , k - 1

3 A Few General Results

This section provides the necessary background for working with the continued fractions of N , where N Z+. For more similar results, see [4]. Most of the results presented here are encountered there.

Theorem 3.1 If is a reduced quadratic irrational,then the continued fraction for is purely periodic.

In order to prove the theorem we need a preliminary lemma:

Lemma 3.2 For any given D there is only a finite number of reduced quadratic irrational associated to it.

Proof of Lemma 3.2 :If is a reduced quadratic irrational, ~ is its conjugate

and

=

P+ Q

D,

then

=

P+ Q

D

>

1, and

-

1

<

~

=

P- Q

D

<

0

2

(1)

The conditions > 1 and ~ > -1 imply that + ~ > 0, or

2P Q

> 0, and since

Q > 0 we conclude that P > 0. Also from ~ < 0 and Q> 0 it follows that

0 < P< D. The inequality > 1 implies that P + D > Q and, thus

Q < 2 D. Once D is fixed there is onlya finite number of positive integers

P and Q such that P < D and Q < 2 D, which proves the assumption.

Proof of Theorem 3.1:

As

is

a

reduced

quadratic

irrational

it

can

be

uniquely

expressed

as

P+ Q

D,

where

P, D, Q

are

positive

integers.

We

can

express

in

the

form

=

a1

+

1 1

,

where a1 is the largest integer less than , and where

1

=

1 -a1

=

P1+ Q1

D

>

1

is again a reduced quadratic irrational associated with D. Repeating step by step the process we convert into a continued fraction such that for every n,

n-1

=

an

+

1 n

(2)

where = 0, 1, ... are all the quadratic irrationals associated with D and where a1, a2, .. are the partial quotients of the continued fraction expansion. From the above lemma we have that we must arrive to a reduced quadratic irrational which has occured before, so that k = l, for 0 k < l. As

k

= ak+1 +

1 k+1

= l

= al+1 +

, 1

l+1

and since ak+1 and al+1 are the greatest integers less than k = l, we conclude that ak+1 = al+1. It then follows that k+1 = l+1. Thus we have that from lth partial quotient, the continued fraction for is periodic.

We show next that k = l for 0 < k < l implies k-1 = l-1, k-2 = l-2, . . . , 0 = l-k. Let ~k = ~l be the conjugates of the equal complete quotients k and l. Then, it follows that

k

=

-

1 ~k

=

-

1 ~l

=

l

3

If k = 0, then by taking conjugates in (2), we obtain

~k-1

= ak +

1 ~k

and

~l-1

= al +

1 ~l

(3)

and thus (4)

k

= ak +

1 k-1

and

l

= al +

1 l-1

Since k-1, l-1 are reduced, we have that

0

<

-~k-1

=

1 k-1

<

1

and

0

<

-~l-1

=

1 l-1

<

1

Thus, ak and al in (4) are the largest integers less than k, l, respectively;from k = l we get that ak = al. Thus, from equation (3) we get that k-1 = l-1. Continuing this process we get that k-2 = l-2, ? ? ? , 0 = l-k. As for each n, an is the greatest integer less than n, we get that a0 = ak-l, a1 = ak-l+1, ? ? ?. Thus the continued fraction of is purely periodic and we can write = [a0, a1, ? ? ? , al-k-1]. This completes the proof of the theorem.

Corollary 3.3 For any N , positive integer which is not a perfect square, the continued fraction of N is simply periodic, more precisely

N = [a1, a2, a3, ? ? ? , an, 2a1], for some n

Proof:

Let a1 be thegreatest integer less than N . Then N + a1 > 1 and its

conjugate, - N + a1 lies between -1 and 0. Thus, N + a1 is a reduced

quadratic irrational with the greatest integer less than it equal to 2a1. We

can apply Theorem 3.1:

N + a1 = [2a1, a2, ? ? ? , an] for some n

which is equivalent to N + a1 = [2a1, a2, a3, ? ? ? an, 2a1]

4

consequently

N = [a1, a2, a3, ? ? ? , an, 2a1] where a1 > 0

Theorem 3.4 If for a reduced quadratic integer = [a1, a2, ? ? ? , an] we de-

note by = [an, an-1, ? ? ? , a1] the continued fraction for with period re-

versed,

then

-

1

=

~

is

the

conjugate

root

of

the

equation

satisfied

by

.

Proof:

We

know

that

if

pn qn

=

[a1, a2, ? ? ? , an]

then

pn

=

anpn-1 + pn-2,

thus

pn pn-1

=

an +

1

pn-1

,

for

any

n.

pn-2

Thus,

we get recursively that

p2 p1

= [a2, a1],...,

pn pn-1

=

[an, an-1, ? ? ? , a1]

=

. p~n

q~n

Similarly, we get that

qn qn-1

=

[an, an-1, ? ? ? , a2]

=

, p~n-1

q~n-1

where

by

p~n q~n

and

p~n-1 q~n-1

we

understand

the

nth

and

the

(n - 1)th

convergents

of the continued fraction [an, an-1, ? ? ? , a1]. Since the fractions are already

reduced we get that

p~n = pn, p~n-1 = qn, q~n = pn-1 and q~n-1 = qn-1 (5)

We also have the recurrences

= and = pn+pn-1 qn +qn-1

p~n+p~n-1 q~n+q~n-1

(6)

According to (5) we notice that

= pn+qn pn-1 +qn-1

And,

thus

from

(6)

we

get

that

and

-

1

satisfy

the

same

quadratic

equa-

tion.

We

conclude

that

-

1

=

~

where

=

[an, an-1, ? ? ? , a1]

.

Lemma3.5 Except for the term 2a1 the periodic part of the continued fraction of N is symmetrical.

5

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