Download Integral Sequences of Infinite Length Whose Terms Are Relatively

Advances in Pure Mathematics, 2013, 3, 24-28
http://dx.doi.org/10.4236/apm.2013.31005 Published Online January 2013 (http://www.scirp.org/journal/apm)
Integral Sequences of Infinite Length Whose
Terms Are Relatively Prime
Kazuyuki Hatada
Department of Mathematics, Faculty of Education, Gifu University, Gifu, Japan
Email: [email protected]
Received September 7, 2012; revised October 10, 2012; accepted November 13, 2012
ABSTRACT

m
n

It is given in Weil and Rosenlicht ([1], p. 15) that G.C.D. c 2  1, c 2  1  1 (resp. 2) for all non-negative integers m
and n with m  n if c is any even (resp. odd) integer. In the present paper we generalize this. Our purpose is to give
other integral sequences
 yn n 1

such that G.C.D.  ym , yn   1 for all positive integers m and n with m  n . Roughly
speaking we show the following 1) and 2). 1) There are infinitely many polynomial sequences
 f n  X n 1  Z  X 

such that G.C.D.  f m  a  , f n  a    1 for all positive integers m and n with m  n and infinitely many rational inte-
gers a. 2) There are polynomial sequences
 gn  X , Y n 1  Z  X , Y 

such that G.C.D.  g m  a, b  , g n  a, b    1 for all
positive integers m and n with m  n and arbitrary (rational or odd) integers a and b with G.C.D.  a, b   1 . Main
results of the present paper are Theorems 1 and 2, and Corollaries 3, 4 and 5.
Keywords: Relatively Prime; Integral Sequences of Infinite Length; Sets of Infinitely Many Prime Numbers
1. Introduction
The numbers Fn  22  1 n  0,1, 2,3, are called Fermat numbers. Fermat conjectured that Fn were all prime
numbers. One has F0  3 , F1  5 , F2  17 , F3  257 ,
F4  65537 and F5  641 6700417 . By now, no Fermat prime has been found except for F j  j  0,1, 2,3, 4  .
In Euclid’s books was given the proof of existence of
infinitely many prime numbers. By proving G.C.D.
 Fm , Fn   1 if m  n , Pólya gave another proof of that,
cf. ([2], Theorem 16, p. 14) and ([3], exercise (viii), p. 7).
Weil and Rosenlicht
([1], p. 15) considered not only Fn
n
but also c 2  1 for any rational integer c.
Let n be any positive integer, and let  n be any
primitive n-th root of unity. Let
S  n     C  is a primitive n-th root of unity . Then
the number of S  n     n  where  denotes the
Euler function. Let  n T  denote the n-th cyclotomic
polynomial over Q. Namely,  n T  denotes the polynomial  Q T  of the minimum degree whose roots contain  n and whose leading coefficient is 1. One has that
 n T  does not depend on choice of  n in S  n , that
 n T   Z T  and that  n T    T   , (see
n
 S  n 
e.g. [4-8]). Below in this paper we write   n, T    n T .
Copyright © 2013 SciRes.
We let G.C.D. denote “greatest
common divisor” as
n1
usual. One has  2n , T  T 2  1. Then exercise IV.3
in [1] asserts G.C.D.  2n , c ,  2m , c  1 (resp. 2)


 
 

for all positive integers m and n with m  n if c is even
(resp. odd).
We generalize this. Let p denote any odd prime number, and let v denote any rational integer. In Theorem 2 in
Section 3 below we show that
G.C.D.  p n , v ,  p m , v  1 (resp. p) for all positive integers m and n with m  n if v is not congruent
modulo p to 1 (resp. if v is congruent modulo p to 1). Our
first proof of Theorem 2 uses Elementary Number Theory. Our second proof of Theorem 2 uses Algebraic
Number Theory and Theory of Cyclotomic Fields. In
Corollary 5 in Section 4 we also show that
G.C.D.  2n p, v ,  2m p, v  1 for all positive integers m and n with m  n and all rational integers v. In
Corollary 4 in Section 3 we study also
G.C.D.  p n q, v ,  p m q, v where p and q are arbitrary odd prime numbers with p  q . The case of
G.C.D.  2 p n , v ,  2 p m , v is reduced to Theorem 2
 
 

 
 

 
 
 

 

  2 p , v     p , v  for any non-negative inu
u
since
teger u. Cf. Corollary 3 in Section 3.
APM
25
K. HATADA
prime number dividing xm in Theorem 1.
Corollary 1. We have pm  pn if m  n . There are
infinitely many prime numbers.
In Section 2 (resp. 4) we consider
hn  X , Y   X 2  X 2 Y 2
n 1
n1
2n1
Y2
n
 resp. H  X , Y   X
n
Y2
n
.
n1

3. On Φ p n , v
In Theorem 1 in Section 2 we show
G.C.D.  hn  a, b  , hm  a, b    1 for all positive integers m
and n with m  n and all rational integers a and b with
G.C.D.  a, b   1 . In Theorem 3 in Section 4 we show
G.C.D.  H n  a, b  , H m  a, b    1 (resp. 2) for all positive
integers m and n with m  n and all rational integers a
and b with ab  0  mod 2  (resp. ab  1 mod 2  ) and
G.C.D.  a, b   1 . The case  b  1 of Theorem 3 gives
a proof of Exercise IV.3 in [1].
n
2. On a 2  a 2
n 1
b2
n 1
 b2
n
n 1
a2
n1

n
n
 a2 b2  b2
n1
n
 a2  a2 b2
n1
n
 b2
n
a
2
n 1
n
 a 2 b2
n 1
 b2
n



mial in Z T  whose leading coefficient is 1. One has
 p n , T   S pn T    , (see e.g. [4-8]). Let m and


 
n be arbitrary positive integers with 1  m  n. Since
there are no common roots of  p m , T and  p n , T
 
a
2n
a b b




j 1
 a 2  ab  b 2   a 2  a 2 b 2
j 1
j
j 1
j
 

 


n1
n
n1
n
 a2 b2  b2 , a2
n
n
 G.C.D. 2a 2 b 2 , a 2
n1
n
n1


 b2 .
n
n
 a 2 b2  b2
n
 a 2 b2  b2
n 1


G.C.D.  x , a
n
G.C.D. 2a 2 b 2 , a 2
Hence
2n1
m
n 1
n
n
 a 2 b2  b2
n
n
 a2 b2  b2
n1
 1
 1
n1

.

 



 

in
 
  in Q T   1 . We have
pm
 

m
1  T p

and

 pm ,T
 T
p nm
pm



n
 1  T p  1



n
1 T p 1

in Z T  if n  m  1 . We have  p m , v
Z if v  Z and n  m  1 . One has


,

 p n , T   p, T p
n1
 v
pn

 1 in

p 1
and   p, T    T j , see e.g. [4-8]. We give
j 0
for all ra-
tional integers 1  m  n . Namely G.C.D.  xm , xn 1   1
for all rational integers 1  m  n .
In Euclid’s books was given the proof of the classical
well known theorem that there are infinitely many prime
numbers. Theorem 1 above gives another proof of this
theorem. For each positive integer m, let pm denote a
Copyright © 2013 SciRes.
in Q T   1 .

T
.
n1

proven.
Note
From G.C.D.  a, b   1 , factoring a and b into products
of prime numbers, we have
n


d m , n T   Z . Since the leading coefficient of
 p m , T  Z T  is 1, d m , n T   1 . Proposition 1 is
n1
n 1
n
n
Hence xm  a 2  a 2 b 2  b 2  for all integers


1  m  n . We have also
G.C.D. a 2

Z T  . If deg d m , n T   1 , this contradicts
2n 1
n
n
We have
Proposition 1. G.C.D.  p m , T ,  p n , T
in
Z T   1, if 1  m  n .
Proof. We have  p m , T and  p n , T are polynomials in Z T  whose leading coefficients are 1, (see
e.g. [4-8]). Use Gauss Lemma for polynomials over the
quotient ring of a factorial ring, (see e.g. ([5], pp. 181-182)).
By applying it to Z T  and Q T  , Z T  is factorial.
G.C.D.  p m , T ,  p n , T
2n

 
m
We may put d m, n T   G.C.D.  p m , T ,  p n , T
and
2n1

in C . Recall  p n , T  Irr  p n , Q , T . It is a polyno-
 
n1
n1
Let p be any odd prime number and let n be any positive
integer. Let  p n denote a primitive p n -th root of unity
in C , G.C.D.  p , T ,  p , T
 n = 1, 2, 3,
Recall hn  X , Y   X 2  X 2 Y 2  Y 2 . We show first
Theorem 1. Let a and b be arbitrary rational integers

with G.C.D.  a, b   1 . Let  xn n 1 denote the sequence
given by xn  hn  a, b  for all positive integers n. Then
we have G.C.D.  xm , xn   1 for all positive integers m
and n with m  n .
Proof. We have
n
  n = 1, 2, 3,
Theorem 2. Let p be any odd prime number, and let v
be any rational integer. Then we have the following.
Case 1 that v is not congruent modulo p to 1:
 
 
G.C.D.  p m , v ,  p n , v
  1
for all rational integers m and n with 1  m  n .
Case 2 that v is congruent modulo p to 1:
APM
26
K. HATADA
 
 
G.C.D.  p m , v ,  p n , v
  p
for all rational integers m and n with 1  m  n .
We give two proofs. The first one uses Elementary
Number Theory. The second one uses (local and global)
Algebraic Number Theory and Theory of Cyclotomic
Fields for which cf. [4-9].
m
Proof 1. We have 1  m  n  1 . Put   v p  1. We
have
vp
n1
 vp
 vp
p
and

 
m n1m


 p n , v   p, v p
m
n1
p n1m
   1
p n1m
 1 mod  
    p,1 mod   p  mod  .
 
 
Since   p , v   , G.C.D.    p , v  ,   p , v   divides
G.C.D.    p , v  ,  . Hence
G.C.D.    p , v  ,   p , v    1 or p .
(1)
There is a rational integer y with  p n , v   y  p.
We have G.C.D.  p n , v ,  G.C.D.  p,   1 or p.
 
m
n
m
n
n
m
In Case 2: We have




 p m , v   p m ,1  mod p   p  mod p 
 0  mod p  .
 
G.C.D.    p , v  ,   p , v    p . Case 2 of Theorem 2
We have also  p n , v  0  mod p  . Hence
n
m
is proven.
In
Case 1: Let   1 be any divisor of  . Then
n1
v p  1 mod    1 mod   . We have



 p n , v   p, v p
n1
    p,1 mod    p  mod  
.
We shall show  does not divide p . Assume it were
truem that  p . Then we would have   p . We have
v p  1  0  mod   . Therefore
 v  mod p  
pm
 1 mod p  and p would not divide v.
It follows that  v  mod p    1 mod p  . The order of
v  mod p  divides p m and p  1 . Hence
v  mod p   1 mod p  , which is a contradiction. Hence
we have   p and  does not divide p . Hence 
does not divide  p n , v . Hence we get

p 1





m
G.C.D.  p n , v , v p  1  1 . Since
  v  1 , we have
G.C.D.    p , v  ,   p , v    1

 pm , v


 p m  T   

Sp 

 p n  T    .
 p m , T   S
and
 p n , T   S
 
m
(resp.   S p n ) arbitrarily. Let B
Take
denote the ring of the algebraic integers in Q  p n . Let
vZ .
In Case 1: Now assume that there is such a prime ideal
P of B that satisfies v    P and v    P. Write
v     v  1  1    and v     v  1  1    . We
have     P and  1   1   P. Let a  Z . We
have  pn  papn
nm
  1pn ap
m

pB  P
p 1 p n1

if 1  m  n . Case 1 of


. We have  1   1  B  P   P . Hence
P   P and P pZ . We have 1    P since
  S  p n  . From v    P , we have 1  v  P. Since
v  Z , we have 1  v  pZ , namely, v  1 mod p  .
This result implies the following. If v is not congruent
modulo p to 1, there is no prime ideal J with v    J
and v    J . Since
  p m , v    S pm  v   
 
and


 p n , v   S
 pn   v   ,


the greatest common divisor ideal of  p n , v B and
 p m , v B is B if v is not congruent modulo p to 1.
Therefore Case 1 of Theorem 2 is proven.
In Case 2: Let v  1 mod p  . Let   S p n . Let P
denote the unique prime ideal in B lying above pZ, and
let BP denote the localization of B at P. Let Bˆ P denote the completion of BP with respect to the P-adic
(non-Archimedean) absolute value. We use local and
global Algebraic Number Theory, cf. [5,9]. We have
P  1    B and 1     v  1  1    in B. We have
v 1
 0 . Hence
 v  1  1    1    in Bˆ P since ord P
1 
we get  v    Bˆ P  1    Bˆ P using
v    v  1  1   . Then we have
ˆ
 p n , v Bˆ  
n v    B


 


 
 S p
P
 1   
Theorem 2 is proven.
We give another proof of Theorem 2.
Proof 2. Let 1  m  n. Recall
Copyright © 2013 SciRes.

which is a primitive p n -th
root of unity since p does not divide 1  ap n  m . So
 1  is a primitive p n -th root of unity. By the theory
of cyclotomic fields (cf. [4-8]),  1   1  B is a
unique prime ideal P  of B lying above pZ, and
pm
n

n m
 
 p 1 p
P
n1
Bˆ P  pBˆ P .
In the same way we have


 p m , v Bˆ P   S
 pm   v   BP
 1   
ˆ
 p 1 p m1
Bˆ P  pBˆ P
APM
27
K. HATADA
 
 
using Q  p m  Q  pn . Here we use (1) in Proof 1
above. Therefore we get
 
 
G.C.D.  p n , v ,  p m , v
  p
if v  1 mod p  .
Case 2 of Theorem 2 is proven.
For each positive integer m , let qm denote a prime
number dividing  p m , v in Case 1 of Theorem 2.
Corollary 2. We have qm  qn if m  n . There are
infinitely many prime numbers.
EXAMPLE of Theorem 2. Let p  5 and let v be a
rational integer which is notm congruent
modulo 5m to 1.
5
25m
35m
1

v

v

v
 v 45 and
Then n we have
that
5
25n
35n
45n
1 v  v
v
v
are relatively prime for all
rational integers m and n with 0  m  n .
We give some computations.




  5, 2   31,  52 , 2  601 1801,


We have p  1 mod q  and
 




(We used “Scientific WorkPlace”, Version 5.5, MacKichan Software, 19307 8th Avenue NE, Suite C, Poulsbo,
WA 98370, USA, for the computations).
Corollary 3 of Theorem 2. Let p be any odd prime
number, and let v be any rational integer. Then we
have the following.
Case 1 that v is not congruent modulo p to 1 :
 
 
  1
 
 
G.C.D.  2 p , v ,  2 p , v
n
  p
 
 

G.C.D.  p q, v ,  p q, v
m
n


  1
for all rational integers m and n with 1  m  n .
Case 2 that v q  1 mod p  and v  1 mod p  :
 
 
G.C.D.  p m q, v ,  p n q, v
  1
for all rational integers m and n with 1  m  n .
Case 3 that v q  1 mod p  and that v is not congruent modulo p to 1:
Copyright © 2013 SciRes.
   p ,T 
u
for any positive integer u. Hence



 p u q, v   p u , v q
   p , v   Z.
u
In Case 1, we have
 
 
  1
 
 
  p
 
 
G.C.D.  p m , v q ,  p n , v q
from Theorem 2.
In Case 2: We have
G.C.D.  p m , v q ,  p n , v q
  p
G.C.D.  p m , v ,  p n , v
from Theorem 2. Hence it follows that
 
 
G.C.D.  p m q, v ,  p n q, v
  1 .
In Case 3: From  v  mod p    1 mod p  , the order
of v  mod p  divides q and p  1 . Since v is not congruent modulo p to 1, the order of v  mod p  is q.
Hence q  p  1 . From Theorem 2, we have
q
 
 
  p
 
 
  1 .
and
G.C.D.  p m , v ,  p n , v
Here we use
for all rational integers m and n with 1  m  n .
Proof. By ([6], p. 280),  2 p u , T   p u , T for
any positive integer u. Then by Theorem 2, Corollary 3
follows.
Corollary 4 of Theorem 2. Let p and q be arbitrary
odd prime numbers with p  q , and let v be any rational integer. Then we have the following.
Case 1 that v q is not congruent modulo p to 1:


G.C.D.  p m , v q ,  p n , v q
for all rational integers m and n with 1  m  n .
Case 2 that v is congruent modulo p to 1 :
m

 p u q, T   p u , T q
 53 , 2  229668251 5519485418336288303251.
G.C.D.  2 p m , v ,  2 p n , v
or p
and
  5, 2   11,  52 , 2  251 4051,

  1
for all rational integers m and n with 1  m  n .
Proof. From ([6], p. 280) we have
 5 , 2  269089806001 4710883168879506001,
3
 
G.C.D.  p m q, v ,  p n q, v



   p , v   Z.
G.C.D.    p q, v  ,   p q, v    1
 p u q, v   p u , v q
u
m
n
or p.
It follows that
From Corollary 4 of Theorem 2 we obtain:
Let p and q be arbitrary odd prime numbers with
p  q , and let v be any rational integer. If p is not
congruent modulo q to 1,
 
 
G.C.D.  p m q, v ,  p n q, v
  1
for all rational integers m and n with 1  m  n .
4. Proof of Exercise IV.3 in [1]
Let us quote the exercise.
Exercise IV.3 in [1]. “If a, m, n are positive integers,
m
n
and m  n , show that the G.C.D. of a 2  1 and a 2  1
odd. (Hint. use the fact
is 1 or 2n according as a is even or
m
that a 2  1 is a multiple of a 2  1 for n  m ). From
APM
28
K. HATADA
this deduce the existence of infinitely many primes.”
We give a proof of this in somewhat generalized form.
Namely we show
Theorem 3. Let a and b be arbitrary positive rational
.D.  an,1 b   1 . Define
integers with G.C
n 1
for any positive n  Z . Let
Hn  X ,Y   X 2  Y 2
m and n be arbitrary rational integers with 0  m  n .
Write  m, n  G.C.D.  H m 1  a, b  , H n 1  a, b   . Then we
have:
 m, n  1 if ab  0  mod 2  ;
n

n
X 2 Y 2  X 2
n1
Y2
 X
n1
2n1
  1
 
  2
by Theorem 3. If v is odd,
 
G.C.D.  2m , v p ,  2n , v p
and
 
 
G.C.D.  2m , v ,  2n , v

Y 2
n1



X 2 Y 2   X Y  X 2 Y 2
n
n
j 0
X
2m
Y2
 X
m
2n
Y 2
n

j
j


 G.C.D.  2a
Hence G.C.D.  a
n
n
n
2n
n




u 1
.
u
 b2
m

2n
2n
u
 T
u 1
2u 1

1
 1.
5. Acknowledgements
for all integers
n
   2 ,T 
1
 T 2 T 2
  G.C.D. a  b , 2b  .
, a  b   for all integers
, a 2  b2
2m
n
u
 2u  3, T   2u , T 3
0  m  n . We have also
 n  G.C.D. a 2  b 2 , a 2  b 2
   2 , v  Z.
In the case of p  3, Corollary 5 is derived also from
Theorem 1. For we have
 T 32
n 1
  2
by Theorem 3. Then we have Corollary 5 using
and
Hence
 
 2u p , v   2u , v p
 m, n  2 if ab  1 mod 2  .
Proof. We have
 
G.C.D.  2m , v p ,  2n , v p
2n
2n
The author concludes that the topic of the present paper
relates to Algebraic Number Theory and Theory of Cyclotomic Fields. He would like to thank the referee for
valuable suggestions for the important improvement of
this paper.
REFERENCES
2n
n
[1]
A. Weil and M. Rosenlicht, “Number Theory for Beginners,” Springer Verlag, New York, 1979.
doi:10.1007/978-1-4612-9957-8
[2]
G. H. Hardy and E. M. Wright, “An Introduction to the
Theory of Numbers,” 4th Edition, Oxford University Press,
Ely House, London, 1971.
[3]
A. Baker, “A Concise Introduction to the Theory of Numbers,” Cambridge University Press, Cambridge, 1984.
doi:10.1017/CBO9781139171601
are even,
[4]
Corollary 5 of Theorem 3. Let p be any odd prime
number, and let v be any rational integer. Then we have
B. J. Birch, “Cyclotomic Fields and Kummer Extensions,”
In: J. W. S. Cassels and A. Fröhlich, Eds., Algebraic Number Theory, Academic Press, London, 1967, pp. 85-93.
[5]
S. Lang, “Algebraic Number Theory,” Addison-Wesley
Publishing Company, Massachusetts, 1970.
[6]
S. Lang, “Algebra,” 3rd Edition, Springer Verlag, New
York, 2002. doi:10.1007/978-1-4613-0041-0
[7]
E. Weiss, “Algebraic Number Theory,” 2nd Edition, Chelsea Publishing Company, New York, 1976.
[8]
H. Weyl, “Algebraic Theory of Numbers,” Princeton University Press, Princeton, 1940.
[9]
J.-P. Serre, “Local Fields,” Springer Verlag, New York,
1979.
number p divides a .
0  m  n . Assume that a prime
n
n
Then p does not divide a 2  b 2 since

G.C.D.  a, b   1 . Use  n  G.C.D. 2a 2 , a 2  b 2
n
n
n
.
Therefore
 n  1 or 2 . We have  n  1 if a is even;
n
n
a 2n  b 2n is even and  n  2 if a and b are odd;
a 2  b 2 is odd and  n  1 if a is odd and b is even.
Recall  m, n  n .  m, n  1 if ab is even.  m, n  2 if
m
ab is odd, since both a 2  b 2
and 2  m, n .
 
m
n
and a 2  b 2
 
G.C.D.  2m p, v ,  2n p, v
n
  1
for all rational integers m and n with 1  m  n .
Proof. By ([6], p. 280),



   2 ,T 
for any positive integer u. We have   2 , T   T
 2u p , T   2u , T p
u
u
If v is even,
Copyright © 2013 SciRes.
2u 1
 1.
APM