Integral Sequences of Infinite Length Whose Terms Are Relatively Prime ()
1. Introduction
The numbers 
are called Fermat numbers. Fermat conjectured that Fn were all prime numbers. One has
, 
, 
, 
, 
and
. By now, no Fermat prime has been found except for
. In Euclid’s books was given the proof of existence of infinitely many prime numbers. By proving G.C.D. 
if
, 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 
but also 
for any rational integer c.
Let n be any positive integer, and let 
be any primitive n-th root of unity. Let
. Then the number of 
where 
denotes the Euler function. Let 
denote the n-th cyclotomic polynomial over Q. Namely, 
denotes the polynomial
of the minimum degree whose roots contain 
and whose leading coefficient is 1. One has that 
does not depend on choice of 
in
, that
and that 
(see e.g. [4-8]). Below in this paper we write
. We let G.C.D. denote “greatest common divisor” as usual. One has 
Then exercise IV.3 in [1] asserts 
(resp. 2)
for all positive integers m and n with 
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
(resp. p) for all positive integers m and n with 
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 
for all positive integers m and n with 
and all rational integers v. In Corollary 4 in Section 3 we study also 
where p and q are arbitrary odd prime numbers with
. The case of
is reduced to Theorem 2 since 
for any non-negative integer u. Cf. Corollary 3 in Section 3.
In Section 2 (resp. 4) we consider
In Theorem 1 in Section 2 we show 
for all positive integers m and n with 
and all rational integers a and b with
. In Theorem 3 in Section 4 we show 
(resp. 2) for all positive integers m and n with 
and all rational integers a and b with 
(resp.
) and
. The case 
of Theorem 3 gives a proof of Exercise IV.3 in [1].
2. On 
Recall
. We show first Theorem 1. Let a and b be arbitrary rational integers with
. Let 
denote the sequence given by 
for all positive integers n. Then we have 
for all positive integers m and n with
.
Proof. We have
and
Hence 
for all integers 
. We have also
From
, factoring a and b into products of prime numbers, we have
Hence 
for all rational integers
. Namely 
for all rational integers
.
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 
denote a prime number dividing 
in Theorem 1.
Corollary 1. We have 
if
. There are infinitely many prime numbers.
3. On 
Let p be any odd prime number and let n be any positive integer. Let 
denote a primitive 
-th root of unity in
. Recall
. It is a polynomial in 
whose leading coefficient is 1. One has
, (see e.g. [4-8]). Let m and n be arbitrary positive integers with 
Since there are no common roots of 
and 
in
, 
in
.
We have Proposition 1. 
in 
if
.
Proof. We have 
and 
are polynomials in 
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 
and
, 
is factorial.
We may put 
in
. If deg
, this contradicts 
. We have
. Since the leading coefficient of
is 1,
. Proposition 1 is proven.
Note
and
in 
if
. We have 
in 
if 
and
. One has
and
, see e.g. [4-8]. We give 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:
for all rational integers m and n with
.
Case 2 that v is congruent modulo 
to 1:
for all rational integers m and n with
.
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].
Proof 1. We have
. Put 
We have
and
There is a rational integer 
with 
We have 
or p.
Since 
divides
. Hence
or
. (1)
In Case 2: We have
We have also 
Hence
. Case 2 of Theorem 2 is proven.
In Case 1: Let 
be any divisor of
. Then 
We have
We shall show 
does not divide
. Assume it were true that
. Then we would have
. We have 
Therefore
and 
would not divide v.
It follows that
. The order of 
divides 
and
. Hence 
, which is a contradiction. Hence we have 
and 
does not divide
. Hence 
does not divide
. Hence we get
. Since
, we have 
if
. Case 1 of Theorem 2 is proven.
We give another proof of Theorem 2.
Proof 2. Let 
Recall
and
.
Take 
(resp.
) arbitrarily. Let B denote the ring of the algebraic integers in
. Let
.
In Case 1: Now assume that there is such a prime ideal P of B that satisfies 
and 
Write 
and
. We have 
and 
Let 
We have 
which is a primitive 
-th root of unity since p does not divide 
So 
is a primitive 
-th root of unity. By the theory of cyclotomic fields (cf. [4-8]), 
is a unique prime ideal 
of B lying above pZ, and
. We have
. Hence
and 
We have 
since 
. From
, we have 
Since 
we have 
namely, 
This result implies the following. If v is not congruent modulo p to 1, there is no prime ideal J with 
and 
Since
and
the greatest common divisor ideal of 
and 
is B if v is not congruent modulo p to 1. Therefore Case 1 of Theorem 2 is proven.
In Case 2: Let
. Let
. Let P denote the unique prime ideal in B lying above pZ, and let 
denote the localization of B at P. Let 
denote the completion of 
with respect to the P-adic (non-Archimedean) absolute value. We use local and global Algebraic Number Theory, cf. [5,9]. We have 
and 
in B. We have
in 
since
. Hence we get 
using 
Then we have
In the same way we have
using 
Here we use (1) in Proof 1 above. Therefore we get
if
. Case 2 of Theorem 2 is proven.
For each positive integer
, let 
denote a prime number dividing 
in Case 1 of Theorem 2.
Corollary 2. We have 
if
. There are infinitely many prime numbers.
EXAMPLE of Theorem 2. Let 
and let 
be a rational integer which is not congruent modulo 5 to 1. Then we have that 
and 
are relatively prime for all rational integers 
and 
with
.
We give some computations.
(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 
be any odd prime number, and let 
be any rational integer. Then we have the following.
Case 1 that v is not congruent modulo 
to
:
for all rational integers 
and 
with
.
Case 2 that v is congruent modulo 
to
:
for all rational integers m and n with
.
Proof. By ([6], p. 280), 
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
, and let 
be any rational integer. Then we have the following.
Case 1 that 
is not congruent modulo 
to 1:
for all rational integers m and n with
.
Case 2 that 
and
:
for all rational integers m and n with
.
Case 3 that 
and that v is not congruent modulo p to 1:
We have 
and
or p for all rational integers m and n with
.
Proof. From ([6], p. 280) we have
for any positive integer u. Hence
In Case 1, we have
from Theorem 2.
In Case 2: We have
and
from Theorem 2. Hence it follows that
.
In Case 3: From
, the order of 
divides q and
. Since v is not congruent modulo p to 1, the order of 
is q. Hence 
From Theorem 2, we have
and
.
Here we use
It follows that 
or p.
From Corollary 4 of Theorem 2 we obtain:
Let 
and q be arbitrary odd prime numbers with
, and let 
be any rational integer. If p is not congruent modulo q to 1,
for all rational integers m and n with
.
4. Proof of Exercise IV.3 in [1]
Let us quote the exercise.
Exercise IV.3 in [1]. “If a, m, n are positive integersand
, show that the G.C.D. of 
and 
is 1 or 2 according as a is even or odd. (Hint. use the fact that 
is a multiple of 
for
). From 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 integers with
. Define 
for any positive
. Let m and n be arbitrary rational integers with
. Write
. Then we have:
Proof. We have
and
.
Hence 
for all integers
. We have also
Hence 
for all integers
. Assume that a prime number p divides
. Then 
does not divide 
since
. Use
.
Therefore
. We have 
if 
is even; 
is even and 
if a and b are odd; 
is odd and 
if a is odd and b is even. Recall
. 
if 
is even. 
if
is odd, since both 
and 
are even, and
.
Corollary 5 of Theorem 3. Let p be any odd prime number, and let v be any rational integer. Then we have
for all rational integers 
and 
with
.
Proof. By ([6], p. 280),
for any positive integer u. We have 
If v is even,
by Theorem 3. If v is odd,
and
by Theorem 3. Then we have Corollary 5 using
In the case of 
Corollary 5 is derived also from Theorem 1. For we have
5. Acknowledgements
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.