The numbers are called Fermat numbers. Fermat conjectured that F_n 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.

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. 

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

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.