[ 15 ] The LMFDB Collaboration, The L-functions and Modular Forms Database, http://www.lmfdb.org, 2013, [On-line; accessed 17 September 2020]. [ 16 ] L. J. Mordell, Diophantine Equations, Academic Press, London and
New York 1969.
[ 17 ] The On-Line Encyclopedia of Integer Sequences, http://www.oeis.org, [On-line; accessed 17 September 2020].
[ 18 ] P. Ribenboim and W. L. McDaniel, The squares in Lucas sequences, J. Number Theory, 58 (1996), 104-123.
[ 19 ] J. H. Rickert, Simultaneous rational approximations and related dio-phantine equations, J. Math. Cambridge Philos. Soc., 113 (1993), 461-472.
[ 20 ] Sage Mathematics Software (Version 9.1), 2020, http://www.sagemath.org. [ 21 ] J. H. Silverman, Friendly Introduction to Number Theory, Addison
Wesley, 2012.
[ 22 ] T. Takagi, Lectures on Elementary Number Theory(2nd ed), Kyoritsu P Shuppan, 1971.
[ 23 ] M. Waldschmidts, A lower bound for linear forms in logarithms, Acta Arith., 37 (1980), 257-283.
[ 24 ] Wolfram Mathworld, Pentagonal Triangular Number, https://mathworld. wolfram.com/PentagonalTriangularNumber.html, [On-line; accessed 17 September 2020].
Rational Values of Powers of Trigonometric
Functions
Genki Shibukawa
Department of Mathematics, Graduate School of Science, Kobe University,
1-1, Rokkodai, Nada-ku, Kobe, 657-8501, JAPAN
e-mail : [email protected]
Received September 30 2020, Revised October 3 2020.
Abstract
We extend the theorem by Olmsted (1945) and Carlitz-Thomas (1963) on rational values of trigonometric functions to powers of trigonometric functions.
2010 Mathematics Subject Classification : 11J72, 11R18, 33B10
1
Introduction
Throughout the paper, we denote the ring of rational numbers byQ, the ring of real numbers byR, the set of positive rational numbers by Q>0 and a mth
root of unity by ζm:= e
2π√−1
m . Olmsted [4] and Carlitz-Thomas [2] determined
all rational values of trigonometric functions.
Theorem 1 (Olmsted (1945), Carlitz-Thomas (1963)). If θ∈ Q, then the only possible rational values of the trigonometric functions are:
sin (πθ), cos (πθ) = 0, ±1
2,±1 ; tan (πθ) = 0, ±1. By this Theorem 1 and well-known facts
cos (πθ)2= 1 + cos (2πθ)
2 , tan (πθ)
2
= 1
cos (πθ)2 − 1, we have the following result immediately.
Corollary 2. If θ∈ Q and cos (πθ)2∈ Q, then the only possible values of the trigonometric functions are:
sin (πθ), cos (πθ) = 0,±12,±√1 2,± √ 3 2 ,±1 ; tan (πθ) = 0, ± 1 √ 3, ±1, ± √ 3. 1
Rational Values of Powers of Trigonometric
Functions
Genki Shibukawa
Department of Mathematics, Graduate School of Science, Kobe University,
1-1, Rokkodai, Nada-ku, Kobe, 657-8501, JAPAN
e-mail : [email protected]
Received September 30 2020, Revised October 3 2020.
Abstract
We extend the theorem by Olmsted (1945) and Carlitz-Thomas (1963) on rational values of trigonometric functions to powers of trigonometric functions.
2010 Mathematics Subject Classification : 11J72, 11R18, 33B10
1
Introduction
Throughout the paper, we denote the ring of rational numbers by Q, the ring of real numbers byR, the set of positive rational numbers by Q>0 and a mth
root of unity by ζm:= e
2π√−1
m . Olmsted [4] and Carlitz-Thomas [2] determined
all rational values of trigonometric functions.
Theorem 1 (Olmsted (1945), Carlitz-Thomas (1963)). If θ∈ Q, then the only possible rational values of the trigonometric functions are:
sin (πθ), cos (πθ) = 0,±1
2,±1 ; tan (πθ) = 0, ±1. By this Theorem 1 and well-known facts
cos (πθ)2=1 + cos (2πθ)
2 , tan (πθ)
2
= 1
cos (πθ)2− 1, we have the following result immediately.
Corollary 2. If θ∈ Q and cos (πθ)2∈ Q, then the only possible values of the trigonometric functions are:
sin (πθ), cos (πθ) = 0,±12,±√1 2, ± √ 3 2 ,±1 ; tan (πθ) = 0, ± 1 √ 3, ±1, ± √ 3. 1
In this note, we prove Theorem 3 and Theorem 4, which are generalizations of Theorem 1 and Corollary 2.
Theorem 3. If N≥ 3 and α is a positive rational number such that αN1, . . . , αN−1N ∈
Q, then for any positive integer m, we have
N
√
α ∈ Q(ζm).
In particular, there is no θ∈ Q such that cos (πθ), cos (πθ)2, . . . , cos (πθ)N−1 ∈ Q and cos (πθ)N
∈ Q (resp. tan (πθ), tan (πθ)2, . . . , tan (πθ)N−1 ∈ Q and tan (πθ)N ∈ Q).
Theorem 4. If there exists a positive integer n and θ∈ Q such that cos (πθ)n∈ Q (resp. tan (πθ)n
∈ Q), then the only possible values of the trigonometric functions are: sin (πθ), cos (πθ) = 0, ±1 2, ±1 (n : odd) 0, ±√1 2, ± 1 2, ± √ 3 2 , ±1 (n : even) . (1) resp. tan (πθ) = 0, ±1 (n : odd) 0, ±√1 3, ±1, ± √ 3 (n : even). (2)
We note that one can easily verify the above theorems using the fundamental properties of Galois theory as in the following sections. But we could find no proof of these facts in print, and hence it will be of some interest to write down the proofs of these facts.
2
Preliminaries
To prove Theorem 3 and Theorem 4, we list some fundamental facts of the cyclotomic fields and Kummer extension in this section. First we mention the Galois group Gal(Q(ζn)/Q) (see [3]).
Lemma 5. The degree of the cyclotomic extensionQ(ζn) overQ is [Q(ζn) :
Q] = φ(n) := |{1 ≤ a ≤ n | gcd(a, n) = 1}| and its Galois group Gal(Q(ζn)/Q)
is (Z/nZ)× Gal(Q(ζ n)/Q) ↷ Q(ζn) → Q(ζn) ∈ ∈ ∈ ∈ c → τc ↷ ζn → τc(ζn) := ζnc . In particular Gal(Q(ζn)/Q) is an abelian extension, and its subfields L ⊃ Q
are Galois and abelian extensions overQ.
Under the following, let α be a positive rational number such that
αn1, . . . , αn−1n ∈ Q (3)
and K :=Q(√nα, ζ
n).
Proposition 6. (1) The binomial polynomial xn− α is irreducible over Q and
[Q(√nα) :Q] = n.
(2) For any n≥ 2, we have √nα
∈ Q(ζn).
Proof. (1) We consider the following decomposition of the binomial polynomial: xn− α = n i=1 (x− √nαζi n) = fI(x)fJ(x) (4)
where I, J are subsets of [n] :={1, 2, . . . , n} such that
[n] = I J, I = ∅, J = ∅, I ∩ J = ∅ (5) and fI(x) := i∈I (x− √n αζni) = x|I|+· · · + (−1)|I|α |I| n i∈I ζni ∈ Q[x], fJ(x) := j∈J (x−√n αζnj) = x|J|+· · · + (−1)mα |J| n j∈J ζnj ∈ Q[x].
If xn− α is reducible over Q, then there exist I, J ⊂ [n] such that satisfy the
condition (5) and fI(x), fJ(x) are rational coefficient polynomials. Since the
constant term of fJ(x) is real, the product j∈Jζnj is real and
j∈Jζnj = 1. Thus we have j∈J ζnj =±1.
From the assumption α|J|n ∈ Q, the constant term of fJ(x)
(−1)mαm n j∈J ζnj =±(−1)mα m n is irrational. It is a contradiction. (2) Assume √nα
∈ Q(ζn). Then we have the contradiction
n = [Q(√n α) :Q] ≤ [Q(ζn) :Q] = φ(n) = n p|n 1−1p < n.
In this note, we prove Theorem 3 and Theorem 4, which are generalizations of Theorem 1 and Corollary 2.
Theorem 3. If N≥ 3 and α is a positive rational number such that αN1, . . . , αN−1N ∈
Q, then for any positive integer m, we have
N
√
α ∈ Q(ζm).
In particular, there is no θ∈ Q such that cos (πθ), cos (πθ)2, . . . , cos (πθ)N−1 ∈ Q and cos (πθ)N
∈ Q (resp. tan (πθ), tan (πθ)2, . . . , tan (πθ)N−1 ∈ Q and tan (πθ)N ∈ Q).
Theorem 4. If there exists a positive integer n and θ∈ Q such that cos (πθ)n∈ Q (resp. tan (πθ)n
∈ Q), then the only possible values of the trigonometric functions are: sin (πθ), cos (πθ) = 0, ±1 2, ±1 (n : odd) 0, ±√1 2, ± 1 2, ± √ 3 2 , ±1 (n : even) . (1) resp. tan (πθ) = 0, ±1 (n : odd) 0, ±√1 3, ±1, ± √ 3 (n : even). (2)
We note that one can easily verify the above theorems using the fundamental properties of Galois theory as in the following sections. But we could find no proof of these facts in print, and hence it will be of some interest to write down the proofs of these facts.
2
Preliminaries
To prove Theorem 3 and Theorem 4, we list some fundamental facts of the cyclotomic fields and Kummer extension in this section. First we mention the Galois group Gal(Q(ζn)/Q) (see [3]).
Lemma 5. The degree of the cyclotomic extension Q(ζn) overQ is [Q(ζn) :
Q] = φ(n) := |{1 ≤ a ≤ n | gcd(a, n) = 1}| and its Galois group Gal(Q(ζn)/Q)
is (Z/nZ)× Gal(Q(ζ n)/Q) ↷ Q(ζn) → Q(ζn) ∈ ∈ ∈ ∈ c → τc ↷ ζn → τc(ζn) := ζnc . In particular Gal(Q(ζn)/Q) is an abelian extension, and its subfields L ⊃ Q
are Galois and abelian extensions over Q.
Under the following, let α be a positive rational number such that
αn1, . . . , αn−1n ∈ Q (3)
and K :=Q(√nα, ζ
n).
Proposition 6. (1) The binomial polynomial xn− α is irreducible over Q and
[Q(√nα) :Q] = n.
(2) For any n≥ 2, we have √nα
∈ Q(ζn).
Proof. (1) We consider the following decomposition of the binomial polynomial: xn− α = n i=1 (x− √nαζi n) = fI(x)fJ(x) (4)
where I, J are subsets of [n] :={1, 2, . . . , n} such that
[n] = I J, I = ∅, J = ∅, I ∩ J = ∅ (5) and fI(x) := i∈I (x−√n αζni) = x|I|+· · · + (−1)|I|α |I| n i∈I ζni ∈ Q[x], fJ(x) := j∈J (x−√n αζnj) = x|J|+· · · + (−1)mα |J| n j∈J ζnj ∈ Q[x].
If xn− α is reducible over Q, then there exist I, J ⊂ [n] such that satisfy the
condition (5) and fI(x), fJ(x) are rational coefficient polynomials. Since the
constant term of fJ(x) is real, the product j∈Jζnj is real and
j∈Jζnj = 1. Thus we have j∈J ζnj =±1.
From the assumption α|J|n ∈ Q, the constant term of fJ(x)
(−1)mαm n j∈J ζnj =±(−1)mα m n is irrational. It is a contradiction. (2) Assume √nα
∈ Q(ζn). Then we have the contradiction
n = [Q(√n α) :Q] ≤ [Q(ζn) :Q] = φ(n) = n p|n 1−1p < n.
Lemma 7. (1) If n = p is a odd prime, then the binomial type polynomial xp−α
is irreducible overQ(ζp) and [K :Q(ζp)] = p. Its Galois group Gal(K/Q) is
Z/pZ ⋊ (Z/pZ)× Gal(K/Q) ↷ K → K ∈ ∈ ∈ ∈ (1, 1) → σ ↷ √pα → σ(√pα) := ζ p√pα ζn → σ(ζp) := ζp (0, c) → τc ↷ √pα → τc(√pα) := √pα ζp → τc(ζp) := ζpc .
In particular, τcσ = σcτc and for n ≥ 3 the Galois group Gal(K/Q) is
non-abelian.
(2) For any n≥ 3, the Galois group Gal(K/Q) is non-abelian. Proof. (1) We consider the factorization (4) xp
− α = fI(x)fJ(x) again. By
gcd(|I|, p) = 1 and gcd(|J|, p) = 1, Q(α|I|p ) andQ(α|J|p ) contain √pα. Hence,
from Proposition 6 (2), α|I|p and α|J|p are not contained in Q(ζ
p). Therefore
the binomial polynomial xp− α is irreducible over Q(ζ
p) and [K :Q(ζp)] = p.
(2) When a odd prime p divides n, K contains a non-abelian Galois extension Q(√pα, ζ
p) overQ, so the Galois group Gal(K/Q) is non-abelian. If n = 2m
(m ≥ 2), then K contains a non-abelian Galois extension Q(√4α, ζ
4) over Q
and Gal(K/Q) is also non-abelian.
Remark 8. Lemma 7 (1) is not true in general. For example, when n = 8 and α = 2 the polynomial x8
− 2 is reducible over Q(ζ8) even though 2
1
8, . . . , 278 are
irrational. In fact
x8− 2 = (x4−√2)(x4+√2) = (x4− ζ8− ζ8−1)(x4+ ζ8+ ζ8−1).
3
Proof of Theorem 3
Assume there exists N ≥ 3, α ∈ Q>0 and a positive integer m such that
αN1, . . . , α N−1 N ∈ Q and N √ α∈ Q(ζm). Then Q(N√ α)⊂ Q(ζm).
Although Q(N√α) is not a Galois extension over Q, K is a Galois extension
overQ and
K⊂ Q(ζm, ζN)⊂ Q(ζmN).
Further the field K is a subfield ofQ(ζmN) and the Galois group Gal(K/Q) is
a normal subgroup of Gal(Q(ζmN)/Q):
Gal(K/Q) ◁ Gal(Q(ζmN)/Q) (Z/mNZ)×.
However the Galois group Gal(K/Q) is non-abelian. It is a contradiction. Then for any positive integer m,
N
√
α ∈ Q(ζm). (6)
For the above N ≥ 3 and positive rational number α ∈ Q>0, assume there
exists θ∈ Q such that
cos (πθ) = N√
α. By θ∈ Q, there exists a positive integer m such that
N√α = cos (πθ)
∈ Q(ζm).
For N ≥ 3, it is contrary to (6). For tan (πθ), one can prove similarly.
Remark 9. From Theorem 3, there is no cyclotomic field over Q containing nth root of a positive rational number α with √nα
∈ Q, for any n ≥ 3. On the other hand, from Gauss sum’s formulas [1]
m −1 k=0 ζmk2 = 1 +√−1 2 (1 + (− √ −1)m)√m = (1 +√−1)√m (m≡ 0 mod 4) √m (m ≡ 1 mod 4) 0 (m≡ 2 mod 4) √ −1√m (m≡ 3 mod 4) , ζ4= √ −1, ζ8+ ζ8−1= √ 2,
there exists a positive integer m such that√α∈ Q(ζm), for any α∈ Q.
4
Proof of Theorem 4
Since the proof of (2) is similar to (1), we only prove (1). The cases of n = 1 and n = 2 are Theorem 1 and Corollary 2 respectively. For n≥ 3, the following three cases are possible:
1) cos (πθ)∈ Q and cos (πθ)2∈ Q, 2) cos (πθ) ∈ Q and cos (πθ)2∈ Q, 3) cos (πθ) ∈ Q and cos (πθ)2 ∈ Q.
In the case of 1), from Theorem 1 and Corollary 2, the possible values of cos (πθ) (or sin (πθ)) are 0,±1
2, ±1. Similarly, for the case of 2), the possible values of
cos (πθ) are ±√1
2, ±
√
3
2 . Finally, the case of 3) is impossible from Theorem 3
(1). Then we obtain the conclusion (1).
Acknowledgement
We would like to thank Professor Takashi Taniguchi (Kobe University) for his comments on cyclotomic and Kummer extensions. We also thank the referee for his helpful comments.
Lemma 7. (1) If n = p is a odd prime, then the binomial type polynomial xp−α
is irreducible overQ(ζp) and [K :Q(ζp)] = p. Its Galois group Gal(K/Q) is
Z/pZ ⋊ (Z/pZ)× Gal(K/Q) ↷ K → K ∈ ∈ ∈ ∈ (1, 1) → σ ↷ √pα → σ(√pα) := ζ p√pα ζn → σ(ζp) := ζp (0, c) → τc ↷ √pα → τc(√pα) := √pα ζp → τc(ζp) := ζpc .
In particular, τcσ = σcτc and for n ≥ 3 the Galois group Gal(K/Q) is
non-abelian.
(2) For any n≥ 3, the Galois group Gal(K/Q) is non-abelian. Proof. (1) We consider the factorization (4) xp
− α = fI(x)fJ(x) again. By
gcd(|I|, p) = 1 and gcd(|J|, p) = 1, Q(α|I|p ) andQ(α|J|p ) contain √pα. Hence,
from Proposition 6 (2), α|I|p and α|J|p are not contained in Q(ζ
p). Therefore
the binomial polynomial xp− α is irreducible over Q(ζ
p) and [K :Q(ζp)] = p.
(2) When a odd prime p divides n, K contains a non-abelian Galois extension Q(√pα, ζ
p) over Q, so the Galois group Gal(K/Q) is non-abelian. If n = 2m
(m ≥ 2), then K contains a non-abelian Galois extension Q(√4α, ζ
4) over Q
and Gal(K/Q) is also non-abelian.
Remark 8. Lemma 7 (1) is not true in general. For example, when n = 8 and α = 2 the polynomial x8
− 2 is reducible over Q(ζ8) even though 2
1
8, . . . , 278 are
irrational. In fact
x8− 2 = (x4−√2)(x4+√2) = (x4− ζ8− ζ8−1)(x4+ ζ8+ ζ8−1).
3
Proof of Theorem 3
Assume there exists N ≥ 3, α ∈ Q>0 and a positive integer m such that
αN1, . . . , α N−1 N ∈ Q and N √ α∈ Q(ζm). Then Q(N√ α)⊂ Q(ζm).
Although Q(N√α) is not a Galois extension over Q, K is a Galois extension
overQ and
K⊂ Q(ζm, ζN)⊂ Q(ζmN).
Further the field K is a subfield ofQ(ζmN) and the Galois group Gal(K/Q) is
a normal subgroup of Gal(Q(ζmN)/Q):
Gal(K/Q) ◁ Gal(Q(ζmN)/Q) (Z/mNZ)×.
However the Galois group Gal(K/Q) is non-abelian. It is a contradiction. Then for any positive integer m,
N
√
α ∈ Q(ζm). (6)
For the above N ≥ 3 and positive rational number α ∈ Q>0, assume there
exists θ∈ Q such that
cos (πθ) = N√
α. By θ∈ Q, there exists a positive integer m such that
N√α = cos (πθ)
∈ Q(ζm).
For N≥ 3, it is contrary to (6). For tan (πθ), one can prove similarly.
Remark 9. From Theorem 3, there is no cyclotomic field over Q containing nth root of a positive rational number α with √nα
∈ Q, for any n ≥ 3. On the other hand, from Gauss sum’s formulas [1]
m −1 k=0 ζmk2= 1 +√−1 2 (1 + (− √ −1)m)√m = (1 +√−1)√m (m≡ 0 mod 4) √m (m ≡ 1 mod 4) 0 (m≡ 2 mod 4) √ −1√m (m≡ 3 mod 4) , ζ4= √ −1, ζ8+ ζ8−1 = √ 2,
there exists a positive integer m such that√α∈ Q(ζm), for any α∈ Q.
4
Proof of Theorem 4
Since the proof of (2) is similar to (1), we only prove (1). The cases of n = 1 and n = 2 are Theorem 1 and Corollary 2 respectively. For n≥ 3, the following three cases are possible:
1) cos (πθ)∈ Q and cos (πθ)2∈ Q, 2) cos (πθ) ∈ Q and cos (πθ)2∈ Q, 3) cos (πθ) ∈ Q and cos (πθ)2 ∈ Q.
In the case of 1), from Theorem 1 and Corollary 2, the possible values of cos (πθ) (or sin (πθ)) are 0,±1
2,±1. Similarly, for the case of 2), the possible values of
cos (πθ) are ±√1
2, ±
√
3
2 . Finally, the case of 3) is impossible from Theorem 3
(1). Then we obtain the conclusion (1).
Acknowledgement
We would like to thank Professor Takashi Taniguchi (Kobe University) for his comments on cyclotomic and Kummer extensions. We also thank the referee for his helpful comments.
References
[1] B. C. Berndt, R. J. Evans and K. S. Williams, Gauss and Jacobi Sums, John Wiley, 1998.
[2] L. Carlitz and J. M. Thomas, Rational tabulated values of trigonomet-ric functions, Amer. Math. Monthly 69 (1962), 789–793.
[3] S. Lang, Algebra , Springer GTM 211 (Revised 3rd ed.), 2002. [4] J. M. H. Olmsted, Rational values of trigonometric functions, Amer.
Math. Monthly 52-9 (1945) 507–508.
Axiomatic Method of Measure
and Integration (V).
Definition and Existence Theorem
of the RS-measure
(Yoshifumi Ito, “RS-integral and LS-integral”, Chapter 3)
By
Yoshifumi Ito
Professor Emeritus, Tokushima University
209-15 Kamifukuman Hachiman-cho
Tokushima 770-8073, JAPAN
e-mail address : [email protected]
(Received September 30, 2020)
Abstract
In this paper, we define the RS-measure on Rd, (d
≥ 1) by prescribing the complete system of axioms. Then we prove the existence theorem of the RS-measure and determine all the RS-measures. This is a new result.
2000 Mathematics Subject Classification. Primary 28Axx.
Introduction
This paper is the part V of the series of papers on the axiomatic method of measure and integration on the Euclidean space. As for the details, we refer to Ito [6], [14]. Further we refer to Ito [1]∼[5], [7]∼[13] and [15]∼ [21].
In this paper, we study the definition of the Riemann-Stieltjes measure on the d-dimensional Euclidean space and prove its existence theorem and study
