Geometric Transcendental Thue

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Introduction The Proof of Thue Theorem The Effectiveness Issue of Thue Theorem Acknowledgments Q & A

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Pondering the Effectiveness of Thue Theorem inTranscendental Number Theory

Qing Zou1, Jun Steed Huang2

1SuQian College, Yangzhou University399 South Huanghe, Jiangsu 223800, P.R. China

2SuQian College, Jiangsu University399 South Huanghe, Jiangsu 223800, P.R. China

July 24, 2015

Qing Zou1 , Jun Steed Huang2 Pondering the Effectiveness of Thue Theorem in Transcendental Number Theory

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The original Paper published on ICCSCM2015 Malaysia on May 8.

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.. Content

...1 IntroductionBeginningThe Statement of Thue Theorem

...2 The Proof of Thue TheoremLemmasThe Proof of Thue Theorem

...3 The Effectiveness Issue of Thue TheoremIneffectivenessEffectiveness

...4 Acknowledgments

...5 Q & A


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Introduction The Proof of Thue Theorem The Effectiveness Issue of Thue Theorem Acknowledgments Q & ABeginning The Statement of Thue Theorem

.. Why We Need to Consider the Thue Theorem

Thue Theorem is an important theorem in Transcendental NumberTheory because it plays an important role in Diophantineapproximation. With the improvement in the status ofTranscendental Number Theory, the researches about Diophantineapproximation keep on increasing.


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.. Our Steps

We first review the original form of Thue Theorem that he gave in1908. Then, two equivalent forms, which are easier to understand,will be shown in this paper. By showing different forms of the sametheorem, one of the most basic issues in Transcendental NumberTheory, i.e. the effectiveness issue of a theorem is commented indetails. Bearing in mind that the Diophantine equation often needsto be solved in engineering of quantum device etc.


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.. The Original Form of Thue Theorem

.Theorem 1. (Thue Theorem)..

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Suppose a, b, r ∈ N, r ⩾ 3,m ∈ Z,m , 0. Then the Diophantineequation

axr − byr = m (1)

cannot have an infinite number of solutions.


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.. Another Form of Thue Theorem

Now, let’s present this theorem in a somewhat different form,looking from possible solutions perspective.

.Theorem 1’. (Thue Theorem)..

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There exists a constant C = C(a, b, r,m) such that any x, y ∈ Zsatisfying (1) must also satisfy the inequality

|x|+ |y| ⩽ C.


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.. Another Form of Thue Theorem

Now, let’s present this theorem in a somewhat different form,looking from possible solutions perspective.

.Theorem 1’. (Thue Theorem)..

......

There exists a constant C = C(a, b, r,m) such that any x, y ∈ Zsatisfying (1) must also satisfy the inequality

|x|+ |y| ⩽ C.


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Introduction The Proof of Thue Theorem The Effectiveness Issue of Thue Theorem Acknowledgments Q & ALemmas The Proof of Thue Theorem

.. Beginning

In order to prove Thue Theorem, we would like to introduce tworesults Thue obtained in his early years.


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.. The First Lemma

.Lemma 1...

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Let n ∈ N, r is an integer that is larger than 2,

Un(z) = zn+n∑

k=1

Ckn(rn + 1)(r(n − 1) + 1) · · · (r(n − k + 1) + 1)

(r − 1)(2r − 1) · · · (kr − 1) zn−k,

Wn(z) = znUn(1

z).

Then we hold the reciprocal polynomial equation

Wn+1(z)Un(z) −Wn(z)Un+1(z) (2)

=2(r + 1)(2r + 1) · · · (nr + 1)

(r − 1)(2r − 1) · · · (nr − 1) (z − 1)2n+1.


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.. The Second Lemma

.Lemma 2...

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Suppose

θ ∈ R − Q; p, p0 ∈ Z; q, q0 ∈ N; pq0 , p0q. (3)

Then

1 ⩽ |pq0 − p0q| =∣∣∣(θq0 − p0) q − (θq − p) q0

∣∣∣ (4)

⩽ |θq0 − p0| q + |θq − p| q0.


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.. The Second Lemma

.Lemma 2...

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Further more, suppose γ, δ ∈ (0, 1), γ + δ = 1, and w > 0; and letφ(x) and ϕ(x) be increasing functions. Also, φ(x) and ϕ(x) areinverse to one another. If

|θq0 − p0| < q−ω0 , φ(q0) ⩽ q ⩽ γqω0 , (5)

then |θq0 − p0| q < γ, and by (4) and (5), we can obtain

|θq − p| ⩾ δq0⩾δ

ϕ(q),

∣∣∣∣∣θ − pq

∣∣∣∣∣ > δ

qϕ(q). (6)


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.. The Proof

.The Proof..

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If X,Y ∈ Z is a solution of (1), then, for α = n√

a/b , we can obtain∣∣∣∣∣θ − YX

∣∣∣∣∣∣∣∣∣∣∣αr−1 Y

X+ · · ·+ Yr−1

Xr−1

∣∣∣∣∣∣ =∣∣∣m/b∣∣∣

|Xr | , (7)

from which we can get that the rational numberYX

is an

approximation of order r to the algebraic number α. Suppose

X1, Y1 ∈ Z is a solution of (1).Taking the rational numberY1

X1as

pq

,

and used (2) to construct an infinite sequence of goodapproximations to α.


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.. The Proof

.The Proof..

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Because pnqn+1 , pn+1qn, it follows that for any other solutionX2, Y2 ∈ Z(|X2| > |X1|) of (1), now carry it into Lemma 2, withpq=

Y2

X2. But when |X2| is larger than |X1|, we get a contradiction

between (6) and (7); hence, |X2| cannot be arbitrarily large, andthis means that (1) cannot have an infinite number of solutions.


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Introduction The Proof of Thue Theorem The Effectiveness Issue of Thue Theorem Acknowledgments Q & AIneffectiveness Effectiveness

.. Ineffectiveness

.Theorem 1”. (Thue Theorem)..

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There exists constants C and c which can computed by givena, b,m, such that either all of the solutions (X, Y) ∈ Z2 of (1) satisfythe condition |X| < C, or else there exists a solution (X1, Y1) ∈ Z2for which |X1| > c, and in the latter case any other solution(X2,Y2) ∈ Z2 satisfies

|X2| ⩽ |X1|C.


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.. Ineffectiveness

According to this form of Thue Theorem, we can obviously get thatall the integer solution of (1) satisfy the inequality |X|+ |Y | ≤ C0 ,but different from the case of C and c, C0 cannot be expressedexplicitly by given a, b,m, since this constant also depends on |X1|,about which we know nothing. For this reason, unfortunately wehave to say that is an ineffective constant, and so Thue Theorem isineffective as well. Although it effectively tells us there are finitesolutions of (1), it does not lead to a method to find the exactsolutions. We are not here to criticize the Thue Theorem, we justhope to see more practical theorems that get us immediatesolutions.


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.. Effectiveness.Condition...

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Let r be an odd prime, a, b, α, β, µ ∈ N satisfied

aαr − bβr = µ, (8)

and(4aαr)r−2 > µ2r−2rr2(r−1)−1(aαrb−1β−r)2r−4+2/r. (9)

It is not hard to prove that if p, q, k ∈ N satisfied the inequality∣∣∣aqr − bpr∣∣∣ ⩽ k, (10)

thenq < θkϑ, (11)

where θ and ϑ are constants that depend and only depend ona, b, α, β, µ and r.


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.. Effectiveness

We can see that under the condition shown above, the ThueTheorem will become effective.


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.. Acknowledgments

This work was partially supported by the 2014 Suqian Science andTechnology Project (“The Counter-example in MathematicalAnalysis”), the Suqian Excellence Expert Allowance2013.18.14.5.9.2, GenieView subsidiary M2M Technology2014.331.7.49 Project.We are also grateful to following individuals for their immeasurablehelps and contributions to this research work: Runping Ye,Jiangyong Gu, Qihong Yu from our university.


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Thanks go to American Mathematical Society for offering of the text book "Iwasawa Theory and Modular Forms".

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.. Q&A

Q&A


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