Geometric Transcendental Thue

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Introduction The Proof of Thue Theorem The Effectiveness Issue of Thue Theorem Acknowledgments Q & A
.
......
Pondering the Effectiveness of Thue Theorem in
Transcendental Number Theory
Qing Zou1, Jun Steed Huang2
1SuQian College, Yangzhou University
399 South Huanghe, Jiangsu 223800, P.R. China
2SuQian College, Jiangsu University
399 South Huanghe, Jiangsu 223800, P.R. China
July 24, 2015
Qing Zou1, Jun Steed Huang2 Pondering the Effectiveness of Thue Theorem in Transcendental Num

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.. Content
...1 Introduction
Beginning
The Statement of Thue Theorem
...2 The Proof of Thue Theorem
Lemmas
The Proof of Thue Theorem
...3 The Effectiveness Issue of Thue Theorem
Ineffectiveness
Effectiveness
...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 Number
Theory because it plays an important role in Diophantine
approximation. With the improvement in the status of
Transcendental Number Theory, the researches about Diophantine
approximation keep on increasing.

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.. Our Steps
We ﬁrst review the original form of Thue Theorem that he gave in
1908. Then, two equivalent forms, which are easier to understand,
will be shown in this paper. By showing different forms of the same
theorem, one of the most basic issues in Transcendental Number
Theory, i.e. the effectiveness issue of a theorem is commented in
details. Bearing in mind that the Diophantine equation often needs
to be solved in engineering of quantum device etc.

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.. The Original Form of Thue Theorem
.
Theorem 1. (Thue Theorem)
..
......
Suppose a, b, r ∈ N, r ⩾ 3, m ∈ Z, m 0. Then the Diophantine
equation
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.

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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 ∈ Z
satisfying (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 two
results Thue obtained in his early years.

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.. The First Lemma
.
Lemma 1.
..
......
Let n ∈ N, r is an integer that is larger than 2,
Un(z) = zn
+
n∑
k=1
Ck
n
(rn + 1)(r(n − 1) + 1) · · · (r(n − k + 1) + 1)
(r − 1)(2r − 1) · · · (kr − 1)
zn−k
,
Wn(z) = zn
Un(
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.
..
......
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.
..
......
Further more, suppose γ, δ ∈ (0, 1), γ + δ = 1, and w > 0; and let
φ(x) and ϕ(x) be increasing functions. Also, φ(x) and ϕ(x) are
inverse to one another. If
|θq0 − p0| < q−ω
0 , φ(q0) ⩽ q ⩽ γqω
0 , (5)
then |θq0 − p0| q < γ, and by (4) and (5), we can obtain
|θq − p| ⩾
δ
q0
⩾
δ
ϕ(q)
, θ −
p
q
>
δ
qϕ(q)
. (6)

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.. The Proof
.
The Proof
..
......
If X, Y ∈ Z is a solution of (1), then, for α = n
√
a/b , we can obtain
θ −
Y
X
αr−1 Y
X
+ · · · +
Yr−1
Xr−1
=
m/b
|Xr|
, (7)
from which we can get that the rational number
Y
X
is an
approximation of order r to the algebraic number α. Suppose
X1, Y1 ∈ Z is a solution of (1).Taking the rational number
Y1
X1
as
p
q
,
and used (2) to construct an inﬁnite sequence of good
approximations to α.

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.. The Proof
.
The Proof
..
......
Because pnqn+1 pn+1qn, it follows that for any other solution
X2, Y2 ∈ Z(|X2| > |X1|) of (1), now carry it into Lemma 2, with
p
q
=
Y2
X2
. But when |X2| is larger than |X1|, we get a contradiction
between (6) and (7); hence, |X2| cannot be arbitrarily large, and
this 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)
..
......
There exists constants C and c which can computed by given
a, b, m, such that either all of the solutions (X, Y) ∈ Z2 of (1) satisfy
the condition |X| < C, or else there exists a solution (X1, Y1) ∈ Z2
for 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 that
all the integer solution of (1) satisfy the inequality |X| + |Y| ≤ C0 ,
but different from the case of C and c, C0 cannot be expressed
explicitly by given a, b, m, since this constant also depends on |X1|,
about which we know nothing. For this reason, unfortunately we
have to say that is an ineffective constant, and so Thue Theorem is
ineffective as well. Although it effectively tells us there are ﬁnite
solutions of (1), it does not lead to a method to ﬁnd the exact
solutions. We are not here to criticize the Thue Theorem, we just
hope to see more practical theorems that get us immediate
solutions.

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.. Effectiveness
.
Condition.
..
......
Let r be an odd prime, a, b, α, β, µ ∈ N satisﬁed
aαr
− bβr
= µ, (8)
and
(4aαr
)r−2
> µ2r−2
rr2(r−1)−1
(aαr
b−1
β−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)
then
q < θkϑ
, (11)
where θ and ϑ are constants that depend and only depend on
a, b, α, β, µ and r.

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.. Effectiveness
We can see that under the condition shown above, the Thue
Theorem will become effective.

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.. Acknowledgments
This work was partially supported by the 2014 Suqian Science and
Technology Project (“The Counter-example in Mathematical
Analysis”), the Suqian Excellence Expert Allowance
2013.18.14.5.9.2, GenieView subsidiary M2M Technology
2014.331.7.49 Project.
We are also grateful to following individuals for their immeasurable
helps and contributions to this research work: Runping Ye,
Jiangyong Gu, Qihong Yu from our university.

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.. Q&A
Q&A

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