Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 893045, 10 pages
http://dx.doi.org/10.1155/2013/893045
Research Article
A Non-NP-Complete Algorithm for a Quasi-Fixed
Polynomial Problem
Yi-Chou Chen1 and Hang-Chin Lai2
1
2
Department of General Education, National Army Academy, Taoyuan 320, Taiwan
Department of Mathematics, National Tsing Hua University, Hsinchu 300, Taiwan
Correspondence should be addressed to Hang-Chin Lai; laihc@mx.nthu.edu.tw
Received 15 January 2013; Accepted 24 February 2013
Academic Editor: Jen-Chih Yao
Copyright © 2013 Y.-C. Chen and H.-C. Lai. This is an open access article distributed under the Creative Commons Attribution
License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly
cited.
𝑠
Let 𝐹 : R × R → R be a real-valued polynomial function of the form 𝐹(𝑥, 𝑦) = ∑𝑖=0 𝑓𝑖 (𝑥)𝑦𝑖 , with degree of 𝑦 in 𝐹(𝑥, 𝑦) = 𝑠 ≥
1, 𝑥 ∈ R. An irreducible real-valued polynomial function 𝑝(𝑥) and a nonnegative integer 𝑚 are given to find a polynomial function
𝑦(𝑥) ∈ R[𝑥] satisfying the following expression: 𝐹(𝑥, 𝑦(𝑥)) = 𝑐𝑝𝑚 (𝑥) for some constant 𝑐 ∈ R. The constant 𝑐 is dependent on
the solution 𝑦(𝑥), namely, a quasi-fixed (polynomial) solution of the polynomial-like equation (∗). In this paper, we will provide a
non-NP-complete algorithm to solve all quasi-fixed solutions if the equation (∗) has only a finite number of quasi-fixed solutions.
1. Introduction and Preliminaries
Lenstra [1] found that a polynomial function 𝐹(𝑥, 𝑦) ∈
Q(𝛼)[𝑥, 𝑦](where 𝛼 is an algebraic number) can be solved
to a polynomial function 𝑦 = 𝑦(𝑥) ∈ Q(𝛼)[𝑥] such that
𝐹(𝑥, 𝑦(𝑥)) = 0. It can thus be derived to find a polynomial
𝑦 = 𝑦(𝑥) such that there exists an 𝑥 ∈ Q(𝛼)[𝑥] as a fixed
point of the polynomial function 𝐹(𝑥, 𝑦(𝑥)); that is,
𝐹 (𝑥, 𝑦 (𝑥)) = 𝑥
(1)
has a polynomial solution 𝑦(𝑥) ∈ Q(𝛼)[𝑥].
Furthermore, Tung [2, 3] extended (1) to solve 𝑦(𝑥) for
the following equation:
𝐹 (𝑥, 𝑦 (𝑥)) = 𝑐𝑥𝑚 , for 𝑥 ∈ K (field) , 𝑚 ∈ N,
(2)
where 𝑐 is a real constant depending on the polynomial
solution 𝑦(𝑥) and the given positive integer 𝑚 ∈ N.
Recently, Lai and Chen [4] extended the expression (2) to
solve 𝑦(𝑥) to satisfy the polynomial equation as the following
form:
𝐹 (𝑥, 𝑦 (𝑥)) = 𝑐𝑝𝑚 (𝑥) , 𝑥 ∈ R,
(3)
where 𝑝(⋅) is an irreducible polynomial in 𝑥 ∈ R and the
polynomial functions 𝐹(𝑥, 𝑦) : R × R → R are written by
𝑠
𝐹 (𝑥, 𝑦) =∑𝑓𝑖 (𝑥) 𝑦𝑖 ,
𝑖=0
(4)
with degree 𝑦 in 𝐹 (𝑥, 𝑦) denoted by deg𝑦 𝐹 = 𝑠 > 1.
Remark 1. Recall that a polynomial function 𝑦 = 𝑦(𝑥)
satisfying (3) is called a quasi-fixed (polynomial) solution
corresponding to the real value 𝑎. This number 𝑎 is called a
quasi-fixed (polynomial) value corresponding to the polynomial solutions 𝑦 = 𝑦(𝑥).
Note that the equation may not be solvable; if it is solvable,
the number of all solutions may be infinitely many, or finitely
many quasi-fixed solutions If this equation has infinitely
many quasi-fixed solutions, by [4, Theorem 3.5], we have
(1) 𝑓𝑠 (𝑥) must be 𝑎𝑝𝑟 (𝑥) for some 𝑎 ∈ R, 𝑟 ∈ N,
(2) those solutions must assume a fixed form like
𝑓 (𝑥)
− 𝑠−1
+ 𝜆𝑝𝑘 (𝑥) for any 𝜆 ∈ R.
𝑠𝑓𝑠 (𝑥)
(5)
Moreover, if this equation has only finitely many quasi-fixed
solutions, by [4, Corollary 3.3], the bound of all solutions is
at most deg𝑦 𝐹 + 2.
2
Abstract and Applied Analysis
In this paper, we deal with the case of finitely many solutions and provide a non-NP-complete algorithm to obtain all
representations of quasi-fixed solutions.
Remark 2. An algorithm is called “NP-complete” if the
algorithm is computed in an exponential time of the input
size “n”; that is, the computing time in the algorithm does
not exceed 𝑂(𝑒𝑛 ). Otherwise, the algorithm is called “nonNP-complete”; that is, the algorithm can be computed in a
polynomial time not exceeding 𝑂(𝑛𝑘 ) for some fixed real
number 𝑘.
The remainder of the paper is organized as follows:
Section 2 introduces an example for the practical algorithm.
Section 3 describes the main algorithm and its processes.
Section 4 proves that the main algorithm is indeed non-NPcomplete. We have given two example problems in Section 5.
and we have
𝑧𝑝 (𝑥) + 2
𝑦={
𝑧𝑝 (𝑥) − 2
={
2 (mod 𝑝 (𝑥))
−2 (mod 𝑝 (𝑥)) ,
(12)
with an indeterminate 𝑧. To determine 𝑧, we substitute this 𝑦
(in (12)) into 𝐹(𝑥, 𝑦), then (8) becomes
(𝑃1 ) {
(1) 𝐹 (𝑥, 𝑦) = 𝐹 (𝑥, 𝑧𝑝 (𝑥) + 2) = 𝑎𝑝2 (𝑥)
(2) 𝐹 (𝑥, 𝑦) = 𝐹 (𝑥, 𝑧𝑝 (𝑥) − 2) = 𝑎𝑝2 (𝑥) .
(13)
This yields
(1) 𝑝2 (𝑥) [𝑝 (𝑥) 𝑧2 + 4𝑧 − 𝑝 (𝑥)] = 𝑎𝑝2 (𝑥)
(14)
(𝑃2 ) {
(2) 𝑝2 (𝑥) [𝑝 (𝑥) 𝑧2 − 4𝑧 − 𝑝 (𝑥)] = 𝑎𝑝2 (𝑥) .
Both sides of (14) divide by 𝑝2 (𝑥) to become
2. Relative Algorithm and Some Examples
Let 𝐹(𝑥, 𝑦) ∈ R[𝑥, 𝑦] and an irreducible polynomial function
𝑝(𝑥) ∈ R[𝑥]. To complete our algorithm, we need an
existing algorithm. In [2], Tung has established the following
algorithm.
Algorithm 3 (Tung [5]). Consider the following technique.
Input. Given is a polynomial 𝐹(𝑥, 𝑦) ∈ R[𝑥, 𝑦].
(𝑃3 ) {
(1) 𝑝 (𝑥) 𝑧2 + 4𝑧 − 𝑝 (𝑥) = 𝑎
(2) 𝑝 (𝑥) 𝑧2 − 4𝑧 − 𝑝 (𝑥) = 𝑎.
(15)
According to Algorithm 3, we obtain 𝑧 = 1 and 𝑧 = −1,
so the solutions (12) of problem (𝑃) are
𝑦={
−𝑝 (𝑥) + 2
𝑧𝑝 (𝑥) + 2
𝑝 (𝑥) + 2
={
or {
𝑧𝑝 (𝑥) − 2
𝑝 (𝑥) − 2
−𝑝 (𝑥) − 2.
(16)
In Section 3, we provide a non-NP-complete algorithm to
satisfy (3).
Output. All solutions 𝑦 = 𝑦(𝑥) satisfy
𝐹 (𝑥, 𝑦 (𝑥)) = 𝑎, for some 𝑎 ∈ R,
(6)
in polynomial time of 𝑤𝜇, where 𝑤 is the computer memory
of all coefficients of 𝐹(𝑥, 𝑦) and 𝜇 = deg𝑥 𝐹(𝑥, 𝑦) +
deg𝑦 𝐹(𝑥, 𝑦). Moreover, the number of all solutions is at most
deg𝑦 𝐹(𝑥, 𝑦).
We provide a simple example to explain our aim for
solving the quasi-fixed polynomial solutions related to
Algorithm 5 (in Section 3) as follows.
3. Computing Procedure
In this paper, we may assume that the number of all solutions
of (3) is finite and then constitute an algorithm of the approximate solutions for the quasi-fixed polynomial equation (3) as
follows.
Algorithm 5. Consider the following technique.
Input. Given a polynomial 𝐹(𝑥, 𝑦) ∈ R[𝑥, 𝑦], 𝑝(𝑥) ∈ R[𝑥]
irreducible and 𝑚 ∈ N.
Example 4. Let
𝐹 (𝑥, 𝑦) = 𝑝 (𝑥) [𝑦2 − 𝑝2 (𝑥) − 4]
(7)
2
and 𝑝(𝑥) = 𝑥 + 𝑥 + 1. Then deg𝑦 𝐹 = 𝑠 = 2. Solve all quasifixed solutions of
𝐹 (𝑥, 𝑦) = 𝑎𝑝2 (𝑥) .
2
2
𝑝 (𝑥) [𝑦 − 𝑝 (𝑥) − 4] = 𝑎𝑝 (𝑥) ,
𝐹 (𝑥, 𝑦 (𝑥)) = 𝑎𝑝𝑚 (𝑥)
for some 𝑎 ∈ R,
(17)
(8)
by a non-NP-complete algorithm.
The following definitions are given by [6, page 421].
(9)
Definition 6. (i) The content of 𝐹(𝑥, 𝑦) in (4), denoted by
cont𝑦 𝐹, is defined by the greatest common divisor (g.c.d):
By (7) and (8), we obtain
2
Output. All solutions 𝑦 = 𝑦(𝑥) satisfy
gcd (𝑓𝑠 (𝑥) , 𝑓𝑠−1 (𝑥) , . . . , 𝑓0 (𝑥)) = cont𝑦 𝐹.
divides both sides by 𝑝(𝑥), then it becomes
𝑦2 − 𝑝2 (𝑥) − 4 = 𝑎𝑝 (𝑥) .
(10)
𝑦2 − 4 = 0 mod 𝑝 (𝑥) ,
(11)
It follows that
(18)
(ii) The primitive part pp𝑦 𝐹 of 𝐹(𝑥, 𝑦) is 𝐹(𝑥, 𝑦)/cont𝑦 𝐹 ∈
R[𝑥, 𝑦], and 𝐹(𝑥, 𝑦) is primitive if cont𝑦 𝐹 = 1.
From (i) and (ii), we have
𝐹 (𝑥, 𝑦) = cont𝑦 𝐹 ⋅ pp𝑦 𝐹.
(19)
Abstract and Applied Analysis
3
To complete the following algorithm, we need the following elementary property.
Lemma 7. Let 𝐹(𝑥, 𝑦) ∈ R[𝑥, 𝑦] and 𝑝(𝑥) be an irreducible
polynomial in R[𝑥]. Consider the module equation
𝐹 (𝑥, 𝑦) = 0 (mod 𝑝 (𝑥)) .
(20)
The number of all solutions 𝑦 = 𝑦(𝑥) (mod 𝑝(𝑥)) is thus at
most deg𝑦 𝐹.
According to Lemma 7, the solution number does not exceed
deg𝑦 𝐹, thus we may assume that 𝑦 = 𝑎0 (𝑥) is a solution of
(28) with deg 𝑎0 (𝑥) < deg 𝑝(𝑥); please note that the choice
of 𝑎0 (𝑥) may be larger than 1 and we may define a solution
set 𝑇0 (𝐹) which collects such 𝑎0 (𝑥) by setting as the following
form:
𝑎0 (𝑥) ∈ 𝑇0 (𝐹) = {𝑎10 (𝑥) , 𝑎20 (𝑥) , . . . , 𝑎𝑟00 (𝑥)} .
(29)
Consequently, the expression (23) can go to Step 2.
Assumption. Throughout this algorithm, for any 𝐹(𝑥, 𝑦) ∈
R[𝑥, 𝑦], one can solve all solutions 𝑦 = 𝑔(𝑥) (mod 𝑝(𝑥)) of
the equation
Step 2. For each 𝑎0 (𝑥) ∈ 𝑇0 (𝐹) in (Step 1, (1.2.2)), we can
replace 𝑦 by 𝑧𝑝(𝑥) + 𝑎0 (𝑥) in (23), and then (23) becomes
𝐹 (𝑥, 𝑔 (𝑥)) = 0 mod 𝑝 (𝑥) .
𝐹 (𝑥, 𝑧𝑝 (𝑥) + 𝑎0 (𝑥)) = 𝑐𝑝𝑚 (𝑥) .
(21)
The procedure of our main algorithm is described below.
Procedures of Main Algorithm (Algorithm 5)
Step 0. If cont𝑦 𝐹 | 𝑝𝑚 (𝑥) and cont𝑦 𝐹 = 1, then let 𝐹1 (𝑥, 𝑦) =
𝐹(𝑥, 𝑦) and move to (Step 1, (1.2.2)).
We let 𝐻1 (𝑥, 𝑧) = 𝐹(𝑥, 𝑧𝑝(𝑥) + 𝑎0 (𝑥)). Thus it follows from
Definition 6(ii) that there exists
𝐹2 (𝑥, 𝑦) = pp𝑦 𝐻1 ∈ R [𝑥, 𝑦]
𝐹 (𝑥, 𝑦) = (cont𝑦 𝐹) 𝐹1 (𝑥, 𝑦) ,
(22)
𝑚
𝐹 (𝑥, 𝑦) = 𝑐𝑝 (𝑥) ,
(32)
Equation (30) then becomes
where 𝐹1 (𝑥, 𝑦) ∈ R[𝑥, 𝑦] is a primitive polynomial.
𝐻1 (𝑥, 𝑦) = 𝑐𝑝𝑚 (𝑥) .
Step 1.1. If cont𝑦 𝐹 ∤ 𝑝𝑚 (𝑥), we would have the problem
(31)
which is a primitive polynomial function so that
𝐻1 (𝑥, 𝑦) = (cont𝑦 𝐻1 ) 𝐹2 (𝑥, 𝑦) .
Step 1. For convenience, we let
(30)
(33)
Next, we will prove that 𝑝(𝑥) | cont𝑦 𝐻1 .
(23)
to deduce that cont𝑦 𝐹 | 𝑐 ⋅ 𝑝𝑚 (𝑥). But cont𝑦 𝐹 ∤ 𝑝𝑚 (𝑥), then
𝑐 = 0. Consequently, (23) becomes
Step 2.1. If cont𝑦 𝐻1 ∤ 𝑝𝑚 (𝑥), (32) and (33) can be extrapolated as to cont𝑦 𝐻1 | 𝑐 ⋅ 𝑝𝑚 (𝑥). Since cont𝑦 𝐻1 ∤ 𝑝𝑚 (𝑥), then
𝑐 = 0. Consequently, (33) becomes
𝐹 (𝑥, 𝑦) = 0.
𝐻1 (𝑥, 𝑦) = 0.
(24)
We can then solve all solutions 𝑦(𝑥) for 𝐹(𝑥, 𝑦) = 0 to get a
solution set
𝑌0 = {𝑦 (𝑥) : 𝐹 (𝑥, 𝑦 (𝑥)) = 0} .
(25)
We can then solve all solutions 𝑦(𝑥) for 𝐻1 (𝑥, 𝑦) = 0 to get a
solution set
𝑌1 = {𝑦 (𝑥) 𝑝 (𝑥)
+ 𝑎0 (𝑥) : 𝐻1 (𝑥, 𝑦 (𝑥)) = 0, 𝑎0 (𝑥) ∈ 𝑇0 (𝐹)} .
Step 1.2. If cont𝑦 𝐹 | 𝑝𝑚 (𝑥) and cont𝑦 𝐹 ≠ 1, then cont𝑦 𝐹 =
𝑝ℓ1 (𝑥) with ℓ1 ≤ 𝑚. In this case, (23) becomes
𝐹1 (𝑥, 𝑦) = 𝑐𝑝𝑚−ℓ1 (𝑥) .
(26)
𝑊0 = {𝑦 (𝑥) : 𝐹1 (𝑥, 𝑦 (𝑥)) = 𝑐} .
𝐹1 (𝑥, 𝑦) = 0 (mod 𝑝 (𝑥)) .
(28)
(36)
Step 2.2.1. In case ℓ2 = 𝑚, (36) becomes
𝐹2 (𝑥, 𝑦) = 𝑐
(27)
Step 1.2.2. If ℓ1 < 𝑚, then 𝑚 − ℓ1 > 0 and we can divide both
sides of (26) by 𝑝(𝑥); consequently,
(35)
Step 2.2. If cont𝑦 𝐻1 | 𝑝𝑚 (𝑥), then cont𝑦 𝐻1 = 𝑝ℓ2 (𝑥) with
ℓ2 ≤ 𝑚. In this case, (33) becomes
𝐹2 (𝑥, 𝑦) = 𝑐𝑝𝑚−ℓ2 (𝑥) .
Step 1.2.1. In case ℓ1 = 𝑚, then (26) becomes 𝐹1 (𝑥, 𝑦) = 𝑐
which can be solved by Algorithm 3 to obtain all solutions
for the equation 𝐹1 (𝑥, 𝑦) = 𝑐 to get 𝑦 = 𝑦(𝑥) and obtain a set
(34)
(37)
which can be solved by Algorithm 3 to get 𝑦 = 𝑦(𝑥) and
obtain a set
𝑊1 = {𝑦 (𝑥) 𝑝 (𝑥)
+ 𝑎0 (𝑥) : 𝐹2 (𝑥, 𝑦 (𝑥)) = 𝑐, 𝑎0 (𝑥) ∈ 𝑇0 (𝐹)} .
(38)
4
Abstract and Applied Analysis
Step 2.2.2. If ℓ2 < 𝑚, then 𝑚 − ℓ2 < 0 and we can divide both
sides of (36) by 𝑝(𝑥); consequently,
𝐹2 (𝑥, 𝑦) = 0 (mod 𝑝 (𝑥)) .
(39)
By the same reasoning used in [Step 1.2.2], we can solve 𝑦 =
𝑎1 (𝑥) with deg 𝑎1 (𝑥) < deg 𝑝(𝑥) and obtain the set
𝑎1 (𝑥) ∈ 𝑉1 = {𝑎11 (𝑥) , 𝑎21 (𝑥) , . . . , 𝑎𝑟11 (𝑥)} .
(40)
Step 3.2.1. If ℓ3 = 𝑚, then (48) becomes
𝐹3 (𝑥, 𝑦) = 𝑐
(49)
and, by Algorithm 3, we compute 𝑦 = 𝑦(𝑥) and collect
𝑦 (𝑥) 𝑝2 (𝑥) + 𝑎1 (𝑥) 𝑝 (𝑥) + 𝑎0 (𝑥)
(50)
in 𝑊2 for all 𝑎1 (𝑥)𝑝(𝑥) + 𝑎0 (𝑥) ∈ 𝑇1 (𝐹). Hence, we output
𝑊2 = {𝑦 (𝑥) 𝑝2 (𝑥) + 𝑎1 (𝑥) 𝑝 (𝑥)+ 𝑎0 (𝑥) : 𝐹3 (𝑥, 𝑦 (𝑥)) = 𝑐,
We let
𝑎1 (𝑥) 𝑝 (𝑥) + 𝑎0 (𝑥) ∈ 𝑇1 (𝐹) } .
𝑇1 (𝐹) = {𝑎1 (𝑥) 𝑝 (𝑥) + 𝑎0 (𝑥) : 𝑎1 (𝑥) ∈ 𝑉1 , 𝑎0 (𝑥) ∈ 𝑇0 (𝐹)} ,
(41)
(51)
Consequently, expression (33) can go to Step 3.
Step 3. For each 𝑎1 (𝑥)𝑝(𝑥) + 𝑎0 (𝑥) ∈ 𝑇1 (𝐹), we can replace 𝑦
by 𝑧𝑝2 (𝑥) + 𝑎1 (𝑥)𝑝(𝑥) + 𝑎0 (𝑥) in (23). It then becomes
𝐹 (𝑥, 𝑧𝑝2 (𝑥) + 𝑎1 (𝑥) 𝑝 (𝑥) + 𝑎0 (𝑥)) = 𝑐𝑝𝑚 (𝑥) .
(42)
Moreover, we let 𝐻2 (𝑥, 𝑧) = 𝐹(𝑥, 𝑧𝑝2 (𝑥) + 𝑎1 (𝑥)𝑝(𝑥) + 𝑎0 (𝑥))
and
𝐻2 (𝑥, 𝑦) = (cont𝑦 𝐻2 ) 𝐹3 (𝑥, 𝑦)
(43)
for a primitive polynomial 𝐹3 (𝑥, 𝑦) = pp𝑦 𝐻2 ∈ R[𝑥, 𝑦], so
that (42) becomes
𝐻2 (𝑥, 𝑦) = 𝑐𝑝𝑚 (𝑥) .
(44)
Next, we will prove that 𝑝2 (𝑥) | cont𝑦 𝐻2 .
𝑚
Step 3.1. If cont𝑦 𝐻2 ∤ 𝑝 (𝑥), and (43) and (44) are extrapolated, we have
cont𝑦 𝐻2 | 𝑐 ⋅ 𝑝𝑚 (𝑥) .
(45)
Since cont𝑦 𝐻2 ∤ 𝑝𝑚 (𝑥), we have 𝑐 = 0.
Thus (44) becomes
𝐻2 (𝑥, 𝑦) = 0.
𝐹3 (𝑥, 𝑦) = 0 (mod𝑝 (𝑥)) .
(46)
𝑌2 = {𝑦 (𝑥) 𝑝2 (𝑥) + 𝑎1 (𝑥) 𝑝 (𝑥) + 𝑎0 (𝑥) : 𝐻2 (𝑥, 𝑦 (𝑥)) = 0,
𝑎1 (𝑥) 𝑝 (𝑥) + 𝑎0 (𝑥) ∈ 𝑇1 (𝐹) } .
(47)
Step 3.2. If cont𝑦 𝐻2 | 𝑝𝑚 (𝑥), then cont𝑦 𝐻2 = 𝑝ℓ3 (𝑥) for some
positive integer ℓ3 ≤ 𝑚 and (44) becomes
(48)
(52)
By the same reasoning used in (Step 1.2.2), we can solve 𝑦 =
𝑎2 (𝑥) with deg 𝑎2 (𝑥) < deg 𝑝(𝑥) and obtain the set
𝑎2 (𝑥) ∈ 𝑉2 = {𝑎12 (𝑥) , 𝑎22 (𝑥) , . . . , 𝑎𝑟22 (𝑥)} .
(53)
We let
𝑇2 (𝐹) = {𝑎2 (𝑥) 𝑝2 (𝑥) + 𝑎1 (𝑥) 𝑝 (𝑥) + 𝑎0 (𝑥)
: 𝑎2 (𝑥) ∈ 𝑉2 , 𝑎1 (𝑥) 𝑝 (𝑥) + 𝑎0 (𝑥) ∈ 𝑇1 (𝐹) } ,
(54)
thus expression (44) can go to Step 4.
Continuing this process, we get the (𝑘 − 1)-th step and a
sequence {ℓ1 , ℓ2 , . . . , ℓ𝑘−1 } and go to the next 𝑘-th step.
𝑖
Step 𝑘. For each ∑𝑘−2
𝑖=0 𝑎𝑖 (𝑥)𝑝 (𝑥) ∈ 𝑇𝑘−2 (𝐹), we can replace 𝑦
𝑘−1
by 𝑧𝑝 (𝑥) + ⋅ ⋅ ⋅ + 𝑎0 (𝑥) in (23), and then (23) becomes
𝐹 (𝑥, 𝑧𝑝𝑘−1 (𝑥) + ⋅ ⋅ ⋅ + 𝑎0 (𝑥)) = 𝑐𝑝𝑚 (𝑥) .
Hence, we solve all solutions 𝑦(𝑥) such that 𝐻2 (𝑥, 𝑦) = 0 to
obtain a solution set
𝐹3 (𝑥, 𝑦) = 𝑐𝑝𝑚−ℓ3 (𝑥) .
Step 3.2.2. If ℓ3 < 𝑚, we use 𝑝(𝑥) to divide both sides of (48)
to obtain
(55)
We let 𝐻𝑘−1 (𝑥, 𝑧) = 𝐹(𝑥, 𝑧𝑝𝑘−1 (𝑥) + ⋅ ⋅ ⋅ + 𝑎0 (𝑥)) and
𝐻𝑘−1 (𝑥, 𝑦) = (cont𝑦 𝐻𝑘−1 ) 𝐹𝑘 (𝑥, 𝑦)
(56)
for a primitive polynomial 𝐹𝑘 (𝑥, 𝑦) ∈ R[𝑥, 𝑦], then (55)
becomes
𝐻𝑘−1 (𝑥, 𝑦) = 𝑐𝑝𝑚 (𝑥) .
(57)
Note that in Section 4, Lemma 9, we will prove that 𝑝𝑘−1 (𝑥) |
cont𝑦 𝐻𝑘−1 .
Step k.1. If cont𝑦 𝐻𝑘−1 ∤ 𝑝𝑚 (𝑥), by (56) and (57), we have
cont𝑦 𝐻𝑘−1 | 𝑐 ⋅ 𝑝𝑚 (𝑥) ,
since cont𝑦 𝐻𝑘−1 ∤ 𝑝𝑚 (𝑥), we have 𝑐 = 0.
(58)
Abstract and Applied Analysis
5
Thus (57) becomes
𝐻𝑘−1 (𝑥, 𝑦) = 0.
(59)
Hence, we solve all solutions 𝑦(𝑥) such that 𝐻𝑘−1 (𝑥, 𝑦) = 0 to
obtain a solution set
𝑘−2
𝑌𝑘−1 = {𝑦 (𝑥) 𝑝𝑘−1 + ∑ 𝑎𝑖 (𝑥) 𝑝𝑖 (𝑥) : 𝐻𝑘−1 (𝑥, 𝑦 (𝑥)) = 0,
𝑖=0
𝑘−2
∑ 𝑎𝑖 (𝑥) 𝑝𝑖 (𝑥) ∈ 𝑇𝑘−2 (𝐹)} .
𝑖=0
(60)
Step k.2. If cont𝑦 𝐻𝑘−1 | 𝑝𝑚 (𝑥), then cont𝑦 𝐻𝑘−1 = 𝑝ℓ𝑘 (𝑥) for
some positive integer ℓ𝑘 ≤ 𝑚 and (57) becomes
𝐹𝑘 (𝑥, 𝑦) = 𝑐𝑝𝑚−ℓ𝑘 (𝑥) .
(61)
(68)
by Corollary 11 and let
𝐻𝑚 (𝑥, 𝑦) = 𝑝𝑚 (𝑥) 𝐹𝑚+1 (𝑥, 𝑦) .
(69)
(62)
𝐹𝑚+1 (𝑥, 𝑦 (𝑥)) = 𝑐.
(63)
Moreover, by Algorithm 3, we can get a solution set 𝑌 such
that the bound of all solutions is “𝑠⋅|𝑇𝑚−1 (𝐹)|.” So the solution
set of (3) is contained in
By Algorithm 3, we compute 𝑦 = 𝑦(𝑥) and obtain
𝑦 (𝑥) 𝑝𝑘−1 (𝑥) + 𝑎𝑘−2 (𝑥) 𝑝𝑘−2 (𝑥) + ⋅ ⋅ ⋅ + 𝑎0 (𝑥)
𝑝𝑚 (𝑥) | 𝐻𝑚 (𝑥, 𝑦)
Thus, the possible solutions of 𝐻𝑚 (𝑥, 𝑦) = 𝑐𝑝𝑚 (𝑥) are those
solutions 𝑦(𝑥) satisfying
Step k.2.1. If ℓ𝑘 = 𝑚, then (61) becomes
𝐹𝑘 (𝑥, 𝑦) = 𝑐.
Remark 8. (i) If 𝑉𝑖 = 0, then 𝑉𝑖+1 = 0 and if 𝑉𝑖+1 = 0, then
𝑇𝑖+1 (𝐹) = 0 for 𝑖 ≥ 0. This means that if 𝑉𝑖 = 0, then the work
finishes at step 𝑖 + 1 for 𝑖 ≥ 0.
(ii) This algorithm is not an infinite loop. Since ℓ𝑘 < 𝑚,
and by Corollary 13, we have ℓ𝑘 < ℓ𝑘+1 for 𝑘 = 1, 2, . . ., thus
this algorithm will terminate at the (𝑚 + 1)-th Step. We can
thus say that this algorithm is a finite loop algorithm.
(iii) We may assume that the cardinal number |𝑊𝑖 | < ∞,
for 𝑖 = 1, 2, . . .; however, if |𝑊𝑖 | = ∞, then Algorithm 5
has an infinite number of solutions, which contradicts our
assumption.
Remark 8(ii) shows that the maximum number of steps
in this algorithm is 𝑚 + 1. Moreover, if the (𝑚 + 1)-th Step
happens, then we obtain
in 𝑊𝑘−1 , for all 𝑎𝑘−2 (𝑥)𝑝𝑘−2 (𝑥) + ⋅ ⋅ ⋅ + 𝑎0 (𝑥) ∈ 𝑇𝑘−2 (𝐹). Hence,
we output
𝑊𝑘−1 = {𝑦 (𝑥) 𝑝𝑘−1 (𝑥) + ⋅ ⋅ ⋅ + 𝑎0 (𝑥) : 𝐹𝑘 (𝑥, 𝑦 (𝑥)) = 𝑐,
𝑘−2
𝑚−1
𝑚−1
𝑖=0
𝑖=0
(70)
𝑇 = ( ⋃ 𝑌𝑖 ) ⋃ ( ⋃ 𝑊𝑖 ) ⋃ 𝑌.
Finally, we check each element 𝑢(𝑥) ∈ 𝑇 to determine
whether or not
𝐹 (𝑥, 𝑢 (𝑥)) = 𝑐𝑝𝑚 (𝑥) .
∑ 𝑎𝑖 (𝑥) 𝑝𝑖 (𝑥) ∈ 𝑇𝑘−2 (𝐹)} .
𝑖=0
(64)
(71)
(72)
The arithmetic computing time to solve each element 𝑢(𝑥)
in 𝑇 only requires non-NP-complete computing time and to
check that each 𝑢(𝑥) ∈ 𝑇 satisfies
Step k.2.2. If ℓ𝑘 < 𝑚, we use 𝑝(𝑥) to divide both sides of (61)
to obtain
𝐹 (𝑥, 𝑢 (𝑥)) = 𝑐𝑝𝑚 (𝑥) .
𝐹𝑘 (𝑥, 𝑦) = 0 (mod 𝑝 (𝑥)) .
Non-NP-complete computing time is also needed given the
cardinal number of 𝑇:
(65)
By the same reasoning used in (Step 1.2.2), we can solve 𝑦 =
𝑎𝑘−1 (𝑥) with deg 𝑎𝑘−1 (𝑥) < deg 𝑝(𝑥) and obtain the set
𝑎𝑘−1 (𝑥) ∈ 𝑉𝑘−1 = {𝑎1𝑘−1 (𝑥) , 𝑎2𝑘−1 (𝑥) , . . . , 𝑎𝑟𝑘−1
(𝑥)} . (66)
𝑘−1
We let
(73)
𝑚−2
𝑚−2
𝑖=0
𝑖=0
󵄨 󵄨 󵄨 󵄨
󵄨
󵄨
󵄨
󵄨
|𝑇| ≤ 󵄨󵄨󵄨𝑊0 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑌0 󵄨󵄨󵄨 + |𝑌| + ∑ 󵄨󵄨󵄨𝑊𝑖+1 󵄨󵄨󵄨 + ∑ 󵄨󵄨󵄨𝑌𝑖+1 󵄨󵄨󵄨
(By Algorithm 3 and Lemma 7)
(74)
𝑚−2
𝑘−2
󵄨
󵄨
󵄨
󵄨
≤ 2𝑠 + 𝑠 ⋅ 󵄨󵄨󵄨𝑇𝑚−1 (𝐹)󵄨󵄨󵄨 + 2 ∑ 𝑠 ⋅ 󵄨󵄨󵄨𝑇𝑖 (𝐹)󵄨󵄨󵄨 ,
𝑖
𝑇𝑘−1 (𝐹) = { ∑ 𝑎𝑖 (𝑥) 𝑝 (𝑥) : 𝑎𝑘−1 (𝑥) ∈ 𝑉𝑘−1 ,
𝑖=0
𝑘−2
𝑖=0
(67)
∑ 𝑎𝑖 (𝑥) 𝑝𝑖 (𝑥) ∈ 𝑇𝑘−2 (𝐹)},
𝑖=0
thus expression (57) can go to the next step.
As explained in Remark 8, this algorithm is not an infinite
loop.
where |𝑇𝑖 (𝐹)|, |𝑊𝑖 |, and |𝑌𝑖 | denote the cardinal number of
𝑇𝑖 (𝐹), 𝑊𝑖 , and 𝑌𝑖 . The remaining work “Checks the bound of
|𝑇𝑖 (𝐹)|, for each 𝑖 = 1, 2, . . . , 𝑚.” In Section 4, Theorem 16, we
prove that
󵄨
󵄨󵄨
(75)
󵄨󵄨𝑇𝑖 (𝐹)󵄨󵄨󵄨 ≤ 𝑠
for 𝑖 = 1, 2, . . . , 𝑚 − 1.
6
Abstract and Applied Analysis
(where 𝑄𝑖 (𝑥, 𝑧)
By (74) and (75), we have the cardinal number
= 𝑧𝑖 𝑝𝑖−1 (𝑥) + 𝑖𝑧𝑖−1 𝑝𝑖−2 (𝑥) 𝑎 (𝑥) + ⋅ ⋅ ⋅ + 𝑖𝑧𝑎𝑖−1 (𝑥))
𝑚−2
󵄨
󵄨
󵄨
󵄨
|𝑇| ≤ 2𝑠 + 𝑠 ⋅ 󵄨󵄨󵄨𝑇𝑚−1 (𝐹)󵄨󵄨󵄨 + 2 ∑ 𝑠 ⋅ 󵄨󵄨󵄨𝑇𝑖 (𝐹)󵄨󵄨󵄨
= 𝑝 (𝑥) (𝑓𝑠 (𝑥) 𝑄𝑠 (𝑥, 𝑧) + 𝑓𝑠−1 (𝑥) 𝑄𝑠−1 (𝑥, 𝑧)
𝑖=0
𝑚−2
(76)
≤ 2𝑠 + 𝑠2 + 2 ∑ 𝑠2
+ (𝑓𝑠 (𝑥) 𝑎𝑠 (𝑥) + 𝑓𝑠−1 (𝑥) 𝑎𝑠−1 (𝑥) + ⋅ ⋅ ⋅ + 𝑓0 (𝑥))
𝑖=0
= 2𝑠 + (2𝑚 − 1) 𝑠2 .
= 𝑝 (𝑥) (𝑓𝑠 (𝑥) 𝑄𝑠 (𝑥, 𝑧) + 𝑓𝑠−1 (𝑥) 𝑄𝑠−1 (𝑥, 𝑧)
That is, we have to check at most “2𝑠 + (2𝑚 − 1)𝑠2 ” solutions
𝑢(𝑥) to satisfy
𝐹 (𝑥, 𝑢 (𝑥)) = 𝑐𝑝𝑚 (𝑥) .
(77)
Checking each solution only takes non-NP-complete time,
thus this algorithm needs non-NP-complete time. That
is, Algorithm 5 is indeed a non-NP-complete time algorithm. Note that, for each 𝑘-th Step, whether 𝑦(𝑥)𝑝𝑘−1 (𝑥) +
𝑎𝑘−2 𝑝𝑘−2 (𝑥) + ⋅ ⋅ ⋅ + 𝑎0 (𝑥) belongs to 𝑌𝑘−1 or 𝑊𝑘−1 , the upper
bound of all elements in 𝑇 may not be as large; in fact,
|𝑇| ≤ 2𝑠 + 𝑠2 + (𝑚 − 1) 𝑠2 = 2𝑠 + 𝑚𝑠2 .
(78)
In the next Section 4, we will complete all necessary
properties for this algorithm.
For convenience, we describe some interesting properties of
the above algorithm.
Lemma 9. Let 𝐹(𝑥, 𝑦) ∈ R[𝑥, 𝑦], 𝑝(𝑥), 𝑎(𝑥) ∈ R[𝑥], and 𝑧
be a parameter, if 𝑝(𝑥) | 𝐹(𝑥, 𝑎(𝑥)), then 𝑝(𝑥) | 𝐹(𝑥, 𝑧𝑝(𝑥) +
𝑎(𝑥)).
Proof. Let 𝐹(𝑥, 𝑦) = 𝑓𝑠 (𝑥)𝑦𝑠 + 𝑓𝑠−1 (𝑥)𝑦𝑠−1 + ⋅ ⋅ ⋅ + 𝑓0 (𝑥). Then
Thus, 𝑝(𝑥) | 𝐹(𝑥, 𝑎(𝑥)) ⇒ 𝑝(𝑥) | 𝐹(𝑥, 𝑧𝑝(𝑥) + 𝑎(𝑥)).
The definitions of 𝐻𝑖 (𝑥, 𝑦), cont𝑦 𝐻𝑖 , 𝐹𝑖+1 (𝑥, 𝑦), and 𝑎𝑖 (𝑥)
for 𝑖 = 1, 2, . . . , 𝑚 + 1 in Algorithm 5 are
𝐻𝑘 (𝑥, 𝑧) = 𝐹 (𝑥, 𝑧𝑝𝑘 (𝑥) + ⋅ ⋅ ⋅ + 𝑎0 (𝑥)) ,
(80)
𝐻𝑘 (𝑥, 𝑦) = (cont𝑦 𝐻𝑘 ) 𝐹𝑘+1 (𝑥, 𝑦) ,
(81)
for a primitive polynomial 𝐹𝑘+1 (𝑥, 𝑦) ∈ R[𝑥, 𝑦] and 𝑝(𝑥) |
𝐹𝑘+1 (𝑥, 𝑎𝑘 (𝑥)) for 𝑘 = 0, 1, . . . , 𝑚.
Next, we prove some properties about 𝐻𝑘 (𝑥, 𝑦) for 𝑘 =
1, 2, . . . , 𝑚 + 1.
Lemma 10. Consider 𝑝(𝑥)cont𝑦 𝐻𝑘 | cont𝑦 𝐻𝑘+1 , for 1 ≤ 𝑘 ≤
𝑚.
Proof. From (80), we have
= 𝐹 (𝑥, (𝑧𝑝 (𝑥) + 𝑎𝑘 (𝑥)) 𝑝𝑘 (𝑥) + ⋅ ⋅ ⋅ + 𝑎0 (𝑥))
= 𝐻𝑘 (𝑥, 𝑧𝑝 (𝑥) + 𝑎𝑘 (𝑥))
(by (80))
= (cont𝑧 𝐻𝑘 ) 𝐹𝑘+1 (𝑥, 𝑧𝑝 (𝑥)+𝑎𝑘 (𝑥))
+ 𝑓𝑠−1 (𝑥) (𝑧𝑝 (𝑥) + 𝑎 (𝑥))
(by (81)) .
(82)
By definition of 𝑎𝑘 (𝑥), we have 𝑝(𝑥) | 𝐹𝑘+1 (𝑥, 𝑎𝑘 (𝑥)). From
Lemma 9, we obtain 𝑝(𝑥) | 𝐹𝑘+1 (𝑥, 𝑧𝑝(𝑥) + 𝑎𝑘 (𝑥)). That is,
(83)
and, by (81), we obtain 𝑝(𝑥)cont𝑦 𝐻𝑘 | cont𝑦 𝐻𝑘+1 .
𝑠
= 𝑓𝑠 (𝑥) (𝑧𝑝 (𝑥) + 𝑎 (𝑥))
𝑠−1
+ ⋅ ⋅ ⋅ + 𝑓0 (𝑥)
= 𝑓𝑠 (𝑥) (𝑧𝑠 𝑝𝑠 (𝑥) + 𝑠𝑧𝑠−1 𝑝𝑠−1 𝑎 (𝑥) + ⋅ ⋅ ⋅ + 𝑎𝑠 (𝑥))
+ 𝑓𝑠−1 (𝑥) (𝑧𝑠−1 𝑝𝑠−1 (𝑥) + (𝑠 − 1)
𝑝
(79)
𝑝 (𝑥) cont𝑦 𝐻𝑘 | 𝐻𝑘+1 (𝑥, 𝑧) ,
𝐹 (𝑥, 𝑧𝑝 (𝑥) + 𝑎 (𝑥))
𝑧
+ ⋅ ⋅ ⋅ + 𝑓1 (𝑥) 𝑄1 (𝑥, 𝑧)) + 𝐹 (𝑥, 𝑎 (𝑥)) .
𝐻𝑘+1 (𝑥, 𝑧) = 𝐹 (𝑥, 𝑧𝑝𝑘+1 (𝑥) + 𝑎𝑘 (𝑥) 𝑝𝑘 (𝑥) + ⋅ ⋅ ⋅ + 𝑎0 (𝑥))
4. Main Theorems
𝑠−2 𝑠−2
+ ⋅ ⋅ ⋅ + 𝑓1 (𝑥) 𝑄1 (𝑥, 𝑧))
𝑠−1
𝑎 (𝑥) + ⋅ ⋅ ⋅ + 𝑎
(𝑥)) + ⋅ ⋅ ⋅ + 𝑓0 (𝑥)
= 𝑓𝑠 (𝑥) (𝑄𝑠 (𝑥, 𝑧) 𝑝 (𝑥) + 𝑎𝑠 (𝑥)) + 𝑓𝑠−1 (𝑥)
× (𝑄𝑠−1 (𝑥, 𝑧) 𝑝 (𝑥) + 𝑎𝑠−1 (𝑥)) + ⋅ ⋅ ⋅ + 𝑓0 (𝑥)
By Lemma 9, 𝑝(𝑥) | 𝐻1 (𝑥, 𝑧), and the definition of
cont𝑦 𝐻1 , it follows that 𝑝(𝑥) | cont𝑦 𝐻1 . Moreover, by
Lemma 10, we can easily obtain the following corollaries.
Corollary 11. Consider 𝑝𝑖 (𝑥) | cont𝑦 𝐻𝑖 , for 𝑖 ≥ 1.
To complete the main theorem, we give the following
definitions.
Definition 12. (i) The function 𝑓(𝑥) is said to have 𝑝(𝑥)power ℓ, denoted by 𝛿𝑝 (𝑓(𝑥)) = ℓ, if 𝑝(𝑥)ℓ | 𝑓(𝑥) and
𝑝(𝑥)ℓ+1 ∤ 𝑓(𝑥).
Abstract and Applied Analysis
7
(ii) We say that 𝐹(𝑥, 𝑦) has 𝑝(𝑥)-power ℓ, if cont𝑦 𝐹 has
𝑝(𝑥)-power ℓ.
The following corollary is an immediate result of
Lemma 10 and Definition 12.
The only possible solution 𝑎0 (𝑥) with 0 ≤ deg 𝑎0 (𝑥) <
deg 𝑝(𝑥) and 𝑝(𝑥)cont𝑦 𝐹(𝑥, 𝑧) | 𝐹(𝑥, 𝑎0 (𝑥)) is
Corollary 13. Consider ℓ𝑖 < ℓ𝑖+1 , for 𝑖 ≥ 1.
That is, |𝑇0 (𝐹)| ≤ 1.
(2) Next, we assume that |𝑇𝑘−1 (𝐹)| ≤ 1, thus we deduce
that |𝑇𝑘 (𝐹)| ≤ 1.
Let
Proof. Since 𝛿𝑝 (𝐻𝑖 ) = ℓ𝑖+1 and by Lemma 10, we have
𝑝 (𝑥) cont𝑦 𝐻𝑘 | cont𝑦 𝐻𝑘+1 .
−1
𝑎0 (𝑥) = (𝑔10 (𝑥)) (𝑔00 (𝑥)) mod 𝑝 (𝑥) .
(84)
V𝑘 = 𝛿𝑝 (𝐹 (𝑥, 𝑧𝑝𝑘 (𝑥) + 𝑏𝑘−1 (𝑥)))
It follows that
1 + 𝛿𝑝 (𝐻𝑖 ) = 𝛿𝑝 (𝑝 (𝑥) 𝐻𝑖 ) ≤ 𝛿𝑝 (𝐻𝑖+1 ) ,
(85)
so 1 + ℓ𝑖 ≤ ℓ𝑖+1 ; hence, ℓ𝑖 < ℓ𝑖+1 .
Next, we obtain some properties of 𝑝(𝑥)-power as follows.
Lemma 14. Let 𝑓(𝑥), 𝑔(𝑥), ℎ(𝑥) ∈ R[𝑥] and 𝐹(𝑥, 𝑦) is defined
in (4). Then
(i) 𝛿𝑝 (𝑓𝑔) = 𝛿𝑝 (𝑓) + 𝛿𝑝 (𝑔),
(ii) 𝛿𝑝 (𝑓 + 𝑔) ≥ min{𝛿𝑝 (𝑓), 𝛿𝑝 (𝑔)}, and
(iii) 𝛿𝑝 (𝐹) = min0≤𝑖≤𝑠 {𝛿𝑝 (𝑓𝑖 )}.
(iv) If ℎ = 𝑓 + 𝑔 and 𝛿𝑝 (ℎ) < 𝛿𝑝 (𝑔), then 𝛿𝑝 (ℎ) = 𝛿𝑝 (𝑓).
For each 𝑘 ∈ N, we rewrite 𝑇𝑘 (𝐹) defined in the
procedures of Algorithm 5 as the following definitions.
for some 𝑏𝑘−1 (𝑥) ∈ 𝑇𝑘−1 (𝐹) ,
𝑇𝑘 (𝐹) = {𝑏𝑘 (𝑥) ∈ R [𝑥] : 𝑏𝑘 (𝑥) = 𝑎 (𝑥) 𝑝𝑘 (𝑥) + 𝑏𝑘−1 (𝑥)
satisfies (𝐷1) and (𝐷2) } ,
(89)
and since |𝑇𝑘−1 (𝐹)| ≤ 1 by assumption, 𝑏𝑘−1 (𝑥) is uniquely
decided, as is V𝑘 . The remaining task is to prove that 𝑎𝑘 is
also uniquely determined. As 𝑏𝑘 (𝑥) ∈ 𝑇𝑘 (𝐹) with 𝑏𝑘 (𝑥) =
𝑎𝑘 (𝑥)𝑝𝑘 (𝑥) + 𝑏𝑘−1 (𝑥), 𝑎𝑘 (𝑥) satisfies
0 ≤ deg 𝑎𝑘 (𝑥) < deg 𝑝 (𝑥)
From the above definitions, we obtain a bound of the
cardinal number |𝑇𝑘 (𝐹)|.
Theorem 16. For each 𝑘 ∈ N, then |𝑇𝑘 (𝐹)| ≤ deg𝑦 𝐹 (|⋅| means
the cardinal number).
Proof. This theorem will be proven by induction on deg𝑦 𝐹.
(i) When deg𝑦 𝐹 = 1, claim that |𝑇𝑘 (𝐹)| ≤ 1 for any 𝑘 ∈ N.
For this purpose, we consider
𝐹 (𝑥, 𝑦) = 𝑓1 (𝑥) 𝑦 + 𝑓0 (𝑥) for some 𝑓1 (𝑥) , 𝑓0 (𝑥) ∈ R [𝑥] .
(86)
(1) If 𝑘 = 0, and let V0 = 𝛿𝑝 (𝐹(𝑥, 𝑧)) and 𝐹(𝑥, 𝑦) =
𝑓1 (𝑥)𝑦 + 𝑓0 (𝑥) = 𝑝V0 (𝑥)(𝑔10 (𝑥)𝑦 + 𝑔00 (𝑥)) for some coprime
polynomials 𝑔10 (𝑥), 𝑔00 (𝑥) ∈ R[𝑥].
(90)
such that
𝑝 (𝑥) cont𝑦 𝐹 (𝑥, 𝑧𝑝𝑘 (𝑥) + 𝑏𝑘−1 (𝑥)) |
𝐹 (𝑥, 𝑎𝑘 (𝑥) 𝑝𝑘 (𝑥) + 𝑏𝑘−1 (𝑥))
(1) 𝑇0 (𝐹) = {𝑎(𝑥) ∈ R[𝑥] : 𝑝(𝑥)cont𝑧 𝐹(𝑥, 𝑧) | 𝐹(𝑥, 𝑎(𝑥))
with 0 ≤ deg 𝑎(𝑥) < deg 𝑝(𝑥)},
(D1): 0 ≤ deg 𝑎(𝑥) < deg 𝑝(𝑥) and 𝑏𝑘−1 (𝑥) ∈ 𝑇𝑘−1 (𝐹),
(D2): 𝑝(𝑥)cont𝑧 𝐹(𝑥, 𝑧𝑝𝑘 (𝑥) + 𝑏(𝑥)) | 𝐹(𝑥, 𝑎(𝑥)𝑝𝑘 (𝑥) +
𝑏(𝑥)).
(88)
and 𝐹(𝑥, 𝑧𝑝𝑘 (𝑥) + 𝑏𝑘−1 (𝑥)) = 𝑝V𝑘 (𝑥)(𝑔1𝑘 (𝑥)𝑧 + 𝑔0𝑘 (𝑥)) for some
𝑔1𝑘 (𝑥), 𝑔0𝑘 (𝑥) ∈ R[𝑥]. Consider
Definition 15. Let 𝐹(𝑥, 𝑦) ∈ R[𝑥, 𝑦], an irreducible polynomial 𝑝(𝑥) ∈ R[𝑥], and 𝑧 be a parameter. We denote the sets
𝑇𝑖 (𝐹), 𝑖 ∈ N defined in Algorithm 5 as follows:
(2) 𝑇𝑘 (𝐹) = {𝑏𝑘 (𝑥) ∈ R[𝑥] : 𝑏𝑘 (𝑥) = 𝑎(𝑥)𝑝𝑘 (𝑥) + 𝑏𝑘−1 (𝑥)
satisfies (𝐷1) and (𝐷2)} where for 𝑘 ≥ 1, and the (𝐷1)
and (𝐷2) properties are given by the following:
(87)
(91)
for some 𝑏𝑘−1 (𝑥) ∈ 𝑇𝑘−1 (𝐹) ,
then
𝑎𝑘 (𝑥) 𝑔1𝑘 (𝑥) + 𝑔0𝑘 (𝑥) = 0 mod 𝑝 (𝑥) ,
(92)
it follows that 𝑎𝑘 (𝑥) must be
−1
𝑎𝑘 (𝑥) = (𝑔1𝑘 (𝑥)) (𝑔0𝑘 (𝑥)) mod 𝑝 (𝑥) .
(93)
Consequently, |𝑇𝑘 (𝐹)| ≤ 1.
By (1) and (2), we get that |𝑇𝑘 (𝐹)| ≤ 1 = deg𝑦 𝐹 for any
𝑘 ∈ N.
(ii) We will prove this theorem by mathematical induction. Assume that
󵄨
󵄨󵄨
󵄨󵄨𝑇𝑘 (𝐹1 )󵄨󵄨󵄨 ≤ deg𝑦 𝐹1 ,
(94)
for any 𝐹1 ∈ R[𝑥] with deg𝑦 𝐹1 ≤ 𝑛 − 1, we want to show
󵄨
󵄨󵄨
󵄨󵄨𝑇𝑘 (𝐹)󵄨󵄨󵄨 ≤ deg𝑦 𝐹,
(95)
for any 𝐹 ∈ R[𝑥] with deg𝑦 𝐹 = 𝑛.
Let 𝑢 = max{𝛿𝑝 (𝐹(𝑥, 𝑐(𝑥))) : 𝑐(𝑥) ∈ 𝑇𝑘 (𝐹)} and choose
𝑑(𝑥) ∈ 𝑇𝑘 (𝐹) with
𝛿𝑝 (𝐹 (𝑥, 𝑑 (𝑥))) = 𝑢.
(96)
8
Abstract and Applied Analysis
Indeed,
and by (D2) in Definition 15, we have
𝛿𝑝 (𝐹 (𝑥, 𝑑 (𝑥))) ≥ 𝛿𝑝 (𝐹 (𝑥, 𝑐 (𝑥)))
for any 𝑐 (𝑥) ∈ 𝑇𝑘 (𝐹) .
(97)
By the division rule, let 𝐹(𝑥, 𝑦) be divided by 𝑦 − 𝑑(𝑥), giving
us
𝑘
𝛿𝑝 (𝐹 (𝑥, 𝑐 (𝑥))) > 𝛿𝑝 (𝐹 (𝑥, 𝑧(𝑝 (𝑥)) + 𝑏1 (𝑥))) .
By (97) and (103), we have
𝛿𝑝 𝐹 (𝑥, 𝑑 (𝑥)) ≥ 𝛿𝑝 (𝐹 (𝑥, 𝑐 (𝑥)))
𝑘
𝐹 (𝑥, 𝑦) = (𝑦 − 𝑑 (𝑥)) 𝐹1 (𝑥, 𝑦) + 𝐹 (𝑥, 𝑑 (𝑥)) ,
> 𝛿𝑝 (𝐹 (𝑥, 𝑧(𝑝 (𝑥)) + 𝑏1 (𝑥))) .
(98)
𝑘
𝛿𝑝 (𝐹 (𝑥, 𝑧(𝑝 (𝑥)) + 𝑏1 (𝑥)))
(99)
For any 𝑐(𝑥) ∈ 𝑇𝑘 (𝐹) − {𝑑(𝑥)}, by the definition of 𝑇𝑘 (𝐹),
we have
𝑘
= 𝛿𝑝 ((𝑧(𝑝 (𝑥)) + 𝑏1 (𝑥) − 𝑑 (𝑥))
𝑘
× 𝐹1 (𝑥, 𝑧(𝑝 (𝑥)) + 𝑏1 (𝑥)) + 𝐹 (𝑥, 𝑑 (𝑥)))
(by (98))
𝑘
𝑐 (𝑥) = 𝑎1 (𝑥) (𝑝 (𝑥)) + 𝑏1 (𝑥) ,
𝑘
(104)
Moreover, we have
for some 𝐹1 (𝑥, 𝑦) ∈ R[𝑥, 𝑦] and deg𝑦 𝐹1 (𝑥, 𝑦) = 𝑛 − 1.
We want to show that
if 𝑐 (𝑥) ∈ 𝑇𝑘 (𝐹) − {𝑑 (𝑥)} , then 𝑐 (𝑥) ∈ 𝑇𝑘 (𝐹1 ) .
(103)
(100)
𝑑 (𝑥) = 𝑎2 (𝑥) (𝑝 (𝑥)) + 𝑏2 (𝑥) ,
for some 𝑏1 (𝑥), 𝑏2 (𝑥) ∈ 𝑇𝑘−1 (𝐹) and 0 ≤ deg 𝑎1 (𝑥),
deg 𝑎2 (𝑥) < deg 𝑝(𝑥). Since the elements 𝑑(𝑥) and 𝑐(𝑥) are
distinct, so 𝑎1 (𝑥) ≠ 𝑎2 (𝑥) or 𝑏1 (𝑥) ≠ 𝑏2 (𝑥),
(a) If 𝑏1 (𝑥) ≠ 𝑏2 (𝑥), by the definition of 𝑏1 (𝑥) and 𝑏2 (𝑥), we
have deg 𝑏1 (𝑥), deg 𝑏2 (𝑥) < deg 𝑝𝑘 (𝑥), so
𝛿𝑝 (𝑏1 (𝑥) − 𝑏2 (𝑥)) < 𝑘 ≤ 𝛿𝑝 ((𝑎1 (𝑥) − 𝑎2 (𝑥))𝑝𝑘 (𝑥)) .
(101)
𝑘
= 𝛿𝑝 ((𝑧(𝑝 (𝑥)) + 𝑏1 (𝑥) − 𝑑 (𝑥))
𝑘
× 𝐹1 (𝑥, 𝑧(𝑝 (𝑥)) + 𝑏1 (𝑥)))
(by (104) and Lemma 14, (iv))
𝑘
= 𝛿𝑝 (𝑧(𝑝 (𝑥)) + 𝑏1 (𝑥) − 𝑑 (𝑥))
𝑘
+ 𝛿𝑝 (𝐹1 (𝑥, 𝑧(𝑝 (𝑥)) + 𝑏1 (𝑥)))
(by Lemma 14, (i)) ,
(105)
𝛿𝑝 (𝐹 (𝑥, 𝑐 (𝑥)))
Therefore, one has the following:
𝑘
(i) 𝛿𝑝 (𝑐(𝑥) − 𝑑(𝑥)) = 𝛿𝑝 ((𝑎1 (𝑥) − 𝑎2 (𝑥))𝑝 (𝑥) + (𝑏1 (𝑥) −
𝑏2 (𝑥))) = 𝛿𝑝 (𝑏1 (𝑥) − 𝑏2 (𝑥)) (by Lemma 14, (iv)),
(ii) 𝛿𝑝 (𝑏1 (𝑥) − 𝑑(𝑥)) = 𝛿𝑝 ((−𝑎2 (𝑥))𝑝𝑘 (𝑥) + (𝑏1 (𝑥) −
𝑏2 (𝑥))) = 𝛿𝑝 (𝑏1 (𝑥) − 𝑏2 (𝑥)) (by Lemma 14, (iv)),
(iii) 𝛿𝑝 (𝑧(𝑝(𝑥))𝑘 + 𝑏1 (𝑥) − 𝑑(𝑥)) = 𝛿𝑝 ((𝑧 − 𝑎2 (𝑥))𝑝𝑘 (𝑥) +
(𝑏1 (𝑥) − 𝑏2 (𝑥))) = 𝛿𝑝 (𝑏1 (𝑥) − 𝑏2 (𝑥)) (by Lemma 14,
(iv)).
(b) if 𝑏1 (𝑥) = 𝑏2 (𝑥), then 𝑎1 (𝑥) ≠ 𝑎2 (𝑥), since 0 ≤ deg 𝑎1 (𝑥),
deg 𝑎2 (𝑥) < deg 𝑝(𝑥), so
(i) 𝛿𝑝 (𝑐(𝑥) − 𝑑(𝑥)) = 𝛿𝑝 ((𝑎1 (𝑥) − 𝑎2 (𝑥))𝑝𝑘 (𝑥))
𝛿𝑝 (𝑎1 (𝑥) − 𝑎2 (𝑥)) + 𝑘 = 𝑘 (by Lemma 14, (i)),
=
= 𝛿𝑝 ((𝑐 (𝑥) − 𝑑 (𝑥)) 𝐹1 (𝑥, 𝑐 (𝑥)) + 𝐹 (𝑥, 𝑑 (𝑥)))
= 𝛿𝑝 ((𝑐 (𝑥) − 𝑑 (𝑥)) 𝐹1 (𝑥, 𝑐 (𝑥)))
= 𝛿𝑝 (𝑐 (𝑥) − 𝑑 (𝑥)) + 𝛿𝑝 (𝐹1 (𝑥, 𝑐 (𝑥)))
(by Lemma 14, (i)) .
From expression (106), we obtain
𝛿𝑝 (𝑐 (𝑥) − 𝑑 (𝑥)) + 𝛿𝑝 (𝐹1 (𝑥, 𝑐 (𝑥)))
= 𝛿𝑝 (𝐹 (𝑥, 𝑐 (𝑥)))
𝑘
(ii) 𝛿𝑝 (𝑏1 (𝑥) − 𝑑(𝑥)) = 𝛿𝑝 ((−𝑎2 (𝑥))𝑝𝑘 (𝑥)) = 𝛿𝑝 (𝑎2 (𝑥)) +
𝑘 = 𝑘 (by Lemma 14, (i)),
> 𝛿𝑝 (𝐹 (𝑥, 𝑧(𝑝 (𝑥)) + 𝑏1 (𝑥)))
(iii) 𝛿𝑝 (𝑧(𝑝(𝑥))𝑘 + 𝑏1 (𝑥) − 𝑑(𝑥)) = 𝛿𝑝 ((𝑧 − 𝑎2 (𝑥))𝑝𝑘 (𝑥) +
(𝑏1 (𝑥)−𝑏2 (𝑥))) = 𝛿𝑝 (𝑧−𝑎2 (𝑥))+𝑘 = 𝑘 (by Lemma 14,
(i)).
= 𝛿𝑝 (𝑧(𝑝(𝑥)) + 𝑏1 (𝑥) − 𝑑 (𝑥))
By the above equations, we obtain
𝛿𝑝 (𝑐 (𝑥) − 𝑑 (𝑥)) = 𝛿𝑝 (𝑏1 (𝑥) − 𝑑 (𝑥))
𝑘
= 𝛿𝑝 (𝑧(𝑝 (𝑥)) + 𝑏1 (𝑥) − 𝑑 (𝑥)) .
(102)
(106)
(by (104) and Lemma 14, (iv) )
(by (104))
(107)
𝑘
𝑘
(by (105)) .
+ 𝛿𝑝 (𝐹1 (𝑥, 𝑧(𝑝 (𝑥)) + 𝑏1 (𝑥)))
From expression (102), that is, 𝛿𝑝 (𝑐(𝑥)−𝑑(𝑥)) = 𝛿𝑝 (𝑧(𝑝(𝑥))𝑘 +
𝑏1 (𝑥) − 𝑑(𝑥)) and by canceling 𝛿𝑝 (𝑐(𝑥) − 𝑑(𝑥)) to both sides
of (107), it follows that
𝑘
𝛿𝑝 (𝐹1 (𝑥, 𝑐 (𝑥))) > 𝛿𝑝 (𝐹1 (𝑥, 𝑧(𝑝 (𝑥)) + 𝑏1 (𝑥))) . (108)
Abstract and Applied Analysis
9
That is, 𝑐(𝑥) ∈ 𝑇𝑘 (𝐹1 ). Then we have 𝑇𝑘 (𝐹) ⊆ 𝑇𝑘 (𝐹1 ) ⋃{𝑑(𝑥)}.
Therefore,
󵄨󵄨
󵄨
󵄨 󵄨
󵄨󵄨𝑇𝑘 (𝐹)󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨𝑇𝑘 (𝐹1 )󵄨󵄨󵄨 + 1 ≤ (𝑛 − 1) + 1 = 𝑛.
(109)
By induction, the proof is completed.
A quasi-fixed polynomial problem can be employed to many
aspects of actuarial science, risk management, game theory,
and so forth. We give an example from game theory as
follows.
Example 1. Let 𝐴 be an insurance company and 𝐵 a customer.
They make a game as follows.
The companies 𝐴 and 𝐵 enter into an 𝑠-year contract that
within with an interest rate 𝑥 = 𝑖. Assume each year 𝐵 pays
a total of 𝑓𝑘 (𝑖) at 𝑘-th year to the insurance company 𝐴, 0 ≤
𝑘 ≤ 𝑠 and 𝐴 prepare the asset 𝑆 for 𝐵 at the starting time of
this contract. Assume the live inflation rate is 𝑦 = 1 + 𝑟 with
𝑟 = 𝑟(𝑖) dependent on the interest 𝑖. The total amount that 𝐵
pays at the s-th year of this contract obeys the law of (year)
motion about a quasi-fixed polynomial function:
𝐹 (𝑖, 𝑦 (𝑖)) = 𝑓𝑠 (𝑖) (1 + 𝑟)𝑠 + 𝑓𝑠−1 (𝑖) (1 + 𝑟)𝑠−1 + ⋅ ⋅ ⋅ + 𝑓0 (𝑖) .
(110)
Now 𝑥 = 𝑖, 𝑦 = 𝑦(𝑥). 𝑝(𝑥) = 1 + 𝑥 + 𝑥2 is an irreducible
polynomial. When the quasi-fixed polynomial equation
𝐹 (𝑖, 𝑦 (𝑖)) = 𝑆(1 + 𝑖 + 𝑖 ) , 𝑚 ∈ N,
Question. How many dollars do 𝐵 pay to 𝐴 at the beginning of
this contract and what is the relationship between the interest
rate and disaster occurrence in the policy?
Answer. 𝐴 pays 𝑎𝑘 dollars if the disaster occurs at the end of
𝑘-th year, 𝑘 = 1, 2, . . . , 𝑠. Note that
5. Applications (As a Game Problem)
2 𝑚
contract, we assume that 𝐴 gives a bonus interest rate, 𝑖 + 𝑖2
to 𝐵.
(1) the probability of the disaster occurring in the 𝑘-th
year is 𝑞(1 − 𝑞)𝑘 ,
(2) according to time factor, the value of 𝑎𝑘 dollars at the
end of 𝑘-th year is equal to the value of 𝑎𝑘 (1 + 𝑖)𝑠−𝑘
dollars at the end of 𝑠-th year where 𝑠 ≥ 𝑘, so the
amount that 𝐴 is expected to pay to 𝐵 in the 𝑘-th year
is
𝑘
𝑎𝑘 (1 + 𝑖)𝑠−𝑘 𝑞(1 − 𝑞) .
Then the value of total amount that 𝐴 pays 𝐵 at the end of 𝑠-th
year is
𝐹 (𝑖, 𝑞) = 𝑎0 (1 + 𝑖)𝑠 𝑞 + 𝑎1 (1 + 𝑖)𝑠−1 𝑞 (1 − 𝑞) + ⋅ ⋅ ⋅ + 𝑎𝑠 . (114)
This can be rewritten as
(i)
𝐹 (𝑖, 𝑞) = 𝑓𝑠 (𝑖) 𝑞𝑠 + 𝑓𝑠−1 (𝑖) 𝑞𝑠−1 + ⋅ ⋅ ⋅ + 𝑓0 (𝑖) .
then 𝐵 can obtain a premium
𝑎(1 + 𝑖 + 𝑖2 )
𝑠+1
𝑆(1 + 𝑖 + 𝑖 ) .
(112)
The second example concerns actuarial science.
Example 2. Traditionally, the disaster occurrence rate of all
policies issued by general insurance companies is fixed. The
actuary can use a variety of accounting systems to calculate the relationship between the interest rate and disaster
occurrence for the policy. But, in general, measurement is
very time consuming, and it would be preferable to use a
simple mathematical calculation to estimate the relationship
between the interest rate and disaster occurrence. In the
following, we consider an 𝑠-year pure endowment contract.
An insurance company 𝐴 signs a contract to some
company 𝐵 to the effect that if, in 𝑘-th year, a disaster occurs,
𝐴 will pay an amount of money, 𝑎𝑘 , to 𝐵, that 𝑎𝑘 represents
the compensation paid by 𝐴 to 𝐵 at the end of the k-th year,
0 ≤ 𝑘 ≤ 𝑠 and the disaster occurrence rate is 𝑞. In calculating
the insurance fee 𝐵 should pay to 𝐴 at the beginning of the
𝑠+1
.
(116)
To balance each other, we set the values (i) and (ii) as
equal. Thus, the relation equation becomes
𝐹 (𝑖, 𝑞) = 𝑎(1 + 𝑖 + 𝑖2 )
2 𝑚
(115)
Moreover, we assume that the total amount that 𝐵 pays to 𝐴
at the beginning of this contract is 𝑎. Then, by translation of
the time factor, the value of this total amount that 𝐵 pays to
𝐴 by the end of the contract can be said to be
(ii)
(111)
(113)
for some 𝑎 ∈ R,
(117)
where 𝑞 and 𝑖, respectively, denote the disaster occurrence
rate and the interest rate.
We can let 𝑥 = 𝑖, 𝑦 = 𝑞, 𝑝(𝑥) = 1 + 𝑥 + 𝑥2 , and 𝑚 = 𝑠 + 1,
then the above problem can be rewritten as
𝐹 (𝑥, 𝑞) = 𝑎𝑝𝑚 (𝑥) for some 𝑎 ∈ R.
(118)
This example explains a quasi-fixed (polynomial) problem
concerning actuarial science.
References
[1] A. K. Lenstra, “Factoring multivariate polynomials over algebraic number fields,” SIAM Journal on Computing, vol. 16, no. 3,
pp. 591–598, 1987.
[2] S. P. Tung, “Near solutions of polynomial equations,” Acta
Arithmetica, vol. 123, no. 2, pp. 163–181, 2006.
[3] S. P. Tung, “Algorithms for near solutions to polynomial
equations,” Journal of Symbolic Computation, vol. 44, no. 10, pp.
1410–1424, 2009.
10
[4] H.-C. Lai and Y.-C. Chen, “A quasi-fixed polynomial problem
for a polynomial function,” Journal of Nonlinear and Convex
Analysis, vol. 11, no. 1, pp. 101–114, 2010.
[5] S. P. Tung, “Approximate solutions of polynomial equations,”
Journal of Symbolic Computation, vol. 33, no. 2, pp. 239–254,
2002.
[6] J. von zur Gathen and J. Gerhard, Modern Computer Algebra,
Cambridge University Press, Cambridge, UK, 2nd edition,
2003.
Abstract and Applied Analysis
Advances in
Operations Research
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Advances in
Decision Sciences
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Mathematical Problems
in Engineering
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Journal of
Algebra
Hindawi Publishing Corporation
http://www.hindawi.com
Probability and Statistics
Volume 2014
The Scientific
World Journal
Hindawi Publishing Corporation
http://www.hindawi.com
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
International Journal of
Differential Equations
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Volume 2014
Submit your manuscripts at
http://www.hindawi.com
International Journal of
Advances in
Combinatorics
Hindawi Publishing Corporation
http://www.hindawi.com
Mathematical Physics
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Journal of
Complex Analysis
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
International
Journal of
Mathematics and
Mathematical
Sciences
Journal of
Hindawi Publishing Corporation
http://www.hindawi.com
Stochastic Analysis
Abstract and
Applied Analysis
Hindawi Publishing Corporation
http://www.hindawi.com
Hindawi Publishing Corporation
http://www.hindawi.com
International Journal of
Mathematics
Volume 2014
Volume 2014
Discrete Dynamics in
Nature and Society
Volume 2014
Volume 2014
Journal of
Journal of
Discrete Mathematics
Journal of
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Applied Mathematics
Journal of
Function Spaces
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Optimization
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014