A Non-NP-Complete Algorithm for a Quasi-Fixed Polynomial Problem

Abstract

Let 𝐹∶ℝ×ℝ→ℝ be a real-valued polynomial function of the form 𝐹(𝑥,𝑦)=∑^𝑠_𝑖=0𝑓_𝑖(𝑥)𝑦^𝑖, 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))=𝑐𝑝^𝑚(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, 𝐹𝑥,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: 𝐹𝑥,𝑦(𝑥)=𝑐𝑥^𝑚,for𝑥∈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: 𝐹𝑥,𝑦(𝑥)=𝑐𝑝^𝑚(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 𝐹𝑥,𝑦=^𝑠_𝑖=0𝑓_𝑖(𝑥)𝑦^𝑖,withdegree𝑦in𝐹𝑥,𝑦denotedbydeg_𝑦𝐹=𝑠>1. (4)

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)𝑓_𝑠(𝑥) must be 𝑎𝑝^𝑟(𝑥) for some 𝑎∈ℝ, 𝑟∈ℕ,(2)those solutions must assume a fixed form like −𝑓_𝑠−1(𝑥)𝑠𝑓_𝑠(𝑥)+𝜆𝑝^𝑘(𝑥)forany𝜆∈ℝ. (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 deg_𝑦𝐹+2.

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(𝑒^𝑛). Otherwise, the algorithm is called “non-NP-complete”; that is, the algorithm can be computed in a polynomial time not exceeding 𝑂(𝑛^𝑘) 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-NP-complete. We have given two example problems in Section 5.

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].
Output. All solutions 𝑦=𝑦(𝑥) satisfy 𝐹𝑥,𝑦(𝑥)=𝑎,forsome𝑎∈ℝ, (6) in polynomial time of 𝑤𝜇, where 𝑤 is the computer memory of all coefficients of 𝐹(𝑥,𝑦) and 𝜇=deg_𝑥𝐹(𝑥,𝑦)+deg_𝑦𝐹(𝑥,𝑦). Moreover, the number of all solutions is at most deg_𝑦𝐹(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.

Example 4. Let 𝐹𝑥,𝑦=𝑝(𝑥)𝑦²−𝑝²(𝑥)−4 (7) and 𝑝(𝑥)=𝑥²+𝑥+1. Then deg_𝑦𝐹=𝑠=2. Solve all quasi-fixed solutions of 𝐹𝑥,𝑦=𝑎𝑝²(𝑥). (8)
By (7) and (8), we obtain 𝑝(𝑥)𝑦²−𝑝²(𝑥)−4=𝑎𝑝²(𝑥), (9) divides both sides by 𝑝(𝑥), then it becomes 𝑦²−𝑝²(𝑥)−4=𝑎𝑝(𝑥). (10)
It follows that 𝑦²−4=0mod𝑝(𝑥), (11) and we have (𝑥)(𝑥)=2(𝑥)(𝑥),𝑦=𝑧𝑝+2𝑧𝑝−2mod𝑝−2mod𝑝 (12) with an indeterminate 𝑧. To determine 𝑧, we substitute this 𝑦 (in (12)) into 𝐹(𝑥,𝑦), then (8) becomes 𝑃₁(1)𝐹𝑥,𝑦=𝐹𝑥,𝑧𝑝(𝑥)+2=𝑎𝑝²(𝑥)(2)𝐹𝑥,𝑦=𝐹𝑥,𝑧𝑝(𝑥)−2=𝑎𝑝²(𝑥). (13)
This yields 𝑃₂(1)𝑝²(𝑥)𝑝(𝑥)𝑧²+4𝑧−𝑝(𝑥)=𝑎𝑝²(𝑥)(2)𝑝²(𝑥)𝑝(𝑥)𝑧²−4𝑧−𝑝(𝑥)=𝑎𝑝²(𝑥). (14)
Both sides of (14) divide by 𝑝²(𝑥) to become 𝑃₃(1)𝑝(𝑥)𝑧²(𝑥)(2)𝑝(𝑥)𝑧+4𝑧−𝑝=𝑎₂−4𝑧−𝑝(𝑥)=𝑎. (15)
According to Algorithm 3, we obtain 𝑧=1 and 𝑧=−1, so the solutions (12) of problem (𝑃) are (𝑥)(𝑥)=𝑝(𝑥)𝑝(𝑥)(𝑥)(𝑥)𝑦=𝑧𝑝+2𝑧𝑝−2+2−2or−𝑝+2−𝑝−2. (16)

In Section 3, we provide a non-NP-complete algorithm to satisfy (3).

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.
Output. All solutions 𝑦=𝑦(𝑥) satisfy 𝐹𝑥,𝑦(𝑥)=𝑎𝑝^𝑚(𝑥)forsome𝑎∈ℝ, (17) by a non-NP-complete algorithm.

The following definitions are given by [6, page 421].

Definition 6. (i) The content of 𝐹(𝑥,𝑦) in (4), denoted by cont_𝑦𝐹, is defined by the greatest common divisor (g.c.d): gcd𝑓_𝑠(𝑥),𝑓_𝑠−1(𝑥),…,𝑓₀(𝑥)=cont_𝑦𝐹. (18)
(ii) The primitive part pp_𝑦𝐹 of 𝐹(𝑥,𝑦) is 𝐹(𝑥,𝑦)/cont_𝑦𝐹∈ℝ[𝑥,𝑦], and 𝐹(𝑥,𝑦) is primitive if cont_𝑦𝐹=1.
From (i) and (ii), we have 𝐹𝑥,𝑦=cont_𝑦𝐹⋅pp_𝑦𝐹. (19)

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 𝐹𝑥,𝑦=0mod𝑝(𝑥). (20)
The number of all solutions 𝑦=𝑦(𝑥)(mod𝑝(𝑥)) is thus at most deg_𝑦𝐹.

Assumption. Throughout this algorithm, for any 𝐹(𝑥,𝑦)∈ℝ[𝑥,𝑦], one can solve all solutions 𝑦=𝑔(𝑥)(mod𝑝(𝑥)) of the equation 𝐹𝑥,𝑔(𝑥)=0mod𝑝(𝑥). (21)

The procedure of our main algorithm is described below.

Procedures of Main Algorithm (Algorithm 5)

Step  0. If cont_𝑦𝐹∣𝑝^𝑚(𝑥) and cont_𝑦𝐹=1, then let 𝐹₁(𝑥,𝑦)=𝐹(𝑥,𝑦) and move to (Step 1, (1.2.2)).

Step  1. For convenience, we let 𝐹𝑥,𝑦=cont_𝑦𝐹𝐹₁𝑥,𝑦, (22) where 𝐹₁(𝑥,𝑦)∈ℝ[𝑥,𝑦] is a primitive polynomial.

Step  1.1. If cont_𝑦𝐹∤𝑝^𝑚(𝑥), we would have the problem 𝐹𝑥,𝑦=𝑐𝑝^𝑚(𝑥), (23) to deduce that cont_𝑦𝐹∣𝑐⋅𝑝^𝑚(𝑥). But cont_𝑦𝐹∤𝑝^𝑚(𝑥), then 𝑐=0. Consequently, (23) becomes 𝐹𝑥,𝑦=0. (24)

We can then solve all solutions 𝑦(𝑥) for 𝐹(𝑥,𝑦)=0 to get a solution set 𝑌₀=𝑦(𝑥)∶𝐹𝑥,𝑦(𝑥)=0. (25)

Step  1.2. If cont_𝑦𝐹∣𝑝^𝑚(𝑥) and cont_𝑦𝐹≠1, then cont_𝑦𝐹=𝑝^ℓ1(𝑥) with ℓ₁≤𝑚. In this case, (23) becomes 𝐹₁𝑥,𝑦=𝑐𝑝^𝑚−ℓ1(𝑥). (26)

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 𝑊₀=𝑦(𝑥)∶𝐹₁𝑥,𝑦(𝑥)=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, 𝐹₁𝑥,𝑦=0mod𝑝(𝑥). (28)

According to Lemma 7, the solution number does not exceed deg_𝑦𝐹, thus we may assume that 𝑦=𝑎₀(𝑥) is a solution of (28) with deg𝑎₀(𝑥)<deg𝑝(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)=𝑎⁰₁(𝑥),𝑎⁰₂(𝑥),…,𝑎⁰_𝑟0(𝑥). (29)

Consequently, the expression (23) can go to Step 2.

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 𝐹𝑥,𝑧𝑝(𝑥)+𝑎₀(𝑥)=𝑐𝑝^𝑚(𝑥). (30)

We let 𝐻₁(𝑥,𝑧)=𝐹(𝑥,𝑧𝑝(𝑥)+𝑎₀(𝑥)). Thus it follows from Definition 6(ii) that there exists 𝐹₂𝑥,𝑦=pp_𝑦𝐻₁∈ℝ𝑥,𝑦 (31) which is a primitive polynomial function so that 𝐻₁𝑥,𝑦=cont_𝑦𝐻₁𝐹₂𝑥,𝑦. (32)

Equation (30) then becomes 𝐻₁𝑥,𝑦=𝑐𝑝^𝑚(𝑥). (33)

Next, we will prove that 𝑝(𝑥)∣cont_𝑦𝐻₁.

Step  2.1. If cont_𝑦𝐻₁∤𝑝^𝑚(𝑥), (32) and (33) can be extrapolated as to cont_𝑦𝐻₁∣𝑐⋅𝑝^𝑚(𝑥). Since cont_𝑦𝐻₁∤𝑝^𝑚(𝑥), then 𝑐=0. Consequently, (33) becomes 𝐻₁𝑥,𝑦=0. (34)

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)∶𝐻₁𝑥,𝑦(𝑥)=0,𝑎₀(𝑥)∈𝑇₀(𝐹). (35)

Step  2.2. If cont_𝑦𝐻₁∣𝑝^𝑚(𝑥), then cont_𝑦𝐻₁=𝑝^ℓ2(𝑥) with ℓ₂≤𝑚. In this case, (33) becomes 𝐹₂𝑥,𝑦=𝑐𝑝^𝑚−ℓ2(𝑥). (36)

Step  2.2.1. In case ℓ₂=𝑚, (36) becomes 𝐹₂𝑥,𝑦=𝑐 (37) which can be solved by Algorithm 3 to get 𝑦=𝑦(𝑥) and obtain a set 𝑊₁=𝑦(𝑥)𝑝(𝑥)+𝑎₀(𝑥)∶𝐹₂𝑥,𝑦(𝑥)=𝑐,𝑎₀(𝑥)∈𝑇₀(𝐹). (38)

Step  2.2.2. If ℓ₂<𝑚, then 𝑚−ℓ₂<0 and we can divide both sides of (36) by 𝑝(𝑥); consequently, 𝐹₂𝑥,𝑦=0mod𝑝(𝑥). (39)

By the same reasoning used in [Step 1.2.2], we can solve 𝑦=𝑎₁(𝑥) with deg𝑎₁(𝑥)<deg𝑝(𝑥) and obtain the set 𝑎₁(𝑥)∈𝑉₁=𝑎¹₁(𝑥),𝑎¹₂(𝑥),…,𝑎¹_𝑟1(𝑥). (40)

We let 𝑇₁(𝐹)=𝑎₁(𝑥)𝑝(𝑥)+𝑎₀(𝑥)∶𝑎₁(𝑥)∈𝑉₁,𝑎₀(𝑥)∈𝑇₀(𝐹), (41)

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 𝐹𝑥,𝑧𝑝²(𝑥)+𝑎₁(𝑥)𝑝(𝑥)+𝑎₀(𝑥)=𝑐𝑝^𝑚(𝑥). (42)

Moreover, we let 𝐻₂(𝑥,𝑧)=𝐹(𝑥,𝑧𝑝²(𝑥)+𝑎₁(𝑥)𝑝(𝑥)+𝑎₀(𝑥)) and 𝐻₂𝑥,𝑦=cont_𝑦𝐻₂𝐹₃𝑥,𝑦 (43) for a primitive polynomial 𝐹₃(𝑥,𝑦)=pp_𝑦𝐻₂∈ℝ[𝑥,𝑦], so that (42) becomes 𝐻₂𝑥,𝑦=𝑐𝑝^𝑚(𝑥). (44)

Next, we will prove that 𝑝²(𝑥)∣cont_𝑦𝐻₂.

Step  3.1. If cont_𝑦𝐻₂∤𝑝^𝑚(𝑥), and (43) and (44) are extrapolated, we have cont_𝑦𝐻₂∣𝑐⋅𝑝^𝑚(𝑥). (45)

Since cont_𝑦𝐻₂∤𝑝^𝑚(𝑥), we have 𝑐=0.

Thus (44) becomes 𝐻₂𝑥,𝑦=0. (46)

Hence, we solve all solutions 𝑦(𝑥) such that 𝐻₂(𝑥,𝑦)=0 to obtain a solution set 𝑌₂=𝑦(𝑥)𝑝²(𝑥)+𝑎₁(𝑥)𝑝(𝑥)+𝑎₀(𝑥)∶𝐻₂(𝑥)𝑎𝑥,𝑦=0,₁(𝑥)𝑝(𝑥)+𝑎₀(𝑥)∈𝑇₁(𝐹). (47)

Step  3.2. If cont_𝑦𝐻₂∣𝑝^𝑚(𝑥), then cont_𝑦𝐻₂=𝑝^ℓ3(𝑥) for some positive integer ℓ₃≤𝑚 and (44) becomes 𝐹₃𝑥,𝑦=𝑐𝑝^𝑚−ℓ3(𝑥). (48)

Step  3.2.1. If ℓ₃=𝑚, then (48) becomes 𝐹₃𝑥,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)𝑎𝑥,𝑦=𝑐,₁(𝑥)𝑝(𝑥)+𝑎₀(𝑥)∈𝑇₁(𝐹). (51)

Step  3.2.2. If ℓ₃<𝑚, we use 𝑝(𝑥) to divide both sides of (48) to obtain 𝐹₃𝑥,𝑦=0mod𝑝(𝑥). (52)

By the same reasoning used in (Step 1.2.2), we can solve 𝑦=𝑎₂(𝑥) with deg𝑎₂(𝑥)<deg𝑝(𝑥) and obtain the set 𝑎₂(𝑥)∈𝑉₂=𝑎²₁(𝑥),𝑎²₂(𝑥),…,𝑎²_𝑟2(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,…,ℓ_𝑘−1} and go to the next𝑘-th step.

Step 𝑘. For each ∑^𝑘−2_𝑖=0𝑎_𝑖(𝑥)𝑝^𝑖(𝑥)∈𝑇_𝑘−2(𝐹), we can replace 𝑦 by 𝑧𝑝^𝑘−1(𝑥)+⋯+𝑎₀(𝑥) in (23), and then (23) becomes 𝐹𝑥,𝑧𝑝^𝑘−1(𝑥)+⋯+𝑎₀(𝑥)=𝑐𝑝^𝑚(𝑥). (55)

We let 𝐻_𝑘−1(𝑥,𝑧)=𝐹(𝑥,𝑧𝑝^𝑘−1(𝑥)+⋯+𝑎₀(𝑥)) and 𝐻_𝑘−1𝑥,𝑦=cont_𝑦𝐻_𝑘−1𝐹_𝑘𝑥,𝑦 (56) for a primitive polynomial 𝐹_𝑘(𝑥,𝑦)∈ℝ[𝑥,𝑦], then (55) becomes 𝐻_𝑘−1𝑥,𝑦=𝑐𝑝^𝑚(𝑥). (57)

Note that in Section 4, Lemma 9, we will prove that 𝑝^𝑘−1(𝑥)∣cont_𝑦𝐻_𝑘−1.

Step  k.1. If cont_𝑦𝐻_𝑘−1∤𝑝^𝑚(𝑥), by (56) and (57), we have cont_𝑦𝐻_𝑘−1∣𝑐⋅𝑝^𝑚(𝑥), (58) since cont_𝑦𝐻_𝑘−1∤𝑝^𝑚(𝑥), we have 𝑐=0.

Thus (57) becomes 𝐻_𝑘−1𝑥,𝑦=0. (59)

Hence, we solve all solutions 𝑦(𝑥) such that 𝐻_𝑘−1(𝑥,𝑦)=0 to obtain a solution set 𝑌_𝑘−1=𝑦(𝑥)𝑝^𝑘−1+^𝑘−2_𝑖=0𝑎_𝑖(𝑥)𝑝^𝑖(𝑥)∶𝐻_𝑘−1𝑥,𝑦(𝑥)=0,_{𝑘−2𝑖=0}𝑎_𝑖(𝑥)𝑝^𝑖(𝑥)∈𝑇_𝑘−2(𝐹). (60)

Step  k.2. If cont_𝑦𝐻_𝑘−1∣𝑝^𝑚(𝑥), then cont_𝑦𝐻_𝑘−1=𝑝^ℓ𝑘(𝑥) for some positive integer ℓ_𝑘≤𝑚 and (57) becomes 𝐹_𝑘𝑥,𝑦=𝑐𝑝^{𝑚−ℓ𝑘}(𝑥). (61)

Step  k.2.1. If ℓ_𝑘=𝑚, then (61) becomes 𝐹_𝑘𝑥,𝑦=𝑐. (62)

By Algorithm 3, we compute 𝑦=𝑦(𝑥) and obtain 𝑦(𝑥)𝑝^𝑘−1(𝑥)+𝑎_𝑘−2(𝑥)𝑝^𝑘−2(𝑥)+⋯+𝑎₀(𝑥) (63) in 𝑊_𝑘−1, for all 𝑎_𝑘−2(𝑥)𝑝^𝑘−2(𝑥)+⋯+𝑎₀(𝑥)∈𝑇_𝑘−2(𝐹). Hence, we output 𝑊_𝑘−1=𝑦(𝑥)𝑝^𝑘−1(𝑥)+⋯+𝑎₀(𝑥)∶𝐹_𝑘𝑥,𝑦(𝑥)=𝑐,_{𝑘−2𝑖=0}𝑎_𝑖(𝑥)𝑝^𝑖(𝑥)∈𝑇_𝑘−2(𝐹). (64)

Step  k.2.2. If ℓ_𝑘<𝑚, we use 𝑝(𝑥) to divide both sides of (61) to obtain 𝐹_𝑘𝑥,𝑦=0mod𝑝(𝑥). (65)

By the same reasoning used in (Step 1.2.2), we can solve 𝑦=𝑎_𝑘−1(𝑥) with deg𝑎_𝑘−1(𝑥)<deg𝑝(𝑥) and obtain the set 𝑎_𝑘−1(𝑥)∈𝑉_𝑘−1=𝑎₁^𝑘−1(𝑥),𝑎₂^𝑘−1(𝑥),…,𝑎_𝑟^𝑘−1_𝑘−1(𝑥). (66)

We let 𝑇_𝑘−1(𝐹)=^𝑘−2_𝑖=0𝑎_𝑖(𝑥)𝑝^𝑖(𝑥)∶𝑎_𝑘−1(𝑥)∈𝑉_𝑘−1,_{𝑘−2𝑖=0}𝑎_𝑖(𝑥)𝑝^𝑖(𝑥)∈𝑇_𝑘−2(𝐹), (67) thus expression (57) can go to the next step.

As explained in Remark 8, this algorithm is not an infinite loop.

Remark 8. (i) If 𝑉_𝑖=∅, then 𝑉_𝑖+1=∅ and if 𝑉_𝑖+1=∅, then 𝑇_𝑖+1(𝐹)=∅ for 𝑖≥0. This means that if 𝑉_𝑖=∅, then the work finishes at step 𝑖+1for 𝑖≥0.
(ii) This algorithm is not an infinite loop. Since ℓ_𝑘<𝑚, and by Corollary 13, we have ℓ_𝑘<ℓ_𝑘+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 |𝑊_𝑖|<∞, for 𝑖=1,2,…; however, if |𝑊_𝑖|=∞, 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 𝑝^𝑚(𝑥)∣𝐻_𝑚𝑥,𝑦 (68) by Corollary 11 and let 𝐻_𝑚𝑥,𝑦=𝑝^𝑚(𝑥)𝐹_𝑚+1𝑥,𝑦. (69)

Thus, the possible solutions of 𝐻_𝑚(𝑥,𝑦)=𝑐𝑝^𝑚(𝑥) are those solutions 𝑦(𝑥) satisfying 𝐹_𝑚+1𝑥,𝑦(𝑥)=𝑐. (70)

Moreover, by Algorithm 3, we can get a solution set 𝑌 such that the bound of all solutions is “𝑠⋅|𝑇_𝑚−1(𝐹)|.” So the solution set of (3) is contained in 𝑇=^𝑚−1_𝑖=0𝑌_𝑖^𝑚−1_𝑖=0𝑊_𝑖𝑌. (71)

Finally, we check each element 𝑢(𝑥)∈𝑇 to determine whether or not 𝐹(𝑥,𝑢(𝑥))=𝑐𝑝^𝑚(𝑥). (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 𝐹(𝑥,𝑢(𝑥))=𝑐𝑝^𝑚(𝑥). (73)

Non-NP-complete computing time is also needed given the cardinal number of 𝑇: |𝑇|≤𝑊₀+𝑌₀+|𝑌|+^𝑚−2_𝑖=0𝑊_𝑖+1+^𝑚−2_𝑖=0𝑌_𝑖+1𝑇ByAlgorithm3andLemma7≤2𝑠+𝑠⋅_𝑚−1(𝐹)+2^𝑚−2_𝑖=0𝑠⋅𝑇_𝑖(𝐹), (74) where |𝑇_𝑖(𝐹)|, |𝑊_𝑖|, and |𝑌_𝑖| denote the cardinal number of 𝑇_𝑖(𝐹), 𝑊_𝑖, and 𝑌_𝑖. The remaining work “Checks the bound of |𝑇_𝑖(𝐹)|, for each 𝑖=1,2,…,𝑚.” In Section 4, Theorem 16, we prove that 𝑇_𝑖(𝐹)≤𝑠 (75) for 𝑖=1,2,…,𝑚−1.

By (74) and (75), we have the cardinal number |𝑇|≤2𝑠+𝑠⋅𝑇_𝑚−1(𝐹)+2^𝑚−2_𝑖=0𝑠⋅𝑇_𝑖(𝐹)≤2𝑠+𝑠²+2^𝑚−2_𝑖=0𝑠²=2𝑠+(2𝑚−1)𝑠². (76)

That is, we have to check at most “2𝑠+(2𝑚−1)𝑠²” solutions 𝑢(𝑥) to satisfy 𝐹(𝑥,𝑢(𝑥))=𝑐𝑝^𝑚(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 each𝑘-th Step, whether 𝑦(𝑥)𝑝^𝑘−1(𝑥)+𝑎_𝑘−2𝑝^𝑘−2(𝑥)+⋯+𝑎₀(𝑥) belongs to 𝑌_𝑘−1 or 𝑊_𝑘−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.

4. Main Theorems

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)=𝑓_𝑠(𝑥)𝑦^𝑠+𝑓_𝑠−1(𝑥)𝑦^𝑠−1+⋯+𝑓₀(𝑥). Then 𝐹𝑥,𝑧𝑝(𝑥)+𝑎(𝑥)=𝑓_𝑠(𝑥)𝑧𝑝(𝑥)+𝑎(𝑥)^𝑠+𝑓_𝑠−1(𝑥)𝑧𝑝(𝑥)+𝑎(𝑥)^𝑠−1+⋯+𝑓₀(𝑥)=𝑓_𝑠(𝑥)𝑧^𝑠𝑝^𝑠(𝑥)+𝑠𝑧^𝑠−1𝑝^𝑠−1𝑎(𝑥)+⋯+𝑎^𝑠(𝑥)+𝑓_𝑠−1(𝑥)𝑧^𝑠−1𝑝^𝑠−1(𝑥)+()𝑧𝑠−1_𝑠−2𝑝^𝑠−2𝑎(𝑥)+⋯+𝑎^𝑠−1(𝑥)+⋯+𝑓₀(𝑥)=𝑓_𝑠(𝑥)𝑄_𝑠(𝑥,𝑧)𝑝(𝑥)+𝑎^𝑠(𝑥)+𝑓_𝑠−1(𝑥)×𝑄_𝑠−1(𝑥,𝑧)𝑝(𝑥)+𝑎^𝑠−1(𝑥)+⋯+𝑓₀(𝑥)where𝑄_𝑖(𝑥,𝑧)=𝑧^𝑖𝑝^𝑖−1(𝑥)+𝑖𝑧^𝑖−1𝑝^𝑖−2(𝑥)𝑎(𝑥)+⋯+𝑖𝑧𝑎^𝑖−1(𝑥(𝑥)𝑓)=𝑝_𝑠(𝑥)𝑄_𝑠(𝑥,𝑧)+𝑓_𝑠−1(𝑥)𝑄_𝑠−1(𝑥,𝑧)+⋯+𝑓₁(𝑥)𝑄₁(+𝑓𝑥,𝑧)_𝑠(𝑥)𝑎^𝑠(𝑥)+𝑓_𝑠−1(𝑥)𝑎^𝑠−1(𝑥)+⋯+𝑓₀(𝑥)=𝑝(𝑥)𝑓_𝑠(𝑥)𝑄_𝑠(𝑥,𝑧)+𝑓_𝑠−1(𝑥)𝑄_𝑠−1(𝑥,𝑧)+⋯+𝑓₁(𝑥)𝑄₁(𝑥,𝑧)+𝐹(𝑥,𝑎(𝑥)). (79)
Thus, 𝑝(𝑥)∣𝐹(𝑥,𝑎(𝑥))⇒𝑝(𝑥)∣𝐹(𝑥,𝑧𝑝(𝑥)+𝑎(𝑥)).

The definitions of 𝐻_𝑖(𝑥,𝑦), cont_𝑦𝐻_𝑖, 𝐹_𝑖+1(𝑥,𝑦), and 𝑎_𝑖(𝑥) for 𝑖=1,2,…,𝑚+1 in Algorithm 5 are 𝐻_𝑘(𝑥,𝑧)=𝐹𝑥,𝑧𝑝^𝑘(𝑥)+⋯+𝑎₀(𝑥),𝐻_𝑘𝑥,𝑦=cont_𝑦𝐻_𝑘𝐹_𝑘+1𝑥,𝑦, (80)(81) for a primitive polynomial 𝐹_𝑘+1(𝑥,𝑦)∈ℝ[𝑥,𝑦] and 𝑝(𝑥)∣𝐹_𝑘+1(𝑥,𝑎_𝑘(𝑥)) for 𝑘=0,1,…,𝑚.

Next, we prove some properties about 𝐻_𝑘(𝑥,𝑦) for 𝑘=1,2,…,𝑚+1.

Lemma 10. Consider 𝑝(𝑥)cont_𝑦𝐻_𝑘∣cont_𝑦𝐻_𝑘+1, for 1≤𝑘≤𝑚.

Proof. From (80), we have 𝐻_𝑘+1(𝑥,𝑧)=𝐹𝑥,𝑧𝑝^𝑘+1(𝑥)+𝑎_𝑘(𝑥)𝑝^𝑘(𝑥)+⋯+𝑎₀(𝑥)=𝐹𝑥,𝑧𝑝(𝑥)+𝑎_𝑘(𝑥)𝑝^𝑘(𝑥)+⋯+𝑎₀(𝑥)=𝐻_𝑘𝑥,𝑧𝑝(𝑥)+𝑎_𝑘(𝑥)()=by80cont_𝑧𝐻_𝑘𝐹_𝑘+1𝑥,𝑧𝑝(𝑥)+𝑎_𝑘(𝑥)by(81). (82)
By definition of 𝑎_𝑘(𝑥), we have 𝑝(𝑥)∣𝐹_𝑘+1(𝑥,𝑎_𝑘(𝑥)). From Lemma 9, we obtain 𝑝(𝑥)∣𝐹_𝑘+1(𝑥,𝑧𝑝(𝑥)+𝑎_𝑘(𝑥)). That is, 𝑝(𝑥)cont_𝑦𝐻_𝑘∣𝐻_𝑘+1(𝑥,𝑧), (83) and, by (81), we obtain 𝑝(𝑥)cont_𝑦𝐻_𝑘∣cont_𝑦𝐻_𝑘+1.

By Lemma 9, 𝑝(𝑥)∣𝐻₁(𝑥,𝑧), and the definition of cont_𝑦𝐻₁, it follows that 𝑝(𝑥)∣cont_𝑦𝐻₁. Moreover, by Lemma 10, we can easily obtain the following corollaries.

Corollary 11. Consider𝑝^𝑖(𝑥)∣cont_𝑦𝐻_𝑖, 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 𝛿_𝑝(𝑓(𝑥))=ℓ, if 𝑝(𝑥)^ℓ∣𝑓(𝑥) and 𝑝(𝑥)^ℓ+1∤𝑓(𝑥).
(ii) We say that 𝐹(𝑥,𝑦) has 𝑝(𝑥)-power ℓ, if cont_𝑦𝐹 has 𝑝(𝑥)-power ℓ.

The following corollary is an immediate result of Lemma 10 and Definition 12.

Corollary 13. Consider ℓ_𝑖<ℓ_𝑖+1, for 𝑖≥1.

Proof . Since 𝛿_𝑝(𝐻_𝑖)=ℓ_𝑖+1 and by Lemma 10, we have 𝑝(𝑥)cont_𝑦𝐻_𝑘∣cont_𝑦𝐻_𝑘+1. (84)
It follows that 1+𝛿_𝑝𝐻_𝑖=𝛿_𝑝𝑝(𝑥)𝐻_𝑖≤𝛿_𝑝𝐻_𝑖+1, (85) so 1+ℓ_𝑖≤ℓ_𝑖+1; hence, ℓ_𝑖<ℓ_𝑖+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)𝛿_𝑝(𝑓𝑔)=𝛿_𝑝(𝑓)+𝛿_𝑝(𝑔), (ii)𝛿_𝑝(𝑓+𝑔)≥min{𝛿_𝑝(𝑓),𝛿_𝑝(𝑔)}, and (iii)𝛿_𝑝(𝐹)=min_{0≤𝑖≤𝑠}{𝛿_𝑝(𝑓_𝑖)}, (iv)If ℎ=𝑓+𝑔 and 𝛿_𝑝(ℎ)<𝛿_𝑝(𝑔), then 𝛿_𝑝(ℎ)=𝛿_𝑝(𝑓).

For each 𝑘∈ℕ, we rewrite 𝑇_𝑘(𝐹) defined in the procedures of Algorithm 5 as the following definitions.

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 𝑇_𝑖(𝐹), 𝑖∈ℕ defined in Algorithm 5 as follows: (1)𝑇₀(𝐹)={𝑎(𝑥)∈ℝ[𝑥]∶𝑝(𝑥)cont_𝑧𝐹(𝑥,𝑧)∣𝐹(𝑥,𝑎(𝑥)) with 0≤deg𝑎(𝑥)<deg𝑝(𝑥)}, (2)𝑇_𝑘(𝐹)={𝑏_𝑘(𝑥)∈ℝ[𝑥]∶𝑏_𝑘(𝑥)=𝑎(𝑥)𝑝^𝑘(𝑥)+𝑏_𝑘−1(𝑥) satisfies (𝐷1) and (𝐷2)} where for 𝑘≥1, and the (𝐷1) and (𝐷2) properties are given by the following:(D1):0≤deg𝑎(𝑥)<deg𝑝(𝑥) and 𝑏_𝑘−1(𝑥)∈𝑇_𝑘−1(𝐹),(D2):𝑝(𝑥)cont_𝑧𝐹(𝑥,𝑧𝑝^𝑘(𝑥)+𝑏(𝑥))∣𝐹(𝑥,𝑎(𝑥)𝑝^𝑘(𝑥)+𝑏(𝑥)).

From the above definitions, we obtain a bound of the cardinal number |𝑇_𝑘(𝐹)|.

Theorem 16. For each 𝑘∈ℕ, then |𝑇_𝑘(𝐹)|≤deg_𝑦𝐹 (|⋅| means the cardinal number).

Proof . This theorem will be proven by induction on deg_𝑦𝐹.
(i) When deg_𝑦𝐹=1, claim that |𝑇_𝑘(𝐹)|≤1 for any 𝑘∈ℕ.
For this purpose, we consider 𝐹𝑥,𝑦=𝑓₁(𝑥)𝑦+𝑓₀(𝑥)forsome𝑓₁(𝑥),𝑓₀(𝑥)∈ℝ[𝑥]. (86)
(1) If 𝑘=0, and let 𝑣₀=𝛿_𝑝(𝐹(𝑥,𝑧)) and 𝐹(𝑥,𝑦)=𝑓₁(𝑥)𝑦+𝑓₀(𝑥)=𝑝^𝑣0(𝑥)(𝑔⁰₁(𝑥)𝑦+𝑔⁰₀(𝑥)) for some coprime polynomials 𝑔⁰₁(𝑥), 𝑔⁰₀(𝑥)∈ℝ[𝑥].
The only possible solution 𝑎₀(𝑥) with 0≤deg𝑎₀(𝑥)<deg𝑝(𝑥) and 𝑝(𝑥)cont_𝑦𝐹(𝑥,𝑧)∣𝐹(𝑥,𝑎₀(𝑥)) is 𝑎₀(𝑥)=𝑔⁰₁(𝑥)⁻¹𝑔⁰₀(𝑥)mod𝑝(𝑥). (87)
That is, |𝑇₀(𝐹)|≤1.
(2) Next, we assume that |𝑇_𝑘−1(𝐹)|≤1, thus we deduce that |𝑇_𝑘(𝐹)|≤1.
Let 𝑣_𝑘=𝛿_𝑝𝐹𝑥,𝑧𝑝^𝑘(𝑥)+𝑏_𝑘−1(𝑥)forsome𝑏_𝑘−1(𝑥)∈𝑇_𝑘−1(𝐹), (88) and 𝐹(𝑥,𝑧𝑝^𝑘(𝑥)+𝑏_𝑘−1(𝑥))=𝑝^𝑣𝑘(𝑥)(𝑔^𝑘₁(𝑥)𝑧+𝑔^𝑘₀(𝑥)) for some 𝑔^𝑘₁(𝑥), 𝑔^𝑘₀(𝑥)∈ℝ[𝑥]. Consider 𝑇_𝑘(𝐹)=𝑏_𝑘(𝑥)∈ℝ[𝑥]∶𝑏_𝑘(𝑥)=𝑎(𝑥)𝑝^𝑘(𝑥)+𝑏_𝑘−1(𝑥)satisfies(𝐷1)and(𝐷2), (89) and since |𝑇_𝑘−1(𝐹)|≤1 by assumption, 𝑏_𝑘−1(𝑥) is uniquely decided, as is 𝑣_𝑘. The remaining task is to prove that 𝑎_𝑘 is also uniquely determined. As 𝑏_𝑘(𝑥)∈𝑇_𝑘(𝐹) with 𝑏_𝑘(𝑥)=𝑎_𝑘(𝑥)𝑝^𝑘(𝑥)+𝑏_𝑘−1(𝑥), 𝑎_𝑘(𝑥) satisfies 0≤deg𝑎_𝑘(𝑥)<deg𝑝(𝑥) (90) such that 𝑝(𝑥)cont_𝑦𝐹𝑥,𝑧𝑝^𝑘(𝑥)+𝑏_𝑘−1(𝑥)∣𝐹𝑥,𝑎_𝑘(𝑥)𝑝^𝑘(𝑥)+𝑏_𝑘−1(𝑥)forsome𝑏_𝑘−1(𝑥)∈𝑇_𝑘−1(𝐹), (91) then 𝑎_𝑘(𝑥)𝑔^𝑘₁(𝑥)+𝑔^𝑘₀(𝑥)=0mod𝑝(𝑥), (92) it follows that 𝑎_𝑘(𝑥) must be 𝑎_𝑘(𝑥)=𝑔^𝑘₁(𝑥)⁻¹𝑔^𝑘₀(𝑥)mod𝑝(𝑥). (93)
Consequently, |𝑇_𝑘(𝐹)|≤1.
By (1) and (2), we get that |𝑇_𝑘(𝐹)|≤1=deg_𝑦𝐹 for any 𝑘∈ℕ.
(ii) We will prove this theorem by mathematical induction. Assume that 𝑇_𝑘𝐹₁≤deg_𝑦𝐹₁, (94) for any 𝐹₁∈ℝ[𝑥] with deg_𝑦𝐹₁≤𝑛−1, we want to show 𝑇_𝑘(𝐹)≤deg_𝑦𝐹, (95) for any 𝐹∈ℝ[𝑥] with deg_𝑦𝐹=𝑛.
Let 𝑢=max{𝛿_𝑝(𝐹(𝑥,𝑐(𝑥)))∶𝑐(𝑥)∈𝑇_𝑘(𝐹)} and choose 𝑑(𝑥)∈𝑇_𝑘(𝐹) with 𝛿_𝑝(𝐹(𝑥,𝑑(𝑥)))=𝑢. (96)
Indeed, 𝛿_𝑝(𝐹(𝑥,𝑑(𝑥)))≥𝛿_𝑝(𝐹(𝑥,𝑐(𝑥)))forany𝑐(𝑥)∈𝑇_𝑘(𝐹). (97)
By the division rule, let 𝐹(𝑥,𝑦) be divided by 𝑦−𝑑(𝑥), giving us 𝐹𝑥,𝑦=𝑦−𝑑(𝑥)𝐹₁𝑥,𝑦+𝐹(𝑥,𝑑(𝑥)), (98) for some 𝐹₁(𝑥,𝑦)∈ℝ[𝑥,𝑦] and deg_𝑦𝐹₁(𝑥,𝑦)=𝑛−1.
We want to show that if𝑐(𝑥)∈𝑇_𝑘(𝐹)−{𝑑(𝑥)},then𝑐(𝑥)∈𝑇_𝑘𝐹₁. (99)
For any 𝑐(𝑥)∈𝑇_𝑘(𝐹)−{𝑑(𝑥)}, by the definition of 𝑇_𝑘(𝐹), we have 𝑐(𝑥)=𝑎₁(𝑥)𝑝(𝑥)^𝑘+𝑏₁(𝑥),𝑑(𝑥)=𝑎₂(𝑥)𝑝(𝑥)^𝑘+𝑏₂(𝑥), (100) for some 𝑏₁(𝑥), 𝑏₂(𝑥)∈𝑇_𝑘−1(𝐹) and 0≤deg𝑎₁(𝑥), deg𝑎₂(𝑥)<deg𝑝(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 deg𝑏₁(𝑥), deg𝑏₂(𝑥)<deg𝑝^𝑘(𝑥), so 𝛿_𝑝𝑏₁(𝑥)−𝑏₂(𝑥)<𝑘≤𝛿_𝑝𝑎₁(𝑥)−𝑎₂(𝑥)𝑝^𝑘(𝑥). (101) Therefore, one has the following:(i)𝛿_𝑝(𝑐(𝑥)−𝑑(𝑥))=𝛿_𝑝((𝑎₁(𝑥)−𝑎₂(𝑥))𝑝^𝑘(𝑥)+(𝑏₁(𝑥)−𝑏₂(𝑥)))=𝛿_𝑝(𝑏₁(𝑥)−𝑏₂(𝑥))(by Lemma 14, (iv)),(ii)𝛿_𝑝(𝑏₁(𝑥)−𝑑(𝑥))=𝛿_𝑝((−𝑎₂(𝑥))𝑝^𝑘(𝑥)+(𝑏₁(𝑥)−𝑏₂(𝑥)))=𝛿_𝑝(𝑏₁(𝑥)−𝑏₂(𝑥))(by Lemma 14, (iv)),(iii)𝛿_𝑝(𝑧(𝑝(𝑥))^𝑘+𝑏₁(𝑥)−𝑑(𝑥))=𝛿_𝑝((𝑧−𝑎₂(𝑥))𝑝^𝑘(𝑥)+(𝑏₁(𝑥)−𝑏₂(𝑥)))=𝛿_𝑝(𝑏₁(𝑥)−𝑏₂(𝑥))(by Lemma 14, (iv)).
(b) if 𝑏₁(𝑥)=𝑏₂(𝑥), then 𝑎₁(𝑥)≠𝑎₂(𝑥), since 0≤deg𝑎₁(𝑥), deg𝑎₂(𝑥)<deg𝑝(𝑥), so (i)𝛿_𝑝(𝑐(𝑥)−𝑑(𝑥))=𝛿_𝑝((𝑎₁(𝑥)−𝑎₂(𝑥))𝑝^𝑘(𝑥))=𝛿_𝑝(𝑎₁(𝑥)−𝑎₂(𝑥))+𝑘=𝑘(by Lemma 14, (i)),(ii)𝛿_𝑝(𝑏₁(𝑥)−𝑑(𝑥))=𝛿_𝑝((−𝑎₂(𝑥))𝑝^𝑘(𝑥))=𝛿_𝑝(𝑎₂(𝑥))+𝑘=𝑘(by Lemma 14, (i)),(iii)𝛿_𝑝(𝑧(𝑝(𝑥))^𝑘+𝑏₁(𝑥)−𝑑(𝑥))=𝛿_𝑝((𝑧−𝑎₂(𝑥))𝑝^𝑘(𝑥)+(𝑏₁(𝑥)−𝑏₂(𝑥)))=𝛿_𝑝(𝑧−𝑎₂(𝑥))+𝑘=𝑘(by Lemma 14, (i)).By the above equations, we obtain 𝛿_𝑝(𝑐(𝑥)−𝑑(𝑥))=𝛿_𝑝𝑏₁(𝑥)−𝑑(𝑥)=𝛿_𝑝𝑧𝑝(𝑥)^𝑘+𝑏₁(𝑥)−𝑑(𝑥). (102) and by (D2) in Definition 15, we have 𝛿_𝑝(𝐹(𝑥,𝑐(𝑥)))>𝛿_𝑝𝐹𝑥,𝑧𝑝(𝑥)^𝑘+𝑏₁(𝑥). (103)
By (97) and (103), we have 𝛿_𝑝𝐹(𝑥,𝑑(𝑥))≥𝛿_𝑝(𝐹(𝑥,𝑐(𝑥)))>𝛿_𝑝𝐹𝑥,𝑧𝑝(𝑥)^𝑘+𝑏₁(𝑥). (104)
Moreover, we have 𝛿_𝑝𝐹𝑥,𝑧𝑝(𝑥)^𝑘+𝑏₁(𝑥)=𝛿_𝑝𝑧𝑝(𝑥)^𝑘+𝑏₁(𝑥)−𝑑(𝑥)×𝐹₁𝑥,𝑧𝑝(𝑥)^𝑘+𝑏₁(𝑥((𝑥())+𝐹𝑥,𝑑))by98=𝛿_𝑝𝑧𝑝(𝑥)^𝑘+𝑏₁(𝑥)−𝑑(𝑥)×𝐹₁𝑥,𝑧𝑝(𝑥)^𝑘+𝑏₁(𝑥)by(104)andLemma14,(iv)=𝛿_𝑝𝑧𝑝(𝑥)^𝑘+𝑏₁(𝑥)−𝑑(𝑥)+𝛿_𝑝𝐹₁𝑥,𝑧𝑝(𝑥)^𝑘+𝑏₁(𝑥(i,𝛿)byLemma14,)_𝑝(𝐹(𝑥,𝑐(𝑥)))=𝛿_𝑝(𝑐(𝑥)−𝑑(𝑥))𝐹₁(𝑥,𝑐(𝑥))+𝐹(𝑥,𝑑(𝑥))=𝛿_𝑝(𝑐(𝑥)−𝑑(𝑥))𝐹₁((𝑥()()𝑥,𝑐))by104andLemma14,iv=𝛿_𝑝(𝑐(𝑥)−𝑑(𝑥))+𝛿_𝑝𝐹₁((𝑥(i).𝑥,𝑐))byLemma14, (105)(106)
From expression (106), we obtain 𝛿_𝑝(𝑐(𝑥)−𝑑(𝑥))+𝛿_𝑝𝐹₁(𝑥,𝑐(𝑥))=𝛿_𝑝(𝐹(𝑥,𝑐(𝑥)))>𝛿_𝑝𝐹𝑥,𝑧𝑝(𝑥)^𝑘+𝑏₁(𝑥)by(104)=𝛿_𝑝𝑧𝑝(𝑥)^𝑘+𝑏₁(𝑥)−𝑑(𝑥)+𝛿_𝑝𝐹₁𝑥,𝑧𝑝(𝑥)^𝑘+𝑏₁(𝑥)by(105). (107)
From expression (102), that is, 𝛿_𝑝(𝑐(𝑥)−𝑑(𝑥))=𝛿_𝑝(𝑧(𝑝(𝑥))^𝑘+𝑏₁(𝑥)−𝑑(𝑥)) and by canceling 𝛿_𝑝(𝑐(𝑥)−𝑑(𝑥)) to both sides of (107), it follows that 𝛿_𝑝𝐹₁(𝑥,𝑐(𝑥))>𝛿_𝑝𝐹₁𝑥,𝑧𝑝(𝑥)^𝑘+𝑏₁(𝑥). (108)
That is, 𝑐(𝑥)∈𝑇_𝑘(𝐹₁). Then we have 𝑇_𝑘(𝐹)⊆𝑇_𝑘(𝐹₁)⋃{𝑑(𝑥)}. Therefore, 𝑇_𝑘(𝐹)≤𝑇_𝑘𝐹₁+1≤(𝑛−1)+1=𝑛. (109)
By induction, the proof is completed.

5. Applications (As a Game Problem)

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 𝑓_𝑘(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 the𝑠-th year of this contract obeys the law of (year) motion about a quasi-fixed polynomial function: 𝐹𝑖,𝑦(𝑖)=𝑓_𝑠(𝑖)(1+𝑟)^𝑠+𝑓_𝑠−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 𝐹𝑖,𝑦(𝑖)=𝑆1+𝑖+𝑖^2𝑚,𝑚∈ℕ, (111) then 𝐵 can obtain a premium 𝑆1+𝑖+𝑖^2𝑚. (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, 𝑎_𝑘, to 𝐵, that 𝑎_𝑘 represents the compensation paid by 𝐴 to 𝐵 at the end of the𝑘-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 contract, we assume that 𝐴 gives a bonus interest rate, 𝑖+𝑖² to 𝐵.

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 𝑎_𝑘 dollars if the disaster occurs at the end of 𝑘-th year, 𝑘=1,2,…,𝑠. Note that(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 𝑎_𝑘 dollars at the end of 𝑘-th year is equal to the value of 𝑎_𝑘(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 𝑎_𝑘(1+𝑖)^𝑠−𝑘𝑞1−𝑞^𝑘. (113)

Then the value of total amount that 𝐴 pays 𝐵 at the end of 𝑠-th year is 𝐹𝑖,𝑞=𝑎₀(1+𝑖)^𝑠𝑞+𝑎₁(1+𝑖)^𝑠−1𝑞1−𝑞+⋯+𝑎_𝑠. (114)

This can be rewritten as (i)𝐹𝑖,𝑞=𝑓_𝑠(𝑖)𝑞^𝑠+𝑓_𝑠−1(𝑖)𝑞^𝑠−1+⋯+𝑓₀(𝑖). (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)𝑎1+𝑖+𝑖^2𝑠+1. (116)

To balance each other, we set the values (i) and (ii) as equal. Thus, the relation equation becomes 𝐹𝑖,𝑞=𝑎1+𝑖+𝑖^2𝑠+1forsome𝑎∈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 𝐹𝑥,𝑞=𝑎𝑝^𝑚(𝑥)forsome𝑎∈ℝ. (118)

This example explains a quasi-fixed (polynomial) problem concerning actuarial science.