Abstract
The main purpose of this paper is to use the mathematical induction and the properties of Lucas polynomials to study the power sum problem of Lucas polynomials. In the end, we obtain an interesting divisible property.
1 Introduction
For any integer n ≥ 0, the famous Fibonacci polynomials {Fn(x)} and Lucas polynomials {Ln(x)} are defined by F0(x) = 0, F1(x) = 1, L0(x) = 2, L1(x) = x and Fn+2(x) = xFn+1(x) + Fn(x), Ln+2(x) = xLn+1(x) + Ln(x). The general terms of Fn(x) and Ln(x) are given by
and
where
It is easy to prove the identities
and
If we take x = 1, then {Fn(x)} becomes Fibonacci sequences {Fn(1)}, and {Ln(x)} becomes Lucas sequences {Ln(1)}.
Since these sequences and polynomials have very important positions in the theory and application of mathematics, many scholars have studied their various properties, and obtained a series of important results, some of which can be found in references [1, 2, 3, 4, 5], and some other related papers can also be found in [6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]. For example, in a private communication with Curtis Cooper, R. S. Melham suggested that it would be interesting to discover an explicit expansion for
as a polynomial in F2n+1. Wiemann and Cooper [1] reported some conjectures of Melham related to the sum
Helmut Prodinger [3] studied the more general summation
For the sum of powers of Lucas numbers, Helmut Prodinger [3] also obtained some similar conclusions.
R. S. Melham [4] also proposed the following two conjectures:
Conjecture A
Letmbe a positive integer. Then the sum
can be expressed as (F2n+1 − 1)2P2m−1(F2n+1), whereP2m−1(x) is a polynomial of degree 2m − 1 with integer coefficients.
Conjecture B
Letm ≥ 0 be an integer. Then the sum
can be expressed as (L2n+1 − 1) Q2m(L2n+1), whereQ2m(x) is a polynomial of degree 2mwith integer coefficients.
Wang Tingting and Zhang Wenpeng [5] studied these problems, and proved several conclusions for
As some applications of Theorem 2 in reference [5], they deduced the following:
Corollary A
Leth ≥ 1 andn ≥ 0 be two integers. Then the sum
can be expressed as (L2h+1(x) − x) Q2n(x, L2h+1(x)), whereQ2n(x, y) is a polynomial in two variablesxandywith integer coefficients and degree 2nofy.
Corollary B
Leth ≥ 1 andn ≥ 0 be two integers. Then the sum
can be expressed as (F2h+1(x) − 1) H2n(x, F2h+1(x)), whereH2n(x, y) is a polynomial in two variablesxandywith integer coefficients and degree 2n of y.
Therefore, Wang Tingting and Zhang Wenpeng [5] solved Conjecture B completely. They also obtained some substantial progress for Conjecture A.
It is easy to see that in the above Conjecture A and Conjecture B, the subscripts of Fn and Ln in sums
If so, what would U(x) look like? How is it different from the form in Conjecture A and Conjecture B ? Obviously, it is hard to guess the exact form of U(x) from (d) of Theorem 1 in reference [5]. Maybe that is why there is no such formal conjecture in [4].
In this paper, we shall use the mathematical induction and the properties of Lucas polynomials to study this problem, and give an exact polynomial U(x) in (3). The result is detailed in the following theorems.
Theorem A
For any positive integersnandhwithh ≥ 2, we have
Theorem B
For any positive integersnandhwithh ≥ 2, we have
Taking x = 1, from Theorem 1 and Theorem 2 we may immediately deduce the following two corollaries:
Corollary A
For any positive integersnandhwithh ≥ 2, we have
Corollary B
For any positive integersnandhwithh ≥ 2, we have
2 Two Lemmas
Lemma A
For any non-negative integernandk, we have the identity
Proof
Let
and
from (2) we have the identity
This completes the proof of Lemma 1. □
Lemma B
For any positive integersnandh, we have the congruence
Proof
From (1) we know that x ∣ L2n+1(x). Because that L1(x) = x, so in order to prove Lemma 2, we only need to prove the congruence
Next we prove (4) by complete induction. It is clear that (4) is true for n = 0. If n = 1, then apply the identity
we have
That is to say, the congruence (4) is true for n = 1.
Suppose that the congruence (4) is true for all integers 0 ≤ n ≤ s. That is,
holds for all integers 0 ≤ n ≤ s.
Then for positive integer n = s + 1, we have the identities
and
from (5) we can deduce the congruence equations as follows
That is to say, the congruence (4) is true for n = s + 1.
Now Lemma 2 follows from (4) and completes the induction. □
3 Proofs of the theorems
In this section, we shall use mathematical induction to complete the proofs of our theorems. Here we only prove Theorem 1. Similarly, we can also deduce Theorem 2 and thus we omit its proving process here. After replacing x by L2m+1(x) in (1), we obtain the following expression with Lemma 1
or
For any positive integer h ≥ 2, we first introduce that the identities
Then, combining (6), (7) and (8) we have
Now we apply (9) and mathematical induction to prove the congruence
If n = 1, then from (9) we have
From Lemma 2 we know that
Combining (11) and (12) we know that the congruence (10) is true for n = 1.
Suppose that (10) is true for all 1 ≤ n ≤ s. That is,
holds for all 1 ≤ n ≤ s.
Then for n = s + 1, from (9) we have
Applying Lemma 2 we have the congruence
If 1 ≤ k ≤ s, then 3 ≤ 2s + 3 − 2k ≤ 2s + 1. From the inductive assumption (13) we have
Combining (14), (15) and (16) we may immediately deduce the congruence
This completes the proof of our theorem by mathematical induction.
Competing interests: The author declare that they have no competing interests.
Acknowledgement
This work is supported by the N. S. F. (Grant No. 11771351) of P. R. China. The author would like to thank the referees for their very helpful and detailed comments, which have significantly improved the presentation of this paper.
References
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© 2018 Xiao, published by De Gruyter
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