The Resultant, the Discriminant, and

the Derivative of Generalized Fibonacci Polynomials

Rigoberto Flórez Department of Mathematical Sciences The Citadel Charleston, SC 29409 USA

                rigo.florez@citadel.edu

Robinson A. Higuita and Alexander Ramírez Instituto de Matemáticas Universidad de Antioquia Medellín Colombia

             robinson.higuita@udea.edu.co
            jalexander.ramirez@udea.edu.co

Abstract

A second-order polynomial sequence is of Fibonacci-type (Lucas-type) if its Binet formula has a structure similar to that for Fibonacci (Lucas) numbers. Known examples of these types of sequences are Fibonacci polynomials, Pell polynomials, Fermat polynomials, Chebyshev polynomials, Morgan-Voyce polynomials, Lucas polynomials, Pell-Lucas polynomials, Fermat-Lucas polynomials, and Chebyshev polynomials.

The resultant of two polynomials is the determinant of the Sylvester matrix and the discriminant of a polynomial p is the resultant of p and its derivative. We study the resultant, the discriminant, and the derivatives of Fibonacci-type polynomials and Lucas-type polynomials as well the resultant of combinations of these two types of polynomials. As a corollary, we give explicit formulas for the resultant, the discriminant, and the derivative for the polynomials mentioned above.

1 Introduction

A second-order polynomial sequence is of Fibonacci-type F_{n} (Lucas-type L_{n}), if its Binet formula has a structure similar to that for Fibonacci (Lucas) numbers. These are known as generalized Fibonacci polynomials GFPs (see, for example [2, 9, 10, 13, 14]). Some known examples are Pell polynomials, Fermat polynomials, Chebyshev polynomials, Morgan-Voyce polynomials, Lucas polynomials, Pell-Lucas polynomials, Fermat-Lucas polynomials, Chebyshev polynomials, Vieta polynomials and Vieta-Lucas polynomials.

The resultant of two polynomials p and q, denoted by Res(p, q), is the determinant of the Sylvester matrix (see (7), [1, 4, 12, 23] or [28, p. 426]). Very often in mathematics we ask the question whether or not two polynomials share a root. In particular, if p and q are two GFPs, we ask whether or not p and q have a common root. Since the resultant of p and q is also the product of p evaluated at each root of q, the resultant of two GFPs can be used to answer this question.

Several authors have been interested in the resultant. The first formula for the resultant of two cyclotomic polynomials was given by Apostol [3]. Recently Bzdega et al. [5] computed the resultant of two cyclotomic polynomials in a short way. Some other papers have been dedicated to the study of the resultant of Chebyshev polynomials [7, 20, 25, 29]. In this paper we deduce simple closed formulas for the resultants of a big family of GFPs. For example, the resultant of Fibonacci polynomials F_{n} and the resultant of Chebyshev polynomials of second kind U_{n} are given by

Res(F_{m}, F_{n}) = 1 if gcd(m, n) = 1, and Res(F_{m}, F_{n}) = 0 otherwise,

and

Res(U_{m}, U_{n}) = (−4)^{(m−1)(n−1)/2} if gcd(m, n) = 1, and Res(U_{m}, U_{n}) = 0 otherwise.

If ν_{2}(n) is the 2-adic valuation of n, then the resultant of Lucas polynomial D_{n}(x) and the resultant of Chebyshev polynomials of the first kind T_{n}(x) are given by

Res(D_{m}, D_{n}) = 2^{δ} if ν_{2}(m) ≠ ν_{2}(n), and Res(D_{m}, D_{n}) = 0 otherwise

and

Res(T_{m}, T_{n}) = (−1)^{\frac{mn}{2}} 2^{(m−1)(n−1)−1}2^{δ} if ν_{2}(m) ≠ ν_{2}(n), and Res(T_{m}, T_{n}) = 0 otherwise.

The resultant for the family of Fibonacci-type polynomials is given in Theorem 1 and the resultant for the family of Lucas-type polynomials is given in Theorem 2. In Tables 1, 2, and 3 we give the resultants for some known polynomials.

Note that the resultant has been used to solve systems of polynomial equations (it encapsulates the solutions) [4, 21, 24, 27]. The resultant can also be used in combination with the elimination theory to answer other different types of questions about the multivariable polynomials.

The discriminant is the resultant of a polynomial and its derivative. If p is a GFP, we ask the question whether or not p has a repeated root. The discriminant helps to answer this question. In this paper we find simple closed formulas for the discriminant of both types of GFPs. For example, the discriminant of Fibonacci polynomials, the Chebyshev polynomials of the second kind, Lucas polynomials, and the Chebyshev polynomials first kind are given by

Disc(F_{n}) = (−1)^{(n−2)(n−1)/2}2^{n−1}n^{n−3}; Disc(U_{n}) = 2^{(n−1)^{2}}n^{n−3},

and

Disc(D_{n}) = (−1)^{n(n−1)/2}2^{n−1}n^{n}; Disc(T_{n}) = 2^{(n−1)^{2}}n^{n}.

The Theorems 4 and 5 give the discriminant of both families of Fibonacci-type polynomials and Lucas-type polynomials. Table 4 shows the discriminant for some known polynomials.

The following formulas generalize the formulas for the derivative of Fibonacci and Lucas polynomials given by several authors [2, 8, 15, 16, 17, 30] to Fibonacci-type polynomials and Lucas-type polynomials (for details see Theorem 6).

F′_{n} = \frac{d′ (nαL_{n} − dF_{n})}{(a − b)^{2}}  and  L′_{n} = \frac{nd′F_{n}}{α} .

2 Main Results

In this section we present the main theorems and corollaries of this paper. Proofs appear elsewhere in the paper. We give some brief definitions (needed to make the theorems readable), all formal definitions are given in Section 3.

For brevity and if there is no ambiguity we present the polynomials without explicit use of “x”. For example, instead of F_{n}(x) and L_{n}(x) we use F_{n} and L_{n}.

We say that a polynomial recurrence relation is of Fibonacci-type if it satisfies that (see also (2))

F_{0} = 0, F_{1} = 1, and F_{n} = dF_{n−1} + gF_{n−2} for n ≥ 2,

where d, and g are fixed non-zero polynomials in ℚ[x]. We say that a polynomial recurrence relation is of Lucas-type if it satisfies that (see also (3))

L_{0} = p_{0}, L_{1} = p_{1}, and L_{n} = dL_{n−1} + gL_{n−2} for n ≥ 2,

where |p_{0}| = 1 or 2 and p_{1}, d = αp_{1}, and g are fixed non-zero polynomials in ℚ[x] with α an integer of the form 2/p_{0}. These are called generalized Fibonacci polynomials (GFPs).

If P is a polynomial, then we use deg(P) and lc(P) to mean the degree of P and the leading coefficient of P. Let

β = lc(d), λ = lc(g), η = deg(d), ω = deg(g), and ρ = Res(g, d).  (1)

Let P and Q be polynomials with a_{n} = lc(P), b_{m} = lc(Q), n = deg(P), and m = deg(Q). If {x_{i}}_{i=1}^{n} and {y_{i}}_{i=1}^{m} are the roots of P and Q in ℂ, respectively, then the resultant of P and Q is given by Res(P, Q) = a_{n}^{m}b_{m}^{n} ∏_{i=1}^{n} ∏_{j=1}^{m}(x_{i} − y_{j}) = a_{n}^{m} ∏_{i=1}^{n} Q(x_{i}) = b_{m}^{n} ∏_{j=1}^{m} P (y_{j}). The discriminant of P, Disc(P ), is defined by (−1)^{n(n−1)/2}a_{n}^{2n−2}Π_{i≠j}(x_{i} − x_{j}).

We use ν_{2}(n) to represent the integer exponent base two of a positive integer n, which is defined to be the largest integer k such that 2^{k} | n (this concept is also known as the 2-adic order or 2-adic valuation of n).

2.1 Theorems of the resultant of GFPs

In this subsection we give simple expressions for the resultant of two GFPs of Fibonacci-type, Res(F_{n}, F_{m}), the resultant of two GFPs of Lucas-type, Res(L_{n}, L_{m}), and the resultant of two equivalent polynomials (Lucas-type and Fibonacci-type), Res(L_{n}, F_{m}).

The proof of Theorem 1 is in Section 4 on page 16, the proof of Theorem 2 is in Section 5 on page 17, and the proof of Theorem 3 is in Section 6 on page 18.

Theorem 1. Let T_{F} = ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{(n−1)(m−1)}{2}} , where n, m ∈ ℤ_{>0}. Then

Res(F_{n}, F_{m}) = \begin{cases}0, & if gcd(m, n) > 1; \\ T_{F}, & otherwise. \end{cases}

Theorem 2. Let T_{L} = α^{−η(n+m)}2^{η gcd(m,n)}((−1)^{ηω}β^{2η−ω}ρ)^{nm/2}, where m, n ∈ ℤ_{>0}. Then

Res(L_{m}, L_{n}) = \begin{cases}0, & if ν_{2}(n) = ν_{2}(m); \\ T_{L}, & if ν_{2}(n) ≠ ν_{2}(m). \end{cases}

Theorem 3. Let T_{LF} = 2^{η gcd(m,n)−η}α^{η(1−m)} ((−1)^{ηω}β^{2η−ω}ρ)^{(n(m−1))/2}, where n, m ∈ ℤ_{>0}. Then

Res(L_{n}, F_{m}) = \begin{cases}0, & if ν_{2}(n) < ν_{2}(m); \\ T_{LF}, & if ν_{2}(n) ≥ ν_{2}(m). \end{cases}

2.2 Theorems of the discriminant of GFPs

As an application of the theorems in the previous subsection we give the discriminants of generalized Fibonacci polynomials. The proofs of Theorems 4 and 5 are in Section 8.2 on page 25.

Theorem 4. If deg(d) = 1, g is a constant, and d′ is the derivative of d, then

Disc(F_{n}) = (−g)^{(n−2)(n−1)/2}2^{n−1}n^{n−3}β^{(n−1)(n−2)}.

Theorem 5. If deg(d) = 1, g is a constant, and d′ is the derivative of d, then

Disc(L_{n}) = (−g)^{n(n−1)/2}2^{n−1}n^{n}α^{2(1−n)}β^{n(n−1)}.

2.3 Theorem of the derivative of GFPs

The Theorem 6 gives formulas for the derivatives of Fibonacci-type polynomials and Lucastype polynomials. Note that the formulas given here are restricted to the special case in which g is a constant. The derivatives of a Lucas-type polynomials is given in term of its equivalent polynomial and the derivative of a Fibonacci-type polynomial is given in terms of Fibonacci-type and its equivalent.

Here we use F′_{n}, L′_{n}, a′, b′ and d′ to mean the derivatives of F_{n}, L_{n}, a, b and d with respect to x, where a and b are given in (4) and (5). The proof of the following theorem is in Section 7.

Theorem 6. If g is a constant, then

(i)

F′_{n} = \frac{d′ (ngF_{n−1} − dF_{n} + nF_{n+1})}{(a − b)^{2}} = \frac{d′ (nαL_{n} − dF_{n})}{(a − b)^{2}}.

(ii)

L′_{n} = \frac{nd′F_{n}}{α} .

2.4 Resultants, discriminants, and derivatives of known GFPs

In this subsection we construct several tables with the resultant, the discriminant and the derivative of some known polynomials (see Table 6).

The Table 1 presents the resultants of some GFPs of Fibonacci-type. The Table 2 presents the resultants of some GFPs of Lucas-type. The Table 3 presents the resultants of two equivalent polynomials (Lucas-type and its equivalent polynomial of Fibonacci-type). The Table 4 gives the discriminants of GFPs of both types. The first half of Table 4 has GFPs of Fibonacci-type and the second half has GFPs of Lucas-type. The Table 5 gives the derivatives for GFPs.

Note that the following property can be used to find the discriminant of a product of GFPs (see [6]). If P and Q are polynomials in ℚ[x], then Disc(P Q) = Disc(P )Disc(Q)Res(P, Q).

Polynomial  gcd(m, n) = 1  gcd(m, n) > 1
Fibonacci  Res(F_{m}, F_{n}) = 1  Res(F_{m}, F_{n}) = 0
Pell  Res(P_{m}, P_{n}) = 2^{(m−1)(n−1)}  Res(P_{m}, P_{n}) = 0
Fermat  Res(Φ_{m}, Φ_{n}) = (−18)^{(m−1)(n−1)/2} Res(Φ_{m}, Φ_{n}) = 0
Chebyshev 2nd kind Res(U_{m}, U_{n}) = (−4)^{(m−1)(n−1)/2} Res(U_{m}, U_{n}) = 0
Morgan-Voyce  Res(B_{m}, B_{n}) = (−1)^{(m−1)(n−1)/2} Res(B_{m}, B_{n}) = 0

Table 1: Resultants of Fibonacci-type polynomials using Theorem 1.

Polynomial  ν_{2}(m) ≠ ν_{2}(n), δ = gcd(m, n)  ν_{2}(m) = ν_{2}(n)
Lucas  Res(D_{m}, D_{n}) = 2^{δ}  Res(D_{m}, D_{n}) = 0
Pell-Lucas-prime  Res(Q′_{m}, Q′_{n}) = 2^{(m−1)(n−1)−1}2^{δ}  Res(Q′_{m}, Q′_{n}) = 0
Fermat-Lucas  Res(ϑ_{m}, ϑ_{n}) = (−1)^{mn/2}18^{mn/2} 2^{δ}  Res(ϑ_{m}, ϑ_{n}) = 0
Chebyshev 1st kind Res(T_{m}, T_{n}) = (−1)^{\frac{mn}{2}} 2^{(m−1)(n−1)−1}2^{δ} Res(T_{m}, T_{n}) = 0
Morgan-Voyce  Res(C_{m}, C_{n}) = (−1)^{\frac{mn}{2}} 2^{δ}  Res(C_{m}, C_{n}) = 0

Table 2: Resultants of Lucas-type polynomials using Theorem 2.

Polynomials  ν_{2}(n) ≥ ν_{2}(m), δ = gcd(n, m)  ν_{2}(n) < ν_{2}(m)
Lucas, Fibonacci  Res(D_{n}, F_{m}) = 2^{δ−1}  Res(D_{n}, F_{m}) = 0
Pell-Lucas-prime, Pell  Res(Q′_{n}, P_{m}) = 2^{(m−1)(n−1)} 2^{δ−1}  Res(Q′_{n}, P_{m}) = 0
Fermat-Lucas, Fermat  Res(ϑ_{n}, Φ_{m}) = (−18)^{n(m−1)/2} 2^{δ−1}  Res(ϑ_{n}, Φ_{m}) = 0
Chebyshev both kinds  Res(T_{n}, U_{m}) = (−1)^{n(m−1)/2}2^{(m−1)(n−1)}2^{δ−1} Res(T_{n}, U_{m}) = 0
Morgan-Voyce both types Res(C_{n}, B_{m}) = (−1)^{n(m−1)/2} 2^{δ−1}  Res(C_{n}, B_{m}) = 0

Table 3: Resultants of two equivalent polynomials using Theorem 3.

Evaluating the derivative of Fibonacci polynomials and the derivative of Lucas polynomials at x = 1 and x = 2 we obtain numerical sequences that appear in [26]. Thus,

\frac{d(F_{n})}{dx}|_{x=1} = \underline{A001629};  \frac{d(F_{n})}{dx}|_{x=2} = \underline{A006645};
\frac{d(D_{n})}{dx}|_{x=1} = \underline{A045925};  \frac{d(D_{n})}{dx}|_{x=2} = \underline{A093967}.

For the sequences generated by the derivatives of the other familiar polynomials studied here see: \underline{A001871}, \underline{A317404}, \underline{A317405}, \underline{A317408}, \underline{A317451}, 3(\underline{A045618}), 2(\underline{A006645}), and 2(\underline{A093967}).

Polynomial  Discriminants of GFPs  The OEIS
Fibonacci  Disc(F_{n}) = (−1)^{(n−2)(n−1)/2}2^{n−1}n^{n−3}  \underline{A317403}
Pell  Disc(P_{n}) = (−1)^{(n−2)(n−1)/2}2^{(n−1)^{2}}n^{n−3} \underline{A317450}
Fermat  Disc(Φ_{n}) = 2^{n(n−1)/2}3^{(n−1)(n−2)}n^{n−3}  \underline{A318184}
Chebyshev 2nd kind Disc(U_{n}) = 2^{(n−1)^{2}}n^{n−3}  \underline{A086804}
Morgan-Voyce  Disc(B_{n}) = 2^{n−1}n^{n−3}  \underline{A127670}
Lucas  Disc(D_{n}) = (−1)^{n(n−1)/2}2^{n−1}n^{n}  \underline{A193678}
Pell-Lucas-prime  Disc(Q′_{n}) = (−1)^{n(n−1)/2}2^{(n−1)^{2}}n^{n}  \underline{A007701}
Fermat-Lucas  Disc(ϑ_{n}) = 2^{(n−1)(n+2)/2}3^{n(n−1)}n^{n}  \underline{A318197}
Chebyshev 1st kind Disc(T_{n}) = 2^{(n−1)^{2}}n^{n}  \underline{A007701}
Morgan-Voyce  Disc(C_{n}) = 2^{n−1}n^{n}  \underline{A193678}

Table 4: Discriminants of GFPs using Theorems 4 and 5.

Fibonacci-type  Derivative  Lucas-Type  Derivative
Fibonacci  \frac{d(F_{n})}{dx} = \frac{nD_{n}−xF_{n}}{4+x^{2}}  Lucas  \frac{d(D_{n})}{dx} = nF_{n}
Pell  \frac{d(P_{n})}{dx} = \frac{nQ_{n}−2xP_{n}}{2(1+x^{2})}  Pell-Lucas-prime  \frac{d(Q_{n})}{dx} = 2nP_{n}
Fermat  \frac{d(Φ_{n})}{dx} = \frac{3(nϑ_{n}−3xΦ_{n})}{−8+9x^{2}} Fermat-Lucas  \frac{d(ϑ_{n})}{dx} = 3nΦ_{m}
Chebyshev 2nd kind \frac{d(U_{n})}{dx} = \frac{2nT_{n}−2xU_{n}}{2(x^{2}−1)}  Chebyshev 1st kind \frac{d(T_{n})}{dx} = nU_{m}
Morgan-Voyce  \frac{d(B_{n})}{dx} = \frac{nC_{n}−(x+2)B_{n}}{x(x+4)} Morgan-Voyce  \frac{d(C_{n})}{dx} = nB_{m}

Table 5: Derivatives of GFPs using Theorem 6.

3 Definitions, background, and basic results

In this section we give formal definitions of the concepts that we are going to use in this paper. Throughout the paper we consider polynomials in ℚ[x].

3.1 Second order polynomial sequences

In this section we introduce the generalized Fibonacci polynomial sequences. This definition gives rise to some known polynomial sequences (see, for example, Table 6 or [9, 10, 14, 22]). The polynomials in this subsection are presented in a formal way (with explicit use of “x”).

For the remaining part of this section we reproduce the definitions given by Flórez et al. [9, 10] for generalized Fibonacci polynomials. We now give the two second-order polynomial recurrence relations in which we divide the generalized Fibonacci polynomials.

F_{0}(x) = 0, F_{1}(x) = 1, and F_{n}(x) = d(x)F_{n−1}(x) + g(x)F_{n−2}(x) for n ≥ 2,  (2)

where d(x), and g(x) are fixed non-zero polynomials in ℚ[x].

We say a polynomial recurrence relation is of Fibonacci-type if it satisfies the relation given in (2), and of Lucas-type if:

L_{0}(x) = p_{0}, L_{1}(x) = p_{1}(x), and L_{n}(x) = d(x)L_{n−1}(x) + g(x)L_{n−2}(x) for n ≥ 2, (3)

where |p_{0}| = 1 or 2 and p_{1}(x), d(x) = αp_{1}(x), and g(x) are fixed non-zero polynomials in ℚ[x] with α an integer of the form 2/p_{0}.

To use similar notation for (2) and (3) on certain occasions we write p_{0} = 0, p_{1}(x) = 1 to indicate the initial conditions of Fibonacci-type polynomials. Some known examples of Fibonacci-type polynomials and Lucas-type polynomials are in Table 6 or in [9, 14, 18, 19, 22].

If G_{n} is either F_{n} or L_{n} for all n ≥ 0 and d^{2}(x) + 4g(x) > 0, then the explicit formula for the recurrence relations in (2) and (3) is given by

G_{n}(x) = t_{1}a^{n}(x) + t_{2}b^{n}(x),

where a(x) and b(x) are the solutions of the quadratic equation associated with the secondorder recurrence relation G_{n}(x). That is, a(x) and b(x) are the solutions of z^{2}−d(x)z−g(x) = 0. If α = 2/p_{0}, then the Binet formula for Fibonacci-type polynomials is stated in (4) and the Binet formula for Lucas-type polynomials is stated in (5) (for details on the construction of the two Binet formulas see [9])

F_{n}(x) = \frac{a^{n}(x) − b^{n}(x)}{a(x) − b(x)}  (4)

and

L_{n}(x) = \frac{a^{n}(x) + b^{n}(x)}{α}.  (5)

Note that for both types of sequence:

a(x) + b(x) = d(x), a(x)b(x) = −g(x), and a(x) − b(x) = \sqrt{d^{2}(x) + 4g(x)},

where d(x) and g(x) are the polynomials defined in (2) and (3).

A sequence of Lucas-type (Fibonacci-type) is equivalent or conjugate to a sequence of Fibonacci-type (Lucas-type), if their recursive sequences are determined by the same polynomials d(x) and g(x). Notice that two equivalent polynomials have the same a(x) and b(x) in their Binet representations. In [9, 10] there are examples of some known equivalent polynomials with their Binet formulas. The polynomials in Table 6 are organized by pairs of equivalent polynomials. For instance, Fibonacci and Lucas, Pell and Pell-Lucas, and so on.

Most of the following conditions were required in the papers that we are citing. Therefore, we require here that gcd(d(x), g(x)) = 1 and deg(g(x)) < deg(d(x)) for both types of sequences. (For instance these conditions hold for polynomial in Table 6.) The conditions in (6) also hold for Lucas-type polynomials;

gcd(p_{0}, p_{1}(x)) = 1, gcd(p_{0}, d(x)) = 1, gcd(p_{0}, g(x)) = 1, and that degree of L_{1} ≥ 1. (6)

Notice that in the definition of Pell-Lucas we have Q_{0}(x) = 2 and Q_{1}(x) = 2x. Thus, the gcd(2, 2x) = 2 ≠ 1. Therefore, Pell-Lucas does not satisfy the extra conditions that we imposed in (6). So, to resolve this inconsistency we use Q′_{n}(x) = Q_{n}(x)/2 instead of Q_{n}(x).

Polynomial  Initial value Initial value Recursive Formula
G_{0}(x) = p_{0}(x) G_{1}(x) = p_{1}(x) G_{n}(x) = d(x)G_{n−1}(x) + g(x)G_{n−2}(x)
Fibonacci  0  1  F_{n}(x) = xF_{n−1}(x) + F_{n−2}(x)
Lucas  2  x  D_{n}(x) = xD_{n−1}(x) + D_{n−2}(x)
Pell  0  1  P_{n}(x) = 2xP_{n−1}(x) + P_{n−2}(x)
Pell-Lucas  2  2x  Q_{n}(x) = 2xQ_{n−1}(x) + Q_{n−2}(x)
Pell-Lucas-prime  1  x  Q′_{n}(x) = 2xQ′_{n−1}(x) + Q′_{n−2}(x)
Fermat  0  1  Φ_{n}(x) = 3xΦ_{n−1}(x) − 2Φ_{n−2}(x)
Fermat-Lucas  2  3x  ϑ_{n}(x) = 3xϑ_{n−1}(x) − 2ϑ_{n−2}(x)
Chebyshev second kind 0  1  U_{n}(x) = 2xU_{n−1}(x) − U_{n−2}(x)
Chebyshev first kind  1  x  T_{n}(x) = 2xT_{n−1}(x) − T_{n−2}(x)
Morgan-Voyce  0  1  B_{n}(x) = (x + 2)B_{n−1}(x) − B_{n−2}(x)
Morgan-Voyce  2  x + 2  C_{n}(x) = (x + 2)C_{n−1}(x) − C_{n−2}(x)
Vieta  0  1  V_{n}(x) = xV_{n−1}(x) − V_{n−2}(x)
Vieta-Lucas  2  x  v_{n}(x) = xv_{n−1}(x) − v_{n−2}(x)

Table 6: Recurrence relation of some GFPs.

3.2 The resultant and the discriminant

In this section we use the Sylvester determinant to define the discriminant of two polynomials. For a complete development of the theory of the resultant of polynomials see [12].

Let P and Q be non-zero polynomials of degree n and m in ℚ[x], with

P = a_{n}x^{n} + a_{n−1}x^{n−1} + ⋯ + a_{1}x + a_{0} and Q = b_{m}x^{m} + b_{m−1}x^{m−1} + ⋯ + b_{1}x + b_{0}.

The resultant of P and Q, denoted by Res(P, Q), is the determinant of Syl(P, Q) (see, for example [1, 4, 20, 28]).

Syl(P, Q) = [ \begin{matrix}a_{n} & a_{n−1} & ⋯ & ⋯ & a_{0} & 0 & ⋯ & 0 \\ 0 & a_{n} & a_{n−1} & ⋯ & ⋯ & a_{0} & ⋯ & 0 \\ ⋮ & ... & … & … & … & ... & ... & ⋮ \\ ⋮ & ... & 0 & a_{n} & ⋯ & & ⋯ & a_{0} \\ b_{m} & b_{m−1} & ⋯ & ⋯ & b_{0} & 0 & ⋯ & 0 \\ 0 & b_{m} & b_{m−1} & ⋯ & ⋯ & b_{0} & ⋯ & 0 \\ ⋮ & ... & … & … & … & ... & ... & ⋮ \\ 0 & ... & 0 & b_{m} & b_{m−1} & ⋯ & ⋯ & b_{0} \end{matrix} ] .  (7)

If P′ is the derivative of P, then the discriminant of P is given by

Disc(P ) = (−1)^{\frac{n(n−1)}{2}} a_{n}^{−1}Res(P, P′).

Note that the discriminant can also be written as a Vandermonde determinant (see [23]).

3.3 Classic results

In this subsection we give some classic properties of the resultant needed to prove the main theorems. Most of the parts of the following lemma can be found in [4, 23].

Let f and h be polynomials, where a_{n} = lc(f), b_{m} = lc(h), n = deg(f) and m = deg(h). Note that if k is a constant, then Res(k, f) = Res(f, k) = k^{deg(f)}. The following lemma summarize some classic result about the resultant (see, for example, Jacobs et al. [20]).

Lemma 7. Let f, h, p, and q be polynomials in ℚ[x]. If n = deg(f), m = deg(h), and a_{n} = lc(f), then

(i) Res(f, h) = (−1)^{nm}Res(h, f),

(ii) Res(f, ph) = Res(f, p)Res(f, h),

(iii) Res(f, p^{k}) = Res(f, p)^{k},

(iv) if G = fq + h and r = deg(G), then Res(f, G) = a_{n}^{r−m}Res(f, h),

(v) Res(f, h) = 0 if and only if f and h have a common divisor of positive degree.

3.4 The GCD of two GFPs and other properties

Most of the results in this subsection are in [9]. Proposition 8 is a result that is in the proof of [9, Proposition 6] therefore its proof is omitted.

In this paper we use ℤ_{≥0} and ℤ_{>0} to mean the set of non-negative integers and positive integers, respectively. Recall that β, λ, η, ω, and ρ are defined in (1) on page 4 and that d, g and α are defined on page 8.

Proposition 8. Let m, n, r, and q be positive integers. If n = mq + r, then there is a polynomial T such that F_{n} = F_{m}T + gF_{mq−1}F_{r}.

Proposition 9. If m, r, and q are positive integers, then if r < m, then there is a polynomial T such that for t = ⌈\frac{q}{2}⌉ we have

L_{mq+r} = \begin{cases}L_{m}T + (−1)^{m(t−1)+t+r}(g)^{(t−1)m+r}L_{m−r}, & if q is odd; \\ L_{m}T + (−1)^{(m+1)t}(g)^{mt}L_{r}, & if q is even. \end{cases}

Proposition 10. If m, q, and r are nonnegative integers with q > 0, then

(i)

F_{mq+r} = \begin{cases}αL_{m}F_{m(q−1)+r} − (−g)^{m}F_{m(q−2)+r}, & if q > 1; \\ αL_{m}F_{r} + (−g)^{r}F_{m−r}, & if q = 1. \end{cases}

αL_{mq+r} = \begin{cases}(a − b)^{2}F_{m}F_{m(q−1)+r} + α(−g)^{m}L_{m(q−2)+r}, & if q > 1; \\ (a − b)^{2}F_{m}F_{r} + α(−g)^{r}L_{m−r}, & if q = 1. \end{cases}

Proof. We prove Part (i), the proof of Part (ii) is similar and it is omitted. If q = 1, then the proof follows from [9, Proposition 3]. We now prove the case in which q > 1. Using the Binet formulas (4) and (5), we obtain

αL_{m}F_{m(q−1)+r} = α\frac{(a^{m} + b^{m})}{α} \frac{(a^{m(q−1)+r} − b^{m(q−1)+r})}{a − b}.

Expanding and simplifying we have

αL_{m}F_{m(q−1)+r} = F_{mq+r} + (ab)^{m} \frac{a^{m(q−2)+r} − b^{m(q−2)+r}}{a − b} = F_{mq+r} + (−g)^{m}F_{m(q−2)+r}.

Solving this equation for F_{mq+r} we have F_{mq+r} = αL_{m}F_{m(q−1)+r} − (−g)^{m}F_{m(q−2)+r}. This completes the proof.

Lemma 11. Let k, n ∈ ℤ_{>0}. Then

(i) deg (F_{k}) = η(k − 1) and lc (F_{k}) = β^{k−1}.

(ii) deg (L_{n}) = ηn and lc(L_{n}) = β^{n}/α.

Proof. We use mathematical induction to prove all parts. We prove Part (i). Let P (k) be the statement:

deg (F_{k}) = η(k − 1) for every k ≥ 1.

The basis step, P (1), is clear, so we suppose that P (k) is true for k = t, where t > 1. Thus, we suppose that deg (F_{t}) = η(t−1) and we prove P (t+1). We know that deg(F_{n}) ≥ deg(F_{n−1}) for n ≥ 1. This, deg (d) > deg (g), and (2) imply

deg (F_{t+1}) = deg (dF_{t}) = deg(d) + deg (F_{t}) = η + η(t − 1) = ηt.

We now prove the second half of Part (i). Let Q(k) be the statement:

lc(F_{k}) = β^{k−1} for every k ≥ 1.

The basis step, Q(1), is clear, so we suppose that Q(k) is true for k = t, where t > 1. Thus, we suppose that lc (F_{t}) = β^{t−1} and we prove Q(t + 1). We know that deg(F_{n}) ≥ deg(F_{n−1}) for n ≥ 1. This, deg (d) > deg (g), and (2) imply lc(F_{t+1}) = lc(d) lc(F_{t}) = lc(d)β^{t−1} = ββ^{t−1} = β^{t}.

We prove Part (ii). Let H(n) be the statement: deg (L_{n}) = ηn for every n > 0. It is easy to see that H(1) is true. Suppose that H(n) is true for some n = k > 1. Thus, suppose that deg (L_{k}) = ηk and we prove H(k + 1). Since L_{k+1} = dL_{k} + gL_{k−1} and deg (d) > deg (g), we have deg (L_{k+1}) = deg (d) + deg (L_{k}) = η + ηk = η(k + 1). This proves the first half of Part (ii).

We now prove the second half of Part (ii). Let N(n) be the statement: lc(L_{n}) = β^{n} for every n > 0. (for simplicity we suppose that α = 1). It is easy to verify that lc(L_{1}) = β^{1}. Suppose that N(n) is true for some n = k > 1. Thus, suppose that lc(L_{k}) = β^{k}. Since L_{k+1} = dL_{k} + gL_{k−1} and deg (d) > deg (g), we have lc(L_{k+1}) = lc(d) lc(L_{k}). This and the inductive hypothesis imply that lc(L_{k+1}) = ββ^{k} = β^{k+1}.

Proposition 12 plays an important role in this paper. This in connection with Lemma 7 Part (v) gives criterions to determine whether or not the resultant of two GFPs is equal to zero (see Corollary 14, Corollary 16, and 18).

Recall that definition of ν_{2}(n) was given in Section 2 on page 4.

Proposition 12 ([9]). If m, n ∈ ℤ_{>0} and δ = gcd(m, n), then

(i) gcd(F_{m}, F_{n}) = 1 if and only if δ = 1.

(ii)

gcd (L_{m}, L_{n}) = \begin{cases}L_{δ}, & if ν_{2}(m) = ν_{2}(n); \\ gcd (L_{δ}, L_{0}) , & otherwise. \end{cases}

(iii)

gcd(L_{n}, F_{m}) = \begin{cases}L_{δ}, & if ν_{2}(m) > ν_{2}(n); \\ 1, & otherwise. \end{cases}

3.5 Some resultants of GFPs of Fibonacci-type polynomials

In this subsection we give some properties of the resultant of two GFPs and some results needed to prove Theorem 1.

Proposition 13. For m and n in ℤ_{≥0} these hold

(i) If n > 0, then Res(g, F_{n}) = ρ^{n−1},

(ii) Res(F_{m}, gF_{n}) = (−1)^{ωη(m−1)}ρ^{m−1}Res(F_{m}, F_{n}),

(iii) Res(L_{m}, gL_{n}) = (−1)^{ωηm}ρ^{m}Res(L_{m}, L_{n}).

Proof. We prove Part (i) using mathematical induction. Let P (n) be the statement:

Res(g, F_{n}) = ρ^{n−1} for every n ≥ 1.

Since F_{1} = 1 the basis step, P (1), is clear. Suppose that P (n) is true for n = k, where k > 1. Thus, suppose that Res(g, F_{k}) = ρ^{k−1}, and we prove P (k + 1). From (2) and Lemma 7 Parts (ii) and (iv) we have

Res(g, F_{k+1}) = Res(g, dF_{k} + gF_{k−1}) = λ^{ηk−(η+η(k−1))}Res(g, d)Res(g, F_{k}).

This and P (k) imply Res(g, F_{n}) = Res(g, d)Res(g, F_{n−1}) = ρ^{n−1}.

We prove Part (ii), the proof of Part (iii) is similar and it is omitted. From Lemma 7 Part (i) and Part (ii), and Lemma 11 Part (i), we have

Res(F_{m}, gF_{n}) = Res(F_{m}, g)Res(F_{m}, F_{n})
= (−1)^{ωη(m−1)}Res(g, F_{m})Res(F_{m}, F_{n})
= (−1)^{ωη(m−1)}ρ^{m−1}Res(F_{m}, F_{n}).

This completes the proof.

The proof of the following corollary is straightforward from Proposition 12 Part (i) and Lemma 7 Part (v).

Corollary 14. Let m, n ∈ ℤ_{>0}. Then gcd (m, n) = 1 if and only if Res(F_{m}, F_{n}) ≠ 0.

Proposition 15. If m, n and q are positive integers, with n > 1 and mq > 1, then

(i) Res(F_{n}, F_{n−1}) = ((−1)^{ωη}β^{2η−ω}ρ)^{(n−2)(n−1)/2},

(ii) Res(F_{m}, F_{mq−1}) = ((−1)^{ηω}β^{2η−ω}ρ)^{(m−1)(mq−2)/2}.

Proof. We prove all parts by mathematical induction. Proof of Part (i). Let Q(n) be the statement:

Res(F_{n}, F_{n−1}) = ((−1)^{ωη}β^{2η−ω}ρ)^{(n−2)(n−1)/2} for every n ≥ 2.

Since F_{1} = 1, the basis step, Q(2), is clear. Suppose that Q(n) is true for n = k − 1, where k > 2. Thus, suppose that Res(F_{k−1}, F_{k−2}) = ((−1)^{ωη}β^{2η−ω}ρ)^{(k−3)(k−2)/2}. We prove Q(k). Using Lemma 11 Part (i) and Lemma 7 Part (i) we get

Res(F_{n}, F_{n−1}) = (−1)^{η^{2}(n−1)(n−2)}Res(F_{n−1}, F_{n})
= Res(F_{n−1}, dF_{n−1} + gF_{n−2}).

This, Lemma 7 Part (iv), Lemma 11 Part (i) and Proposition 13 Part (ii) imply

Res(F_{n}, F_{n−1}) = (β^{n−2})^{η(n−1)−(ω+η(n−3))}Res(F_{n−1}, gF_{n−2})
= (−1)^{ωη(n−2)}β^{(n−2)(2η−ω)}ρ^{n−2}Res(F_{n−1}, F_{n−2}).

Simplifying we have

Res(F_{n}, F_{n−1}) = ((−1)^{ωη}β^{2η−ω}ρ)^{n−2} Res(F_{n−1}, F_{n−2}).

This and Q(k − 1) give

Res(F_{n}, F_{n−1}) = ((−1)^{ωη}β^{2η−ω}ρ)^{n−2} (β^{2η−ω}(−1)^{ωη}ρ)^{\frac{(n−3)(n−2)}{2}}
= ((−1)^{ωη}β^{2η−ω}ρ)^{\frac{(n−2)(n−1)}{2}} .

Proof of Part (ii). The proof is straightforward when m = 1 and q > 1. Therefore, we suppose that m > 1. Let W (q) be the statement: for a fixed integer m we have

Res(F_{m}, F_{mq−1}) = ((−1)^{ηω}β^{2η−ω}ρ)^{(m−1)(mq−2)/2} for every q ≥ 1.

From Part (i) it follows that W (1) is true. Suppose that W (q) is true for q = k, where k > 1. Thus, suppose that Res(F_{m}, F_{mk−1}) = ((−1)^{ηω}β^{2η−ω}ρ)^{(m−1)(mk−2)/2}. We prove W (k + 1). From Proposition 8 we know that there is a polynomial T such that F_{m(k+1)−1} = F_{m}T + gF_{km−1}F_{m−1}. This, Lemma 7 Part (ii) and Part (iv) and Proposition 13 Part (ii) imply

Res(F_{m}, F_{m(k+1)−1}) = Res(F_{m}, F_{m}T + gF_{km−1}F_{m−1})
= (β^{m−1})^{η(mk+m−2)−(ω+η(mk+m−4))}Res(F_{m}, gF_{km−1}F_{m−1})
= β^{(m−1)(2η−ω)}Res(F_{m}, gF_{km−1}F_{m−1})
= (−1)^{ωη(m−1)}β^{(m−1)(2η−ω)}ρ^{m−1}Res(F_{m}, F_{m−1})Res(F_{m}, F_{km−1}).

From this, Part (i), and W (k) we conclude

Res(F_{m}, F_{mk+m−1}) = ((−1)^{ωη}β^{2η−ω}ρ)^{m−1} ((−1)^{ωη}β^{2η−ω}ρ)^{\frac{(m−1)(m−2)}{2}} Res(F_{m}, F_{km−1})
= ((−1)^{ωη}β^{2η−ω}ρ)^{\frac{(m−1)m}{2}} Res(F_{m}, F_{km−1})
= ((−1)^{ωη}β^{2η−ω}ρ)^{\frac{(m−1)m}{2}} ((−1)^{ωη}β^{2η−ω}ρ)^{\frac{(m−1)(mk−2)}{2}} .

Simplifying the last expression, we have

Res(F_{m}, F_{mk+m−1}) = ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{(m−1)(m(k+1)−2)}{2}} .

This completes the proof.

3.6 Some resultants of GFPs of Lucas-type polynomials

In this subsection we give some properties of the resultant of two GFPs of Lucas-type and some results needed to prove Theorem 2.

Recall that a GFP of Lucas-type is a polynomial sequence such that L_{0} ∈ {1, 2}, L_{1} = 2^{−1}L_{0}d, and L_{n} = dL_{n−1} + gL_{n−2} for n > 1.

Note that if we take the particular case of the Lucas-type sequence L_{n} in which L_{0} = 1 and L_{1} = 2^{−1}d, then using the initial conditions we define a new Lucas-type sequence as follows: Let \overline{L_{0}} = 2L_{0}, \overline{L_{1}} = 2L_{1} = d and \overline{L_{n}} = d\overline{L_{n−1}} + g\overline{L_{n−2}} for n > 1. It is easy to verify that \overline{L_{n}} = 2L_{n} for n ≥ 0. Therefore, to find the resultant of a polynomial of Lucas-type L_{n}, it is enough to find the resultant for L_{n} in which L_{0} = 2.

The following corollary is a direct consequence of Proposition 12 Part (ii), Lemma 7 Part (v), and Lemma 11 Part (ii). So, we omit its proof.

Corollary 16. Let m, n ∈ ℤ_{>0}. ν_{2}(n) = ν_{2}(m) if and only if Res(L_{m}, L_{n}) = 0.

Proposition 17. If n ∈ ℤ_{>0} and L_{0} = 2, then

(i) Res(g, L_{n}) = ρ^{n},

(ii)

Res(L_{1}, L_{n}) = \begin{cases}0, & if n is odd; \\ 2^{η} ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{n}{2}} , & if n is even. \end{cases}

Proof. We prove Part (i) by mathematical induction on n. Since Res(g, L_{1}) = Res(g, d) = ρ, it holds that the result is true for n = 1. Suppose that for some integer n = k > 1, Res(g, L_{k}) = ρ^{k} holds. From (3) and Lemma 7 Parts (ii) and (iv) we have

Res(g, L_{k+1}) = Res(g, dL_{k} + gL_{k−1}) = α^{η(k+1)−η(k+1)}Res(g, d)Res(g, L_{k}).

This and the inductive hypothesis imply that

Res(g, L_{k+1}) = Res(g, d)Res(g, L_{k}) = Res(g, d)^{k+1},

which is our claim.

We prove Part (ii) by induction on n. Let Q(n) be the statement:

Res(L_{1}, L_{n}) = \begin{cases}0, & if n is odd; \\ 2^{η} ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{n}{2}} , & if n is even. \end{cases}

Since Res(L_{1}, L_{1}) = 0, Q(1) holds. Note that L_{1} = (p_{0}/2)d = d. This and Lemma 7 Parts (ii) and (iv) imply that

Res(L_{1}, L_{2}) = Res(L_{1}, dL_{1} + gL_{0}) = β^{2η−ω}Res(d, 2g) = 2^{η}(−1)^{ηω}β^{2η−ω}ρ.

This proves Q(2).

Suppose that Q(k − 2) and Q(k − 1) is true and we prove Q(k). Note that if k is odd by Corollary 16 we have that Res(L_{1}, L_{n}) = 0. We suppose that k is even. Lemma 7 Parts (ii) and (iv), L_{1} = d, Lemma 11 Part (ii), and (5) imply

Res(L_{1}, L_{k}) = Res(d, dL_{k−1} + gL_{k−2})
= β^{2η−ω}Res(d, gL_{k−2})
= β^{2η−ω}Res(d, g)Res(d, L_{k−2})
= ((−1)^{ηω}β^{2η−ω}ρ) Res(L_{1}, L_{k−2}).

Note that k − 2 and k have the same parity. This and Q(k − 2) imply that

Res(L_{1}, L_{k}) = ((−1)^{ηω}β^{2η−ω}ρ) 2^{η} ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{k−2}{2}} = 2^{η} ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{k}{2}} .

This proves Q(k).

4 Proof of Theorem 1

In this section we prove the theorem about the resultant of two GFPs of Fibonacci-type (see Section 2).

Proof of Theorem 1. Let A be the set of all i ∈ ℤ_{>0} such that for every j ∈ ℤ_{>0} we have

Res(F_{i}, F_{j}) = \begin{cases}0, & if gcd(m, n) > 1; \\ ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{(i−1)(j−1)}{2}} , & otherwise. \end{cases}  (8)

Since 1 ∈ A, we have that A ≠ ∅. The following claim completes the proof of the Theorem. Claim. A = ℤ_{>0}. Proof of Claim. Suppose B := ℤ_{>0} \ A is a non-empty set. Let n ≠ 1 be the least element of B. So, there is h ∈ ℤ_{>0} such that Res(F_{n}, F_{h}) does not satisfy Property (8) (if m = n, then Res(F_{i}, F_{j}) = 0). Let m be the least element of the non-empty set H = {h ∈ ℤ_{>0} | Res(F_{n}, F_{h}) does not satisfy (8)}. Note that Corollary 14 and (8) imply that gcd(m, n) = 1. We now consider two cases.

Case m < n. Since n is the minimum element of B, m ∈ A. Either m or n is odd, because gcd(m, n) = 1. We know, from Lemma 11 Part (i), that deg(F_{m}) = η(m − 1). This implies that Res(F_{n}, F_{m}) = Res(F_{m}, F_{n}). Since m ∈ A, we have that (8) holds for j ∈ ℤ_{>0}, in particular (8) holds when j = n. That is a contradiction.

Case n < m. The Euclidean algorithm and gcd(m, n) = 1 guarantee that there are q, r ∈ ℤ such that m = nq + r with 0 < r < n. We now can proceed analogously to the proof of Proposition 15 Part (ii). From the Euclidean algorithm, Proposition 8 and Lemma 7 Part (iv) we have

Res(F_{n}, F_{m}) = Res(F_{n}, F_{nq+r})
= Res(F_{n}, F_{n}T + gF_{nq−1}F_{r})
= (β^{n−1})^{η(m−1)−(ω+η(nq−2+r−1))}Res(F_{n}, gF_{nq−1}F_{r}).

This, Lemma 7 Part (ii) and Proposition 13 Part (ii) imply

Res(F_{n}, F_{m}) = β^{(n−1)(2η−ω)}Res(F_{n}, gF_{r})Res(F_{n}, F_{nq−1})
= (−1)^{ωη(n−1)}β^{(n−1)(2η−ω)}ρ^{n−1}((−1)^{ηω}β^{2η−ω}ρ)^{\frac{(n−1)(nq−2)}{2}} Res(F_{n}, F_{r})
= ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{(n−1)nq}{2}} Res(F_{n}, F_{r}).  (9)

Since gcd(n, m) = gcd(n, r) = 1, either n or r is odd. So, Res(F_{n}, F_{r}) = Res(F_{r}, F_{n}). It is easy to verify that r ∈ A, because r < n and n is the minimum element of B. Set j = n, so Res(F_{r}, F_{n}) = ((−1)^{ηω}β^{2η−ω}ρ)^{(n−1)(r−1)/2}. This and (9) imply that

Res(F_{n}, F_{m}) = ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{(m−1)(n−r)}{2}} ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{(r−1)(m−1)}{2}}
= ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{(n−1)(m−1)}{2}} .

That is a contradiction. This implies that A = ℤ_{>0}.

5 Proof of Theorem 2

In this section we prove the theorem about the resultant of two GFPs of Lucas-type (see Section 2).

Proof of Theorem 2. We consider two cases: L_{0} = 2 and L_{0} = 1. We first prove the case L_{0} = 2. Let A = {i ∈ ℤ_{>0} | ∀j ∈ ℤ_{>0}, Property (10) holds for Res(L_{i}, L_{j})}.

Res(L_{i}, L_{j}) = \begin{cases}0, & if ν_{2}(i) = ν_{2}(j); \\ 2^{η gcd(i,j)} ((−1)^{ηω}β^{2η−ω}ρ)^{ij/2} , & if ν_{2}(i) ≠ ν_{2}(j). \end{cases}  (10)

From Proposition 17 Part (ii) we have that i = 1 ∈ A. So, A ≠ ∅. The following claim completes the proof part L_{0} = 2. Claim. A = ℤ_{>0}. Proof of Claim. Suppose B := ℤ_{>0}\A is a non-empty set. Let n ≠ 1 be the least element of B. So, there is h ∈ ℤ_{>0} such that Res(L_{n}, L_{h}) does not satisfy Property (10). Let m be the least element of the non-empty set H = {h ∈ ℤ_{>0} | Res(L_{n}, L_{h}) does not satisfy Property (10)}. Note that if ν_{2}(n) = ν_{2}(m), then Res(L_{n}, L_{m}) = 0 (by Corollary 16). That is a contradiction by the definition of H. Therefore, we have that ν_{2}(n) ≠ ν_{2}(m). So, n ≠ m, where at least one of them is even. This implies that Res(L_{m}, L_{n}) = (−1)^{η^{2}mn}Res(L_{n}, L_{m}) = Res(L_{n}, L_{m}). Therefore, Res(L_{m}, L_{n}) does not satisfy (10). So, m ∉ A. Since n ≠ m is the least element of B, we have m > n. From the Euclidean algorithms we know that there are q, r ∈ ℤ_{≥0} such that m = nq + r with 0 ≤ r < n.

We now proceed by cases over q.

Case q odd Suppose that q = 2t − 1. Note that t = ⌈q/2⌉ and that (m − n + r)/2 = (t − 1)n + r. Since ν_{2}(n) ≠ ν_{2}(m), r ≠ 0 and n(n − r) is even. This, Proposition 9 for odd case, and Lemma 7 Part (ii) and Part (iv) imply that Res(L_{n}, L_{m}) equals

Res(L_{n}, L_{nq+r}) = Res(L_{n}, L_{n}T + (−1)^{t(n+1)+r−n}g^{(t−1)n+r}L_{n−r})
= Res(L_{n}, L_{n}T + (−1)^{t(n+1)+r−n}g^{\frac{m−n+r}{2}} L_{n−r})
= (β^{n})^{ηm− \frac{ω(m−n+r)}{2} −η(n−r)}Res(L_{n}, (−1)^{t(n+1)+r−n}g^{\frac{m−n+r}{2}} L_{n−r})
= β^{\frac{n(m−n+r)(2η−ω)}{2}} Res(L_{n}, (−1)^{t(n+1)+r−n}g^{\frac{m−n+r}{2}} L_{n−r}).

Note that Res(L_{n}, (−1)^{t(n+1)+r−n}) = (−1)^{ηn(t(n+1)+r−n)} = (−1)^{ηn(r−n)} = 1. This, Lemma 7 Parts (iii) and (iv) and Proposition 17 Part (i) imply

Res(L_{n}, L_{m}) = β^{\frac{n(m−n+r)(2η−ω)}{2}} Res(L_{n}, g^{\frac{m−n+r}{2}} L_{n−r})
= β^{\frac{n(m−n+r)(2η−ω)}{2}} Res(L_{n}, g)^{\frac{n−m+r}{2}} Res(L_{n}, L_{n−r})
= β^{\frac{n(m−n+r)(2η−ω)}{2}} ((−1)^{ηnω}ρ^{n})^{\frac{m−n+r}{2}} Res(L_{n}, L_{n−r}).

Thus,

Res(L_{n}, L_{m}) = ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{(m−n+r)}{2} n} Res(L_{n−r}, L_{n}).  (11)

Since n(n − r) is even, we have that Res(L_{n−r}, L_{n}) = Res(L_{n}, L_{n−r}). This and m > n − r imply that Res(L_{n}, L_{n−r}) satisfies (10). From this and (11) we have

Res(L_{n}, L_{m}) = ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{(m−n+r)}{2} n} 2^{η gcd(n−r,n)} ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{n(n−r)}{2}}
= 2^{η gcd(n,m)} ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{nm}{2}} .

Thus, Res(L_{n}, L_{m}) satisfies (10). That is a contradiction.

Case q is even Let q = 2t. Note that t = ⌈q/2⌉. Using Proposition 9 for the even case, Lemma 7 Parts (ii) and (iv) and following a similarly procedure as in the proof of the case q = 2t+1 we obtain Res(L_{n}, L_{m}) = β^{(2η−ω)(m−r)n/2}Res(L_{n}, (−1)^{(n+1)t})Res(L_{n}, g^{nt}L_{r}). This, the fact that Res(L_{n}, (−1)^{(n+1)t}) = 1 and following a similar procedure as in the proof of the case q = 2t+1 we obtain that Res(L_{n}, L_{m}) = ((−1)^{ηω}β^{2η−ω}ρ)^{(m−r)n/2}Res(L_{r}, L_{n}). Since r < n, we have r ∉ B. Therefore, r ∈ A. This implies that

Res(L_{n}, L_{m}) = ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{(m−r)n}{2}} Res(L_{r}, L_{n})
= ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{(m−r)n}{2}} ((−1)^{ηω}2^{η gcd(r,n)}β^{2η−ω}ρ)^{\frac{nr}{2}}
= 2^{η gcd(n,m)} ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{mn}{2}} .

Thus, Res(L_{n}, L_{m}) satisfies (10). That is a contradiction. This completes the proof that A = ℤ_{>0}.

We now prove the case L_{0} = 1. It is easy to see that \overline{L_{n}} = 2L_{n} is a GFP sequence of Lucas-type, where \overline{L_{0}} = 2. This and the previous case imply

Res(\overline{L_{m}}, \overline{L_{n}}) = \begin{cases}0, & if ν_{2}(n) = ν_{2}(m); \\ 2^{η gcd(m,n)} ((−1)^{ηω}β^{2η−ω}ρ)^{\frac{nm}{2}} , & if ν_{2}(n) ≠ ν_{2}(m). \end{cases}

Since Res(\overline{L_{m}}, \overline{L_{n}}) = Res(2L_{m}, 2L_{n}), we have

Res(\overline{L_{m}}, \overline{L_{n}}) = Res(2, L_{n})Res(L_{m}, 2)Res(L_{m}, L_{n}) = 2^{(n+m)η}Res(L_{m}, L_{n}).

Therefore, Res(L_{m}, L_{n}) = 2^{−(n+m)η}Res(\overline{L_{m}}, \overline{L_{n}}), completing the proof.

6 Proof of Theorem 3

In this section we prove the theorem about the resultant of two equivalent polynomials (see Section 2).

The following corollary is a consequence of Proposition 12 Part (iii). So, we omit its proof.

Corollary 18. Let m, n ∈ ℤ_{>0}. ν_{2}(n) < ν_{2}(m) if and only if Res(L_{n}, F_{m}) = 0.

Proof of Theorem 3. We consider two cases: α = 1 and α = 2. We prove the case α = 1, the case α = 2 is similar and it is omitted. Let

A = {i ∈ ℤ_{>0} | ∀j ∈ ℤ_{>0}, Property (12) holds for Res(L_{j}, F_{i})}.

Res(L_{j}, F_{i}) = \begin{cases}0, & if ν_{2}(j) < ν_{2}(i); \\ 2^{η gcd(i,j)−η} ((−1)^{ηω}β^{2η−ω}ρ)^{(j(i−1))/2} , & if ν_{2}(j) ≥ ν_{2}(i). \end{cases}  (12)

Since Res(L_{j}, 1) = 1, we have i = 1 ∈ A. So, A ≠ ∅. The following claim completes the proof. Claim. A = ℤ_{>0}. Proof of Claim. Suppose B := ℤ_{>0} \ A is a non-empty set. Let m ≠ 1 be the least element of B. So, there is h ∈ ℤ_{>0} such that Res(L_{h}, F_{m}) does not satisfy (12). Let n be the least element of the non-empty set H = {h ∈ ℤ_{>0} | Res(L_{h}, F_{m}) does not satisfy (12)}.

If ν_{2}(n) < ν_{2}(m), then by Corollary 18 it holds Res(L_{n}, F_{m}) = 0. This and (12) imply that m ∈ A. That is a contradiction. Let us suppose that ν_{2}(n) ≥ ν_{2}(m).

We now analyze cases on m.

Case m = n. Note that Res(L_{n}, F_{n}) = Res(F_{n}, L_{n}). From Proposition 10 Part (i) with r = q = α = 1 (if α ≠ 1 is similar) we have L_{n} = F_{n+1} + gF_{n−1} = dF_{n} + 2gF_{n−1}. Therefore, Res(F_{n}, L_{n}) equals

Res(F_{n}, dF_{n} + 2gF_{n−1}) = (β^{n−1})^{ηn−(ω+η(n−2))}Res(F_{n}, 2gF_{n−1})
= β^{(n−1)(2−ω)}2^{η(n−1)}Res(F_{n}, gF_{n−1}).

By Lemma 7 Part (ii) and Proposition 15 Part (i), we have

Res(F_{n}, dF_{n} + 2gF_{n−1}) = 2^{η(n−1)}(−1)^{ωη(n−1)}(β^{(n−1)(2η−ω)})ρ^{n−1}((−1)^{ωη}β^{2η−ω}ρ)^{\frac{(n−2)(n−1)}{2}} .
= 2^{η(n−1)}((−1)^{ωη}β^{2η−ω}ρ)^{\frac{n(n−1)}{2}} .

So, Res(L_{n}, F_{n}) satisfies Property (12). That is contradiction. Therefore m ≠ n.

Case m > n. From the Euclidean algorithm we know that m = nq + r for 0 ≤ r < n. We consider two sub-cases on q.

Sub-case q = 1. Note that 0 < r < n. So, m = n+r and Res(L_{n}, F_{m}) = Res(L_{n}, F_{n+r}). This and Proposition 10 Part (i) imply that Res(L_{n}, F_{m}) = Res(L_{n}, αL_{n}F_{r} + (−g)^{r}F_{n−r}). Therefore,

Res(L_{n}, F_{m}) = β^{n(η(n+r−1)−(ωr+η(n−r−1)))}Res(L_{n}, (−g)^{r}F_{n−r})
= β^{n(2η−ω)r}Res(L_{n}, (−g)^{r}F_{n−r})
= β^{n(2η−ω)r}Res(L_{n}, (−g)^{r})Res(L_{n}, F_{n−r}).

This and Lemma 7 Part (iii) imply

Res(L_{n}, F_{m}) = β^{n(2η−ω)r} (Res(L_{n}, −g))^{r} Res(L_{n}, F_{n−r})
= β^{n(2η−ω)r} (Res(L_{n}, −1))^{r} (Res(L_{n}, g))^{r} Res(L_{n}, F_{n−r})
= β^{n(2η−ω)r}(−1)^{rηn}(−1)^{ηnωr}Res(g, L_{n})^{r}Res(L_{n}, F_{n−r}).

Therefore,

Res(L_{n}, F_{m}) = (−1)^{ηnr(ω+1)}β^{n(2η−ω)r}ρ^{nr}Res(L_{n}, F_{n−r}).  (13)

Since m > n − r is the least element of B, we have n − r ∈ A. Therefore, it holds Res(L_{n}, F_{n−r}) = 2^{η(gcd(n,n−r)−1)} ((−1)^{ηω}β^{2η−w}ρ)^{n(n−r−1)/2}. This and (13) (after simplifications) imply that

Res(L_{n}, F_{m}) = 2^{η(gcd(n,n−r)−1)} [β^{2η−ω}(−1)^{ηω}ρ]^{\frac{n(n+r−1)}{2}} .

Therefore, Res(L_{n}, F_{m}) satisfies Property (12). That is a contradiction. Thus, m ≠ n + r.

Sub-case q > 1. Since Res(L_{n}, F_{m}) = Res(L_{n}, F_{nq+r}), by Proposition 10 Part (i) we have Res(L_{n}, F_{m}) = Res(L_{n}, αL_{n}F_{n(q−1)+r} − (−g)^{n}F_{n(q−2)+r}). This, Lemma 7 Part (iv) and the fact that n(η(nq + r − 1) − (ωn + η(n(q − 2) + r − 1))) = (2η − ω)n^{2}, imply that

Res(L_{n}, F_{m}) = β^{n(2η−ω)n}Res(L_{n}, −(−g)^{n}F_{n(q−2)+r}).

So, by Lemma 7 we have

Res(L_{n}, F_{m}) = β^{(2η−ω)n^{2}}Res(L_{n}, −1)Res(L_{n}, (−g)^{n})Res(L_{n}, F_{n(q−2)+r})
= β^{(2η−ω)n^{2}}(−1)^{ηn} (Res(L_{n}, −g))^{n} Res(L_{n}, F_{n(q−2)+r})
= β^{(2η−ω)n^{2}}(−1)^{ηn}(−1)^{ηn^{2}} (Res(L_{n}, g))^{n} Res(L_{n}, F_{n(q−2)+r})
= β^{(2η−ω)n^{2}}(−1)^{ηn(n+1)} (Res(L_{n}, g))^{n} Res(L_{n}, F_{n(q−2)+r})
= β^{(2η−ω)n^{2}} (Res(L_{n}, g))^{n} Res(L_{n}, F_{n(q−2)+r}).

Since m is the least element of B and n(q − 2) + r < m, we have that n(q − 2) + r ∈ A. This and ν_{2}(n) ≥ ν_{2}(m) = ν_{2}(n(q − 2) + r) imply that

Res(L_{n}, F_{n(q−2)+r}) = 2^{η(gcd(n,n(q−2)+r)−1)} [(−1)^{ηω}β^{2η−ω}ρ]^{\frac{n(n(q−2)+r−1)}{2}} .

Note that gcd(n, n(q − 2) + r) = gcd(n, nq + r). Therefore,

Res(L_{n}, F_{m}) = β^{(2η−ω)n^{2}}(−1)^{ηnω}ρ^{n^{2}}2^{η(gcd(n,n(q−2)+r)−1)} [(−1)^{ηω}β^{2η−ω}ρ]^{\frac{n(n(q−2)+r−1)}{2}}
= β^{\frac{(2η−ω)n(nq+r−1)}{2}} (−1)^{\frac{ηωn(nq+r−1)}{2}} ρ^{\frac{n(nq+r−1)}{2}} 2^{η(gcd(n,nq+r)−1)}
= 2^{η(gcd(n,nq+r)−1)} [(−1)^{ηω}β^{2η−ω}ρ]^{\frac{n(nq+r−1)}{2}} .

From this we conclude that Res(L_{n}, F_{m}) = 2^{η(gcd(n,m)−1)} [β^{2η−ω}(−1)^{ηω}ρ]^{\frac{n(m−1)}{2}} . Therefore, m ∈ A. That is is a contradiction.

Case m < n. From the Euclidean algorithm we know that n = mq + r for 0 ≤ r < m. There are two sub-cases to consider, q = 1 and q > 1. However, we prove only the case in which q = 1, the other case is similar and it is omitted.

Sub-case q = 1. In this case r ≠ 0, for r = 0 see the case m = n. Therefore, Res(L_{n}, F_{m}) = Res(L_{m+r}, F_{m}). This, Proposition 10 Part (ii) and Lemma 7 Parts (i) imply that

Res(L_{n}, F_{m}) = Res( ((a − b)^{2}/α) F_{m}F_{r} + (−g)^{r}L_{m−r}, F_{m})
= (−1)^{η^{2}(m+r)(m−1)}Res(F_{m}, \frac{(a − b)^{2}}{α} F_{m}F_{r} + (−g)^{r}L_{m−r}).

Note that (m ± r)(m − 1) and r(m − 1) are even (it is clear if m is odd), if m is even, then 1 ≤ ν_{2}(m) ≤ ν_{2}(n = m + r). So, both n and r are even. Therefore, we have (−1)^{η^{2}(m+r)(m−1)} = (−1)^{η(m−1)r} = (−1)^{η(m−1)ωr} = 1.

From Lemma 7 Parts (ii), (iii), and (iv) we have

Res(L_{n}, F_{m}) = (β^{m−1})^{η(m+r)−(η(m−r)+ωr)}Res(F_{m}, (−g)^{r}L_{m−r})
= β^{(m−1)(2η−ω)r}Res(F_{m}, (−g)^{r})Res(F_{m}, L_{m−r})
= β^{(m−1)(2η−ω)r}Res(F_{m}, (−g)^{r})Res(F_{m}, L_{m−r})
= β^{(m−1)(2η−ω)r}Res(F_{m}, (−1)^{r})Res(F_{m}, g^{r})Res(F_{m}, L_{m−r})
= β^{(m−1)(2η−ω)r}(−1)^{η(m−1)r}(−1)^{η(m−1)ωr}Res(g, F_{m})^{r}Res(F_{m}, L_{m−r})
= β^{(m−1)(2η−ω)r}ρ^{r(m−1)}(−1)^{η^{2}(m−1)(m−r)}Res(L_{m−r}, F_{m})
= (−1)^{η(m−1)(1+ω)r}β^{(m−1)(2η−ω)r}ρ^{r(m−1)}Res(L_{m−r}, F_{m}).

Since n = m + r is the least element of H, we have that (m − r) ∉ H. Therefore, Res(L_{m−r}, F_{m}) satisfies (12). So,

Res(L_{n}, F_{m}) = β^{(m−1)(2η−ω)r}ρ^{r(m−1)}2^{η(gcd(m−r,m)−1)} (β^{2η−ω}ρ)^{(m−r)(m−1)/2} .

Since 0 < r < m, the gcd(m − r, m) = 1. This (after some simplifications) implies that Res(L_{n}, F_{m}) = ((−1)^{ηω}β^{2η−ω}ρ)^{(m−1)(m+r)/2}. Therefore, m ∈ A. That is a contradiction. This completes the proof of the claim.

7 Proof of Theorem 6

In this section we prove a theorem that gives closed formulas for the derivatives of GFPs.

Recall that from (4) and (5) we have d = a + b, b = −g/a, where d and g are the polynomials defined in (2) and (3). This implies that a − b = a + ga^{−1}. Here F′_{n}, L′_{n}, a′, b′ and d′ are the derivatives of F_{n}, L_{n}, a, b and d with respect to x.

Proof of Theorem 6. We prove Part (i). From Binet formula (4) and b = −g/a we have

F_{n} = (a^{n} − (−g)^{n}a^{−n}) / (a − (−g)a^{−1}) .

Differentiating F_{n} with respect to x, using a − b = a + ga^{−1} and simplifying we have

F′_{n} = \frac{na′ (a^{n−1} + (−g)^{n}a^{−n−1})}{(a − b)^{2}} − \frac{a′(1 − ga^{−2}) (a^{n} − (−g)^{n}a^{−n})}{(a − b)^{2}}.  (14)

Since d = a + b, and b = −g/a, we have a′ + b′ = d′, and b′ = ga^{−2}a′. These imply that

a′ = \frac{ad′}{a + ga^{−1}} and 1 − ga^{−2} = \frac{d}{a}.

Substituting these results in (14) and simplifying we have

F′_{n} = \frac{nad′ (a^{n−1} + (−g)^{n}a^{−n−1})}{(a − b)^{2}} − \frac{d · d′}{(1 + ga^{−1})^{2}} \frac{(a^{n} − (−g)^{n}a^{−n})}{(a − b)}.

Thus,

F′_{n} = \frac{nd′ (a^{n} + b^{n})}{(a − b)^{2}} − \frac{d · d′}{(a − b)^{2}} \frac{(a^{n} − (−g)^{n}a^{−n})}{(a − b)}.

It is known that (see, for example [11]) a^{n} + b^{n} = gF_{n−1} + F_{n+1}. So,

F′_{n} = \frac{nd′ (gF_{n−1} + F_{n+1}) − d · d′F_{n}}{(a − b)^{2}} .

This completes the proof of Part (i).

We now prove Part (ii). From [11] we know that L_{n} = (gF_{n−1} + F_{n+1}) /α. Differentiating L_{n} with respect to x, we have (recall that g is constant) L′_{n} = (gF′_{n−1} + F′_{n+1}) /α. This and Part (i) imply that

L′_{n} = \frac{gd′ ((n − 1)αL_{n−1} − dF_{n−1})}{α(a − b)^{2}} + \frac{d′ ((n + 1)αL_{n+1} − dF_{n+1})}{α(a − b)^{2}}.

Simplifying we have

L′_{n} = \frac{d′}{α(a − b)^{2}} ((n − 1)αgL_{n−1} + (n + 1)αL_{n+1} − dα\frac{gF_{n−1} + F_{n+1}}{α} ) .

This and L_{n} = (gF_{n−1} + F_{n+1}) /α imply that

L′_{n} = \frac{d′ ((n − 1)gL_{n−1} + (n + 1)L_{n+1} − dL_{n})}{(a − b)^{2}}.

Therefore,

L′_{n} = \frac{d′ (n(gL_{n−1} + L_{n+1}) + (L_{n+1} − gL_{n−1}) − dL_{n})}{(a − b)^{2}}.  (15)

From [11] we know that

gL_{n−1} + L_{n+1} = (a − b)^{2}F_{n}/α, L_{n+1} − gL_{n−1} = αL_{n}L_{1}, and αL_{1} − d = 0.

Substituting these identities in (15) completes the proof.

8 Proof of Theorems 4 and 5

In this section we prove Theorems 4 and 5. We first prove some basic lemmas needed for the main proof.

8.1 Basic lemmas for the discriminant

Recall that one of the expressions for the discriminant of a polynomial f is given by Disc(f) = (−1)^{n(n−1)/2}a^{−1}Res(f, f′), where a = lc(f), n = deg(f) and f′ the derivative of f. Lemma 19. For n ∈ ℤ_{≥0} we have

F_{n} mod d^{2} + 4g ≡ \begin{cases}n(−g)^{(n−1)/2}, & if n is odd; \\ (−1)^{(n+2)/2} (ndg^{(n−2)/2}) /2, & if n is even. \end{cases}

Proof. We use mathematical induction. Let S(k) be statement:

F_{k} mod d^{2} + 4g ≡ \begin{cases}(−1)^{(k−1)/2}kg^{(k−1)/2}, & if k is odd; \\ (−1)^{(k+2)/2} (kdg^{(k−2)/2}) /2, & if k is even. \end{cases}

Since F_{1} = 1 and F_{2} = d, we have S(1) and S(2) are true. Suppose that the statement is true for some k = n − 1 and k = n. Thus, suppose that S(n − 1) and S(n) are true and we prove S(n + 1). We consider two cases on the parity of n.

Case n even. Recall that F_{n+1} = dF_{n} + gF_{n−1}. This and S(n − 1) and S(n) (with n even and n−1 odd) imply that F_{n+1} ≡ (−1)^{(n+2)/2} (nd^{2}g^{(n−2)/2}/2)+(n−1)(−g)^{(n−2)/2}g mod d^{2} + 4g. Simplifying

F_{n+1} ≡ (−1)^{(n+2)/2} \frac{nd^{2}g^{(n−2)/2}}{2} + (2n − (n + 1))(−1)^{(n−2)/2}g^{n/2} mod d^{2} + 4g.

It is easy to see that

(−1)^{(n+2)/2} \frac{nd^{2}g^{(n−2)/2}}{2} + 2n(−1)^{(n−2)/2}g^{n/2} = (−1)^{(n+2)/2} \frac{ng^{(n−2)/2}}{2} (d^{2} + 4g) .

Thus,

F_{n+1} ≡ (−1)^{(n+2)/2} \frac{ng^{(n−2)/2}}{2} (d^{2} + 4g) + (n + 1)(−g)^{n/2} mod d^{2} + 4g.

This implies that F_{n+1} ≡ (n + 1)(−g)^{n/2} mod d^{2} + 4g.

Case n odd. S(n − 1) and S(n) (with n odd and n − 1 even) and F_{n+1} = dF_{n} + gF_{n−1}, imply that

F_{n+1} ≡ n(−g)^{(n−1)/2}d + (−1)^{(n+1)/2} (\frac{(n − 1)dg^{(n−3)/2}}{2} ) g mod d^{2} + 4g
≡ dg^{(n−1)/2} (\frac{(−1)^{(n−1)/2}2n − (−1)^{(n−1)/2}(n − 1)}{2}) mod d^{2} + 4g
≡ \frac{(−1)^{(n+3)/2}(n + 1)dg^{(n−1)/2}}{2} mod d^{2} + 4g.

This completes the proof.

Lemma 20. If n ∈ ℤ_{≥0}, then Res((a − b)^{2}, F_{n}) = (β^{2η−ω}ρ)^{(n−1)}n^{2η}.

Proof. From [11] we know that (a − b)^{2} = d^{2} + 4g. This and Lemma 19 imply that there is a polynomial T such that

F_{n} = \begin{cases}(a − b)^{2}T + n(−g)^{(n−1)/2}, & if n is odd; \\ (a − b)^{2}T + (−1)^{(n+2)/2}2^{−1}dg^{(n−2)/2}n, & if n is even. \end{cases}  (16)

Using Lemma 7 Parts (i), (iii) and (iv) and simplifying we have

Res(d^{2} +4g, g^{m}) = Res(d^{2} +4g, g)^{m} = Res(d^{2} +4g, g)^{m} = (λ^{2η−2η}Res(g, d)^{2})^{m} = ρ^{2m}. (17)

To find Res((a − b)^{2}, F_{n}) we consider two cases, depending on the parity of n. Case n is even. From (16) we have

Res((a − b)^{2}, F_{n}) = Res((a − b)^{2}, (a − b)^{2}T + (−1)^{(n+2)/2}2^{−1}dg^{(n−2)/2}n).

This and Lemma 7 Parts (i), (ii) and (iv) imply that

Res((a − b)^{2}, F_{n}) = β^{(2η−ω)(n−2)}Res((a − b)^{2}, (−1)^{(n+2)/2}2^{−1}n)Res((a − b)^{2}, dg^{(n−2)/2})
= β^{(2η−ω)(n−2)}(2^{−1}n)^{2η}Res((a − b)^{2}, d)Res((a − b)^{2}, g^{(n−2)/2})
= β^{(2η−ω)(n−2)}(2^{−1}n)^{2η}Res(d, d^{2} + 4g)Res((a − b)^{2}, g^{(n−2)/2}).

Using similar analysis as in (17) we have Res((a − b)^{2}, g^{(n−2)/2}) = ρ^{n−2}. It is easy to see that Res((a − b)^{2}, d) = Res(d, d^{2} + 4g) = β^{2n−ω}2^{η}. Therefore,

Res((a − b)^{2}, F_{n}) = β^{(2η−ω)(n−1)}n^{2η}ρρ^{n−2} = (β^{2η−ω}ρ)^{(n−1)}n^{2η}.

Case n is odd. From (16) we have

Res((a − b)^{2}, F_{n}) = Res((a − b)^{2}, (a − b)^{2}T + (−1)^{(n+2)/2}2^{−1}dg^{(n−2)/2}n).

This and Lemma 7 Parts (i), (ii) and (iv) imply that

Res((a − b)^{2}, F_{n}) = (β^{2})^{η(n−1)−ω(n−1)/2}Res(d^{2} + 4g, n(−g)^{(n−1)/2})
= β^{(2η−ω)(n−1)}n^{2η}Res(d^{2} + 4g, g^{(n−1)/2})
= (β^{2η−ω}ρ)^{(n−1)}n^{2η}.

This completes the proof.

8.2 Proof of main theorems about discriminant

Proof of Theorem 4. From Theorem 6 we have

Res(F_{n}, (a − b)^{2}F′_{n}) = Res(F_{n}, d′(nαL_{n} − dF_{n})).

Since deg(d) = 1, we have that d′ is a constant. (Recall that when F_{n} and L_{n} are together in a resultant, they are equivalent.) Therefore,

Res(F_{n}, (a − b)^{2}F′_{n}) = (d′)^{n−1}Res(F_{n}, nαL_{n} − dF_{n}).

Since deg(F_{n}) = η(n−1) and deg(L_{n}) = ηn, we have Res(F_{n}, nαL_{n}−dF_{n}) = Res(F_{n}, nαL_{n}). So, Res(F_{n}, (a − b)^{2}F′_{n}) = (αd′n)^{n−1}Res(F_{n}, L_{n}) = (αd′n)^{n−1}Res(L_{n}, F_{n}). This and Theorem 3 imply that

Res(F_{n}, (a − b)^{2}F′_{n}) = (αd′n)^{n−1}2^{n−1}α^{1−n}(β^{2}ρ)^{n(n−1)/2} = (2d′n)^{n−1}(β^{2}ρ)^{n(n−1)/2}. (18)

On the other hand, from Lemma 20 and the fact that deg(a − b)^{2} is even we have

Res(F_{n}, (a − b)^{2}F′_{n}) = Res(F_{n}, (a − b)^{2})Res(F_{n}, F′_{n}) = n^{2}(β^{2}ρ)^{(n−1)}Res(F_{n}, F′_{n}).

This and (18) imply that

Res(F_{n}, F′_{n}) = \frac{(2d′n)^{n−1}(β^{2}ρ)^{n(n−1)/2}}{n^{2}(β^{2}ρ)^{(n−1)}} = n^{n−3}(2d′)^{n−1}(β^{2}ρ)^{(n−1)(n−2)/2}.

Therefore,

Disc(F_{n}) = (−1)^{\frac{(n−1)(n−2)}{2}} β^{1−n}Res(F_{n}, F′_{n}) = β^{1−n}n^{n−3}(2d′)^{n−1}(−β^{2}ρ)^{(n−1)(n−2)/2}.

This completes the proof.

Proof of Theorem 5. From the definition of the discriminant we have

Disc(L_{n}) = (−1)^{n(n−1)/2}αβ^{−n}Res(L_{n}, L′_{n}).

This and Theorem 6 imply that Disc(L_{n}) = (−1)^{n(n−1)/2}αβ^{−n}Res(L_{n}, (nd′F_{n})/α). Since (nd′)/α is a constant, Disc(L_{n}) = (−1)^{n(n−1)/2}αβ^{−n}(nd′/α)^{n}Res(L_{n}, F_{n}). This and Theorem 3 imply that

Disc(L_{n}) = (−1)^{\frac{n(n−1)}{2}} αβ^{−n}(\frac{nd′}{α} )^{n}2^{n−1}α^{1−n}(β^{2}ρ)^{(n(n−1))/2}
= β^{n(n−2)}(nd′)^{n}2^{n−1}α^{2−2n}( − ρ)^{(n(n−1))/2}.

Completing the proof.

Theorems 4 and 5 generalize as follows.

If deg(d) = m, g is a constant and d′ is the derivative of d, then

Disc(F_{n}) = (−ρ)^{(n−2)(n−1)/2}2^{n−1}n^{n−3}β^{(n−1)(n−3)}Res(F_{n}, d′)^{n−1}.

If deg(d) = m, g is a constant and d′ is the derivative of d, then

Disc(L_{n}) = (−ρ)^{n(n−1)/2} 2^{n−1}n^{n}α^{2−2n}β^{n(n−2)}Res(L_{n}, d′)^{n}.

Open question. In this paper we did not investigate the case deg(g) ≥ deg(d). This property is satisfied by Jacobsthal polynomials.

9 Acknowledgments

The first author was partially supported by Grant No. 344524, 2018 from the Citadel Foundation, Charleston, SC. The second and third author would like to thank the Universidad de Antioquia for the support received from this institution. This work is dedicated to our mutual friend, Gilberto Garcia Pulgarín, who many years ago imparted to us an understanding of the significance of the resultant.

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2010 Mathematics Subject Classification: Primary 11B39; Secondary 11B83. Keywords: resultant, discriminant, derivative, Fibonacci polynomial, Lucas polynomial, polynomial sequence.

(Concerned with sequences \underline{A006645}, \underline{A001629}, \underline{A001871}, \underline{A006645}, \underline{A007701}, \underline{A045618}, \underline{A045925}, \underline{A086804}, \underline{A093967}, \underline{A127670} \underline{A193678}, \underline{A317404}, \underline{A317405}, \underline{A317408}, and \underline{A317451}.)

Received August 22 2018; revised versions received August 23 2018; March 28 2019; June 15 2019. Published in Journal of Integer Sequences, June 28 2019.

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