ON LINEAR FORMS WITH COEFFICIENTS IN ℕ ζ(1 + ℕ)

The detailed version, Part 3

Leonid A. Gutnik

In memory of Professor Alexander Buchstab

Through the shade of sleep, Through the cloudy strife Of Death and of Life P.B. Shelley.

Content

§0. Foreword. §1. ”Exercise.” §2. Sextette comes to the foreground.

§0. Foreword

At the end of the second part, we have come to the necessity of studying the zeros of the polynomial

(3.0.1,)  D^{∧}(z, η) = (η + 1)^{3}(η + γ_{1})^{3} − 2^{3}(1 + γ_{1})^{3}zη^{3}.

where γ_{1} = \frac{1−δ_{0}}{1+δ_{0}} for δ_{0} = \frac{1}{Δ} and Δ ∈ ℕ + 1. I do this here in the paragraph 1. Then I give, as compensation for the heavy paragraph 1, an easy paragraph 2.

Acknowledgment. I express my deep thanks to Professor B.Z.Moroz, to Professor I.I. Piatetski-Shapiro, to Professor A.G.Aleksandrov, to Professor P.Bundshuh, and to Professor S.G.Gindikin for the support of my work.

Typeset by AMS-TEX

§1. ”Exercise.”

Let z = −r^{3} exp(3iϕ), where 1 ≤ r, −\frac{π}{3} < ϕ ≤ \frac{π}{3} . Let further ψ = ϕ+ \frac{2kπ}{3} , where k = −1, 0, 1. Then

(3.1.0) D^{∧}(z, η) = ∏_{k=−1}^{1} ((η + 1)(η + γ_{1}) + 2r exp (iϕ + \frac{2ikπ}{3} ) (1 + γ_{1})η)

So, we have to study the zeros of the polynomial

(3.1.1)  D^{∨}(z, η) = (η + 1)(η + γ_{1}) + 2r exp(iψ)(1 + γ_{1})η

for 1 ≤ r and −π < ψ ≤ π. We shall study the zeros of the polynomial (3.1.1) under more general conditions

(3.1.2)  −1 < δ_{0} < 1, 1 ≤ r, γ_{1} = (1 − δ_{0})/(1 + δ_{0}), −π ≤ ψ ≤ π.

Putting η = θ(1 + γ_{1}), θ^{∨} = − exp(−iψ)θ, we get the equalities

(3.1.3)  D^{∨}(z, η) = (1 + γ_{1})^{2}D_{1}^{∨}(z, θ),

where

(3.1.4) D_{1}^{∨}(z, θ) = θ^{2} + 2 (\frac{1}{2} + r exp(iψ)) θ + \frac{γ_{1}}{(1 + γ_{1})^{2}} = exp(2iψ)D_{2}^{∨}(z, θ^{∨})

with

(3.1.5)  D_{2}^{∨}(z, θ^{∨}) = (θ^{∨})^{2} − 2 (\frac{exp(−iψ)}{2} + r) θ^{∨} + \frac{γ_{1} exp(− 2iψ)}{(1 + γ_{1})^{2}}.

Let

(3.1.6)  D_{0}(r, ψ, δ_{0}) = r^{2} + r exp(−iψ) + (\frac{δ_{0}}{2} )^{2} exp(− 2iψ).

If we put t = cos(ψ), then

(3.1.7)  Im(D_{0}(r, ψ, δ_{0})) = − (\frac{(δ_{0})^{2}}{2} t + r) sin(ψ),

(3.1.8)  Re (D_{0}(r, ψ, δ_{0})) = \frac{(δ_{0})^{2}t^{2}}{2} + rt + r^{2} − (\frac{δ_{0}}{2} )^{2} .

ON LINEAR FORMS WITH COEFFICIENTS IN ℕζ(1 + ℕ) 3

Since r ≥ 1 ≥ (δ_{0})^{2}, it follows that

(3.1.9)  min_{−π≤ψ≤π} Re (D_{0}(r, ψ, δ_{0})) = r^{2} − r + (\frac{δ_{0}}{2} )^{2} ≥ 0;

and the equality Re(D_{0}(r, ψ, γ)) = 0 is reached only if r = 1, δ_{0} = 0, ψ = ±π; this is the only case when the equality D(r, ψ, δ_{0}) = 0 holds. Clearly, for t = cos(ψ), the following equality will hold:

(3.1.10)  |D_{0}(r, ψ, δ_{0})|^{2} = r^{4} + r^{2} + (\frac{δ_{0}}{2} )^{4} +

2r^{2} (\frac{δ_{0}}{2} )^{2} (2t^{2} − 1) + 2r(r^{2} + (\frac{δ_{0}}{2} )^{2})t =

r^{2} + (r^{2} − (\frac{δ_{0}}{2} )^{2})^{2} + 2r (r^{2} + (\frac{δ_{0}}{2} )^{2}) t + r^{2}(δ_{0})^{2}t^{2}.

Below, in order to extract the square root w from the complex number z = x + iy, we use the well known formulas

(3.1.11)  w = ε_{1} (\sqrt{\frac{|z| + x}{2}} + i sign (y)\sqrt{\frac{|z| − x}{2}}) ,

where ε_{1}^{2} = 1, z ∉ (−∞, 0), and

(3.1.12)  w = ε_{1}ε_{2}i\sqrt{|z|},

where ε_{1}^{2} = 1, ε_{2}^{2} = 1, z ∈ (−∞, 0). By fixed ε_{1} and ε_{2}, formulas (3.1.11) - (3.1.12) determine the defined on ℂ branch w_{0}(ε_{1}, ε_{2}, z) of multivalued function z^{1/2}. This branch is regular in the domain ℂ\(−∞, 0] and has the continuous restrictions on each of the sets ℂ\(−∞, 0) and {z ∈ ℂ: ε_{2}Im(z) ≥ 0}. In view of (3.1.11),

(3.1.13)  Re(w_{0}(1, ε_{2}, z)) ≥ 0,

where z ∈ ℂ\(−∞, 0), and the equality in (3.1.13) for this case is fulfilled only if z = 0. According to the Riemann - Schwarz principle of symmetry, since w_{0}(ε_{1}, ε_{2}, z) ≥ 0 for z > 0, it follows that

(3.1.14)  w(ε, 1, z) = \overline{w(ε, 1, z̄)},

where ε^{2} = 1, z ∈ ℂ\(−∞, 0). We are interested now in the behavior of D_{0}(r, ψ, δ_{0}) when ψ changes in [−π, π]. According to (3.1.9), when ψ changes in [−π, π], the value of

D_{0}(r, ψ, δ_{0})

remains in the set ℂ\(−∞, 0), where the function

w(1, 1, z) = w(1, −1, z)

is continuous. Therefore, the value

(3.1.15)  w_{0}(1, 1, D_{0}(r, ψ, δ_{0})) = w_{0}(1, −1, D_{0}(r, ψ, δ_{0}))

in this case is continuous on the set

(3.1.16)  −π ≤ ψ ≤ π, 1 ≤ r.

It will further be denoted by R_{0}(r, ψ, δ_{0}). Since

R_{0}(r, −π, δ_{0}) = R_{0}(r, π, δ_{0}),

it follows that the function R_{0}(r, ψ, δ_{0}), as function from ψ, admits 2π−periodical continuation on ℝ. Since

(3.1.17)  D_{0}(r, 0, δ_{0}) = r^{2} + r + (δ_{0}/2)^{2} > 0

and

D_{0}(r, ψ, δ_{0}) = \overline{D_{0}(r, −ψ, δ_{0})},

it follows from (3.1.14) that

(3.1.18)  R_{0}(r, ψ, δ_{0}) = \overline{R_{0}(r, −ψ, δ_{0})}.

In (3.1.6), we considered the value D_{0}(r, ψ, δ_{0}) only for |δ_{0}| < 1, but it may also be considered for δ_{0} = 1, and in this case, according to (3.1.6),

D_{0}(r, ψ, 1) = (r + \frac{exp(−iψ)}{2})^{2} ;

the inequality (3.1.9) is also fulfilled for δ_{0} = 1, and the value

(3.1.19)  w_{0}(1, 1, D_{0}(r, ψ, 1)) = w_{0}(1, −1, D_{0}(r, ψ, 1))

in this case is continuous on the set (3.1.16). It will further be denoted by R_{0}(r, ψ, 1). Since

(3.1.20)  R_{0}(r, −π, 1) = R_{0}(r, π, 1),

it follows that the function R_{0}(r, ψ, 1), as function from ψ, admits 2π−periodical continuation on ℝ and (3.1.18) also holds for δ_{0} = 1. Since

(3.1.21)  Re (r + \frac{exp(−iψ)}{2}) = r + \frac{cos(ψ)}{2} > 0,

if r ≥ 1, in view of (3.1.13), it follows that

(3.1.22)  R_{0}(r, ψ, 1) = r + \frac{exp(−iψ)}{2}.

According to (3.1.22), for r ≥ 1, |δ_{0}| < 1 and k = 0, 1, the values

(3.1.23)  θ_{k}^{∨}(r, ψ, δ_{0}) = R_{0}(r, ψ, 1) + (−1)^{k}R_{0}(r, ψ, δ_{0})

compose the set of all the roots of the polynomial (3.1.5), the values

(3.1.24)  θ_{k}^{∧}(r, ψ, δ_{0}) = −e^{iψ}(R_{0}(r, ψ, 1) + (−1)^{k}R_{0}(r, ψ, δ_{0}))

compose the set of all the roots of the polynomial (3.1.4) and, according to (3.1.3), the values

(3.1.25)  η_{k}^{∧}(r, ψ, δ_{0}) = −(1 + γ_{1})e^{iψ}(R_{0}(r, ψ, 1) + (−1)^{k}R_{0}(r, ψ, δ_{0}))

compose the set of all the roots of the polynomial (3.1.1). It is clear that the values (3.1.25), as functions from ψ, are 2π−periodical. Clearly, the six numbers

(3.1.26)  η_{k,l}^{∗}(δ_{0}) = η_{k}^{∧}(1, (2l − 1)π/3, δ_{0}),

where k = 0, 1 and l = 0, 1, 2, compose the set of all the roots of the polynomial (3.1.1) for z = 1. It is clear that

(3.1.27)  η_{k,2}^{∗}(δ_{0}) ∈ ℝ, η_{k,1}^{∗}(δ_{0}) = \overline{η_{k,0}^{∗}(δ_{0})}.

Let, as in [8] and [9], (2.4.55),

(3.1.28)  h^{∼}(η) = (η^{2} − 1)^{3}(1 − δ_{0})^{−3d_{1}}2^{−6}η^{3d_{1}}.

Let further

F_{0}^{∨}(η) = ∏_{l=0}^{2} ∏_{k=0}^{1}(η − h^{∼}(η_{k,l}^{∗}))

and 𝔇_{0}^{∨} be the discriminant relatively to η of the polynomial F_{0}^{∨}(η). The goal of this paragraph is the proof of the following

Theorem 3.1.1. If Δ = 13, then the following inequality holds:

(3.1.29)  𝔇_{0} ≠ 0.

To prove this theorem we shall first prove the following Theorem 3.1.2. If Δ = 13, then the following inequalities hold:

|h^{∼}(η_{0,l}^{∗})| > |h^{∼}(η_{1,l}^{∗})|,

where l = 0, 1;

0 = |h^{∼}(η_{0,2}^{∗})| < |h^{∼}(η_{1,2}^{∗})|.

Then we shall prove

Theorem 3.1.3. If Δ = 13, then h^{∼}(η_{k,l}^{∗}) ∉ ℝ, where k = 0, 1 and l = 0, 1. Theorem 3.1.1 follows from (3.1.27), Theorem 3.1.3 and Theorem 3.1.2.

Lemma 3.1.1. The value |D_{0}(r, ψ, δ_{0})| is increasing and concave (convex upwards) function from t = cos(ψ).

Proof. According to (3.1.10),

|D_{0}(r, ψ, δ_{0})| = \sqrt{at^{2} + bt + c},

where

a = r^{2}(δ_{0})^{2}, b = 2r (r^{2} + (\frac{δ_{0}}{2} )^{2}) , c = r^{2} + (r^{2} − (\frac{δ_{0}}{2} )^{2})^{2} .

In view of

\frac{∂}{∂t}\sqrt{at^{2} + bt + c} = \frac{2at + b}{2\sqrt{at^{2} + bt + c}}

(\frac{∂}{∂t})^{2} \sqrt{at^{2} + bt + c} = \frac{−(2at + b)^{2} + 4a(at^{2} + bt + c)}{4\sqrt{(at^{2} + bt + c)^{3}}} =

\frac{4ac − b^{2}}{4\sqrt{(at^{2} + bt + c)^{3}}}

and

4ac − b^{2} = 4r^{4}(δ_{0})^{2}+

4r^{2} ((δ_{0})^{2} (r^{2} − (\frac{δ_{0}}{2} )^{2})^{2} − (r^{2} + (\frac{δ_{0}}{2} )^{2})^{2}) =

−4r^{2}(1 − (δ_{0})^{2}) (r^{2} − (\frac{δ_{0}}{2} )^{2})^{2} ≤ 0,

the lemma is proved.

Let ε^{3} = ε; according to (3.1.24),

(3.1.30)  η_{k}^{∧}(r, ψ, δ_{0}) + ε = −\frac{2 exp(iψ)}{1 + δ_{0}}×

(r + \frac{exp(−iψ)}{2} − ε\frac{1 + δ_{0}}{2} exp(−iψ) + (−1)^{k}R_{0}(r, ψ, δ_{0})) ,

where ε^{3} = ε, k = 0, 1. Let

(3.1.31)  R_{ε}^{∗}(r, ψ, δ_{0}) = r + \frac{exp(−iψ)}{2} − ε\frac{1 + δ_{0}}{2} exp(−iψ),

where ε^{3} = ε.

Lemma 3.1.2. The values

|η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2},

where k = 0, 1, r ≥ 1, ψ ∈ [0, π], r − 1 + π − ψ > 0, ε^{3} = ε, are nonnegative zeros of the polynomial

(3.1.32)  x^{2} − xp_{1+ε}(r, ψ, δ_{0}) + q_{1+ε}(r, ψ, δ_{0}),

where ε^{3} = ε, and

(3.1.33)  p_{1+ε}(r, ψ, δ_{0}) = 8(|R_{ε}^{∗}(r, ψ, δ_{0})|^{2} + |R_{0}(r, ψ, δ_{0})|^{2})/(1 + δ_{0})^{2},

(3.1.34)  q_{1+ε}(r, ψ, δ_{0}) = |D^{∨}(z, −ε)|^{2}.

Proof. According to theorem about the sum of the squares of the diagonales of the parallelogramm, the following equality holds:

(3.1.35)  ∑_{k=0}^{1} |η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2} =

8(|R_{ε}^{∗}(r, ψ, δ_{0})|^{2} + |R_{0}(r, ψ, δ_{0})|^{2})/(1 + δ_{0})^{2}.

According to (3.1.1),

(3.1.36)  ∏_{k=0}^{1} |η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2} = |D^{∨}(z, −ε)|^{2}.

The lemma is proved.

Lemma 3.1.3. Let ε^{2} = ε. With increasing ψ ∈ (0, π), the value |η_{0}^{∧}(r, ψ, δ_{0})+ε| monotonically decreases, but the value

|η_{1}^{∧}(r, ψ, δ_{0}) + ε|

monotonically increases and

(3.1.37)  |η_{1}^{∧}(r, ψ, δ_{0}) + ε| < |η_{0}^{∧}(r, ψ, δ_{0}) + ε|,

where ψ ∈ [0, π);

(3.1.38)  |η_{1}^{∧}(r, π, δ_{0}) + ε| ≤ \frac{1 − δ_{0}}{1 + |δ_{0}|} + ε ≤

\frac{1 + |δ_{0}|}{1 + δ_{0}} + ε ≤ |η_{0}^{∧}(r, π, δ_{0}) + ε|;

moreover, the first and the last inequalities (3.1.38) become equalities only if r = 1, and the middle inequality (3.1.38) becomes equality only if δ_{0} = 0; consequently,

(3.1.39)  |η_{1}^{∧}(r, π, δ_{0}) + ε| = |η_{0}^{∧}(r, π, δ_{0}) + ε| = 1 + ε

only if r = 1 and δ_{0} = 0.

Proof. According to Lemma 3.1.2, if ε^{3} = ε, then

(3.1.40)  |η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2} = \frac{p_{1+ε}}{2} ± \sqrt{\frac{p_{1+ε}^{2}}{4} − q_{1+ε}},

where k = 0, 1, ψ ∈ [0, π]. According to (3.1.31), (3.1.33) and Lemma 3.1.2 ,

(3.1.41)  p_{1} = 8(|R_{0}^{∗}(r, ψ, δ_{0})|^{2} + |R_{0}(r, ψ, δ_{0})|^{2})/(1 + δ_{0})^{2} =

8 (r^{2} + rt + \frac{1}{4} + |D_{0}(r, ψ, δ_{0})|)/(1 + δ_{0})^{2} ,

(3.1.42)  p_{2} = 8(|R_{1}^{∗}(r, ψ, δ_{0})|^{2} + |R_{0}(r, ψ, δ_{0})|^{2})/(1 + δ_{0})^{2} =

8 (r^{2} − rδ_{0}t + \frac{(δ_{0})^{2}}{4} + |D_{0}(r, ψ, δ_{0})|) /(1 + δ_{0})^{2}.

In view of (3.1.34) and (3.1.3),

(3.1.43)  q_{1}(r, ψ, δ_{0}) = (\frac{1 − δ_{0}}{1 + δ_{0}} )^{2} , q_{2}(r, ψ, δ_{0}) = (\frac{4r}{1 + δ_{0}} )^{2} .

In view of (3.1.41) and Lemma 3.1.1, the value p_{1} increases with increasing t = cos(ψ) ∈ (−1, 1). According to the assertion of the Lemma 3.1.1 and the proof of this Lemma,

\frac{∂}{∂t}|D_{0}(r, ψ, δ_{0})||_{t=cos(ψ)} > \frac{∂}{∂t}|D_{0}(r, ψ, δ_{0})||_{t=1} =

\frac{2a + b}{2|D_{0}(r, 0, δ_{0})|},

where

a = r^{2}(δ_{0})^{2}, b = 2r(r^{2} + (δ_{0}/2)^{2}), 0 < ψ ≤ π.

Therefore, according to (3.1.42),

\frac{∂}{∂t}p_{2} = \frac{8(−rδ_{0} + \frac{∂}{∂t}|D_{0}(r, ψ, δ_{0})|)}{(1 + δ_{0})^{2}}

and

−rδ_{0} + \frac{∂}{∂t}|D_{0}(r, ψ, δ_{0} >

−rδ_{0} + \frac{r^{2}(δ_{0})^{2} + r(r^{2} + (δ_{0}/2)^{2})}{r^{2} + r + (δ_{0}/2)^{2}} = r×

\frac{−δ_{0}(r^{2} + r + (δ_{0}/2)^{2}) + r(δ_{0})^{2} + rδ_{0} + (r − δ_{0}/2)^{2}}{r^{2} + r + (δ_{0}/2)^{2}} =

\frac{r(1 − δ_{0})(r − δ_{0}/2)^{2}}{r^{2} + r + (δ_{0}/2)^{2}} > 0.

Consequently, the value of p_{2} increases with increasing t = cos(ψ) ∈ (−1, 1). The value of

Δ_{1+ε}^{∗}(r, ψ, δ_{0}) = (p_{1+ε}/2)^{2} − q_{1+ε}

10 LEONID A. GUTNIK

for some ε ∈ {0, 1} is equal to 0 if and only if p_{1+ε}/2 = (q_{1+ε})^{1/2}, i.e. if

(3.1.44)  r^{2} + rt + \frac{1}{4} + |D_{0}(r, ψ, δ_{0})| = (1 − δ_{0}^{2})/4,

when ε = 0, and

(3.1.45)  r^{2} − δ_{0}rt + (\frac{δ_{0}}{2} )^{2} + |D_{0}(r, ψ, δ_{0})| = r(1 + δ_{0}),

when ε = 1. The equation (3.1.44) is equivalent to the system consisting of the inequality

(3.1.46)  r^{2} + rt + δ_{0}^{2}/4 ≤ 0

and the equation

(3.1.47)  (r^{2} + δ_{0}^{2}/4)^{2} + 2r(r^{2} + δ_{0}^{2}/4)t + r^{2}t^{2} =
= r^{2} + (r^{2} − δ_{0}^{2}/4)^{2} + 2r(r^{2} + δ_{0}^{2}/4)t + r^{2}δ_{0}^{2}t^{2}.

The solutions of the equation (3.1.47) are the roots of the polynomial

(3.1.48)  r^{2}(1 − δ_{0}^{2})(1 − t^{2}).

The equation (3.1.45) is equivalent to the system consisting of the inequality

(3.1.49)  (r − \frac{δ_{0}}{2} )^{2} − r − δ_{0}rt ≤ 0,

and the equation

(3.1.50)  (r − \frac{δ_{0}}{2} )^{2} ((r − \frac{δ_{0}}{2} )^{2} − 2r) + r^{2}−

2 ((r − \frac{δ_{0}}{2} )^{2} − r) δ_{0}rt + (δ_{0}rt)^{2} =

= r^{2} + (r^{2} − δ_{0}^{2}/4)^{2} + 2r(r^{2} + δ_{0}^{2}/4)t + r^{2}δ_{0}^{2}t^{2}.

The solutions of the equation (3.1.50) are the roots of the polynomial

(3.1.51)  2r(1 + δ_{0})(r − δ/2)^{2}(1 + t).

Since, for t = 1, the inequality (3.1.46) is not fulfilled, it follows, according to (3.1.48) and (3.1.51), that the value

Δ_{1+ε}^{∗}(r, ψ, δ_{0})

is not equal to 0, when ε^{2} = ε and t ∈ (−1, 1]; according to (3.1.30) and (3.1.17), if ε^{3} = ε and k^{2} = k, then

(3.1.52)  η_{k}^{∧}(r, 0, δ_{0}) + ε = −\frac{2 (r + \frac{1}{2} − ε\frac{1+δ_{0}}{2} + (−1)^{k}\sqrt{r^{2} + r + (\frac{δ_{0}}{2} )^{2}})}{1 + δ_{0}},

(3.1.53)  |η_{0}^{∧}(r, 0, δ_{0}) + ε| = \frac{2 (r + \frac{1}{2} − ε\frac{1+δ_{0}}{2} + \sqrt{r^{2} + r + (\frac{δ_{0}}{2} )^{2}})}{1 + δ_{0}} <

|η_{1}^{∧}(r, 0, δ_{0}) + ε| = \frac{2 |r + \frac{1}{2} − ε\frac{1+δ_{0}}{2} − \sqrt{r^{2} + r + (\frac{δ_{0}}{2} )^{2}}|}{1 + δ_{0}}.

Therefore, (3.1.37) is fulfilled and, if

ε^{2} = ε,

then instead (3.1.40), one may write a more exact equality

(3.1.54)  |η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2} = \frac{p_{1+ε}}{2} + (−1)^{k}\sqrt{(\frac{p_{1+ε}}{2} )^{2} − q_{1+ε}};

consequently, for ε ∈ {0, 1} with increasing t = cos(ψ) ∈ (−1, 1), the value

|η_{0}^{∧}(r, ψ, δ_{0}) + ε|

monotonically increases and, in view of (3.1.43), the value

|η_{1}^{∧}(r, ψ, δ_{0}) + ε|

monotonically decreases. According to (3.1.30), if

ε^{3} = ε

and k^{2} = k, then

(3.1.55)  η_{k}^{∧}(r, π, δ_{0}) + ε = \frac{2 (r − \frac{1}{2} + ε\frac{1+δ_{0}}{2} + (−1)^{k}\sqrt{r^{2} − r + (\frac{δ_{0}}{2} )^{2}})}{1 + δ_{0}}.

If ε^{2} = ε, then the following from (3.1.55) and (3.1.54) relations |η_{0}^{∧}(r, π, δ_{0}) + ε| = η_{0}^{∧}(r, π, δ_{0}) + ε =

\frac{2 (r − \frac{1}{2} + \sqrt{r^{2} − r + (\frac{δ_{0}}{2} )^{2}})}{1 + δ_{0}} + ε ≥ \frac{1 + |δ_{0}|}{1 + δ_{0}} + ε,

together with (3.1.43), entail the inequality

|η_{1}^{∧}(r, π, δ_{0})| + ε ≤ \frac{1 − δ_{0}}{1 + |δ_{0}|} + ε =

\frac{1 − |δ_{0}|}{1 + δ_{0}} + ε

and therefore also entail (3.1.38) and (3.1.39). The lemma is proved.

Corollary 1. If −1 < δ_{0} < 1, r ≥ 1, ψ ∈ (0, π), then

(3.1.56)  |η_{0}^{∧}(r, ψ, δ_{0})| > |η_{0}^{∧}(r, π, δ_{0})| = η_{0}^{∧}(r, π, δ_{0}) =

\frac{2}{1 + δ_{0}} (r − 1/2 + (r^{2} − r + δ_{0}^{2}/4)^{1/2}) ≥

\frac{2}{1 + δ_{0}} (r − 1/2 − (r^{2} − r + δ_{0}^{2}/4)^{1/2}) =

η_{1}^{∧}(r, π, δ_{0}) = |η_{1}^{∧}(r, π, δ_{0})| > |η_{1}^{∧}(r, ψ, δ_{0})|, |η_{1}^{∧}(r, ψ, δ_{1}) + 1| ≤ |η_{1}^{∧}(r, ψ, δ_{1})| + 1 < |η_{1}^{∧}(r, π, δ_{0})| + 1 = η_{1}^{∧}(r, π, δ_{0}) + 1 =

\frac{2}{1 + δ_{0}} (r − 1/2 − (r^{2} − r + δ_{0}^{2}/4)^{1/2}) + 1 =

\frac{2}{1 + δ_{0}} (r + δ_{0}/2 − (r^{2} − r + δ_{0}^{2}/4)^{1/2}) ≤

\frac{2}{1 + δ_{0}} (r + δ_{0}/2 + (r^{2} − r + δ_{0}^{2}/4)^{1/2}) =

η_{0}^{∧}(r, π, δ_{0}) + 1 = |η_{0}^{∧}(r, π, δ_{0}) + 1| =

q_{2}(r, π, δ_{0})/|η_{1}^{∧}(r, π, δ_{0}) + 1| = \frac{4r}{(1 + δ_{0})|η_{1}^{∧}(r, π, δ_{0}) + 1|} <

\frac{4r}{(1 + δ_{0})|η_{1}^{∧}(r, ψ, δ_{0}) + 1|} = q_{2}(r, ψ, δ_{0})/|η_{1}^{∧}(r, ψ, δ_{0}) + 1| =

|η_{0}^{∧}(r, ψ, δ_{0}) + 1|.

Proof. Since, according to the Lemma, for ε ∈ {0, 1} with increasing ψ ∈ (0, π), the value |η_{0}^{∧}(r, ψ, δ_{0}) + ε| monotonically decreases, but the value

|η_{1}^{∧}(r, ψ, δ_{0}) + ε|

monotonically increases, it follows that the first, the seventh, the ninth and the eighteenth relations of the assertions of the corollary 1 hold; the second, the third, the fifth, the sixth, the tenth, the twelfth the fourteenth and the fifteenth relations of the assertions of the corollary 1 follow directly from (3.1.55); the fourth, the eighth, the thirteenth, the sixteenth and twentieth relations of the assertions of the corollary 1 are obvious; the seventeenth and the nineteenth relations of the assertions of the corollary 1 follow from (3.1.43). The corollary is proved.

Corollary 2. If r ≥ 1, −1 < δ_{0} < 1, ψ ∈ [0, π], then

(3.1.57) |η_{0}^{∧}(r, ψ, δ_{0})| ≥ |η_{0}^{∧}(r, π, δ_{0})| = \frac{2}{1 + δ_{0}} (r − 1/2 + (r^{2} − r + δ_{0}^{2}/4)^{1/2}) ≥

\frac{1 + |δ_{0}|}{1 + δ_{0}} ≥ \frac{1 − |δ_{0}|}{1 + δ_{0}} ≥

|η_{1}^{∧}(r, π, δ_{0})| ≥ |η_{1}^{∧}(r, ψ, δ_{0})|;

|η_{1}^{∧}(r, ψ, δ_{0}) + 1| ≤ |η_{1}^{∧}(r, ψ, δ_{0})| + 1 ≤

\frac{2 − |δ_{0}| + δ_{0}}{1 + δ_{0}} ;

|η_{0}^{∧}(r, ψ, δ_{0}) + 1| = \frac{4r}{((1 + δ_{0})|η_{1}^{∧}(r, ψ, δ_{0}) + 1|)} ≥ \frac{4r}{2 − |δ_{0}| + δ_{0}} .

Proof. The first relation (3.1.57) follows from the first relation (3.1.56); the second relation (3.1.57) coincides with the first relation (3.1.56); since r ≥ 1, the third relation (3.1.57) is obvious; the fourth and seventh relations (3.1.57) are also obvious; since

(1 + |δ_{0}|)(1 − |δ_{0}|) = (1 − (δ_{0})^{2}),

the fifth relation (3.1.57) follows from the second relation (3.1.57), from the third relation (3.1.57) and from (3.1.43); the sixth relation (3.1.57) follows from the seventh relation (3.1.56); the eighth relation (3.1.57) follows from the fifth relation (3.1.57); the ninth relation (3.1.57) follows from (3.1.43); the tenth relation (3.1.57) follows from the eighth relation (3.1.57). The corollary is proved.

Since

(3.1.58)  |η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2} + |η_{k}^{∧}(r, ψ, δ_{0}) + 1|^{2} = 2|η_{k}^{∧}(r, ψ, δ_{0})|^{2} + 2,

in view of (3.1.53) and (3.1.43), it follows that

(3.1.59)  |η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2} = 2 + 2|η_{k}^{∧}(r, ψ, δ_{0})|^{2} − |η_{k}^{∧}(r, ψ, δ_{0}) + 1|^{2} =

= \frac{2p_{1} − p_{2}}{2} +

(−1)^{k}2\sqrt{(\frac{p_{1}}{2} )^{2} − q_{1}} − (−1)^{k}\sqrt{(\frac{p_{2}}{2} )^{2} − q_{2}}.

Since, according to (3.1.48) and (3.1.51), the value

Δ_{1+ε}^{∗}(r, ψ, δ_{0})

is not equal to 0, when ε^{2} = ε and t ∈ (−1, 1], it follows that

(3.1.60)  \frac{∂}{∂t}(|η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2}) =

(−1)^{k} (\frac{∂}{∂t} \frac{p_{1+ε}}{2} ) |η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2}/\sqrt{(\frac{p_{1+ε}}{2} )^{2} − q_{1+ε}},

where ε^{2} = ε, k^{2} = k, |δ_{0}| < 1, ψ ∈ [0, π). Let

(3.1.61)  s = \sqrt{\frac{(r − 1)^{2}}{4} + r cos^{2} (\frac{ψ}{2} )},

where ψ ∈ [0, π]. Then

(3.1.62)  \frac{r − 1}{2} ≤ s ≤ \frac{r + 1}{2},

(3.1.63)  t = cos(ψ) = \frac{4s^{2} − r^{2} − 1}{2r}

Lemma 3.1.4. The following inequalities hold:

(3.1.64)  |η_{0}^{∧}(r, ψ, δ_{0}) − 1| > |η_{1}^{∧}(r, ψ, δ_{0}) − 1|,

ON LINEAR FORMS WITH COEFFICIENTS IN ℕζ(1 + ℕ) 15

where r > 1 + \frac{δ_{0}}{2} , \frac{r−1}{2} ≤ s ≤ \frac{r+1}{2};

(3.1.65)  |η_{0}^{∧}(r, ψ, δ_{0}) − 1| > |η_{1}^{∧}(r, ψ, δ_{0}) − 1|,

where 1 ≤ r ≤ 1 + δ_{0}/2, δ_{0}/4 < s ≤ (r + 1)/2;

(3.1.66)  |η_{0}^{∧}(r, ψ, δ_{0}) − 1| = |η_{1}^{∧}(r, ψ, δ_{0}) − 1|,

where 1 ≤ r ≤ 1 + δ_{0}/2, s = δ_{0}/4;

(3.1.67)  |η_{0}^{∧}(r, ψ, δ_{0}) − 1| < |η_{1}^{∧}(r, ψ, δ_{0}) − 1|,

where 1 ≤ r ≤ 1 + δ_{0}/2, (r − 1)/2 ≤ s < δ_{0}/4;

Proof. According to (3.1.31),

R_{−1}^{∗}(r, ψ, δ_{0}) = r + e^{−iψ}(2 + δ_{0})/2.

Therefore,

(3.1.68)  |R_{−1}^{∗}(r, ψ, δ_{0})|^{2} = r^{2} + (2 + δ_{0})^{2}/4 + r(2 + δ_{0}) cos(ψ) =
r^{2} + (2 + δ_{0})^{2}/4 + r(2 + δ_{0})(4s^{2} − r^{2} − 1)/2r =

2(2 + δ_{0})s^{2} − δ_{0} \frac{r^{2} − 1}{2} + (\frac{δ_{0}}{2} )^{2}

decreases with increasing ψ ∈ [0, π]. According to (3.1.10) and (3.1.61),

(3.1.69)  |R_{0}(r, ψ, δ_{0})|^{4} = |D_{0}(r, ψ, δ_{0})|^{2} =

r^{2} + (r^{2} − (\frac{δ_{0}}{2} )^{2})^{2} + 2r (r^{2} + (\frac{δ_{0}}{2} )^{2}) t + r^{2}(δ_{0})^{2}t^{2} =

r^{2}(δ_{0})^{2} (\frac{4s^{2} − r^{2} − 1}{2r})^{2} +

2r (r^{2} + (\frac{δ_{0}}{2} )^{2}) \frac{4s^{2} − r^{2} − 1}{2r} + r^{2} + (r^{2} − (\frac{δ_{0}}{2} )^{2})^{2} =

4(δ_{0})^{2}s^{4} + (4r^{2} − (δ_{0})^{2}(2r^{2} + 1))s^{2} + r^{2}(r^{2} − 1) (\frac{δ_{0}}{2} )^{2} + (\frac{δ_{0}}{2} )^{4} .

16 LEONID A. GUTNIK

According to (3.1.33), (3.1.68) and (3.1.69),

(3.1.70)  p_{0} = 8\frac{|R_{−1}^{∗}(r, ψ, δ_{0})|^{2} + |R_{0}(r, ψ, δ_{0})|^{2}}{(1 + δ_{0})^{2}} =

\frac{8}{(1 + δ_{0})^{2}} (2(2 + δ_{0})s^{2} − δ_{0} \frac{r^{2} − 1}{2} + (\frac{δ_{0}}{2} )^{2}) + \frac{8}{(1 + δ_{0})^{2}} ×

\sqrt{4(δ_{0})^{2}s^{4} + (4r^{2} − (δ_{0})^{2}(2r^{2} + 1))s^{2} + r^{2}(r^{2} − 1) (\frac{δ_{0}}{2} )^{2} + (\frac{δ_{0}}{2} )^{4} .}

According to (3.1.34), (3.1.1) and (3.1.63),

(3.1.71)  q_{0}(r, ψ, δ_{0}) = |D^{∨}(z, 1)|^{2} =

16\frac{r^{2} + 1 + 2rcos(ψ)}{(1 + δ_{0})^{2}} =

\frac{64s^{2}}{(1 + δ_{0})^{2}} .

The value

Δ_{0}^{∗}(r, ψ, δ_{0}) = (p_{0}/2)^{2} − q_{0}

is equal to 0 if and only if p_{0}/2 = (q_{0})^{1/2}, i.e. if

(3.1.72)  2s(1 + δ_{0}) = 2(2 + δ_{0})s^{2} − δ_{0} \frac{r^{2} − 1}{2} + (\frac{δ_{0}}{2} )^{2} +
\sqrt{4(δ_{0})^{2}s^{4} + (4r^{2} − (δ_{0})^{2}(2r^{2} + 1))s^{2} + r^{2}(r^{2} − 1) (\frac{δ_{0}}{2} )^{2} + (\frac{δ_{0}}{2} )^{4} .}

The equation (3.1.72) is equivalent to the system consisting of the inequality

(3.1.73)  2(2 + δ_{0})s^{2} − δ_{0} \frac{r^{2} − 1}{2} + (\frac{δ_{0}}{2} )^{2} −
2s(1 + δ_{0}) ≤ 0

and the equation

(3.1.74)  P_{0}(s) = 0,

where

P_{0}(s) = a_{0}s^{4} + b_{0}s^{3} + c_{0}s^{2} + d_{0}s + r_{0},
a_{0} = 4(2 + δ_{0})^{2} − 4(δ_{0})^{2} = 16(1 + δ_{0}),
b_{0} = (1 + δ_{0})(−16 − 8δ_{0}),
c_{0} = 4(2 + δ_{0})(−δ_{0}(r^{2} − 1)/2 + (δ_{0}/2)^{2})+
4(1 + δ_{0})^{2} − (4r^{2} − (δ_{0})^{2}(2r^{2} + 1)) =
−4r^{2}(δ_{0} + 1) + 4(1 + δ_{0})^{2} + 4δ_{0} + 5δ_{0}^{2} + δ_{0}^{3} =
(δ_{0} + 1)×
(−4(r^{2} − 1) + δ_{0}^{2} + 8δ_{0}),
d_{0} = −4(1 + δ_{0})(−δ_{0}(r^{2} − 1)/2 + (δ_{0}/2)^{2}),
r_{0} = (δ_{0}/2)^{2}(r^{2} − 1 − δ_{0}/2)^{2}−
(δ_{0}/2)^{2}r^{2}(r^{2} − 1) − (δ_{0}/2)^{4} =
−(δ_{0}/2)^{2}(1 + δ_{0})(r^{2} − 1).

Therefore,

P (s) = (16s^{2} − 8sδ_{0} + δ_{0}^{2})×
(s^{2} − s − (r^{2} − 1)/4)(1 + δ_{0}) =
(1 + δ_{0})(4s − δ_{0})^{2}(s − (r + 1)/2)(s + (r − 1)/2).

The left part of (3.1.73) for s = (r + 1)/2 ≥ 1 is equal to the positive number

2(2 + δ_{0})s^{2} − δ_{0}(r^{2} − 1)/2 + (δ_{0}/2)^{2}−
2s(1 + δ_{0}) = 2(s^{2} − s)(1 + δ_{0})+
(r + 1)^{2}/2 − δ_{0}(r^{2} − 1)/2 + (δ_{0}/2)^{2};

therefore, s = (r+1)/2 is not the solution of the system consisting of the inequality (3.1.73) and the equation (3.1.74). According to (3.1.62) and (3.1.63), the equality s = −(r − 1)/2 is only possible when when the equalities r = 1 and ψ = π are fulfilled at the same time. According to (3.1.30) and (3.1.6),

η_{k}^{∧}(r, π, δ_{0}) − 1 = (2/(1 + δ_{0}))×
(r − 1 − δ_{0}/2 + (−1)^{k}\sqrt{r^{2} − r + (δ_{0}/2)^{2}}),

where k = 0, 1. Therefore,

(3.1.75)  |η_{0}^{∧}(r, π, δ_{0}) − 1| > |η_{1}^{∧}(r, π, δ_{0}) − 1|,

if r > 1 + δ_{0}/2;

(3.1.76)  |η_{0}^{∧}(r, π, δ_{0}) − 1| = |η_{1}^{∧}(r, π, δ_{0}) − 1|,

if r = 1 + δ_{0}/2;

(3.1.77)  |η_{0}^{∧}(r, π, δ_{0}) − 1| < |η_{1}^{∧}(r, π, δ_{0}) − 1|,

if 1 ≤ r < 1 + δ_{0}/2. In view of (3.1.77), if 1 ≤ r < 1 + δ_{0}/2, then the number s = −(r − 1)/2 is not the solution of the system consisting of the inequalities (3.1.62) and (3.1.73) and the equation (3.1.74). For s = (r − 1)/2, the inequality (3.1.64) follows from (3.1.75). For s = (r−1)/2, i.e. for r = 1+δ_{0}/2, the inequality (3.1.66) follows from (3.1.76). For s = (r − 1)/2, the inequality (3.1.67) follows from (3.1.77). Now we consider the number

(3.1.78)  s = δ_{0}/4

as a possible solution of the system consisting of the inequalities (3.1.62) and (3.1.73) and the equation (3.1.74). As before, we consider only r ≥ 1. According to (3.1.62), the equality (3.1.78) is only possible if

(3.1.79)  r ≤ 1 + δ_{0}/2.

Therefore, according to (3.1.75),the inequality (3.1.64) is fulfilled.

The case when equality is achieved in (3.1.79) and s = δ_{0}/4, was considered above. It remains to consider the case when

(3.1.80)  r − 1 < δ_{0}/2 = 2s.

But (3.1.73) for s = δ_{0}/4 turns into inequality

δ_{0}((δ_{0})^{2}/4 − r^{2}) < 0.

So, if (3.1.79) is fulfilled, then s = δ_{0}/4 is the solution of the system consisting of inequalities (3.1.62) and (3.1.73) and equation (3.1.74); consequently, (3.1.66) holds. From the equalities

(3.1.81)  η_{k}^{∧}(r, 0, δ_{0}) − 1 = (2/(1 + δ_{0}))×

(r + 1 + δ_{0}/2 + (−1)^{k}\sqrt{r^{2} + r + (δ_{0}/2)^{2}}),

where k = 0, 1; from the continuity of the sum

∑_{k=0}^{1} |η_{k}^{∧}(r, 0, δ_{0}) − 1|

as function of s; from (3.1.77); and from the necessity of (3.1.79) for s = δ_{0}/4 to satisfy the system consisting of inequalities (3.1.62) and (3.1.73) and equation (3.1.74), follow (3.1.65) and (3.1.67). The lemma is proved.

Proof of Theorem 3.1.2. Since now r = 1, it follows, according to (3.1.61), that

s = cos^{2} (\frac{ψ}{2} ) ,

where ψ = (2l − 1)π/3 and l = 0, 1, 2. If l = 0, 1, then s = 1/\sqrt{3} > 1/52 = δ_{0}/4. From Lemma 3.1.4 it follows that |η_{0,l}^{∗} − 1| > |η_{1,l}^{∗} − 1|. This inequalty and Lemma 3.1.3 have as consequence the first assertion of Theorem 3.1.2. If l = 2, then s = 0 < δ_{0}/4 = 1/52; the second assertion of Theorem 3.1.2 follows from Lemma 3.1.4, or directly from (3.1.55) with ε = −1. Theorem is proved.

For l = 0, 1, 2 we denote by F_{l}^{∗}(η) the polynomial (3.1.1) as polynomial of η with r = 1, ψ = (2l − 1)π/3 and Δ = 13. It is clear that

(3.1.82)  ∏_{l=0}^{2} F_{l}^{∗}(η) = D^{∧}(1, η) = (η + 1)^{3}(η + γ_{1})^{3} − 2^{3}(1 + γ_{1})^{3}η^{3}

with γ = 6/7.

Lemma 3.1.5. For l = 0, 1, the polynomial F_{l}^{∗}(η) is irreducible over the field K_{0} = ℚ[exp(2πi/3)].

Proof. Clearly, it is sufficient to consider the case l = 1. Since the degree of the polynomial F_{1}^{∗}(η) is equal to 2, the contrary means that it has a zero in the field K_{0}; then there is a ∈ K_{0} such that for D_{0}(1, π/3, δ_{0}) from (3.1.6) the following equality holds:

D_{0}(1, π/3, δ_{0}) = 1 − exp(2iπ/3) − (δ_{0}/2)^{2}(1 + exp(2iπ/3)) =
1 − (δ_{0}/2)^{2} − exp(2iπ/3)(1 + (δ_{0}/2)^{2}) = a^{2}.

Then

(3.1.83)  Nm_{K_{0}/ℚ}(D_{0}(1, π/3, δ_{0}) = 3 + (δ_{0}/2)^{4}) = \frac{48Δ^{4} + 1}{16Δ^{4}}

(Nm_{K_{0}/ℚ}(a))^{2} = 3 + (δ_{0}/2)^{4}.

Therefore,

16Δ^{4}(Nm_{K_{0}/ℚ}(a))^{2} = 48Δ^{4} + 1 ≡ −1 mod 11,

but −1 is a non-quadratic residue mod 11. This contradiction proves the lemma. The lemma is proved.

Let F_{l}^{∼}(η) = ∏_{k=0}^{1}(η − h^{∼}(η_{k,l}^{∗})).

Lemma 3.1.6. Let Δ = 13. Then, for l = 0, 1, the polynomial F_{l}^{∼}(η) is irreducible over the field K_{0} = ℚ[exp(2πi/3)], and the polynomial

F_{0}(η) = ∏_{l=0}^{1} ∏_{k=0}^{1}(η − h^{∼}(η_{k,l}^{∗}))

is irreducible over the field ℚ.

Proof. According to Lemma 3.1.5, for k = 0, 1, l = 0, 1, the field K_{l+1} = K_{0}[η_{k,l}^{∗}] is the quadratic extension over the field K_{0}. In view of (3.1.1),

(3.1.84)  exp(−i(2l − 1)2π/3) = \frac{(η_{k,l}^{∗} + 1)(η_{k,l}^{∗} + γ_{1})}{2(1 + γ_{1})η_{k,l}^{∗}},

where k = 0, 1, l = 0, 1. Therefore,

(3.1.85)  K_{l+1} = K_{0}[η_{k,l}^{∗}] = ℚ[η_{k,l}^{∗}]

According to (3.1.82), if Δ = 13, then the numbers η_{k,l}^{∗} with k = 0, 1, l = 0, 1 are 5ℤ_{K_{l+1}} - integer numbers of the field K_{l+1} (the ideal 5ℤ_{K_{l+1}}, as it will be shown below, is not prime ). For l = −1, 0, 1, let 𝔒_{l+1} be the ring of all 5ℤ_{K_{l+1}} - integer numbers of the field K_{l+1}; let further K_{3} = ℚ and 𝔒_{3}, the ring of all 5ℤ_{K_{3}} - integer numbers of the field K_{3}.

Since the polynomial η^{2} + η + 1 is irreducible over the field ℤ/5ℤ, it follows that the ideal 5ℤ_{K_{0}} is the prime ideal of the second degree.

Below, for the prime ideal 𝔭 of the number field K, we denote by v_{𝔭} the additive 𝔭−adic valuation of the field K. According to (3.1.83), if Δ = 13, then

v_{5ℤ_{K_{0}}} (Nm_{K_{0}/ℚ}(D_{0}(1, π/3, δ_{0})) = 0.

Therefore, for any k = 0, 1, l = 0, 1, the numbers 1, η_{k,l}^{∗} form the 5𝔒_{0}−integer basis of the extension K_{l+1}/K_{0} and the four numbers

(3.1.86)  (η_{k,l}^{∗})^{α} exp (2πβ \frac{2l − 1}{3}i) ,

where α = 0, 1, β = 0, 1, form 5𝔒_{3}−integer basis of the extension K_{l+1}/K_{3}. Further we have

(3.1.87)  F_{l}^{∗}(η) = (η + \frac{6}{7}) (η + 1) + \frac{26}{7}η exp (π \frac{2l − 1}{3}i) ≡

(η + 3)(η + 1) + 3η exp (π \frac{2l − 1}{3}i) ≡

η^{2} + (3 exp (π \frac{2l − 1}{3}i) − 1) η + 3 ≡

η^{2} + (2 exp (4π \frac{2l − 1}{3}i) − 1) η + 3 ≡

η^{2} + (−2 exp (2π \frac{2l − 1}{3}i) + 2) η − 2 ≡

(η + exp (2π \frac{2l − 1}{3}i)) (η + 2 exp (2π \frac{2l − 1}{3}i) + 2) ≡

(η + exp (2π \frac{2l − 1}{3}i)) (η − 2 exp (4π \frac{2l − 1}{3}i)) mod 5𝔒_{0}[η],

where l = 0, 1; therefore,

Nm_{K_{l+1}/K_{0}}(η_{k,l}^{∗} + exp (2l − 1)2iπ/3) =
F_{l}^{∗}(− exp((2l − 1)2iπ/3)) ∈ 5𝔒_{0},
Nm_{K_{l+1}/K_{0}}(η_{k,l}^{∗} − exp((2l − 1)4iπ/3)) =
F_{l}^{∗}(exp (2l − 1)4iπ/3) ∈ 5𝔒_{0},

and for each l=0,1, by transition from the field K_{0} to the field K_{l+1}, the prime ideal 5𝔒_{0} turns into the product

(3.1.88)  5𝔒_{l+1} = 𝔭_{1,l+1}𝔭_{2,l+1}

of two prime ideals

(3.1.89)  𝔭_{1,l} = (η_{0,l}^{∗} + exp (2π \frac{2l − 1}{3}i)) 𝔒_{l+1} + 5𝔒_{l+1}

and

(3.1.90)  𝔭_{2,l} = (η_{0,l}^{∗} − 2 exp (4π \frac{2l − 1}{3}i)) 𝔒_{l+1} + 5𝔒_{l+1}.

According to (3.1.87),

(3.1.91) η^{2} ≡ (2 exp((2l − 1)2iπ/3) − 2)η + 2 mod (5𝔒_{0}[η] + F_{l}^{∗}(η)𝔒_{0}[η]),

where l = 0, 1,

(3.1.92)  η^{2} exp (2π \frac{2l − 1}{3}i) ≡

(−2 + exp (2π \frac{2l − 1}{3}i))η + 2 exp (2π \frac{2l − 1}{3}i) mod 5𝔒_{0}[η] + F_{l}^{∗}(η)𝔒_{0}[η],

where l = 0, 1.

Let 𝔐 denote the 𝔒_{0}[η]−module consisting of all columns with four elements contained in the ring 𝔒_{0}[η]. For l = 0, 1, according to (3.1.87), (3.1.91) and (3.1.92),

η ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ exp((2l − 1)2iπ/3))η \end{matrix} ) ≡

( \begin{matrix}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 2 & 0 & 3 & 2 \\ 0 & 2 & 3 & 1 \end{matrix} ) ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ η exp((2l − 1)2iπ/3)) \end{matrix} ) mod (5𝔐 + F_{l}^{∗}(η)𝔐),

(3.1.93)  η^{2} ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ η exp((2l − 1)2iπ/3)) \end{matrix} ) ≡

( \begin{matrix}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 2 & 0 & 3 & 2 \\ 0 & 2 & 3 & 1 \end{matrix} )^{2} ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ η exp((2l − 1)2iπ/3)) \end{matrix} ) ≡

( \begin{matrix}2 & 0 & 3 & 2 \\ 0 & 2 & 3 & 1 \\ 1 & 4 & 2 & 3 \\ 1 & 2 & 2 & 4 \end{matrix} ) ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ η exp((2l − 1)2iπ/3)) \end{matrix} ) mod (5𝔐 + F_{l}^{∗}(η)𝔐),

η^{4} ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ η exp((2l − 1)2iπ/3)) \end{matrix} ) ≡

( \begin{matrix}2 & 0 & 3 & 2 \\ 0 & 2 & 3 & 1 \\ 1 & 4 & 2 & 3 \\ 1 & 2 & 2 & 4 \end{matrix} )^{2} ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ η exp((2l − 1)2iπ/3)) \end{matrix} ) ≡

( \begin{matrix}4 & 1 & 1 & 1 \\ 4 & 3 & 4 & 0 \\ 2 & 2 & 0 & 4 \\ 3 & 0 & 1 & 1 \end{matrix} ) ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ η exp((2l − 1)2iπ/3)) \end{matrix} ) mod (5𝔐 + F_{l}^{∗}(η)𝔐),

η^{8} ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ η exp((2l − 1)2iπ/3)) \end{matrix} ) ≡

( \begin{matrix}4 & 1 & 1 & 1 \\ 4 & 3 & 4 & 0 \\ 2 & 2 & 0 & 4 \\ 3 & 0 & 1 & 1 \end{matrix} )^{2} ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ η exp((2l − 1)2iπ/3)) \end{matrix} ) ≡

( \begin{matrix}0 & 4 & 4 & 4 \\ 1 & 1 & 1 & 0 \\ 3 & 3 & 4 & 1 \\ 2 & 0 & 4 & 3 \end{matrix} ) ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ η exp((2l − 1)2iπ/3)) \end{matrix} ) mod (5𝔐 + F_{l}^{∗}(η)𝔐),

(3.1.94)  η^{12} ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ η exp((2l − 1)2iπ/3)) \end{matrix} ) ≡ ( \begin{matrix}4 & 1 & 1 & 1 \\ 4 & 3 & 4 & 0 \\ 2 & 2 & 0 & 4 \\ 3 & 0 & 1 & 1 \end{matrix} ) ×

( \begin{matrix}0 & 4 & 4 & 4 \\ 1 & 1 & 1 & 0 \\ 3 & 3 & 4 & 1 \\ 2 & 0 & 4 & 3 \end{matrix} ) ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ η exp((2l − 1)2iπ/3)) \end{matrix} ) ≡

( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} ) ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ η exp((2l − 1)2iπ/3)) \end{matrix} ) mod (5𝔐 + F_{l}^{∗}(η)𝔐).

Let me explain the relation (3.1.94). According to (3.1.88),

(3.1.95)  𝔒_{l+1}/5𝔒_{l+1} = 𝔒_{l+1}/𝔭_{1,l+1} ⊕ 𝔒_{l+1}/𝔭_{2,l+1}.

In view of (3.1.89),

η_{0,l}^{∗} ≡ − exp (2π \frac{2l − 1}{3}i) mod 𝔭_{1,l+1}

and, therefore, the order of η_{0,l}^{∗} + 𝔭_{1,l+1} in the multiplicative group of the field 𝔒_{l+1}/𝔭_{1,l+1} is equal to 6; in view of (3.1.90),

η_{0,l}^{∗} ≡ 2 exp (4π \frac{2l − 1}{3}i) mod 𝔭_{2,l+1}

and, therefore, the order of η_{0,l}^{∗} + 𝔭_{2,l+1} in the multiplicative group of the field 𝔒_{l+1}/𝔭_{2,l+1} is equal to 12.

For l = 0, 1, according to (3.1.93) and (3.1.94),

(3.1.96)  (η^{2} − 1)η^{12} ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ exp((2l − 1)2iπ/3))η \end{matrix} ) ≡

( \begin{matrix}1 & 0 & 3 & 2 \\ 0 & 1 & 3 & 1 \\ 1 & 4 & 1 & 3 \\ 1 & 2 & 2 & 3 \end{matrix} ) ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ exp((2l − 1)2iπ/3))η \end{matrix} ) mod (5𝔐 + F_{l}^{∗}(η)𝔐),

(η^{2} − 1)^{2}η^{24} ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ exp((2l − 1)2iπ/3))η \end{matrix} ) ≡

( \begin{matrix}1 & 0 & 3 & 2 \\ 0 & 1 & 3 & 1 \\ 1 & 4 & 1 & 3 \\ 1 & 2 & 2 & 3 \end{matrix} )^{2} ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ exp((2l − 1)2iπ/3))η \end{matrix} ) ≡

( \begin{matrix}1 & 1 & 0 & 2 \\ 4 & 0 & 3 & 3 \\ 0 & 4 & 2 & 3 \\ 1 & 1 & 2 & 4 \end{matrix} ) ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ exp((2l − 1)2iπ/3))η \end{matrix} ) mod (5𝔐 + F_{l}^{∗}(η)𝔐),

(3.1.97)  (η^{2} − 1)^{3}η^{36} ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ exp((2l − 1)2iπ/3))η \end{matrix} ) ≡

( \begin{matrix}1 & 1 & 0 & 2 \\ 4 & 0 & 3 & 3 \\ 0 & 4 & 2 & 3 \\ 1 & 1 & 2 & 4 \end{matrix} ) ×

( \begin{matrix}1 & 0 & 3 & 2 \\ 0 & 1 & 3 & 1 \\ 1 & 4 & 1 & 3 \\ 1 & 2 & 2 & 3 \end{matrix} ) ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ exp((2l − 1)2iπ/3))η \end{matrix} ) ≡

( \begin{matrix}3 & 0 & 0 & 4 \\ 0 & 3 & 1 & 1 \\ 0 & 3 & 0 & 4 \\ 2 & 2 & 1 & 1 \end{matrix} ) ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ exp((2l − 1)2iπ/3))η \end{matrix} ) mod (5𝔐 + F_{l}^{∗}(η)𝔐),

(3.1.98)  (η^{2} − 1)^{6}η^{72} ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ exp((2l − 1)2iπ/3))η \end{matrix} ) ≡

( \begin{matrix}3 & 0 & 0 & 4 \\ 0 & 3 & 1 & 1 \\ 0 & 3 & 0 & 4 \\ 2 & 2 & 1 & 1 \end{matrix} )^{2} ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ η exp((2l − 1)2iπ/3)) \end{matrix} ) ≡

( \begin{matrix}2 & 3 & 4 & 1 \\ 2 & 4 & 4 & 3 \\ 3 & 2 & 2 & 2 \\ 3 & 1 & 3 & 0 \end{matrix} )^{2} ( \begin{matrix}1 \\ exp((2l − 1)2iπ/3)) \\ η \\ η exp((2l − 1)2iπ/3)) \end{matrix} ) mod (5𝔐 + F_{l}^{∗}(η)𝔐).

In correspondence to (3.1.28), let

(3.1.99)  h_{1}^{∼}(η) = (η^{2} − 1)^{3}η^{3}6 = h^{∼}(η)(12/13)^{36}2^{6}.

From (3.1.97) and (3.1.98), it follows that

(3.1.100)  h_{1}^{∼}(η_{k,l}^{∗}) ≡ 3 + 4 exp (2π \frac{2l − 1}{3}i) η_{k,l}^{∗} mod 5𝔒_{l+1}.

(3.1.101)  (h_{1}^{∼}(η_{k,l}^{∗}))^{2} ≡ 2 + 3 exp (2π \frac{2l − 1}{3}i) +

(4 + exp (2π \frac{2l − 1}{3}i)) η_{k,l}^{∗} mod 5𝔒_{l+1},

26 LEONID A. GUTNIK

where k = 0, 1, l = 0, 1; therefore, h_{1}^{∼}(η_{k,l}^{∗}) ∉ K_{0} and, since [K_{l+1} : K_{0}] = 2 for any l = 0, 1, it follows, according to (3.1.99), that

(3.1.102)  K_{0}[h^{∼}(η_{k,l}^{∗})] = K_{l+1},

where l = 0, 1. Consequently, the polynomial

(3.1.103)  F_{l}(η)^{∼} = ∏_{k=0}^{1}(η − h^{∼}(η^{∗}(k, l)))

is irreducible over the field K_{0}. Moreover, according to (3.1.100) and (3.1.101),

h_{1}^{∼}(η^{∗}(k, l)) + 2 = a_{1} + b_{1} exp (2π \frac{2l − 1}{3}i) +

c_{1}η^{∗}(k, l) + d_{1} exp (2π \frac{2l − 1}{3}i) η^{∗}(k, l),

h_{1}^{∼}(η^{∗}(k, l)) + (h_{1}^{∼}(η^{∗}(k, l))^{2} = a_{2}+

b_{2} exp (2π \frac{2l − 1}{3}i) + c_{2}η^{∗}(k, l) + d_{2} exp (2π \frac{2l − 1}{3}i) η^{∗}(k, l),

where k = 0, 1, l = 0, 1,

a_{1} ∈ 𝔒_{3}, b_{1} ∈ 𝔒_{3}, c_{1} ∈ 𝔒_{3}, d_{1} ∈ 𝔒_{3},
a_{2} ∈ 𝔒_{3}, b_{2} ∈ 𝔒_{3}, c_{2} ∈ 𝔒_{3}, d_{2} ∈ 𝔒_{3},
a_{1} ≡ b_{1} ≡ c_{1} ≡ 0 mod 5𝔒_{3}, d_{1} ≡ 4 mod 5𝔒_{3}
a_{2} ≡ d_{2} ≡ 0 mod 5𝔒_{3},
b_{2} ≡ 3 mod 5𝔒_{3}, c_{2} ≡ 4 mod 5𝔒_{3}.

Therefore, the system of numbers

1, h_{1}^{∼}(η^{∗}(k, l)) + 2, h_{1}^{∼}(η^{∗}(k, l)) + h_{1}^{∼}(η^{∗}(k, l))^{2}

is linearly independent over ℚ. In fact, the contrary means that there exist α, β, γ in ℚ, not all of which are equal to zero, such that

α + βa_{1} + γa_{2} = 0,

βb_{1} + γb_{2} = 0,

βc_{1} + γc_{2} = 0,
βd_{1} + γd_{2} = 0;

but this is impossible, because

det (( \begin{matrix}1 & 0 & 0 \\ a_{1} & b_{1} & d_{1} \\ a_{2} & b_{2} & d_{2} \end{matrix} )) ≡ 3 mod 5𝔒_{3}.

Therefore,

[ℚ(h_{1}^{∼}(η^{∗}(k, l))) : ℚ] ≥ 3

and, since 4/[ℚ(h_{1}^{∼}(η^{∗}(k, l))) : ℚ] ∈ ℕ, it follows that

[ℚ(h_{1}^{∼}(η^{∗}(k, l))) : ℚ] = 4.

Consequently, according to (3.1.99),

[ℚ(h^{∼}(η^{∗}(k, l))) : ℚ] = 4,

where k = 0, 1, l = 0, 1, and the polynomial

F_{0}(η) = ∏_{l=0}^{1} ∏_{k=0}^{1}(η − h^{∼}(η_{k,l}^{∗}))

is irreducible over the field ℚ. The lemma is proved.

Corollary. The assertion of the Theorem 3.1.3 holds.

Proof. Since

ℚ(h^{∼}(η^{∗}(k, l))) = K_{l+1},

where k = 0, 1, ‘l = 0, 1, it follows that, for k = 0, 1, l = 0, 1, h^{∼}(η^{∗}(k, l) ∈ ℝ is impossible. The corollary is proved.

So, Theorem 3.1.1 is also proved.

As in the first part of this paper, we shall work with the set Ω^{∗} = {z ∈ ℂ: |z| ≥ 1, −\frac{3π}{2} < arg(z) < \frac{π}{2} \}. For any ε > 0, let D_{ε}^{∼} denote the domain \{z ∈ ℂ: |z| < 1 + ε, Re (z) > 1\}. Clearly, D_{ε}^{∼} lies in the circle |z−1| < \sqrt{ε}, and | arg(z)| < arctan(\sqrt{ε}) for any z ∈ D_{ε}. Let further D_{ε}^{∧} be the closure of D_{ε}^{∼}. For any k = 0, 1, l = 0, 1, 2, and z ∈ D_{ε}^{∧}, let

η_{k,l}^{∗∗}(z) = η_{k}^{∧}(|z|, argz + (2l − 1)π/3, δ_{0}),

where, as above, δ_{0} = 1/13. Let further

F_{0}^{∧}(η; z) = ∏_{l=0}^{2} ∏_{k=0}^{1}(η − h^{∼}(η_{k,l}^{∗∗}(z)))

and 𝔇_{0}^{∧}(z) be the discriminant relatively to η of the polynomial F_{0}^{∧}(η; z). Then 𝔇_{0}^{∧}(z) and

η_{k,l}^{∗∗}(z)

with any k = 0, 1, l = 0, 1, 2 are continuos functions in D_{ε}^{∧}; moreover,

η_{k,l}^{∗∗}(1) = η_{k,l}^{∗},

where k = 0, 1, l = 0, 1, 2, and, according to Theorem 3.1.1, 𝔇_{0}^{∧}(1) = 𝔇_{0} ≠ 0. Therefore, there exist ε > 0 such that 𝔇_{0}^{∧}(z) ≠ 0 for any z ∈ D_{ε}^{∧}; we fix this ε for further considerations.

According to (3.1.43) and (3.1.71),

h^{∼}(η_{k,l}^{∗∗}(z)) ≠ 0

for any z ∈ D_{ε} and any k = 0, 1, l = 0, 1. Let ε_{1} ∈ (0, ε), ε_{3} ∈ (ε_{1}, ε). Let further

D_{ε_{1},ε_{3}}^{∼,∼} = {z ∈ ℂ: |z| < 1 + ε_{3}, Re (z) > 1 + ε_{1}},
D_{ε_{1},ε_{3}}^{∧,∧} = {z ∈ ℂ: |z| ≤ 1 + ε_{3}, Re (z) ≥ 1 + ε_{1}}.

I wish to apply Lemma 2.4.5 of Part 2 of this work.

For the role of the compact set F of this lemma, I take the set D_{ε_{1},ε_{3}}^{∧∧}, which is a compact subset of the domain D_{ε}^{∼} ⊂ ℂ\{1}. The polynomial D^{∼}(z, w) in (2.4.3) is connected to D^{∧}(z, η) from (3.1.0) by means of the equality (2.4.52) and therefore, according to the substitution (2.4.51), Lemma 3.1.3 and Lemma 3.1.5, has only simple roots. It will play the role of the polynomial D^{∼}(z, λ) in (2.4.45) of part 2. The independent of z polynomials (2.4.4) and (2.4.5) will play the role of the polynomials P^{∼}(z, λ) and Q^{∼}(z, λ) of Lemma 2.4.5. According to (2.4.55) and Theorem 3.1.1, the map (2.4.46) is injective on the set of all the roots of the polynomial

D^{∼}(z, λ)

(i.e. the polynomial (2.4.3)). The matrix B^{∼}(z) of Lemma 2.4.5 will now play the role of the matrix B^{∼}(z) from (2.4.6). The matrix (2.4.16) will play the role of the matrix A(z; ν^{−1}) from (2.4.44); according to (2.4.10) and (2.4.11), the following equality holds: A(z, 0) = A^{∼}(z), with A^{∼}(z) from (2.4.15); therefore, the condition (2.4.47) is also fulfilled. So, all the conditions of Lemma 2.4.5 are fulfilled. Consequently, according to the assertion of Lemma 2.4.5, for every z ∈ F, all the roots of the polynomial (2.4.20) are simple and there exists a matrix C = C(z) ∈ Mat_{6}(ℂ^{F}) satisfying to the following conditions:

1) the relations (2.4.21) are fulfilled;

2) for every z ∈ F, the matrix (C(z))^{−1}A(z, 0)C(z) is diagonal;

3) for every z ∈ F, no element in the first row of the matrix C(z) is equal to sero.

Therefore, we can now apply Lemma 2.4.4 of the Part 2 of this work. As above, the set D_{ε_{1},ε_{3}}^{∧∧} plays the role of the compact F, the matrix (2.4.16) plays the role of the matrix A(z; ν^{−1}) from (2.4.44) and (2.4.18), the column X_{k}(z; ν) from (2.4.17) with k = 1, 2, 3, 5 and ν ∈ ℕ + [1/ε_{0}(F)] plays the role of the column X_{ν}(z) from (2.4.19). Consequently, according to the assertion of Lemma 2.4.4, there is ε_{2} ∈ (0, ε_{0}) such that, for any z ∈ F, ν ∈ ℕ + [1/ε_{2}(F)] and k = 1, 2, 3, 5, the first component f_{k}^{∗}(z; ν) of the column X_{k}(z; ν) satisfies the equation

f_{k}^{∗}(z; ν + 6) + ∑_{j=0}^{5} q_{j}^{∗}(z; ν^{−1})f_{k}^{∗}(z; ν + j) = 0,

where q_{j}^{∗}(z; ν^{−1}) ∈ ℌ(F, ε_{2}) for j = 0 , …, 5; moreover the polynomial

λ^{6} + ∑_{j=0}^{5} q_{j}^{∗}(z; 0)λ^{j},

coincides with polynomial (2.4.20), the roots of which, in view of (2.4.55), coincide with the roots of the polynomial

F_{0}^{∨}(η) = ∏_{l=0}^{2} ∏_{k=0}^{1}(η − h^{∼}(η_{k,l}^{∗})),

having, according to Theorem 3.1.1, only simple roots.

So, the functions f_{k}^{∗}(z; ν), for any z ∈ F, ν ∈ ℕ + [1/ε_{2}(F)] and k = 1, 2, 3, 5, satisfy the equation

(3.1.104)  x_{ν+6} + ∑_{j=0}^{5} q_{j}^{∗}(z; ν^{−1})x_{ν+j} = 0.

In view of (2.11) and (2.15) of the Part 1 of this work, x_{ν} = f_{4}^{∗}(z; ν) is the solution of the equation (3.1.104). In view of (2.13) and (2.15) of the Part 1 of this work, x_{ν} = f_{6}^{∗}(z; ν), is also the solution of the equation (3.1.104).

Clearly, the rational functions q_{j}^{∗}(z; ν^{−1}) from (3.1.104) with j = 0 , …, 5 admit the representation in the form

(3.1.105)  q_{j}^{∗}(z; ν^{−1}) = q_{j}(z; ν^{−1})/q_{6}(z; ν^{−1}),

where q_{j}(z; ν^{−1}) ∈ ℚ[z; ν^{−1}] for j = 0 , …, 6 have no common divisors with the exception of different from zero constants, and q_{6}(z; ν^{−1}) is different from zero for any z ∈ F, ν ∈ ℕ + [1/ε_{2}(F)]. Consequently, the equation (3.1.104) is equivalent to the equation

(3.1.106)  q_{6}(z; ν^{−1})x_{ν+6} + ∑_{j=0}^{5} q_{j}(z; ν^{−1})x_{ν+j} = 0,

where z ∈ F, ν ∈ ℕ + [1/ε_{2}(F)]; and x_{ν} = f_{k}^{∗}(z; ν), for k = 1, 2, 4, 6, is the solution of this equation. In view of (2.19) - (2.28) of the first part of this work, the functions f_{k}^{∗}(z; ν), for k = 1, 2, 4, 6 and ν ∈ ℕ, are regular in the domain |z| > 1 and, since the equality (3.1.106) for x_{ν} = f_{k}^{∗}(z; ν) with k = 1, 2, 4, 6, ν ∈ ℕ + [1/ε_{2}(F)] is fulfiiled for all z in the domain

D_{ε_{1},ε_{3}}^{∼,∼} = {z ∈ ℂ: |z| < 1 + ε_{3}, Re (z) > 1 + ε_{1}},

it follows that, according to the uniqueness theorem, it is fulfilled for all z with |z| > 1.

§2. Sextette comes to the foreground.

Lemma 3.2.1. Let r ∈ ℕ + 1,

(3.2.1)  L_{k}(z) = ∑_{y=1}^{∞} z^{y}/y^{k},

where k ∈ ℕ.

Then the functions

(3.2.2)  1, L_{1}(z) , …, L_{r}(z)

form a linear idependent system over ℂ(z).

Proof. Let

(3.2.3)  P_{0}(z) + ∑_{k=1}^{r} P_{k}(z)L_{k}(z) = 0,

where P_{k}(z) ∈ ℂ[z] with k = 0 , …, r. Let further P_{κ}(z) = 0 for κ ∈ ℕ + r. If we direct z ∈ (0, 1) to 1, then we see that

(3.2.4)  P_{1}(1) = 0.

Therefore, P_{1}(z) = (1−z)Q_{1}(z) with Q_{1}(z) ∈ ℂ[z]. Applying the operator δ = z \frac{∂}{∂z} to (3.2.3), we obtain the equality

(δP_{0})(z) + zQ_{1}(z) + ∑_{k=1}^{r}((δP_{k})(z) + P_{k+1}(z))L_{k}(z) = 0.

Repeating previous arguments m times, we obtain, instead of (3.2.4), the equality

(3.2.5)  ∑_{κ=0}^{min(m,r−1)} \binom{m}{κ}((δ)^{m−κ}P_{1+κ})(1) = 0,

where m ∈ ℕ − 1. Let k = κ + 1, where κ = 0 , …, min(m, r − 1). Then

(3.2.6)  ∑_{k=1}^{min(m+1,r)} \binom{m}{k − 1}((δ)^{m−k+1}P_{k})(1) = 0,

where m ∈ ℕ − 1. We fix n ∈ ℕ, such that, for any k ∈ ℕ ∩ [1, r], if P_{k}(z) is not a zero polynomial, then deg(P_{k}(z)) ≤ n; let

P_{k}(z) = ∑_{l=0}^{n} a_{k,l}z^{l},

where a_{k,l} ∈ ℂ for k = 1 , …, r and 1 = 1 , …, n. Then the equality (3.2.6) for m ≥ r may be written in the form

(3.2.7)  ∑_{k=1}^{r} \binom{m}{k − 1} ∑_{l=0}^{n} l^{m−k+1}a_{k,l} =

∑_{l=1}^{n} ∑_{k=1}^{r} \binom{m}{k − 1}l^{m−k+1}a_{k,l} = 0.

Since y_{κ,l}(m) = \binom{m}{k−1}l^{m−k+1}, where k = 1 , …, r and l = 1 , …, n, form a basis for the space of the solutions of the difference equation

(∏_{l=1}^{n}(▽ − l)^{r}y)(m),

where, as usually, the operator ▽ turns y(m) into y(m + 1), it follows from (3.2.7) that a_{k,l}(m) = 0, where k = 1 , …, r and l = 1 , …, n. Therefore, P_{k}(z) = a_{k,0}, where k = 1 , …, r. In (3.2.6), taking successively m = 0 , …, r − 1, we convince that a_{k,0} = 0, where k = 1 , …, r. The lemma is proved.

Corollary. Let r ∈ ℕ + 1. Then the functions

(3.2.8)  1, L_{1}(1/z) , …, L_{r}(1/z)

form a linear idependent system over ℂ(z).

Proof. Let

(3.2.9)  T_{0}(z) + ∑_{k=1}^{r} T_{k}(z)L_{k}(1/z) = 0,

where T_{k}(z) ∈ ℂ[z] with k = 0 , …, r. We fix n ∈ ℕ, such that, for any k ∈ ℕ∩[1, r], if T_{k}(z) is not a zero polynomial, then deg(T_{k}(z)) ≤ n. From (3.2.9), it follows that

P_{0}(ζ) + ∑_{k=1}^{r} P_{k}(ζ)L_{k}(ζ) = 0,

where ζ = 1/z and P_{k}(ζ) = ζ^{n}T_{k}(1/ζ) with k = 1 , …, r. According to Lemma 3.2.1, P_{k}(ζ) = 0 for k = 1 , …, r and, therefore, T_{k}(z) = 0 for k = 1 , …, r . The Corollary is proved.

Since x_{ν} = f_{k}^{∗}(z; ν) with k = 1, 2, 4, 6, ν ∈ ℕ + [1/ε_{2}(F)], |z| > 1 is the solution of the equality (3.1.106), it follows that, in view of (2.19) - (2.28) of the first part of this work and Corollary of Lemma 3.2.1, the six sequences of the polynomials

(3.2.10)  x_{ν} = α^{∗}(z; ν), x_{ν} = β^{∗}(z; ν), x_{ν} = γ^{∗}(z; ν),
x_{ν} = ϕ^{∗}(z; ν), x_{ν} = ψ^{∗}(z; ν), x_{ν} = ξ^{∗}(z; ν)

with ν ∈ ℕ + [1/ε_{2}(F)], |z| > 1 are also the solutions of the equation (3.1.106). I will start the next part by studying the sequences of the polynomials (3.2.10).

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Moscow State Institute of Electronics and Mathematics Bolshoi Trekhsvyatitelskij pereulok, 1-3/12, str.8 Moscow 109028 Russian Federation E-mail: gutnik@gutnik.mccme.ru