Sets with Even Partition Functions and 2-adic Integers, II

N. Baccar^{1}

Université de Sousse ISITCOM Hammam Sousse Dép. de Math Inf. 5 Bis Rue 1 Juin 1955 4011 Hammam Sousse Tunisie

                 naceurbaccar@yahoo.fr

A. Zekraoui Université de Monastir F. S. M. Dép. de Math. Avenue de l’environnement 5000 Monastir Tunisie

                ahlemzekraoui@yahoo.fr

Abstract

For P ∈ 𝔽_{2}[z] with P (0) = 1 and deg(P) ≥ 1, let A = A(P) be the unique subset of ℕ such that ∑_{n≥0} p(A, n)z^{n} ≡ P (z) (mod 2), where p(A, n) is the number of partitions of n with parts in A. Let p be an odd prime number, and let P be irreducible of order p ; i.e., p is the smallest positive integer such that P divides 1 + z^{p} in 𝔽_{2}[z]. N. Baccar proved that the elements of A(P) of the form 2^{k}m, where k ≥ 0 and m is odd, are given by the 2-adic expansion of a zero of some polynomial R_{m} with integer coefficients. Let s_{p} be the order of 2 modulo p, i.e., the smallest positive integer such that 2^{s_{p}} ≡ 1 (mod p). Improving on the method with which R_{m} was obtained explicitly only when s_{p} = \frac{p−1}{2}, here we make explicit R_{m} when s_{p} = \frac{p−1}{3}. For that, we have used the number of points of the elliptic curve x^{3} + ay^{3} = 1 modulo p.

1Research supported DGRST of Tunisia, UR 99/15-18, Faculté des Sciences de Tunis.

1 Introduction.

Let ℕ denote the set of positive integers, and let A = {a_{1}, a_{2}, …} be a non-empty subset of ℕ. For n ∈ ℕ, let p(A, n) be the number of partitions of n with parts in A, i.e., the number of solutions of the diophantine equation

a_{1}x_{1} + a_{2}x_{2} + ⋯ = n  (1)

in non-negative integers x_{1}, x_{2}, . . .. By convention, p(A, 0) = 1 and p(A, n) = 0 for all n < 0. The generating series of p(A, n) is

F_{A}(z) := ∑_{n=0}^{∞} p(A, n)z^{n} = ∏_{a∈A} \frac{1}{1 − z^{a}} .  (2)

Let 𝔽_{2} be the field with two elements and P (z) = 1 + ϵ_{1}z + ⋯ + ϵ_{N}z^{N} ∈ 𝔽_{2} [z], N ≥ 1. J.-L. Nicolas, I. Z. Ruzsa and A. Sárközy [10] proved that there exists a unique set A = A(P) satisfying

F_{A}(z) ≡ P (z) (mod 2),  (3)

which means that

p(A, n) ≡ ϵ_{n} (mod 2) for 1 ≤ n ≤ N  (4)

and p(A, n) is even for all n > N. Indeed, for n = 1,

p(A, 1) = \{ \begin{matrix}1, & if 1 ∈ A; \\ 0, & if 1 ∉ A. \end{matrix}

and so, by (4),

1 ∈ A ⇔ ϵ_{1} = 1.

Further, assume that we Know A_{n−1} = A∩{1, …, n−1}; since there exists only one partition of n containing the part n, then

p(A, n) = p(A_{n−1}, n) + χ(A, n),

where χ(A, .) is the characteristic function of the set A, i.e.,

χ(A, n) = \{ \begin{matrix}1, & if n ∈ A; \\ 0, & if n ∉ A, \end{matrix}

which with (3) allow one to decide whether n belongs to A.

Let p be an odd prime number, and let s_{p} be the order of 2 modulo p, i.e., s_{p} is the smallest positive integer such that p divides 2^{s_{p}} − 1. Let P ∈ 𝔽_{2}[z] be irreducible of order p (ord(P) = p); in other words, p is the smallest positive integer such that P divides 1 + z^{p} in 𝔽_{2}[z]. N. Baccar and F. Ben Saïd [2] determined the sets A(P) for all p such that s_{p} = \frac{p−1}{2}. Moreover, they proved that if k ≥ 0 and m is an odd positive integer, then the elements of A(P) of the form 2^{k}m are given by the 2-adic expansion of some zero of a polynomial R_{m} with integer coefficients. N. Baccar [1] extended this last result to any odd prime number p. Unfortunately, the method used in that paper can make explicit R_{m} only when s_{p} = \frac{p−1}{2}. In this paper, we will improve on the method given by N. Baccar [1], by introducing elliptic curves, to make R_{m} explicit when s_{p} = \frac{p−1}{3}. In Section 2, some properties of the polynomial R_{m} are exposed. In Section 3, we introduce elliptic curves to compute some cardinalities used in Section 4 to make R_{1} explicit, and in Section 5 to get R_{m} explicitly for any odd integer m ≥ 3.

Throughout this paper, p is an odd prime number and P is some irreducible polynomial in 𝔽_{2}[z] of order p. We also denote by s_{p} the order of 2 modulo p. For a ∈ ℤ and b ∈ ℕ, we should write a mod b for the remainder of the euclidean division of a by b.

2 Some results on the polynomial R_{m}

Let p be an odd prime. We denote by (ℤ/pℤ)^{∗} the group of invertible elements modulo p and by < 2 > its subgroup generated by 2. We consider the action ⋆ of < 2 > on the set ℤ/pℤ given by a ⋆ n = an for all a ∈< 2 > and all n ∈ ℤ/pℤ. The quotient set will be denoted by (ℤ/pℤ)/_{<2>} and the orbit of some n ∈ ℤ/pℤ by O(n). So, we can write

ℤ/pℤ = O(1) ∪ O(g) ∪ ⋯ ∪ O(g^{r−1}) ∪ O(p),

where g is some generator of (ℤ/pℤ)^{∗}, r = \frac{p−1}{s_{p}} is the number of invertible orbits of ℤ/pℤ,

O(g^{i}) = {2^{j}g^{i} mod p : 0 ≤ j ≤ s_{p} − 1} , 0 ≤ i ≤ r − 1,  (5)
O(p) = {0} .

Note that for any integer t,

O(g^{t}) = O(g^{t mod r}).  (6)

The orbits O(n) are defined as parts of ℤ/pℤ; however, by extension, they are also considered as parts of ℕ.

If ϕ_{p} is the cyclotomic polynomial over 𝔽_{2} of index p, then

1 + z^{p} = (1 + z)ϕ_{p}(z).

Moreover, one has

ϕ_{p}(z) = P_{0}(z)P_{1}(z) ⋯ P_{r−1}(z),

where P_{0}, P_{1}, … and P_{r−1} are the only distinct irreducible polynomials in 𝔽_{2} [z] of the same degree s_{p} and all of which are of order p. For all l, 0 ≤ l ≤ r − 1, let A_{l} = A(P_{l}) be the set obtained from (3). If m is an odd positive integer, we define the 2−adic integer y_{l}(m) by

y_{l}(m) = χ(A_{l}, m) + 2χ(A_{l}, 2m) + 4χ(A_{l}, 4m) + ⋯ = ∑_{k=0}^{∞} χ(A_{l}, 2^{k}m)2^{k}.  (7)

By computing y_{l}(m) mod 2^{k+1}, one can deduce χ(A_{l}, 2^{j}m) for all j, 0 ≤ j ≤ k, and obtain all the elements of A_{l} of the form 2^{j}m. In [3], some necessary conditions on integers to be in A_{l} were given. For instance:

p^{2}n ∉ A_{l}, ∀ n ∈ ℕ,  (8)

if q is an odd prime in O(1), then qn ∉ A_{l}, ∀n ∈ ℕ.  (9)

Let 𝕂 be some field, and let u(z) = ∑_{j=0}^{n} u_{j}z^{j} and v(z) = ∑_{j=0}^{t} v_{j}z^{j} be polynomials in 𝕂[z]. We denote the resultant of u and v with respect to z by res_{z}(u(z), v(z)), and recall the following well known result

Lemma 1. (i) The resultant res_{z}(u(z), v(z)) is a homogeneous multivariate polynomial with integer coefficients, of degree n + t in the n + t + 2 variables u_{i}, v_{j}. (ii) If u(z) is written as u(z) = u_{n}(z − α_{1})(z − α_{2}) ⋯ (z − α_{n}) in the splitting field of u over 𝕂 then

res_{z}(u(z), v(z)) = u_{n}^{t} ∏_{i=1}^{n} v(α_{i}).  (10)

N. Baccar proved [1] that, for all l, 0 ≤ l ≤ r − 1, the 2-adic integers y_{l}(m) defined by (7) are the zeros of some polynomial R_{m} with integer coefficients and which can be written as the resultant of two polynomials. We mention here that the expressions given in that paper to R_{m}, for m = 1 and m ≥ 3, can be encoded in only one. So that we have

Theorem 2. ([1]) 1) Let m be an odd positive integer such that m ∉ O(p) (i.e., gcd(m, p) = 1), and let δ = δ(m) be the unique integer in {0, 1, …, r − 1} such that m ∈ O(g^{δ}). We define the polynomial A_{m} by

A_{m}(z) = ∑_{h=0}^{r−1} α_{h}(m)B_{h}(z),  (11)

where for all h, 0 ≤ h ≤ r − 1,

α_{h}(m) = ∑_{d | m̃, d∈O(g^{h})} μ(d),  (12)

m̃ = ∏_{q prime q|m} q is the radical of m with ˜1 = 1, μ is the Möbius function and B_{h} is the polynomial

B_{h}(z) = B_{h,m}(z) = ∑_{j=0}^{s_{p}−1} z^{(2^{j}g^{(δ−h) mod r}) mod p}.  (13)

Then, the 2-adic integers y_{0}(m), y_{1}(m), … and y_{r−1}(m) are the zeros of the polynomial R_{m}(y) of ℤ[y] defined by the resultant

R_{m}(y) = res_{z}(ϕ_{p}(z), my + A_{m}(z))  (14)

and we have

R_{m}(y) = m^{p−1}((y − y_{0}(m))(y − y_{1}(m)) ⋯ (y − y_{r−1}(m)))^{s_{p}}.  (15)

2) The 2-adic integers y_{0}(p), y_{1}(p), … and y_{r−1}(p) are the zeros of the polynomial R_{1}(−py − s_{p}); while if m = pm′, m′ ≥ 3 and gcd(m′, p) = 1, then y_{0}(m), y_{1}(m), … and y_{r−1}(m) are the zeros of the polynomial R_{m′}(−py) defined by (14).

3) If m is divisible by p^{2} or by some prime q belonging to O(1) then we extend the definition (14) to R_{m}(y) = m^{p−1}y^{s_{p}}; so that y_{0}(m), y_{1}(m), … and y_{r−1}(m) remain zeros of R_{m} since, from (8) and (9), they all vanish.

Remark 3. Explicit formulas to the polynomials R_{m} defined by (14), when s_{p} = \frac{p−1}{2}, are given in [1]. Moreover in that paper, it is shown that if θ is a certain primitive p-th root of unity over the 2-adic field ℚ_{2}, then for all l, 0 ≤ l ≤ r − 1,

y_{l}(1) = −T_{l},  (16)

where, for all l ∈ ℤ,

T_{l} = T_{l mod 3} = ∑_{k=0}^{s_{p}−1} θ^{2^{k}g^{l}} = ∑_{j∈O(g^{l})} θ^{j}.  (17)

We also mention here that N. Baccar [1] proved that for all m ∈ ℕ,

R_{m}(y) = ∏_{l=0}^{r−1} (my + A_{m}(θ^{g^{l}}))^{s_{p}} .  (18)

3 Orbits and elliptic curves.

From now on, we keep the above notation and assume that the prime number p is such that s_{p} = \frac{p−1}{3} (the first ones up to 1000 are: p = 43, 109, 157, 229, 277, 283, 307, 499, 643, 691, 733, 739, 811, 997 ). So the number of invertible orbits is r = 3 and

ℤ/pℤ = O(1) ∪ O(g) ∪ O(g^{2}) ∪ O(p),  (19)

where g is some generator of the cyclic group (ℤ/pℤ)^{∗}. The order of 2 is s_{p} = \frac{p−1}{3}; if 2 were a square modulo p, its order should divide \frac{p−1}{2}, which is impossible. Hence 2 cannot be a square modulo p, and by Euler criterion, p has to satisfy p ≡ ±3 (mod 8), and, as p ≡ 1 (mod 3), p ≡ 13, 19 (mod 24).

Lemma 4. For all i, 0 ≤ i ≤ 2, let O(g^{i}) be the orbit of g^{i} defined by (5). Then

O(g^{i}) = {−g^{i}, −2g^{i}, …, −2^{s_{p}−1}g^{i}} = {g^{i}, g^{i+3}…. , g^{i+3(s_{p}−1)}} .  (20)

In particular, 2 is a cube modulo p and the sub-group generated by 2 is the sub-group of cubes (generated by g^{3}) and contains −1.

Proof. To get the first equality of (20), it suffices to show that −1 ∈ O(1). This follows from −1 = (\frac{2}{p}) ≡ 2^{\frac{p−1}{2}} (mod p). To prove the second equality of (20), one just use the fact that (cf. (6)) g^{3} ∈ O(1).

Let us define the integers ℓ_{i,j}, 0 ≤ i, j ≤ 2, by

ℓ_{i,j} = |{t : 0 ≤ t ≤ s_{p} − 1, 1 + g^{j+3t} ∈ O(g^{i})}| .  (21)

Remark 5. As shown just above, −1 ∈ O(1), so that there exists one and only one t ∈ {0, 1, …, s_{p} − 1} such that 1 + g^{3t} ∈ O(p). Moreover, for all t ∈ {0, 1, …, s_{p} − 1} and j ∈ {1, 2}, 1 + g^{j+3t} ∉ O(p). Hence the integers ℓ_{i,j} defined by (21) satisfy

∑_{i=0}^{2} ℓ_{i,j} = s_{p} − δ_{0,j},  (22)

where δ_{i,j} is the Kronecker symbol given by

δ_{i,j} = \{ \begin{matrix}1, & if i = j; \\ 0, & otherwise. \end{matrix}

The integers ℓ_{i,j} defined above are cardinalities of some curves. Indeed, let us consider the curve over 𝔽_{p},

C_{i,j} : 1 + g^{j}X^{3} = g^{i}Y^{3}

and denote by c_{i,j} its cardinality, c_{i,j} =| C_{i,j} |. Since −1 is a cube modulo p, it is clear that

c_{j,i} = c_{i,j}.

Using (21) it follows that

ℓ_{i,j} = |{(X^{3}, Y^{3}) : X ≠ 0, Y ≠ 0 and (X, Y) ∈ C_{i,j}}|.

Therefore,

ℓ_{j,i} = ℓ_{i,j}.  (23)

Note that, (X^{3}, Y^{3}) = (X^{′3}, Y^{′3}) if and only if X′ = Xg^{vs_{p}} and Y′ = Y g^{ws_{p}} for some v, w ∈ {0, 1, 2}. Moreover, if i ≠ 0 (resp. j ≠ 0), no point on the curve C_{i,j} can be of the form (0, Y) (resp. (X, 0)). But if i = 0 (resp. j = 0), we obtain three points on the curve C_{i,j} with X = 0 (resp. Y = 0). Consequently, we obtain the relation

c_{i,j} = 9ℓ_{i,j} + 3δ_{i,0} + 3δ_{0,j}.  (24)

Now, let us consider the projective plane cubic curve

E_{i,j} : Z^{3} + g^{j}X^{3} = g^{i}Y^{3}

and e_{i,j} =| E_{i,j} | its cardinality. If i ≠ j, E_{i,j} has no points at infinity; whereas if i = j, it has three points at infinity. Hence

e_{i,j} = c_{i,j} + 3δ_{i,j}.  (25)

By multiplying the equation Z^{3} + gX^{3} = gY^{3} by g^{2}, we get the curve g^{2}Z^{3} + X′^{3} = Y′^{3}. So, by permuting the variables, we deduce that e_{1,1} = e_{2,0}. Similarly, we obtain e_{2,2} = e_{1,0}. Hence, by (25) and (24) we find that

ℓ_{1,1} = ℓ_{2,0},  (26)

ℓ_{2,2} = ℓ_{1,0}.  (27)

Therefore, from (22), it follows that

ℓ_{2,1} = ℓ_{0,0} + 1.  (28)

Furthermore, from (25) and (24) we have for all i, 0 ≤ i ≤ 2,

9ℓ_{i,0} = c_{i,0} − 3δ_{i,0} − 3
= e_{i,0} − 6δ_{i,0} − 3.  (29)

Hence, to get all the numbers ℓ_{i,j}, 0 ≤ i, j ≤ 2, it suffices to know the values of e_{i,0}, 0 ≤ i ≤ 2.

Computation of e_{i,0}, i ∈ {0, 1, 2}. Here, we are interested with the curve E_{i,0} : Z^{3} + X^{3} = g^{i}Y^{3}. By setting X = 9g^{i}z + 2y, Y = 6x and Z = 9g^{i}z − 2y, we get the Weierstrass’s form

zy^{2} = x^{3} − (27/4)g^{2i}z^{3},

which, when divided by z^{3}, gives the form

y^{2} = x^{3} − (27/4)g^{2i}.

Let

y^{2} = x^{3} + αx + β

be the equation of an elliptic curve E defined over 𝔽_{p}. It is well known that the number of points of E is equal to

| E |= p + 1 + ∑_{x∈𝔽_{p}} (\frac{x^{3} + αx + β}{p} ) ,  (30)

where (\frac{.}{p}) is the Legendre’s symbol. For α = 0, the sum ∑_{x∈𝔽_{p}} (\frac{x^{3}+αx+β}{p} ) was investigated by S. A. Katre [8]. He obtained:

Lemma 6. Let p be a prime number such that p ≡ 1 (mod 3). Then there exist a unique L, L ≡ 1 (mod 3) and a unique M up to a sign such that 4p = L^{2} + 27M^{2}. Moreover, if β is an integer ≠ 0 then

∑_{x∈𝔽_{p}} (\frac{x^{3} + β}{p} ) = \{ \begin{matrix}(\frac{β}{p})L, & if 4β is a cube modulo p; \\ −\frac{1}{2}(\frac{β}{p})(L + 9M), & otherwise, where M is chosen uniquely \\ & by (4β)^{\frac{p−1}{3}} ≡ \frac{L+9M}{L−9M} (mod p) . \end{matrix}

Thanks to Lemma 6, we can give the values of e_{i,0} for 0 ≤ i ≤ 2. Computation of e_{0,0}. From (30), since −27 is a cube , by using Lemma 6 with β = −27/4, we obtain

e_{0,0} = p + 1 + (\frac{−27/4}{p}) L
= p + 1 + (\frac{−27}{p}) L
= p + 1 + (\frac{−3}{p})^{3} L.

Since p ≡ 1 (mod 3) then, by the quadratic reciprocity law, −3 is a quadratic residue modulo p. Hence,

e_{0,0} = p + 1 + L.  (31)

Computation of e_{1,0}. If β = −27g^{2}/4 then 4β = −27g^{2} is not a cube modulo p. Hence, by using Lemma 6 again, it follows that

e_{1,0} = p + 1 − \frac{1}{2} (\frac{−27g^{2}/4}{p}) (L + 9M)
= p + 1 − \frac{1}{2}(L + 9M),  (32)

where the sign of M is given by

(−27g^{2})^{(p−1)/3} ≡ (g^{2})^{(p−1)/3} (mod p)
≡ \frac{L + 9M}{L − 9M} (mod p).

Computation of e_{2,0}. Let M be fixed by last congruence, and let β = −27g^{4}/4. Since (g^{4})^{(p−1)/3} ≢ (g^{2})^{(p−1)/3} (mod p), Lemma 6 implies that

e_{2,0} = p + 1 − \frac{1}{2}(L − 9M).  (33)

4 The explicit form of R_{1} when s_{p} = \frac{p−1}{3}.

First, we remark that the polynomial R_{1}(y) given by (14) can also be defined (cf. (15), (16)) by

(R_{1}(y))^{1/s_{p}} = ∏_{i∈{0,1,2}} (y + T_{i}),  (34)

where θ ≠ 1 is some p-th root of unity.

Theorem 7. Let p be an odd prime such that s_{p} = \frac{p−1}{3}. Then the polynomial R_{1} given by (14) or (34) is equal to

R_{1}(y) = (y^{3} − y^{2} − s_{p}y + λ_{p})^{s_{p}} ,

with

λ_{p} = \frac{p(L + 3) − 1}{27},

where L is the unique integer satisfying 4p = L^{2} + 27M^{2} and L ≡ 1 (mod 3).

Remark 8. p(L + 3) ≡ 1 (mod 27) follows easily from the congruences L ≡ 1 (mod 3) and p ≡ L^{2} (mod 27).

Proof of Theorem 7. From (34), we have

R_{1}(y) = ((y + T_{0})(y + T_{1})(y + T_{2}))^{s_{p}}.

This can be written as

R_{1}(y) = (y^{3} + λ′′_{p}y^{2} + λ′_{p}y + λ_{p})^{s_{p}},

with

λ_{p} = T_{0}T_{1}T_{2},  (35)
λ′_{p} = T_{0}T_{1} + T_{0}T_{2} + T_{1}T_{2},  (36)
λ′′_{p} = T_{0} + T_{1} + T_{2}.

We begin by calculating λ′′_{p}. It follows immediately from (17) and (19) that

λ′′_{p} = T_{0} + T_{1} + T_{2}
= ∑_{i=0}^{2} ∑_{k=0}^{s_{p}−1} θ^{2^{k}g^{i}}

= ∑_{j=1}^{p−1} θ^{j}
= −1,  (37)

since cf. Remark 3, θ is a primitive p-th root of unity.

Now, let us prove that λ′_{p} = −s_{p}. From (36) and (17), we have

λ′_{p} = ∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}g+2^{k′}g^{2}} + ∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}g^{2}+2^{k′}} + ∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}g+2^{k′}}.  (38)

To treat the last sum in (38), let us fix k and k′ in {0, 1, …, s_{p} − 1}. We have θ^{2^{k}g+2^{k′}} = θ^{2^{k′}(1+2^{k−k′}g)}. Since 2^{k−k′}g ∈ O(g) then, from the second equality in (20), there exists a unique t ∈ {0, 1, …, s_{p} − 1} such that 2^{k−k′}g = g^{1+3t}. Hence, the last sum in (38) becomes

∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}g+2^{k′}} = ∑_{0≤k′,t≤s_{p}−1} θ^{2^{k′}(1+g^{1+3t})}.  (39)

For the first and second sums in (38), arguing as above, we get

∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}g^{2}+2^{k′}} = ∑_{0≤k,k′≤s_{p}−1} θ^{2^{k′}(1+2^{k−k′}g^{2})}
= ∑_{0≤k′,t≤s_{p}−1} θ^{2^{k′}(1+g^{2+3t})}  (40)

and

∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}g+2^{k′}g^{2}} = ∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}(g+2^{k′−k}g^{2})}
= ∑_{0≤k,t≤s_{p}−1} θ^{2^{k}(g+g^{2+3t})}.  (41)

Now, from (21), (26), (27), (17) and Remark 5, we obtain

∑_{0≤k′,t≤s_{p}−1} θ^{2^{k′}(1+g^{1+3t})} = ∑_{i=0}^{2} ℓ_{i,1}T_{i}
= ℓ_{1,0}T_{0} + ℓ_{2,0}T_{1} + ℓ_{2,1}T_{2}  (42)

and

∑_{0≤k′,t≤s_{p}−1} θ^{2^{k′}(1+g^{2+3t})} = ∑_{i=0}^{2} ℓ_{i,2}T_{i}
= ℓ_{2,0}T_{0} + ℓ_{2,1}T_{1} + ℓ_{1,0}T_{2}.  (43)

On the other hand, since for all v ≥ 0, 0 ≤ i, j ≤ 2, 1 + g^{j+3t} ∈ O(g^{i}) ⟺ g^{v} + g^{v+j+3t} ∈ O(g^{v+i}), again by (21), (26), (17) and Remark 5, we get

∑_{0≤k,t≤s_{p}−1} θ^{2^{k}(g+g^{2+3t})} = ∑_{i=0}^{2} ℓ_{i,1}T_{1+i}
= ℓ_{2,1}T_{0} + ℓ_{1,0}T_{1} + ℓ_{2,0}T_{2}.  (44)

Clearly, from (22) and (28), one can deduce that

ℓ_{1,0} + ℓ_{2,0} + ℓ_{2,1} = s_{p},  (45)

which, by (38)-(44) and (37), gives

λ′_{p} = s_{p} (T_{0} + T_{1} + T_{2})
= s_{p} ∑_{j=1}^{p−1} θ^{j}
= −s_{p}.  (46)

Finally, let us calculate λ_{p}. From (35) and (17), we have

λ_{p} = ∑_{0≤k,k′,k”≤s_{p}−1} θ^{2^{k}+2^{k′}g+2^{k”}g^{2}}

= ∑_{0≤k≤s_{p}−1} θ^{2^{k}} ( ∑_{0≤k′,k”≤s_{p}−1} θ^{2^{k′}g+2^{k”}g^{2}} ) .  (47)

Hence, by (41), (44) and (17), we get

λ_{p} = ∑_{0≤k≤s_{p}−1} θ^{2^{k}} (ℓ_{2,1} ∑_{j∈O(1)} θ^{j} + ℓ_{1,0} ∑_{j∈O(g)} θ^{j} + ℓ_{2,0} ∑_{j∈O(g^{2})} θ^{j})
= ℓ_{2,1} ∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}+2^{k′}} + ℓ_{1,0} ∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}+2^{k′}g} + ℓ_{2,0} ∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}+2^{k′}g^{2}}.

Consequently, from (39), (42), (40) and (43), it happens that

λ_{p} = ℓ_{2,1} ∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}+2^{k′}} + (ℓ_{1,0}^{2} + ℓ_{2,0}^{2})T_{0} + (ℓ_{1,0}ℓ_{2,0} + ℓ_{2,0}ℓ_{2,1})T_{1}
+(ℓ_{1,0}ℓ_{2,1} + ℓ_{1,0}ℓ_{2,0})T_{2}.  (48)

Since 2^{k′−k} ∈ O(1) = O(g^{3}) then, cf. (20), there exists a unique t ∈ {0, 1, …, s_{p} − 1} such that 2^{k′−k} = g^{3t}. Hence

∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}+2^{k′}} = ∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}(1+2^{k′−k})}
= ∑_{0≤k,t≤s_{p}−1} θ^{2^{k}(1+g^{3t})}.

Now, we recall that cf. Remark 5 there exists one and only one t ∈ {0, 1, …, s_{p} − 1} satisfying 1 + g^{3t} ∈ O(p). Consequently, by (21) and (17), we get

∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}+2^{k′}} = ∑_{0≤k,t≤s_{p}−1} θ^{2^{k}(1+g^{3t})}
= ℓ_{0,0}T_{0} + ℓ_{1,0}T_{1} + ℓ_{2,0}T_{2} + s_{p}.  (49)

Hence, (48) gives

λ_{p} = (ℓ_{1,0}^{2} + ℓ_{2,0}^{2} + ℓ_{0,0}ℓ_{2,1})T_{0} + (ℓ_{1,0}ℓ_{2,0} + ℓ_{2,0}ℓ_{2,1} + ℓ_{1,0}ℓ_{2,1})T_{1}
+(ℓ_{1,0}ℓ_{2,0} + ℓ_{1,0}ℓ_{2,1} + ℓ_{2,0}ℓ_{2,1})T_{2} + ℓ_{2,1}s_{p}.  (50)

On the other hand, from (47), by changing the order of summation, λ_{p} can be written as

λ_{p} = ∑_{0≤k′≤s_{p}−1} θ^{2^{k′}g} ( ∑_{0≤k,k”≤s_{p}−1} θ^{2^{k}+2^{k”}g^{2}} )

and we get in the way as above

λ_{p} = (ℓ_{1,0}ℓ_{2,0} + ℓ_{1,0}ℓ_{2,1} + ℓ_{2,0}ℓ_{2,1})T_{0} + (ℓ_{1,0}^{2} + ℓ_{2,0}^{2} + ℓ_{0,0}ℓ_{2,1})T_{1}
+(ℓ_{1,0}ℓ_{2,0} + ℓ_{2,0}ℓ_{2,1} + ℓ_{1,0}ℓ_{2,1})T_{2} + ℓ_{2,1}s_{p}  (51)

Whereas, if we write λ_{p} in the form

λ_{p} = ∑_{0≤k”≤s_{p}−1} θ^{2^{k”}g^{2}} ( ∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}+2^{k′}g}) ,

we get

λ_{p} = (ℓ_{1,0}ℓ_{2,0} + ℓ_{2,0}ℓ_{2,1} + ℓ_{1,0}ℓ_{2,1})T_{0} + (ℓ_{1,0}ℓ_{2,0} + ℓ_{1,0}ℓ_{2,1} + ℓ_{2,0}ℓ_{2,1})T_{1}
+(ℓ_{1,0}^{2} + ℓ_{2,0}^{2} + ℓ_{0,0}ℓ_{2,1})T_{2} + ℓ_{2,1}s_{p}.  (52)

By summing (50), (51) and (52), we obtain

3λ_{p} = (ℓ_{1,0}^{2} + ℓ_{2,0}^{2} + 2ℓ_{1,0}ℓ_{2,0} + 2ℓ_{1,0}ℓ_{2,1} + 2ℓ_{2,0}ℓ_{2,1} + ℓ_{0,0}ℓ_{2,1}) ×
(T_{0} + T_{1} + T_{2}) + 3ℓ_{2,1}s_{p}.

But, according to (37), T_{0} + T_{1} + T_{2} = −1. Hence,

3λ_{p} = −(ℓ_{1,0}^{2} + ℓ_{2,0}^{2} + 2ℓ_{1,0}ℓ_{2,0} + 2ℓ_{1,0}ℓ_{2,1} + 2ℓ_{2,0}ℓ_{2,1} + ℓ_{0,0}ℓ_{2,1}) + 3ℓ_{2,1}s_{p}
= − ((ℓ_{1,0} + ℓ_{2,0})^{2} + ℓ_{1,0}ℓ_{2,1} + ℓ_{2,0}ℓ_{2,1} + ℓ_{2,1}(ℓ_{0,0} + ℓ_{1,0} + ℓ_{2,0})) + 3ℓ_{2,1}s_{p}.

So that, from (45) and (22), we obtain

3λ_{p} = −(s_{p} − ℓ_{2,1})^{2} − ℓ_{2,1}(s_{p} − ℓ_{2,1}) − ℓ_{2,1}(s_{p} − 1) + 3ℓ_{2,1}s_{p},

which gives λ_{p} = \frac{(3s_{p}+1)ℓ_{2,1}−s_{p}^{2}}{3} = \frac{pℓ_{2,1}−s_{p}^{2}}{3} . Finally, using the value of ℓ_{2,1}:

ℓ_{2,1} = \frac{1}{9}(p + 1 + L)  (53)

which follows from (28), (29) and (31), we complete the proof of Theorem 7.

In the following table we give L, g, M and R_{1}(y) when p ≤ 1000.

\begin{matrix}p & L & g & M & R_{1}(y) \\ 43 & -8 & 3 & -2 & (y^{3} − y^{2} − 14y − 8)^{14} \\ 109 & -2 & 6 & 4 & (y^{3} − y^{2} − 36y + 4)^{36} \\ 157 & -14 & 5 & 4 & (y^{3} − y^{2} − 52y − 64)^{52} \\ 229 & 22 & 6 & 4 & (y^{3} − y^{2} − 76y + 212)^{76} \\ 277 & -26 & 5 & -4 & (y^{3} − y^{2} − 92y − 236)^{92} \\ 283 & -32 & 3 & -2 & (y^{3} − y^{2} − 94y − 304)^{94} \\ 307 & 16 & 5 & -6 & (y^{3} − y^{2} − 102y + 216)^{102} \\ 499 & -32 & 7 & -6 & (y^{3} − y^{2} − 166y − 536)^{166} \\ 643 & 40 & 11 & 6 & (y^{3} − y^{2} − 214y + 1024)^{214} \\ 691 & -8 & 3 & 10 & (y^{3} − y^{2} − 230y − 128)^{230} \\ 733 & -50 & 6 & 4 & (y^{3} − y^{2} − 244y − 1276)^{244} \\ 739 & 16 & 3 & 10 & (y^{3} − y^{2} − 246y + 520)^{246} \\ 811 & -56 & 3 & -2 & (y^{3} − y^{2} − 270y − 1592)^{270} \\ 997 & 10 & 7 & -12 & (y^{3} − y^{2} − 332y + 480)^{332} \end{matrix}

xxxx

5 The explicit form of R_{m} when s_{p} = \frac{p−1}{3} and m ≥ 3.

We remind that if m ∈ O(p) or m is divisible by some prime q belonging to O(1), then the polynomial R_{m} is given by Theorem 2, 2) and 3). Let m be an odd integer ≥ 3 such that all its prime divisors are in O(g) ∪ O(g^{2}). For i ∈ {1, 2}, we denote by ω_{i} the arithmetic function which counts the number of distinct prime divisors belonging to O(g^{i}) of an integer, i.e.,

ω_{i}(n) = ∑_{q prime, q∈O(g^{i}), q | n} 1.  (54)

Let the decomposition of m into irreducible factors be

m = q_{1,1}^{γ_{1,1}}q_{1,2}^{γ_{1,2}} ⋯ q_{1,ω_{1}}^{γ_{1,ω_{1}}}q_{2,1}^{γ_{2,2}}q_{2,2}^{γ_{2,2}} ⋯ q_{2,ω_{2}}^{γ_{2,ω_{2}}},  (55)

where ω_{i} = ω_{i}(m), ω = ω(m) = ω_{1} + ω_{2} and q_{i,j} ∈ O(g^{i}).

We shall begin with some result concerning binomial coefficients:

Lemma 9. For all n ∈ ℕ and all j , 0 ≤ j ≤ 2,

∑_{k≥0} \binom{n}{3k + j}(−1)^{k+j} = 2.3^{\frac{n}{2} −1} cos(\frac{nπ}{6} + \frac{2jπ}{3} ).  (56)

Proof. Let z_{1} = e^{(2iπ)/3} and z_{2} = e^{(4iπ)/3} be the two cubic primitive roots of unity, and let f(z) = ∑_{k≥0} a_{k}z^{k} be some convergent power series. Since for all j, 0 ≤ j ≤ 2,

\frac{1 + z_{1}^{n−j} + z_{2}^{n−j}}{3} = \begin{cases}1, if n ≡ j (mod 3); \\ 0, otherwise , \end{cases}

it follows that

\frac{f(z) + \frac{1}{z_{1}^{j}} f(z_{1}z) + \frac{1}{z_{2}^{j}} f(z_{2}z)}{3} = ∑_{k≥0} a_{3k+j}z^{3k+j}.

Hence, defining g_{j}(z), 0 ≤ j ≤ 2, by

g_{j}(z) = ∑_{k≥0} \binom{n}{3k + j}z^{3k+j},

and taking f(z) = (1 + z)^{n}, we get

g_{j}(z) = \frac{f(z) + \frac{1}{z_{1}^{j}} f(z_{1}z) + \frac{1}{z_{2}^{j}} f(z_{2}z)}{3}.

By making the substitution z = −1, we obtain

g_{j}(−1) = ∑_{k≥0} \binom{n}{3k + j}(−1)^{k+j}

= \frac{\frac{1}{z_{1}^{j}} (1 − z_{1})^{n} + \frac{1}{z_{2}^{j}} (1 − z_{2})^{n}}{3}
= \frac{1}{3} \{\frac{1}{z_{1}^{j}} (\frac{3 − i\sqrt{3}}{2})^{n} + \frac{1}{z_{2}^{j}} (\frac{3 + i\sqrt{3}}{2})^{n}\} .

To get (56), we need only transform the right hand-side of the last equality.

Corollary 10. Let m be an odd integer ≥ 3 of the form (55), and let α_{h}(m) be the quantity defined by (12). For all h, 0 ≤ h ≤ 2, we have

α_{h}(m) = η(m) cos ((ω_{2} − ω_{1})\frac{π}{6} + 4h\frac{π}{3} )  (57)

where

η(m) = 2.3^{\frac{ω}{2} −1}.  (58)

Proof. From (12), for all h, 0 ≤ h ≤ 2, we have

α_{h}(m) = ∑_{d | m̃, d∈O(g^{h})} μ(d).

First, let us suppose that ω_{1} ≠ 0 and ω_{2} ≠ 0. By (54) and (55), we obtain that for all h, 0 ≤ h ≤ 2,

α_{h}(m) = ∑_{i_{1}=0}^{ω_{1}}(−1)^{i_{1}} \binom{ω_{1}}{i_{1}} ∑_{\begin{matrix}i_{2}=0 \\ i_{2}≡i_{1}+2h (mod 3) \end{matrix}}^{ω_{2}}(−1)^{i_{2}} \binom{ω_{2}}{i_{2}}.

So that, by (56), we get

α_{h}(m) = ∑_{i_{1}=0}^{ω_{1}}(−1)^{i_{1}} \binom{ω_{1}}{i_{1}}2.3^{\frac{ω_{2}}{2} −1} cos (\frac{ω_{2}π}{6} + \frac{2(i_{1} + 2h)π}{3} )
= 2.3^{\frac{ω_{2}}{2} −1} ∑_{j=0}^{2} cos (\frac{ω_{2}π}{6} + \frac{2(j + 2h)π}{3} ) ∑_{\begin{matrix}i_{1}=0 \\ i_{1}≡j (mod 3) \end{matrix}}^{ω_{1}}(−1)^{i_{1}} \binom{ω_{1}}{i_{1}},

which, by (56) again, gives

α_{h}(m) = 4.3^{\frac{ω}{2} −2} ∑_{j=0}^{2} cos (\frac{ω_{2}π}{6} + \frac{2(j + 2h)π}{3} ) cos (\frac{ω_{1}π}{6} + \frac{2jπ}{3} ) .

Consequently, to get (57), one need only use the elementary trigonometric formulas

cos a cos b = \frac{1}{2} (cos(a + b) + cos(a − b)) for all a and b in ℝ

and

cos c + cos(c + \frac{2π}{3} ) + cos(c + \frac{4π}{3} ) = 0, for all c ∈ ℝ.

In case ω_{1} = or ω_{2} = 0, (57) follows immediately from (56). Theorem 11. Let m be an odd integer ≥ 3 of the form (55). Let η(m) be as defined in (58), and let R_{m} be the polynomial given by (14). Then

R_{m}(y) = (m^{3}y^{3} − \frac{3}{4}pmη^{2}(m)y + ν_{p})^{s_{p}} ,  (59)

with

ν_{p} = \{ \begin{matrix}\frac{1}{8} (−1)^{\frac{ω_{2}−ω_{1}}{2}} pη^{3}(m)L, & if ω_{2} − ω_{1} is even ; \\ \frac{3\sqrt{3}}{8} (−1)^{\frac{ω_{2}−ω_{1}−1}{2}} pη^{3}(m)M, & if ω_{2} − ω_{1} is odd, \end{matrix}  (60)

where L and M are the unique integers satisfying 4p = L^{2} + 27M^{2}, L ≡ 1 (mod 3) and (g^{2})^{(p−1)/3} ≡ \frac{L+9M}{L−9M} (mod p). Proof. From (18), we have

R_{m}(y) = ∏_{l=0}^{2} (my + A_{m}(θ^{g^{l}}))^{s_{p}}
= (m^{3}y^{3} + m^{2}ν′′_{p}y^{2} + mν′_{p}y + ν_{p})^{s_{p}} ,  (61)

where

ν′′_{p} = A_{m}(θ) + A_{m}(θ^{g}) + A_{m}(θ^{g^{2}}),  (62)

ν′_{p} = A_{m}(θ)A_{m}(θ^{g}) + A_{m}(θ)A_{m}(θ^{g^{2}}) + A_{m}(θ^{g})A_{m}(θ^{g^{2}})  (63)

and

ν_{p} = A_{m}(θ)A_{m}(θ^{g})A_{m}(θ^{g^{2}}).  (64)

Recall that cf. Theorem 2, δ is the unique integer in {0, 1, 2} such that m ∈ O(g^{δ}). So that from (11)-(13) and (17), we get for i ∈ {0, 1, 2}

A_{m}(θ^{g^{i}}) = ∑_{h=0}^{2} α_{h}(m)T_{δ−h+i}.  (65)

Computation of ν′′_{p}. From (65) and (62) we deduce that

ν′′_{p} = (α_{0}(m) + α_{1}(m) + α_{2}(m))(T_{0} + T_{1} + T_{2}).

Since gcd(m, p) = 1 and m ≠ 1, it follows immediately from (12) and (19) that

α_{0}(m) + α_{1}(m) + α_{2}(m) = ∑_{d | m̃} μ(d) = 0  (66)

and thus ν′′_{p} = 0. Computation of ν′_{p}. From (61), to prove (59) it suffices to show (60) and that ν′_{p} = \frac{−3}{4}pη^{2}(m). By (63) and (65), we have

ν′_{p} = ∑_{k=0}^{2} ∑_{h=0}^{k} α_{h}(m)α_{k}(m)U(h, k)

with

U(h, h) = ∑_{(i,j)∈{(0,1),(0,2),(1,2)}} T_{δ−h+i}T_{δ−h+j}

and, for h < k,

U(h, k) = ∑_{(i,j)∈{(0,1),(0,2),(1,2)}} (T_{δ−h+i}T_{δ−k+j} + T_{δ−k+i}T_{δ−h+j}) .

Observing that, for 0 ≤ δ, h, k ≤ 2, U(h, k) does not depend on δ and is equal to T_{0}T_{1} + T_{0}T_{2} + T_{1}T_{2} when h = k and to T_{0}T_{1} + T_{0}T_{2} + T_{1}T_{2} + T_{0}^{2} + T_{1}^{2} + T_{2}^{2} when h < k, we obtain

ν′_{p} = β(m)(T_{0}^{2} + T_{1}^{2} + T_{2}^{2}) + β′(m)(T_{0}T_{1} + T_{0}T_{2} + T_{1}T_{2}),  (67)

where

β(m) = α_{0}(m)α_{1}(m) + α_{0}(m)α_{2}(m) + α_{1}(m)α_{2}(m)

and

β′(m) = α_{0}^{2}(m) + α_{1}^{2}(m) + α_{2}^{2}(m) + β(m).

From (57), it is easy to check that

β(m) = −\frac{3}{4}η^{2}(m).

By (66), we find that

β′(m) = (∑_{i=0}^{2} α_{i}(m))^{2} − 2β(m) + β(m)
= −β(m)
= \frac{3}{4}η^{2}(m).

On the other hand, using (17), we get

T_{0}^{2} + T_{1}^{2} + T_{2}^{2} = (∑_{k=0}^{s_{p}−1} θ^{2^{k}})^{2} + (∑_{k=0}^{s_{p}−1} θ^{2^{k}g})^{2} + (∑_{k=0}^{s_{p}−1} θ^{2^{k}g^{2}})^{2}.

The first sum in the last equality is, by (49), equal to

T_{0}^{2} = ∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}+2^{k′}}
= ℓ_{0,0}T_{0} + ℓ_{1,0}T_{1} + ℓ_{2,0}T_{2} + s_{p}.  (68)

Similarly, for the second and third sums, we obtain

T_{1}^{2} = ∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}g+2^{k′}g}
= ℓ_{0,0}T_{1} + ℓ_{1,0}T_{2} + ℓ_{2,0}T_{0} + s_{p}  (69)

and

T_{2}^{2} = ∑_{0≤k,k′≤s_{p}−1} θ^{2^{k}g^{2}+2^{k′}g^{2}}
= ℓ_{0,0}T_{2} + ℓ_{1,0}T_{0} + ℓ_{2,0}T_{1} + s_{p}.  (70)

Consequently,

T_{0}^{2} + T_{1}^{2} + T_{2}^{2} = 3s_{p} + (ℓ_{0,0} + ℓ_{1,0} + ℓ_{2,0})(T_{0} + T_{1} + T_{2}).

So that, by (22) and (37), we get

T_{0}^{2} + T_{1}^{2} + T_{2}^{2} = 3s_{p} − (s_{p} − 1)
= 2s_{p} + 1.  (71)

Therefore, with the use of (67), (36) and the fact that s_{p} = \frac{p−1}{3}, we obtain

ν′_{p} = −\frac{3}{4}pη^{2}(m).

Computation of ν_{p}. By (64) and (65), we obtain

ν_{p} = ∑_{h,k,t∈{0,1,2}} α_{h}(m)α_{k}(m)α_{t}(m)T_{δ−h}T_{δ−k+1}T_{δ−t+2}

and by observing the 27 terms of the expansion of the above sum, we find that

ν_{p} = γ_{1}(m)(T_{0}T_{1}^{2} + T_{1}T_{2}^{2} + T_{2}T_{0}^{2}) + γ_{2}(m)(T_{0}T_{2}^{2} + T_{1}T_{0}^{2} + T_{2}T_{1}^{2})
+γ_{3}(m)(T_{0}^{3} + T_{1}^{3} + T_{2}^{3}) + γ_{4}(m)T_{0}T_{1}T_{2},  (72)

where

γ_{1}(m) = α_{0}^{2}(m)α_{1}(m) + α_{0}(m)α_{2}^{2}(m) + α_{1}^{2}(m)α_{2}(m),  (73)
γ_{2}(m) = α_{0}^{2}(m)α_{2}(m) + α_{0}(m)α_{1}^{2}(m) + α_{1}(m)α_{2}^{2}(m),  (74)
γ_{3}(m) = α_{0}(m)α_{1}(m)α_{2}(m)  (75)

and

γ_{4}(m) = α_{0}^{3}(m) + α_{1}^{3}(m) + α_{2}^{3}(m) + 3γ_{3}(m).  (76)

Using (68)-(70), we get

T_{0}^{3} = ℓ_{0,0}T_{0}^{2} + ℓ_{1,0}T_{0}T_{1} + ℓ_{2,0}T_{0}T_{2} + s_{p}T_{0},
T_{1}^{3} = ℓ_{0,0}T_{1}^{2} + ℓ_{1,0}T_{1}T_{2} + ℓ_{2,0}T_{0}T_{1} + s_{p}T_{1}

and

T_{2}^{3} = ℓ_{0,0}T_{2}^{2} + ℓ_{1,0}T_{0}T_{2} + ℓ_{2,0}T_{1}T_{2} + s_{p}T_{2}.

Therefore,

T_{0}^{3} + T_{1}^{3} + T_{2}^{3} = s_{p}(T_{0} + T_{1} + T_{2}) + ℓ_{0,0}(T_{0}^{2} + T_{1}^{2} + T_{2}^{2})

+(ℓ_{1,0} + ℓ_{2,0})(T_{0}T_{1} + T_{0}T_{2} + T_{1}T_{2}).

So that, from (37), (71) and (36) , we get

T_{0}^{3} + T_{1}^{3} + T_{2}^{3} = −s_{p} + ℓ_{0,0}(2s_{p} + 1) − (ℓ_{1,0} + ℓ_{2,0})s_{p}, which, by (22), gives

T_{0}^{3} + T_{1}^{3} + T_{2}^{3} = ℓ_{0,0}(3s_{p} + 1) − s_{p}^{2}
= pℓ_{0,0} − s_{p}^{2}.  (77)

Similarly, by again using (68)-(70), we obtain

T_{0}^{2}T_{1} + T_{0}T_{2}^{2} + T_{1}^{2}T_{2} = pℓ_{1,0} − s_{p}^{2}  (78)

and

T_{0}^{2}T_{2} + T_{0}T_{1}^{2} + T_{1}T_{2}^{2} = pℓ_{2,0} − s_{p}^{2}.  (79)

Using (73)-(75) and (57), it is easy to show that

γ_{1}(m) = \frac{3}{4}η^{3}(m) cos ((ω_{2} − ω_{1})\frac{π}{2} + \frac{4π}{3} ) ,

γ_{2}(m) = \frac{3}{4}η^{3}(m) cos ((ω_{2} − ω_{1})\frac{π}{2} + \frac{2π}{3} )

and

γ_{3}(m) = \frac{1}{4}η^{3}(m) cos ((ω_{2} − ω_{1})\frac{π}{2} ) .

Since, cf. (66), α_{0}(m) + α_{1}(m) + α_{2}(m) = 0, from (76) we find that

γ_{4}(m) = −γ_{1}(m) − γ_{2}(m) + 3γ_{3}(m).

Therefore,

γ_{4}(m) = \frac{3}{2}η^{3}(m) cos ((ω_{2} − ω_{1})\frac{π}{2} ) .

Note that if w_{2} − w_{1} is even then

γ_{1}(m) = γ_{2}(m) = −\frac{3}{2}γ_{3}(m) = −\frac{1}{4}γ_{4}(m) = −\frac{3}{8}η^{3}(m) (−1)^{\frac{w_{2}−w_{1}}{2}} ;

while if w_{2} − w_{1} is odd then

γ_{1}(m) = −γ_{2}(m) = −\frac{3\sqrt{3}}{8}η^{3}(m)(−1)^{\frac{w_{2}−w_{1}+1}{2}} , γ_{3}(m) = 0, γ_{4}(m) = 0.

For w_{2} − w_{1} even, from (72), (77)-(79) and (28), we get

ν_{p} = \frac{1}{8}(−1)^{\frac{ω_{2}−ω_{1}}{2}} pη^{3}(m) (9ℓ_{2,1} − p − 1) .

For w_{2} − w_{1} odd, from (72) and (77)-(79), we get

ν_{p} = \frac{3\sqrt{3}}{8}(−1)^{\frac{ω_{2}−ω_{1}+1}{2}} pη^{3}(m)(ℓ_{1,0} − ℓ_{2,0}).

By (29), (32) and (33), we have

ℓ_{1,0} − ℓ_{2,0} = −M.

Lastly, for w_{2} − w_{1} even (resp. odd), (60) follows from (53) (resp. the last equality).

Example: p = 43. As an explicit example, let us consider the case p = 43. Then

1 + z^{43} = (1 + z)P_{1}(z)P_{2}(z)P_{3}(z),

where P_{1}(z) = z^{14} + z^{12} + z^{10} + z^{7} + z^{4} + z^{2} + 1, P_{2}(z) = z^{14} + z^{11} + z^{10} + z^{9} + z^{8} + z^{7} + z^{6} + z^{5} + z^{4} + z^{3} + 1 and P_{3}(z) = z^{14} + z^{13} + z^{11} + z^{7} + z^{3} + z + 1 are the only irreducible polynomials over 𝔽_{2}[z] of order 43. For 1 ≤ l ≤ 3, let A(P_{l}) be the unique set defined by (3). For m ≥ 1, let A(P_{l})_{m} denote the set of the elements of A(P_{l}) of the form 2^{k}m. We give bellow the description of the sets A(P_{l})_{1} and A(P_{l})_{3}; 1 ≤ l ≤ 3. Since p = 43 then g = 3 is a generator of the cyclic group (ℤ/43ℤ)^{∗}. Let L and M be the unique integers satisfying 4p = 172 = L^{2} + 27M^{2}, L ≡ 1 (mod 3) and (g^{2})^{(p−1)/3} = (3^{2})^{14} ≡ \frac{L+9M}{L−9M} (mod 43). Hence, L = −8, M = −2, R_{1}(y) = (y^{3} − y^{2} − 14y − 8)^{14} and R_{3}(y) = (27y^{3} − 129y + 86)^{14}. By using the function polrootspadic of PARI, the 2−adic expansions of the zeros of the polynomial R_{1}(y) are 2^{2} + 2^{3} + 2^{6} + 2^{10} + 2^{13} + 2^{17} + 2^{18} + 2^{20} + 2^{22} + 2^{25} + 2^{27} + 2^{29} + 2^{30} + 2^{32} + 2^{33} + 2^{36} + ⋯ 2 + 2^{4} + 2^{6} + 2^{7} + 2^{10} + 2^{15} + 2^{16} + 2^{19} + 2^{20} + 2^{23} + 2^{26} + 2^{27} + 2^{31} + 2^{34} + 2^{35} + ⋯ 1 + 2 + 2^{5} + 2^{6} + 2^{7} + 2^{9} + 2^{10} + 2^{12} + 2^{14} + 2^{20} + 2^{24} + 2^{27} + ⋯ and the 2−adic expansions of the zeros of the polynomial R_{3}(y) are 1+2^{2} +2^{3} +2^{4} +2^{6} +2^{7} +2^{12} +2^{17} +2^{18} +2^{19} +2^{20} +2^{21} +2^{25} +2^{27} +2^{31} +2^{32} +2^{35} +2^{36} +⋯ 1 + 2^{2} + 2^{5} + 2^{7} + 2^{10} + 2^{13} + 2^{14} + 2^{19} + 2^{20} + 2^{22} + 2^{23} + 2^{24} + 2^{25} + 2^{27} + 2^{29} + 2^{34} + ⋯ 2+2^{2} +2^{3} +2^{4} +2^{5} +2^{6} +2^{9} +2^{11} +2^{15} +2^{16} +2^{19} +2^{21} +2^{22} +2^{23} +2^{24} +2^{27} +2^{30} +2^{33} +· · ·. After computing some first few elements of the sets A(P_{l}), we deduce that A(P_{1})_{1} = {2, 2^{4}, 2^{6}, 2^{7}, 2^{10}, 2^{15}, 2^{16}, 2^{19}, 2^{20}, 2^{23}, 2^{26}, 2^{27}, 2^{31}, 2^{34}, 2^{35}, …} A(P_{2})_{1} = {2^{2}, 2^{3}, 2^{6}, 2^{10}, 2^{13}, 2^{17}, 2^{18}, 2^{20}, 2^{22}, 2^{25}, 2^{27}, 2^{29}, 2^{30}, 2^{32}, 2^{33}, 2^{36}, …} A(P_{3})_{1} = {1, 2, 2^{5}, 2^{6}, 2^{7}, 2^{9}, 2^{10}, 2^{12}, 2^{14}, 2^{20}, 2^{24}, 2^{27}, …}. A(P_{1})_{3} = {2.3, 2^{2}.3, 2^{3}.3, 2^{4}.3, 2^{5}.3, 2^{6}.3, 2^{9}.3, 2^{11}.3, 2^{15}.3, 2^{16}.3, 2^{19}.3, 2^{21}.3, 2^{22}.3, 2^{23}.3, 2^{24}.3, …} A(P_{2})_{3} = {3, 2^{2}.3, 2^{3}.3, 2^{4}.3, 2^{6}.3, 2^{7}.3, 2^{12}.3, 2^{17}.3, 2^{18}.3, 2^{19}.3, 2^{20}.3, 2^{21}.3, 2^{25}.3, 2^{27}.3, 2^{31}.3, …} A(P_{3})_{3} = {3, 2^{2}.3, 2^{5}.3, 2^{7}.3, 2^{10}.3, 2^{13}.3, 2^{14}.3, 2^{19}.3, 2^{20}.3, 2^{22}.3, 2^{23}.3, 2^{24}.3, 2^{25}.3, 2^{27}.3, 2^{29}.3, …}.

6 Acknowledgments

We are pleased to thank professors F. Ben Saïd, F. Morain, C. Delaunay and J.- L. Nicolas for valuable comments and helpful discussions. We also would like to thank the referees for their valuable remarks that help to improve the initial version of this paper.

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2000 Mathematics Subject Classification: Primary 11P83; Secondary 11B50, 11D88, 11G20. Keywords: Partitions, periodic sequences, order of a polynomial, cyclotomic polynomials, resultant, 2-adic integers, elliptic curves.

Received July 16 2009; revised version received December 23 2009. Published in Journal of Integer Sequences, December 31 2009.

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