GEORGIAN MATHEMATICAL JOURNAL: Vol. 2, No. 5, 1995, 491-516

ON THE NUMBER OF REPRESENTATIONS OF INTEGERS BY SOME QUADRATIC FORMS IN TEN VARIABLES

G. LOMADZE

Abstract. A method of finding the so-called Liouville’s type formulas for the number of representations of integers by a_{1}(x_{1}^{2} + x_{2}^{2}) + a_{2}(x_{3}^{2} + x_{4}^{2}) + a_{3}(x_{5}^{2} + x_{6}^{2}) + a_{4}(x_{7}^{2} + x_{8}^{2}) + a_{5}(x_{9}^{2} + x_{10}^{2}) quadratic forms is developed.

In the papers [4,5] four classes of entire modular forms of weight 5 for the congruence subgroup Γ_{0}(4N) are constructed. The Fourier coefficients of these modular forms have a simple arithmetical sense. This allows one to get sometimes the so-called Liouville’s type formulas for the number of representations of positive integers by positive quadratic forms in ten variables.

In the present paper we consider positive primitive quadratic forms

f = a_{1}(x_{1}^{2}+x_{2}^{2})+a_{2}(x_{3}^{2}+x_{4}^{2})+a_{3}(x_{5}^{2}+x_{6}^{2})+
+ a_{4}(x_{7}^{2}+x_{8}^{2})+a_{5}(x_{9}^{2}+x_{10}^{2}).  (1)

For the purpose of illustration we obtain a formula for the number of representations of positive integers by the form (1) for a_{1} = ⋯ = a_{4} = 1 a_{5} = 4. In a similar way one can investigate as well other forms of the kind (1). As is well known, Liouville obtained in 1865 the corresponding formula for a_{1} = ⋯ = a_{5} = 1 only.

1. Some known results

1.1. In this paper N, a, d, k, n, q, r, s, λ denote positive integers; b, u, v are odd positive integers; p is a prime number; ν, l are non-negative integers; H, c, g, h, j, m, x, y, α, β, γ, δ are integers; i is an imaginary unit; z, τ are

491

1072-947X/95/0900-0491$7.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1995 Plenum Publishing Corporation

1991 Mathematics Subject Classification. 11E25, 11E20.

Key words and phrases. Quadratic form, number of representations, entire modular form, congruence subgroup, Gaussian sum, Ramanujan’s sum, theta-function, singular series.

492 G. LOMADZE

complex variables (Im τ > 0); e(z) = exp 2πiz; Q = e(τ); (\frac{h}{u}) is the generalized Jacobi symbol. Further, ∑_{hmodq} and ^{∑′}_{hmodq} denote respectively sums in which h runs a complete and a reduced residue system modulo q.

Let

S(h, q) = ∑_{jmodq} e(\frac{hj^{2}}{q} ) (Gaussian sum),  (1.1)
c(h, q) =\begin{matrix}∑ & ′ \\ jmodq \end{matrix} e(\frac{hj^{2}}{q} ) (Ramanujan’s sum),  (1.2)
ϑ_{gh}(z|τ; c, N) =
= ∑_{m≡c (mod N)}(−1)^{h(m−c)/N}e(\frac{1}{2N} (m + \frac{g}{2})^{2}τ) e((m + \frac{g}{2})z) (1.3)

(theta-function with characteristics g, h),

hence

\frac{∂^{n}}{∂z^{n}} ϑ_{gh}(z|τ; c, N) = (πi)^{n} ∑_{m≡c (mod N)}(−1)^{h(m−c)/N}(2m + g)^{n} ×

×e(\frac{1}{2N} (m + \frac{g}{2})^{2}τ) e((m + \frac{g}{2})z).  (1.4)

Put

\begin{matrix}ϑ_{gh}(τ; c, N) = ϑ_{gh}(0|τ; c, N), \\ ϑ_{gh}^{(n)}(τ; c, N) = \frac{∂^{n}}{∂z^{n}} ϑ_{gh}(z|τ; c, N)|_{z=0}. \end{matrix}  (1.5)

It is known (see, for example, [3], p. 112, formulas (2.3) and (2.5)) that

\begin{matrix}ϑ_{g+2j,h}(τ; c, N) = ϑ_{gh}(τ; c + j, N), \\ ϑ_{g+2j,h}^{(n)}(τ; c, N) = ϑ_{gh}^{(n)}(τ; c + j, N), \end{matrix}  (1.6)
\begin{matrix}ϑ_{gh}(τ; c + N_{j}, N) = (−1)^{hj}ϑ_{gh}(τ; c, N), \\ ϑ_{gh}^{(n)}(τ; c + N_{j}, N) = (−1)^{hj}ϑ_{gh}^{(n)}(τ; c, N). \end{matrix}  (1.7)

From (1.3), in particular, according to the notations (1.5), it follows that

ϑ_{gh}(τ; 0, N) = ∑_{m=−∞}^{∞}(−1)^{hm}Q^{(2Nm+g)^{2}/8N},  (1.8)
ϑ_{gh}^{(n)}(τ; 0, N) = (πi)^{n} ∑_{m=−∞}^{∞}(−1)^{hm}(2Nm + g)^{n}Q^{(2Nm+g)^{2}/8N}. (1.9)

From (1.8) and (1.9) it follows that

ϑ_{−g,h}(τ; 0, N)=ϑ_{gh}(τ; 0, N), ϑ_{−g,h}^{(n)}(τ; 0, N)=(−1)^{n}ϑ_{gh}^{(n)}(τ; 0, N). (1.10)

Everywhere in this paper a denote a least common multiple of the coefficients a_{k} of the quadratic form (1) and Δ = ∏_{k=1}^{5} a_{k}^{2} is its determinant.

Denoting by r(n; f) the number of representations of n by the form (1), we get

∏_{k=1}^{5} ϑ_{00}^{2}(τ; 0, a_{k}) = 1 + ∑_{n=1}^{∞} r(n; f)Q^{n}.  (1.11)

Further, put

θ(τ; f) = 1 + ∑_{n=1}^{∞} ρ(n; f)Q^{n},  (1.12)

where

ρ(n; f) = \frac{π^{5}}{4!Δ^{1/2}} n^{4} ∑_{q=1}^{∞} A(q)  (1.13)

(singular series of the problem) and

A(q) = q^{−10}\begin{matrix}∑ & ′ \\ hmodq \end{matrix} e( − \frac{hn}{q}) ∏_{k=1}^{5} S^{2}(a_{k}h, q).  (1.14)

Finally let

Γ_{0}(4N) = \{\frac{ατ + β}{γτ + δ} | αδ − βγ = 1, γ ≡ 0 (mod 4N)\}

(nonhomogeneous congruence subgroup).

1.2. For the convenience of references we quote some known results as the following lemmas.

Lemma 1. If (h, q) = 1, then

S(kh, kq) = kS(h, q).

Lemma 2 (see, for example, [6], p. 13, Lemma 6). If (h, q) = 1, then

S^{2}(h, q) = (\frac{−1}{q})q for q ≡ 1 (mod 2),
= 2i^{h}q for q ≡ 0 (mod 4),
= 0 for q ≡ 2 (mod 4).

Lemma 3 (see, for example, [6], p. 16, Lemma 8). If (h, q) = 1, then

S(h, u) = (\frac{h}{u})i^{(u−1)^{2}/4}u^{1/2}.

Lemma 4 (see, for example [6], p. 177, formula 20). Let q = p^{λ} and p^{ν} ∥h. Then

c(h, q) = 0 for ν < λ − 1,
= −p^{λ−1} for ν = λ − 1,
= p^{λ−1}(p − 1) for ν > λ − 1.

Lemma 5 (see, for example, [2], p. 14, Lemma 10). Let

χ_{p} = 1 + A(p) + A(p)^{2}) + · · · .  (1.15)

Then

∑_{q=1}^{∞} A(q) = ∏_{p} χ_{p}.

Lemma 6 ([1], pp. 811 and 953). The entire modular form F (τ) of weight r for the congruence subgroup Γ_{0}(4N) is identically zero, if in its expansion in the series

F (τ) = ∑_{m=0}^{∞} C_{m}Q^{m},

C_{m} = 0 for all m ≤ \frac{r}{3} N ∏_{p|4N} (1 + \frac{1}{p}).

Lemma 7 ([7], p. 18, Lemma 14). The function θ(τ; f) is an entire modular form of weight 5 and character χ(δ) = sgn δ(\frac{−Δ}{|δ|}) for Γ_{0}(4a).

Lemma 8 ([2], p. 21, the remark to Lemma 18). The function ∏_{k=1}^{5} ϑ_{00}(τ; 2, a_{k}) is an entire modular form of weight 5 and character χ(δ) = sgn δ(\frac{−Δ}{|δ|}) for Γ_{0}(4a).

Lemma 9 ([4], p. 67, Theorem 1^{1}and [5], p. 193, Theorem 1). For a given N the functions

(1) Ψ_{2}(τ; g_{1}, g_{2}; h_{1}, h_{2}; 0, 0; N_{1}, N_{2}) =

= \frac{1}{N_{1}^{2}} ϑ_{g_{1}h_{1}}^{(4)}(τ; 0, 2N_{1})ϑ_{g_{2}h_{2}}(τ; 0, 2N_{2})+
+\frac{1}{N_{2}^{2}} ϑ_{g_{1}h_{1}}(τ; 0, 2N_{1})ϑ_{g_{2}h_{2}}^{(4)}(τ; 0, 2N_{2})−

−\frac{6}{N_{1}N_{2}} ϑ′′_{g_{1}h_{1}}(τ; 0, 2N_{1})ϑ′′_{g_{2}h_{2}}(τ; 0, 2N_{2}), (1.16)

1There is a misprint in the formulation of Theorem 1 [4, p. 67] which can be easily corrected by substituting 5 for 10 and vice versa.

ON THE NUMBER OF REPRESENTATIONS OF INTEGERS 495

where

(a) 2|g_{1}, 2|g_{2}, N_{1}|N, N_{2}|N, 4|N(\frac{h_{1}^{2}}{N_{1}} + \frac{h_{2}^{2}}{N_{2}} ), 4|\frac{g_{1}^{2}}{4N_{1}} + \frac{g_{2}^{2}}{4N_{2}} ,

(b) for all α and δ with αδ ≡ 1 (mod 4N)

(\frac{N_{1}N_{2}}{|δ|} )Ψ_{2}(τ; αg_{1}, αg_{2}; h_{1}, h_{2}; 0, 0; N_{1}, N_{2}) =

= (\frac{Δ}{|δ|})Ψ_{2}(τ; g_{1}, g_{2}; h_{1}, h_{2}; 0, 0; N_{1}, N_{2}),

(2) Ψ_{3}(τ; g_{1}, …, g_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}) =
= \{\frac{1}{N_{1}} ϑ′′_{g_{1}h_{1}}(τ; 0, 2N_{1})ϑ_{g_{2}h_{2}}(τ; 0, 2N_{2})−
−\frac{1}{N_{2}} ϑ_{g_{1}h_{1}}(τ; 0, 2N_{1})ϑ′′_{g_{2}h_{2}}(τ; 0, 2N_{2})\}×
×ϑ′_{g_{3}h_{3}}(τ; 0, 2N_{3})ϑ_{g_{4}h_{4}}(τ; 0, 2N_{4}) (1.17)

and

(3) Ψ_{4}(τ; g_{1}, …, g_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}) =
= ∏_{k=1}^{3} ϑ′_{g_{k}h_{k}}(τ; 0, 2N_{k})ϑ_{g_{4}h_{4}}(τ; 0, 2N_{4}), (1.18)

where

(a) 2|g_{k}, N_{k}|N (k = 1, 2, 3, 4), 4|N ∑_{k=1}^{4} \frac{h_{k}^{2}}{N_{k}} , 4| ∑_{k=1}^{4} \frac{g_{k}^{2}}{4N_{k}},

(b) for all α and δ with αδ ≡ 1 (mod 4N)

(\frac{∏_{k=1}^{4} N_{k}}{|δ|} ) Ψ_{j}(τ; αg_{1}, …, αg_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}) =

= sgn δ(\frac{−Δ}{|δ|}Ψ_{j}(τ; g_{1}, …, g_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}) (j = 3, 4),

are entire modular forms of weight 5 and character χ(δ) = sgn δ(\frac{−Δ}{|δ|}) for Γ_{0}(4N).

2. Summation of the singular series ρ(n; f)

Everywhere in this section α, β, γ denote non-negative integers and m positive odd integers.

Lemma 10. Let n = 2^{α}m, a_{k} = 2^{γ_{k}}b_{k} (k = 1, 2, …, 5), (b_{1}, …, b_{5}) = 1, b = [b_{1}, …, b_{5}], γ_{5} ≥ γ_{4} ≥ γ_{3} ≥ γ_{2} ≥ γ_{1} = 0, γ = ∑_{k=1}^{5} γ_{k}. Then

χ_{2} = 1 + (−1)^{(b_{1}−m)/2} for 0 ≤ α ≤ γ_{2} − 2,
= 1 for α = γ_{2} − 1, α = γ_{2} < γ_{3}, γ_{2} = γ_{3} ≤ α = γ_{4} − 1,
γ_{2} + 1 = γ_{3} ≤ α = γ_{4} − 1, γ_{2} = γ_{3} ≤ α = γ_{4} < γ_{5},

γ_{2} + 1 = γ_{3} ≤ α = γ_{4} < γ_{5},

= 1 + (−1)^{(b_{1}+b_{2})/2}(1 − 2^{γ_{2}−α} · 3) for γ_{2} + 1 ≤ α < γ_{3},

= 1 + (−1)^{(∑_{k=1}^{3} b_{k}−m)/2} · 2^{γ_{2}+γ_{3}−2α−2} for γ_{2} = γ_{3} ≤ α < γ_{4} − 2, γ_{2} + 1 = γ_{3} ≤ α ≤ γ_{4} − 2,

= 1 + (−1)^{(b_{1}+b_{2})/2}(1 − 2^{γ_{2}−γ_{3}+1}) + (−1)^{(∑_{k=1}^{3} b_{k}−m)/2} · 2^{γ_{2}−γ_{3}−2α−2} for γ_{2} + 2 ≤ γ_{3} ≤ α ≤ γ_{4} − 2,

= 1 + (−1)^{(b_{1}+b_{2})/2}(1 − 2^{γ_{2}−γ_{3}+1}) for γ_{2} + 2 ≤ γ_{3} ≤ α = γ_{4} − 1, γ_{2} + 2 ≤ γ_{3} ≤ α = γ_{4} < γ_{5},

= 1 + (−1)^{(b_{1}+b_{2})/2}(1 − 2^{γ_{2}−γ_{3}+1}) + (−1)^{(∑_{k=1}^{4} b_{k})/2} ×

× 2^{γ_{2}+γ_{3}−2γ_{4}}\{\frac{1}{7}(1 − 2^{−3(α−γ_{4}−1)}) − 2^{−3(α−γ_{4})}\}

for γ_{4} + 1 ≤ α < γ_{5} but γ_{4} ≥ γ_{3} ≥ γ_{2} + 2,

= 1 + (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{γ_{2}+γ_{3}−2γ_{4}}\{\frac{1}{7}(1 − 2^{−3(α−γ_{4}−1)}) − 2^{−3(α−γ_{4})}\}

for γ_{4}+1≤α<γ_{5} but γ_{4} ≥γ_{3} =γ_{2} or γ_{4} ≥γ_{3} =γ_{2}+1,

= 1 + (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−4α−4} for α ≥ γ_{5} = γ_{4} ≥ γ_{3} = γ_{2}, α ≥ γ_{5} = γ_{4} ≥ γ_{3} = γ_{2} + 1,

= 1 + (−1)^{(b_{1}+b_{2})/2}(1 − 2^{γ_{2}−γ_{3}+1}) + (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−4α−4} for α ≥ γ_{5} = γ_{4} ≥ γ_{3} ≥ γ_{2} + 2,

= 1 + (−1)^{(b_{1}+b_{2})/2}(1 − 2^{γ_{2}−γ_{3}+1}) + (−1)^{(∑_{k=1}^{4} b_{k})/2} ×

× 2^{γ_{2}+γ_{3}−2γ_{4}}(1 − 2^{−3(γ_{5}−γ_{4}−1)})/7 + (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−4α−4} for α ≥ γ_{5} > γ_{4} ≥ γ_{3} ≥ γ_{2} + 2,

= 1 + (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{γ_{2}+γ_{3}−2γ_{4}}(1 − 2^{−3(γ_{5}−γ_{4}−1)})/7 +

+ (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−4α−4} for α ≥ γ_{5} > γ_{4} ≥ γ_{3} = γ_{2}, α ≥ γ_{5} > γ_{4} ≥ γ_{3} = γ_{2} + 1.

Proof. I. If in (1.14) we put q = 2^{λ} and then instead of h introduce a new letter of summation y defined by the congruence h ≡ by (mod 2^{λ}), then we get

A(2^{λ}) = 2^{−10λ}\begin{matrix}∑ & ′ \\ ymod2^{λ} \end{matrix} e(−2^{α−λ}mby) ∏_{k=1}^{5} S^{2}(2^{γ_{k}}b_{k}by, 2^{λ}).  (2.1)

From (2.1), according to Lemmas 1, 2, and 4 it follows that (1) for λ = 1, γ_{k} + 1 (k = 2, 3, 4, 5)

A(2^{λ}) = 0,  (2.2)

because S^{2}(b_{1}by, 2) = 0, S^{2}(2^{γ_{k}}b_{k}by, 2^{γ_{k}+1}) = 0 (k = 2, 3, 4, 5); (2) for 2 ≤ λ ≤ γ_{2}

A(2^{λ}) = 2^{−10λ}\begin{matrix}∑ & ′ \\ ymod2^{λ} \end{matrix} e(−2^{α−λ}mby)(2i^{b_{i}by} · 2^{λ})2^{8λ} =

= 2^{1−λ}\begin{matrix}∑ & ′ \\ ymod2^{λ} \end{matrix} e(\frac{b_{1}by}{4} − \frac{2^{α}mby}{2^{λ}} ) =

= 2^{1−λ}e(\frac{(b_{1} − 2^{α−λ+2}m)b}{4}) ∑_{y=0}^{2^{λ−1}−1} e(\frac{(2^{λ−2}b_{1} − 2^{α}m)by}{2^{λ−1}} ) =

= \begin{cases}e((b_{1} − 2^{α−λ+2}m)b/4) & if 2^{λ−1}|(2^{λ−2}b_{1} − 2^{α}m)b, \\ 0 & if 2^{λ−1}†(2^{λ−2}b_{1} − 2^{α}m)b, \end{cases}

i.e.,

A(2^{λ}) = (−1)^{b_{1}−m)/2} if λ = α + 2,  (2.3)
= 0 if λ ≠ α + 2;  (2.3_{1})

(3) for γ_{2} + 2 ≤ λ ≤ γ_{3}

A(2^{λ}) = 2^{−10λ}\begin{matrix}∑ & ′ \\ ymod2^{λ} \end{matrix} e(−2^{α−λ}mby)(2i^{b_{i}by} · 2^{λ})2^{2γ_{2}} · 2i^{b_{2}by} ×
× 2^{λ−γ_{2}} · 2^{6λ} = 2^{γ_{2}−2λ+2}(−1)^{(b_{1}+b_{2})/2}c(2^{α}mb, 2^{λ}),

hence

A(2^{λ}) = (−1)^{(b_{2}+b_{2})/2} · 2^{γ_{2}+1−λ} if λ < α + 1,  (2.4)
= −(−1)^{(b_{2}+b_{2})/2} · 2^{γ_{2}−α} if λ = α + 1,  (2.4_{1})
= 0 if λ > α + 1;  (2.4_{2})

(4) for γ_{3} + 2 ≤ λ ≤ γ_{4}

A(2^{λ}) = 2^{−10λ}\begin{matrix}∑ & ′ \\ ymod2^{λ} \end{matrix} e(−2^{α−λ}mby)(2i^{b_{1}by} · 2^{λ})2^{2γ_{2}} ×
× (2i^{b_{2}by} · 2^{λ−γ_{2}})2^{2γ_{3}}(2i^{b_{3}by} · 2^{λ−γ_{3}}) · 2^{4λ} =

= 2^{γ_{2}+γ_{3}−3λ+3}\begin{matrix}∑ & ′ \\ ymod2^{λ} \end{matrix} e(\frac{(b_{1} + b_{2} + b_{3})by}{4} − \frac{2^{α}mby}{2^{λ}} ) =

= 2^{γ_{2}+γ_{3}−3λ+3}e(\frac{1}{4}((b_{1} + b_{2} + b_{3})b − 2^{α−λ+2}mb)) ×

× ∑_{y=0}^{2^{λ−1}−1} e(2^{1−λ}(2^{λ−2}(b_{1} + b_{2} + b_{3}) − 2^{α}m)by),

hence, as in case (2)

A(2^{λ}) =(−1)^{(∑_{k=1}^{3} b_{k}−m)/2} · 2^{γ_{2}+γ_{3}−2α−2} if λ=α+2,  (2.5)
=0 if λ ≠ α + 2;  (2.5_{1})

(5) for γ_{4} + 2 ≤ λ ≤ γ_{5}

A(2^{λ}) = 2^{−10λ}\begin{matrix}∑ & ′ \\ ymod2^{λ} \end{matrix} e(−2^{α−λ}mby)(2i^{b_{1}by} · 2^{λ})(2^{2γ_{2}} · 2i^{b_{2}by} · 2^{λ−γ_{2}}) ×

× (2^{2γ_{3}} · 2i^{b_{3}by} · 2^{λ−γ_{3}})(2^{2γ_{4}} · 2i^{b_{4}by} · 2^{λ−γ_{4}})2^{2λ} =

= 2^{γ_{2}+γ_{3}+γ_{4}−4λ+4}(−1)^{(∑_{k=1}^{4} b_{k})/2}c(2^{α}bm, 2^{λ}),

hence

A(2^{λ}) = (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{γ−γ_{5}−3λ−3} if λ < α + 1,  (2.6)
= −(−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{γ−γ_{5}−3α} if λ = α + 1,  (2.6_{1})
= 0 if λ > α + 1;  (2.6_{2})

(6) for λ > γ_{5} + 1

A(2^{λ}) = 2^{−10λ}\begin{matrix}∑ & ′ \\ ymod2^{λ} \end{matrix} e(−2^{α−λ}mby)(2i^{b_{1}by} · 2^{λ})2^{2γ_{2}} ×

× (2i^{b_{2}by} · 2^{λ−γ_{2}})2^{2γ_{3}}(2i^{b_{3}by} · 2^{λ−γ_{3}}) ×

× (2^{2γ_{4}} · 2i^{b_{4}by} · 2^{λ−γ_{4}})(2^{2γ_{5}} · 2i^{b_{5}by} · 2^{λ−γ_{5}}) =

= 2^{γ−5λ+5}\begin{matrix}∑ & ′ \\ ymod2^{λ} \end{matrix} e (\frac{∑_{k=1}^{5} b_{k}by}{4} − \frac{2^{α}mby}{2^{λ}} ) =

= 2^{γ−5λ+5}e(\frac{1}{4}( ∑_{k=1}^{5} b_{k}b − 2^{α−λ+2}mb)) ×

× ∑_{y=0}^{2^{λ−1}−1} e(2^{1−λ}(2^{λ−2} ∑_{k=1}^{5} b_{k} − 2^{α}m)by),

hence, as in case (2),

A(2^{λ}) = (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−4α−4} if λ = α + 2,  (2.7)
= 0 if λ ≠ α + 2;  (2.7_{1})

II. According to (1.15) and (2.2), we have

χ_{2} = 1 + ∑_{λ=2}^{γ_{2}} A(2^{λ}) + ∑_{λ=γ_{2}+2}^{γ_{3}} A(2^{λ}) + ∑_{λ=γ_{3}+2}^{γ_{4}} A(2^{λ}) +

+ ∑_{λ=γ_{4}+2}^{γ_{5}} A(2^{λ}) + ∑_{λ=γ_{5}+2}^{∞} A(2^{λ}).  (2.8)

Consider the following cases:

(1) Let 0 ≤ α ≤ γ_{2} − 2. Then from (2.8), (2.3), (2.3_{1}), (2.4_{2}), (2.5_{1}), (2.6_{2}), and (2.7_{1}) we get

χ_{2} = 1 + ∑_{λ=2}^{γ_{2}} A(2^{λ}) = 1 + (−1)^{(b_{1}−m)/2}.

(2) Let α = γ_{2} −1 or α = γ_{2} < γ_{3}. Then from (2.8), (2.3_{1}), (2.4_{2}), (2.5_{1}), (2.6_{2}), and (2.7_{1}) we get

χ_{2} = 1.

(3) Let γ_{2} + 1 ≤ α < γ_{3}. Then from (2.8), (2.3_{1}), (2.4), (2.4_{1}), (2.4_{2}), (2.5_{1}), (2.6_{2}), and (2.7_{1}) we get

χ_{2} =1+ ∑_{λ=γ_{2}+2}^{γ_{3}}A(2^{λ})=1+ ∑_{λ=γ_{2}+2}^{α}(−1)^{(b_{1}+b_{2})/2}·2^{γ_{2}+1−λ}−(−1)^{(b_{1}+b_{2})/2}·2^{γ_{2}−α}.

(4) Let γ_{2} = γ_{3} ≤ α ≤ γ_{4} − 2 or γ_{2} + 1 = γ_{3} ≤ α ≤ γ_{4} − 2. Then from (2.8), (2.3_{1}), (2.5), (2.5_{1}), (2.6_{2}), and (2.7_{1}) we get

χ_{2} = 1 + ∑_{λ=γ_{3}+2}^{γ_{4}} A(2^{λ}) = 1 + (−1)^{(∑_{k=1}^{3} b_{k}−m)/2} · 2^{γ_{2}+γ_{3}−2α−2}.

(5) Let γ_{2} + 2 ≤ γ_{3} ≤ α ≤ γ_{4} − 2. Then from (2.8), (2.3_{1}), (2.4), (2.5), (2.5_{1}), (2.6_{2}), and (2.7_{1}) we get

χ_{2} = 1 + ∑_{λ=γ_{2}+2}^{γ_{3}} A(2^{λ}) + ∑_{λ=γ_{3}+2}^{γ_{4}} A(2^{λ}) =

= 1 + ∑_{λ=γ_{2}+2}^{γ_{3}}(−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}+1−λ} + (−1)^{(∑_{k=1}^{3} b_{k}−m)/2} · 2^{γ_{2}+γ_{3}−2α−2}.

(6) Let γ_{2} = γ_{3} ≤ α = γ_{4} − 1 or γ_{2} + 1 = γ_{3} ≤ α = γ_{4} − 1 or γ_{2} = γ_{3} ≤ α = γ_{4} < γ_{5} or γ_{2} + 1 = γ_{3} ≤ α = γ_{4} < γ_{5}. Then from (2.8), (2.3_{1}), (2.5_{1}), (2.6_{2}), and (2.7_{1}) we get

χ_{2} = 1.

(7) Let γ_{2} + 2 ≤ γ_{3} ≤ α = γ_{4} − 1 or γ_{2} + 2 ≤ γ_{3} ≤ α = γ_{4} < γ_{5}. Then from (2.8), (2.3_{1}), (2.4), (2.4_{2}), (2.5_{1}), (2.6_{1}), and (2.7_{1}) we get

χ_{2} = 1 + ∑_{λ=γ_{2}+2}^{γ_{3}} A(2^{λ}) = 1 + ∑_{λ=γ_{2}+2}^{γ_{3}}(−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}+1−λ}.

8) Let γ_{4} + 1 ≤ α < γ_{5}, but γ_{4} ≥ γ_{3} ≥ γ_{2} + 2. Then from (2.8), (2.3_{1}), (2.4), (2.5_{1}), (2.6), (2.6_{1}), and (2.7_{1}) we get

χ_{2} = 1 + ∑_{λ=γ_{2}+2}^{γ_{3}} A(2^{λ}) + ∑_{λ=γ_{4}+2}^{γ_{5}} A(2^{λ}) =

= 1 + ∑_{λ=γ_{2}+2}^{γ_{3}}(−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}+1−λ} +

+ ∑_{λ=γ_{4}+2}^{α}(−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{γ−γ_{5}−3λ+3} − (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{γ−γ_{5}−3α}.

(9) Let γ_{4} + 1 ≤ α < γ_{5}, but γ_{4} ≥ γ_{3} = γ_{2} or γ_{4} ≥ γ_{3} = γ_{2} + 1. Then from (2.8), (2.3_{1}), (2.5_{1}), (2.6), (2.6_{1}), and (2.7_{1}) we get

χ_{2} = 1 + ∑_{λ=γ_{4}+2}^{γ_{5}} A(2^{λ}) + ∑_{λ=γ_{5}+2}^{∞} A(2^{λ}) =

= 1 + ∑_{λ=γ_{4}+2}^{α}(−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{γ−γ_{5}−3λ+3} − (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{γ−γ_{5}−3α}.

(10) Let α ≥ γ_{5} = γ_{4} ≥ γ_{3} = γ_{2} or α ≥ γ_{5} = γ_{4} ≥ γ_{3} = γ_{2} + 1. Then from (2.8), (2.3_{1}), (2.5_{1}), and (2.7_{1}) we get

χ_{2} = 1 + ∑_{λ=γ_{5}+2}^{∞} A(2^{λ}) = 1 + (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−4α−4}.

(11) Let α ≥ γ_{5} = γ_{4} ≥ γ_{3} ≥ γ_{2} + 2. Then from (2.8), (2.3_{1}), (2.4), (2.5_{1}), (2.7), and (2.7_{1}) we get

χ_{2} = 1 + ∑_{λ=γ_{2}+2}^{γ_{3}} A(2^{λ}) + ∑_{λ=γ_{5}+2}^{∞} A(2^{λ}) =

= 1 + ∑_{λ=γ_{2}+2}^{γ_{3}}(−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}+1−λ} + (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−4α−4}.

(12) Let α ≥ γ_{5} > γ_{4} ≥ γ_{3} ≥ γ_{2} + 2. Then from (2.8), (2.3_{1}), (2.4), (2.5_{1}), (2.6), (2.7), and (2.7_{1}) we get

χ_{2} = 1 + ∑_{λ=γ_{2}+2}^{γ_{3}} A(2^{λ}) + ∑_{λ=γ_{4}+2}^{γ_{5}} A(2^{λ}) +

+ ∑_{λ=γ_{5}+2}^{∞} A(2^{λ}) = 1 + ∑_{λ=γ_{2}+2}^{γ_{3}}(−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}+1−λ} +

+ ∑_{λ=γ_{4}+2}^{γ_{5}}(−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{γ−γ_{5}−3λ+3} + (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−4α−4}.

(13) Let α ≥ γ_{5} > γ_{4} ≥ γ_{3} = γ_{2} or α ≥ γ_{5} > γ_{4} ≥ γ_{3} = γ_{2} + 1. Then from (2.8), (2.3_{1}), (2.5_{1}), (2.6), (2.7), and (2.7_{1}) we get

χ_{2} = 1 + ∑_{λ=γ_{4}+2}^{γ_{5}} A(2^{λ}) + ∑_{λ=γ_{5}+2}^{∞} A(2^{λ}) =

=1+ ∑_{λ=γ_{4}+2}^{γ_{5}}(−1)^{(∑_{k=1}^{4} b_{k})/2}·2^{∑_{k=2}^{4} γ_{k}−3λ+3}+(−1)^{(∑_{k=1}^{5} b_{k}−m)/2}·2^{γ−4α−4}.

Calculating the sums in the right-hand sides of the above given equalities, we complete the proof of the lemma.

Lemma 11. Let p > 2, p^{β}∥n, p^{ℓ_{k}}∥a_{k} (k =1, …, 5), max ℓ_{k} =\overline{ℓ}, min ℓ_{k} = \underline{ℓ} = 0, \overline{ℓ} ≥ ℓ′′′ ≥ ℓ′′ ≥ ℓ′ ≥ \underline{ℓ} = 0, ℓ = ∑_{k=1}^{5} ℓ_{k} = \overline{ℓ} + ℓ′′′ + ℓ′′ + ℓ′ + \underline{ℓ}, η(ℓ′) = 1 if 2|ℓ′ and η(ℓ′) = 0 if 2†ℓ′. Then

χ_{p} = (1 − p^{−1})(β + 1) for ℓ′ ≥ β + 1, p ≡ 1 (mod 4),

= (1 + p^{−1}) for ℓ′ ≥ β + 1, p ≡ 3 (mod 4), 2|β,

= 0 for ℓ′ ≥ β + 1, p ≡ 3 (mod 4), 2†β;

= (1 − p^{−1})ℓ′ + (1 + p^{−1})(1 − p^{−(β−ℓ′+1)})

for ℓ′ ≤ β < ℓ′′, p ≡ 1 (mod 4),

= (1 + p^{−1})η(ℓ′) − (−1)^{ℓ′}(1 + p)p^{−(β−ℓ′+2)})

for ℓ′ ≤ β < ℓ′′, p ≡ 3 (mod 4),

= (1 − p^{−1})ℓ′ + (1 + p^{−1}) − p^{−(ℓ′′−ℓ′)}{(1 + p^{−2(β−ℓ′′)−1)})(p + 1)^{−1} +

+ p^{−2(β−ℓ′′)−3}} for ℓ′′ ≤ β < ℓ′′′, p ≡ 1 (mod 4),

= (1 + p^{−1})η(ℓ′) − (−1)^{ℓ′}p^{−(ℓ′′−ℓ′+1)}{(1 + (p − 1)(p^{2} + 1)^{−1} ×

× (1 − (−1)^{β+ℓ′′}p^{−2(β−ℓ′′)}) − (−1)^{β+ℓ′′}p^{−2(β−ℓ′′+1)}}

for ℓ′′ ≤ β < ℓ′′′, p ≡ 3 (mod 4),

= (1 − p^{−1})ℓ′ + (1 + p^{−1}) − p^{−(ℓ′′−ℓ′)}{(1 + p^{−2(ℓ′′′−ℓ′′)−1})(p + 1)^{−1} −

− p^{−2(ℓ′′′−ℓ′′)−1}(1 − p^{−3(β−ℓ′′′)}(p^{2} + p + 1)^{−1}\} − p^{−3(β+1)+ℓ−\overline{ℓ}−1}

for ℓ′′′ ≤ β < \overline{ℓ}, p ≡ 1 (mod 4),

= (1 + p^{−1})η(ℓ′) − (−1)^{ℓ′}p^{−(ℓ′′−ℓ′+1)}{(1 + (p − 1)(p^{2} + 1)^{−1} ×

× (1 − (−1)^{ℓ′′′+ℓ′′}p^{−2(β−ℓ′′)})\} + (−1)^{ℓ−\overline{ℓ}}p^{ℓ−\overline{ℓ}−3ℓ′′′−1} ×

× {(1 − p^{−3(β−ℓ′′′)})(p^{2} + p + 1)^{−1} − p^{−3(β−ℓ′′′+1)}}

for ℓ′′′ ≤ β < \overline{ℓ}, p ≡ 3 (mod 4),

= (1 − p^{−1})ℓ′ + (1 + p^{−1}) − p^{−(ℓ′′−ℓ′)}{(1 + p^{−2(ℓ′′′−ℓ′′)−1})(p + 1)^{−1} −

− p^{−2(ℓ′′′−ℓ′′)−1}(1 − p^{−3(\overline{ℓ}−ℓ′′′)}(p^{2} + p + 1)^{−1}\} +

+ p^{ℓ−4\overline{ℓ}−1}\{(1 − p^{−4(β−\overline{ℓ})})(p^{3} + p^{2} + p + 1)^{−1} − p^{−4(β−\overline{ℓ}+1)}\}

for β ≥ \overline{ℓ}, p ≡ 1 (mod 4),

= (1 + p^{−1})η(ℓ′) − (−1)^{ℓ′}p^{−(ℓ′′−ℓ′+1)}{(1 + (p − 1)(p^{2} + 1)^{−1} ×

× (1 − (−1)^{ℓ′′′+ℓ′′}p^{−2(ℓ′′′−ℓ′′)})\} + (−1)^{ℓ−\overline{ℓ}}p^{ℓ−\overline{ℓ}−3ℓ′′′−1} ×

× \{(1 − p^{−3(\overline{ℓ}−ℓ′′′)})(p^{2} + p + 1)^{−1} − p^{−3(\overline{ℓ}−ℓ′′′)}(1 − (−1)^{\overline{ℓ}+β}p^{−4(β−\overline{ℓ})} ×

× (p − 1)(p^{4} + 1)^{−1}\} + (−1)^{β+ℓ}p^{ℓ−4β−5} for β ≥ \overline{ℓ}, p ≡ 3 (mod 4).

Proof. I. Let 2†q and q = (q, a_{k})q_{k} (k = 1, …, 5). Then from (1.14) and Lemmas 1 and 2 it follows that

A(q) = q^{−10}\begin{matrix}∑ & ′ \\ hmodq \end{matrix} e( − \frac{hn}{q}) ∏_{k=1}^{5}(q, a_{k})^{2}S^{2}(\frac{a_{k}}{(q, a_{k})} h, q_{k}) =

= q^{−10}\begin{matrix}∑ & ′ \\ hmodq \end{matrix} e( − \frac{hn}{q}) ∏_{k=1}^{5}(q, a_{k})^{2}(\frac{−1}{q_{k}})q_{k} =

= q^{−5}\begin{matrix}∑ & ′ \\ hmodq \end{matrix} e( − \frac{hn}{q}) ∏_{k=1}^{5}(q, a_{k})(\frac{−1}{q_{k}}),

where, putting q = p^{λ} and taking into account that (a_{1}, …, a_{5}) = 1, we get

A(p^{λ})=p^{−5λ}\begin{matrix}∑ & ′ \\ hmodp^{λ} \end{matrix}e( − \frac{hn}{p^{λ}})∏_{k=1}^{5}(p^{λ}, a_{k})(\frac{−1}{p^{λ}/(p^{λ}, a_{k})})=

= (\frac{−1}{p})^{λ+min(λ,\overline{ℓ})+min(λ,ℓ′′′)+min(λ,ℓ′′)+min(λ,ℓ′)} ×

× p^{min(λ,\overline{ℓ})+min(λ,ℓ′′′)+min(λ,ℓ′′)+min(λ,ℓ′)}p^{−5λ}c(n, p^{λ}).  (2.9)

It follows from (2.9) and Lemma 4 that

(1) for λ ≤ ℓ′

A(p^{λ}) = (\frac{−1}{p})^{λ}(1 − p^{−1}) if λ < β + 1,  (2.10)
= −(\frac{−1}{p})^{β+1}p^{−1} if λ = β + 1;  (2.10_{1})

(2) for ℓ′ < λ ≤ ℓ′′

A(p^{λ}) = (\frac{−1}{p})^{ℓ′}p^{−(λ−ℓ′)}(1 − p^{−1}) if λ < β + 1,  (2.11)
= −(\frac{−1}{p})^{ℓ′}p^{ℓ′−β−2} if λ = β + 1;  (2.11_{1})

(3) for ℓ′′ < λ ≤ ℓ′′′

A(p^{λ})=(\frac{−1}{p})^{λ+ℓ′′+ℓ′}p^{−2λ+ℓ′′ℓ′}(1−p^{−1}) if λ<β+1,  (2.12)
=−(\frac{−1}{p})^{β+1+ℓ′′+ℓ′}p^{−2β+ℓ′′+ℓ′−3} if λ=β+1;  (2.12_{1})

504 G. LOMADZE

(4) for ℓ′′′ < λ ≤ \overline{ℓ}

A(p^{λ})=(\frac{−1}{p})^{ℓ−\overline{ℓ}}p^{ℓ−\overline{ℓ}−3λ}(1−p^{−1}) if λ<β+1,  (2.13)
=−(\frac{−1}{p})^{ℓ−\overline{ℓ}}p^{ℓ−\overline{ℓ}} if λ=β+1;  (2.13_{1})

(5) for λ > \overline{ℓ}

A(p^{λ}) = (\frac{−1}{p})^{λ+ℓ}p^{ℓ−4λ}(1 − p^{−1}) if λ < β + 1,  (2.14)
= −(\frac{−1}{p})^{β+ℓ+1}p^{ℓ−4β−5} if λ = β + 1.  (2.14_{1})

In all the above-mentioned cases

A(p^{λ}) = 0 if λ > β + 1.  (2.15)

II. According to (1.15) we have

χ_{p} = 1 + ∑_{λ=1}^{ℓ′} A(p^{λ}) + ∑_{λ=ℓ′+1}^{ℓ′′} A(p^{λ}) + ∑_{λ=ℓ′′+1}^{ℓ′′′} A(p^{λ}) +
+ ∑_{λ=ℓ′′′+1}^{\overline{ℓ}} A(p^{λ}) + ∑_{λ=\overline{ℓ}+1}^{∞} A(p^{λ}).  (2.16)

Consider the following cases:

(1) Let = ℓ′ ≥ β + 1. Then from (2.16), (2.10), (2.10_{1}), and (2.15) we get

χ_{p} = 1 + ∑_{λ=1}^{β} (\frac{−1}{p})^{λ}(1 − p^{−1}) − (\frac{−1}{p})^{β+1}p^{−1}.

(2) Let ℓ′ ≤ β < ℓ′′. Then from (2.16), (2.10) (2.11), (2.11_{1}), and (2.15) we get

χ_{p} = 1 + ∑_{λ=1}^{ℓ′} (\frac{−1}{p})^{λ}(1 − p^{−1}) + ∑_{λ=ℓ′+1}^{β} (\frac{−1}{p})^{ℓ′}p^{−(λ−ℓ′)}(1 − p^{−1}) −

−(\frac{−1}{p})^{ℓ′}p^{−(β−ℓ′+1)}.

(3) Let ℓ′′ ≤ β < ℓ′′′. Then from (2.16), (2.10), (2.11), (2.12), (2.12_{1}), and (2.15) we get

χ_{p} = 1 + ∑_{λ=1}^{ℓ′} (\frac{−1}{p})^{λ}(1 − p^{−1}) + ∑_{λ=ℓ′+1}^{ℓ′′} (\frac{−1}{p})^{ℓ′}p^{−(λ−ℓ′)}(1 − p^{−1}) +

+ ∑_{λ=ℓ′′+1}^{β} (\frac{−1}{p})^{λ+ℓ′′+ℓ′}p^{−2λ+ℓ′′+ℓ′}(1−p^{−1})−(\frac{−1}{p})^{β+1+ℓ′′+ℓ′}p^{−2β+ℓ′′+ℓ′+3}.

(4) Let ℓ′′′ ≤ β < \overline{ℓ}. Then from (2.16), (2.10), (2.11), (2.12), (2.13), (2.13_{1}), and (2.15) we get

χ_{p} = 1 + ∑_{λ=1}^{ℓ′} (\frac{−1}{p})^{λ}(1 − p^{−1}) + ∑_{λ=ℓ′+1}^{ℓ′′} (\frac{−1}{p})^{ℓ′}p^{−(λ−ℓ′)}(1 − p^{−1}) +

+ ∑_{λ=ℓ′′+1}^{ℓ′′′} (\frac{−1}{p})^{λ+ℓ′′+ℓ′}p^{−2λ+ℓ′′+ℓ′}(1 − p^{−1}) +

+ ∑_{λ=ℓ′′′+1}^{β} (\frac{−1}{p})^{ℓ−\overline{ℓ}}p^{ℓ−\overline{ℓ}−3λ}(1 − p^{−1}) − (\frac{−1}{p})^{ℓ−\overline{ℓ}}p^{ℓ−\overline{ℓ}−3β−4}.

(5) Let β ≥ \overline{ℓ}. Then from (2.16), (2.10), (2.11), (2.12), (2.13), (2.14), (2.14_{1}), and (2.15) we get

χ_{p} =1+∑_{λ=1}^{ℓ′}(\frac{−1}{p})^{λ}(1 − p^{−1})+ ∑_{λ=ℓ′+1}^{ℓ′′}(\frac{−1}{p})^{ℓ′}(1−p^{−1})p^{−(λ−ℓ′)}+

+ ∑_{λ=ℓ′′+1}^{ℓ′′′}(\frac{−1}{p})^{λ+ℓ′′+ℓ′}p^{ℓ′′+ℓ′−2λ}(1−p^{−1})+ ∑_{λ=ℓ′′′+1}^{\overline{ℓ}}(\frac{−1}{p})^{ℓ−\overline{ℓ}}p^{ℓ−\overline{ℓ}−3λ}(1 − p^{−1})+

+ ∑_{λ=\overline{ℓ}+1}^{β} (\frac{−1}{p})^{λ+ℓ}p^{ℓ−4λ}(1 − p^{−1}) − (\frac{−1}{p})^{β+ℓ+1}p^{ℓ−4β−5}. (2.17)

Calculating the sums in the right-hand sides of the above-mentioned equalities, we complete the proof of the lemma.

Theorem 1. Let n = 2^{α}m = 2^{α}uv, u = ∏_{\begin{matrix}p|n \\ p†2Δ \end{matrix}} p^{β}, v = ∏_{\begin{matrix}p|n \\ p|Δ,p>2 \end{matrix}} p^{β}. Then

ρ(n; f) = \frac{2^{4α+6}v^{4}}{5Δ^{1/2}} χ_{2} ∏_{\begin{matrix}p|Δ \\ p>2 \end{matrix}} χ_{p} ∏_{p|Δ,p>2} (1 − (\frac{−1}{p})p^{−5})^{−1} ∑_{d_{1}d_{2}=u} (\frac{−1}{d_{1}})d_{2}^{4},

506 G. LOMADZE

where the values of χ_{2} and χ_{p} are given in Lemmas 10 and 11. Proof. Let p > 2 p^{β}∥n, p†Δ (i.e., ℓ = 0). In (2.17) putting ℓ = 0, we get

χ_{p} = 1 + ∑_{λ=1}^{β} (\frac{−1}{p})^{λ}p^{−4λ}(1 − p^{−1}) − (\frac{−1}{p})^{β+1}p^{−4β−5} =

= 1 + ∑_{λ=1}^{β} (\frac{−1}{p})^{λ}p^{−4λ} − (\frac{−1}{p})p^{−5} ∑_{λ=1}^{β} (\frac{−1}{p})^{λ−1}p^{−4(λ−1)} −

−(\frac{−1}{p})^{β+1}p^{−4β−5} = ∑_{λ=0}^{β} (\frac{−1}{p})^{λ}p^{−4λ} − (\frac{−1}{p})p^{−5} ∑_{λ=0}^{β} (\frac{−1}{p})^{λ}p^{−4λ} =

= (1 − (\frac{−1}{p})p^{−5}) ∑_{λ=0}^{β} (\frac{−1}{p})^{λ}p^{−4λ} =

= (1 − (\frac{−1}{p})p^{−5}) ∑_{d|p^{β}} (\frac{−1}{d})p^{−4}.  (2.18)

For p†Δn, i.e., for β = 0, from (2.18) we get

χ_{p} = 1 − (\frac{−1}{p})p^{−5}.  (2.19)

Thus it follows from Lemma 5, (2.18), and (2.19) that

∑_{q=1}^{∞} A(q) = χ_{2} ∏_{\begin{matrix}p|Δ \\ p>2 \end{matrix}} χ_{p} ∏_{\begin{matrix}p|n \\ p†2Δ \end{matrix}} χ_{p} ∏_{\begin{matrix}p†Δn \\ p>2 \end{matrix}} χ_{p} =

= χ_{2} ∏_{\begin{matrix}p|Δ \\ p>2 \end{matrix}} χ_{p} ∏_{\begin{matrix}p|n \\ p†2Δ \end{matrix}} \{(1 − (\frac{−1}{p})p^{−5}) ∑_{d|p^{β}} (\frac{−1}{d})d^{−4}\} ∏_{\begin{matrix}p†Δn \\ p>2 \end{matrix}} (1 − (\frac{−1}{p})p^{−5}) =

= χ_{2} ∏_{\begin{matrix}p|Δ \\ p>2 \end{matrix}} χ_{p} ∏_{p>2} (1 − (\frac{−1}{p})p^{−5}) ∏_{\begin{matrix}p|Δ \\ p>2 \end{matrix}} (1 − (\frac{−1}{p})p^{−5})^{−1} ∑_{d|u} (\frac{−1}{d})d^{4} =

=χ_{2}∏_{\begin{matrix}p|Δ \\ p>2 \end{matrix}} χ_{p}L^{−1}(5, −1)∏_{\begin{matrix}p|Δ \\ p>2 \end{matrix}}(1−(\frac{−1}{p})p^{−5})^{−1} \frac{1}{u^{4}} ∑_{d_{1}d_{2}=u}(\frac{−1}{d_{1}})d_{2}^{4}, (2.20)

where L(5, −1) is the Dirichlet L-function and it is well known that

L(5, −1) = \frac{5π^{4}}{2^{6} · 4!}.  (2.21)

Thus the lemma follows from (1.13), (2.20), and (2.21).

ON THE NUMBER OF REPRESENTATIONS OF INTEGERS 507

3. Formulas for r(n; f)

As an example let us consider the quadratic form

f = x_{1}^{2} + ⋯ + x_{8}^{2} + 4(x_{9}^{2} + x_{10}^{2})

in which

a_{1} = ⋯ a_{4} = 1, a_{5} = 4, b_{1} = ⋯ = b_{5} = 1,
γ_{5} = 2, γ_{4} = γ_{3} = γ_{2} = γ_{1} = 0, γ = 2, a = 4, Δ = 2^{4}.

Lemma 12. The function

ψ(τ; f)=ϑ_{00}^{8}(τ; 0, 2)ϑ_{00}^{2}(τ; 0, 8)−θ(τ; f)+ \frac{42}{5·512π^{4}} Ψ_{2}(τ; 8, 8; 0, 0; 0, 0; 4, 4)−

−\frac{544}{5 · 8192π^{4}} Ψ_{2}(τ; 0, 0; 0, 0; 0, 0; 4, 4)+ \frac{3}{5·128π^{4}} Ψ_{2}(τ; 4, 4; 0, 0; 0, 0; 2, 2)+

+\frac{2}{32π^{3}i}Ψ_{3}(τ; 4, 0, 4, 0; 0, 1, 1, 1; 0, 0, 0, 0; 2, 4, 2, 4) +

+\frac{24}{512π^{3}i}Ψ_{4}(τ; 8, 8, 8, 0; 1, 1, 1, 1; 0, 0, 0, 0; 4, 4, 4, 4)  (3.1)

is an entire modular form of weight 5 and character χ(δ) = sgn δ(\frac{−1}{|δ|}) for Γ_{0}(16). Proof. According to Lemmas 7 and 8, the first two summands in (3.1) are entire modular forms of weight 5 and character χ(δ) = sgn δ(\frac{−1}{|δ|}) for Γ_{0}(16).

Put N = 4. Then it is obvious that the following five summands in (3.1) satisfy the condition (a) from Lemma 9.

If αδ ≡ 1 (mod 16), then αδ ≡ 1 (mod 4), i.e.,

α ≡ ±1 (mod 4) and respectively δ ≡ ±1 (mod 4).  (3.2)

In the third, fourth, and fifth summands from (3.1) we have

(\frac{N_{1}N_{2}}{|δ|} ) = 1 and (\frac{Δ}{|δ|}) = 1.  (3.3)

By (1.16), (1.6), (1.7), and (1.10) we have

Ψ_{2}(τ; 8α, 8α; 0, 0; 0, 0; 4, 4)= \frac{1}{8}ϑ_{8α,0}^{(4)}(τ; 0, 8)ϑ_{8α,0}(τ; 0, 8)− \frac{3}{8}ϑ_{8α,0}^{′′2}(τ; 0, 8)=

= \frac{1}{8}ϑ_{±8+8(α∓1),0}^{(4)}(τ; 0, 8)ϑ_{±8+8(α∓1),0}(τ; 0, 8) − \frac{3}{8}ϑ_{±8+8(α∓1),0}^{′′2}(τ; 0, 8) =

= \frac{1}{8}ϑ_{±8,0}^{(4)}(τ; 4(α ∓ 1), 8)ϑ_{±8,0}(τ; 4(α ∓ 1), 8) − \frac{3}{8}ϑ_{±8,0}^{′′2}(τ; 4(α ∓ 1), 8) =

= \frac{1}{8}ϑ_{±8,0}^{(4)}(τ; 0, 8)ϑ_{±8,0}(τ; 0, 8) − \frac{3}{8}ϑ_{±8,0}^{′′2}(τ; 0, 8) = \frac{1}{8}ϑ_{80}^{(4)}(τ; 0, 8) ×

×ϑ_{80}(τ; 0, 8) − \frac{3}{8}ϑ_{80}^{′′2}(τ; 0, 8) = Ψ_{2}(τ; 8, 8; 0, 0; 0, 0; 4, 4)  (3.4)

and similarly

Ψ_{2}(τ; 4α, 4α; 0, 0; 0, 0; 2, 2)= \frac{1}{2}ϑ_{4α,0}^{(4)}(τ; 0, 4)ϑ_{4α,0}(τ; 0, 4)− \frac{3}{2}ϑ_{4α,0}^{′′2}(τ; 0, 4)=

= \frac{1}{2}ϑ_{40}^{(4)}(τ; 0, 4)ϑ_{40}(τ; 0, 4) − \frac{3}{2}ϑ_{40}^{′′2}(τ; 0, 4) =

= Ψ_{2}(τ; 4, 4; 0, 0; 0, 0; 2, 2).  (3.5)

Hence, according to (3.2)–(3.5) the functions Ψ_{2}(τ; 8, 8; 0, 0; 0, 0; 4, 4) and Ψ_{2}(τ; 4, 4; 0, 0; 0, 0; 2, 2) satisfy also the condition (b) from Lemma 9. Thus they are entire modular forms of weight 5 and character χ(δ) = sgn δ(\frac{−1}{|δ|}) for Γ_{0}(16).

Analogously, by (1.17), (1.18), (1.6), (1.7), and (1.10) we obtain

Ψ_{3}(τ; 4α, 0, 4α, 0; 0, 1, 1, 1; 0, 0, 0, 0; 2, 4, 2, 4)=\{\frac{1}{2}ϑ′′_{4α,0}(τ; 0, 4)ϑ_{01}(τ; 0, 8)−

−\frac{1}{4}ϑ_{4α,0}(τ; 0, 4)ϑ′′_{01}(τ; 0, 8)\}ϑ′′_{4α,1}(τ; 0, 4)ϑ_{01}(τ; 0, 8) =

= ±\{\frac{1}{2}ϑ′′_{40}(τ; 0, 4)ϑ_{01}(τ; 0, 8) − \frac{1}{4}ϑ_{40}(τ; 0, 4)ϑ′′_{01}(τ; 0, 8)\}ϑ′_{41}(τ; 0, 4)×

×ϑ_{01}(τ; 0, 8)=±Ψ_{3}(τ; 4, 0, 4, 0; 0, 1, 1, 1; 0, 0, 0, 0; 2, 4, 2, 4), (3.6)
Ψ_{4}(τ; 8α, 8α, 8α, 0; 1, 1, 1, 1; 0, 0, 0, 0; 4, 4, 4, 4) =
= ϑ_{8α,1}^{′3}(τ; 0, 8)ϑ_{01}(τ; 0, 8) = ±ϑ_{81}^{′3}(τ; 0, 8)ϑ_{01}(τ; 0, 8) =
= ±Ψ_{4}(τ; 8, 8, 8, 0; 1, 1, 1, 1; 0, 0, 0, 0; 4, 4, 4, 4).  (3.7)

In ((3.6) and (3.7) we have “+” if α ≡ 1 (mod 4) and “−” if α ≡ −1 (mod 4). Hence the functions Ψ_{3}(τ; 4, 0, 4, 0; 0, 1, 1, 1; 0, 0, 0, 0; 2, 4, 2, 4) and Ψ_{4}(τ; 8, 8, 8, 0; 1, 1, 1, 1; 0, 0, 0, 0; 4, 4, 4, 4) also satisfy the condition (b) from Lemma 9, since

(\frac{∏_{k=1}^{4} N_{k}}{|δ|} ) = 1 and sgn δ(\frac{−1}{|δ|}) = \begin{cases}1 & if δ ≡ 1 (mod 4), \\ −1 & if δ ≡ −1 (mod 4). \end{cases}

Thus, they are entire modular forms of weight 5 and character χ(δ) = sgn δ(\frac{−1}{|δ|}) for Γ_{0}(16).

Theorem 2.

ϑ_{00}^{8}(τ; 0, 2)ϑ_{00}^{2}(τ; 0, 8) = θ(τ; f) − \frac{42}{5 · 512π^{4}} Ψ_{2}(τ; 8, 8; 0, 0; 0, 0; 4, 4) +

+\frac{544}{5 · 8192π^{4}} Ψ_{2}(τ; 0, 0; 0, 0; 0, 0; 4, 4) − \frac{3}{5 · 128π^{4}} Ψ_{2}(τ; 4, 4; 0, 0; 0, 0; 2, 2) −

−\frac{2}{32π^{3}i}Ψ_{3}(τ; 4, 0, 4, 0; 0, 1, 1, 1; 0, 0, 0, 0; 2, 4, 2, 4) −

−\frac{24}{512π^{3}i}Ψ_{4}(τ; 8, 8, 8, 0; 1, 1, 1, 1; 0, 0, 0, 0; 4, 4, 4, 4).  (3.8)

Proof. According to Lemma 6, the function ψ(r; f) will be identically zero if all coefficients by Q^{n} (n ≤ 10) in its expansion by powers of Q are zero.

I. In Theorem 1 and Lemma 10 put n = 2^{α}u, m = u, v = 1, Δ = 2^{4}. Then

ρ(n; f) = \frac{16}{5} ∑_{d_{1}d_{2}=u} (\frac{−1}{d_{1}})d_{2}^{4} for α = 0,

= \frac{224}{5} ∑_{d_{1}d_{2}=u} (\frac{−1}{d_{1}})d_{2}^{4} for α = 1,

= \frac{4}{5}(2^{4α−1} · 9 + (\frac{−1}{u})) ∑_{d_{1}d_{2}=u} (\frac{−1}{d_{1}})d_{2}^{4} for α ≥ 2.

Calculating the values of ρ(n; f) by the above formulas for all n ≤ 10, we get

θ(τ; f) = 1 + \frac{16}{5}Q + \frac{224}{5}Q^{2} + 256Q^{3} + \frac{4612}{5}Q^{4} + \frac{10016}{5}Q^{5} + 3584Q^{6} +
+ 7680Q^{7} + \frac{73732}{5}Q^{8} + \frac{103696}{5}Q^{9} + \frac{140224}{5}Q^{10} + · · · .  (3.9)

It follows from (1.8) that

ϑ_{00}^{8}(τ; 0, 2)ϑ_{00}^{2}(τ; 0, 8) = 1 + 16Q + 112Q^{2} + 448Q^{3} + 1140Q^{4} + 2080Q^{5} +

+3584Q^{6} + 7296Q^{7} + 13876Q^{8} + 20240Q^{9} + 27104Q^{10} + · · · . (3.10)

II. From (1.8) and (1.9) we have

(1) \frac{1}{8}ϑ_{80}^{(4)}(τ; 0, 8)ϑ_{80}(τ; 0, 8) = \frac{1}{8}4096π^{4} ∑_{m_{1}=−∞}^{∞}(2m_{1} + 1)^{4}Q^{(2m_{1}+1)^{2}} ×

× ∑_{m_{2}=−∞}^{∞} Q^{(2m_{1}+1)^{2}} = 512π^{4} · 4Q^{2}(1 + 81Q^{8} + 625Q^{24} + ⋯) ×

×(1 + Q^{8} + Q^{24} + ⋯) = 512π^{4}(4Q^{2} + 328Q^{10} + ⋯),  (3.11)

\frac{3}{8}ϑ_{80}^{′′2}(τ; 0, 8) = \frac{3}{8}4096π^{4} ∑_{m_{1}=−∞}^{∞}(2m_{1} + 1)^{4}Q^{(2m_{1}+1)^{2}} ×

510 G. LOMADZE

× ∑_{m_{2}=−∞}^{∞}(2m_{2} + 1)^{4}Q^{(2m_{1}+1)^{2}} = 3 · 512π^{4} · 4Q^{2} ×

×(1 + 9Q^{8} + 25Q^{24} + ⋯)^{2} = 512π^{4}(12Q^{2} + 216Q^{10} + ⋯); (3.12)

(2) \frac{1}{8}ϑ_{00}^{(4)}(τ; 0, 8)ϑ_{00}(τ; 0, 8) = \frac{1}{8} · 2^{16}π^{4} ∑_{m_{1}=−∞}^{∞} m_{1}^{4}Q^{4m_{1}^{2}} ∑_{m_{2}=−∞}^{∞} Q^{4m_{2}^{2}} =

= 8192π^{4}(2Q^{4} + 4Q^{8} + 32Q^{16} + ⋯),  (3.13)

\frac{3}{8}ϑ_{∞}^{′′2}(τ; 0, 8) = \frac{3}{8} · 2^{16}π^{4} ∑_{m_{1}=−∞}^{∞} m_{1}^{2}Q^{4m_{1}^{2}} ∑_{m_{2}=−∞}^{∞} m_{2}^{2}Q^{4m_{2}^{2}} =

= 8192π^{4}(12Q^{8} + 96Q^{20} + ⋯);  (3.14)

(3) \frac{1}{2}ϑ_{40}^{(4)}(τ; 0, 4)ϑ_{40}(τ; 0, 4) = \frac{1}{2} · 256π^{4} ∑_{m_{1}=−∞}^{∞}(2m_{1} + 1)^{4}Q^{(2m_{1}+1)^{2}/2} ×

× ∑_{m_{2}=−∞}^{∞} Q^{(2m_{2}+1)^{2}/2} =

= 128π^{4}(4Q + 328Q^{5} + 324Q^{9} + 2504Q^{13} + ⋯),  (3.15)

\frac{3}{2}ϑ_{40}^{′′2}(τ; 0, 4) = \frac{3}{2} · 256π^{4} ∑_{m_{1}=−∞}^{∞}(2m_{1} + 1)^{2}Q^{(2m_{1}+1)^{2}/2} ×

× ∑_{m_{2}=−∞}^{∞}(2m_{2} + 1)^{2}Q^{(2m_{2}+1)^{2}/2} =

= 128π^{4}(12Q + 216Q^{5} + 972Q^{9} + 600Q^{13} + ⋯);  (3.16)

(4) \frac{1}{2}ϑ40^{′′}(τ; 0, 4)ϑ_{01}(τ; 0, 8) = \frac{1}{2} · 16π^{2} ∑_{m_{1}=−∞}^{∞}(2m_{1} + 1)^{2}Q^{(2m_{1}+1)^{2}/2} ×

× ∑_{m_{3}=−∞}^{∞}(−1)^{m_{3}}Q^{4m_{3}^{2}} = −8π^{2} · 2Q^{1/2}(1 + 9Q^{4} + 25Q^{12} + ⋯)(1 − 2Q^{4} +

+2Q^{16} − ⋯) = −8π^{2} · 2Q^{1/2}(1 + 7Q^{4} − 18Q^{8} + 25Q^{12} + ⋯), (3.17)

\frac{1}{4}ϑ_{40}(τ; 0, 4)ϑ′′_{01}(τ; 0, 8) = −\frac{1}{4} ∑_{m_{1}=−∞}^{∞} Q^{(2m_{1}+1)^{2}/2} · 256π^{2} ×

× ∑_{m_{3}=−∞}^{∞}(−1)^{m_{3}}m_{3}^{2}Q^{4m_{3}^{2}} = −64π^{2} · 2Q^{1/2}(1 + Q^{4} + Q^{12} + ⋯)(−2Q^{4} +

+8Q^{16} − ⋯) = −64π^{2} · 2Q^{1/2}(−2Q^{4} − 2Q^{8} + 6Q^{16} + ⋯), (3.18)

ϑ′_{41}(τ; 0, 4)ϑ_{01}(τ; 0, 8) = −4πi ∑_{m_{2}=−∞}^{∞}(−1)^{m_{2}}Q^{(2m_{2}+1)^{2}/2} ×

× ∑_{m_{4}=−∞}^{∞}(−1)^{m_{4}}Q^{4m_{4}^{2}} = 4πi · 2Q^{1/2}(1 − 3Q^{4} + 5Q^{12} − ⋯)(1 − 2Q^{4} +

+2Q^{16} − ⋯) = 4πi · 2Q^{1/2}(1 − 5Q^{4} + 6Q^{8} + 5Q^{12} − ⋯), (3.19)

(5) ϑ_{81}^{′3}(τ; 0, 8)ϑ_{01}(τ; 0, 8)=−512π^{3}i ∑_{m_{1},m_{2},m_{3}=−∞}^{∞}(−1)^{m_{1}+m_{2}+m_{3}}(2m_{1}+1) ×

×(2m_{2} + 1)(2m_{3} + 1)Q^{(2m_{1}+1)^{2}+(2m_{2}+1)^{2}+(2m_{3}+1)^{2}} ∑_{m_{4}=−∞}^{∞}(−1)^{m_{4}}Q^{4m_{4}^{2}} =

= −512π^{3}i(8Q^{3} − 16Q^{7} − 72Q^{11} + ⋯).  (3.20)

III. From (3.4), (3.11), and (3.12) it follows that

\frac{1}{512π^{4}} Ψ_{2}(τ; 8, 8; 0, 0; 0, 0; 4, 4) = −8Q^{2} + 112Q^{10} + · · · . (3.21)

From (1.16), (3.13), and (3.14) it follows that

\frac{1}{8192π^{4}} Ψ_{2}(τ; 0, 0; 0, 0; 0, 0; 4, 4) = 2Q^{4} − 8Q^{8} + 31Q^{16} + · · · . (3.22)

From (3.5), (3.15), and (3.16) it follows that

\frac{1}{128π^{4}} Ψ_{2}(τ; 4, 4; 0, 0; 0, 0; 2, 2)=−8Q + 112Q^{5}+1904Q^{13}−· · · . (3.23)

From (3.6), (3.17)–(3.19) it follows that

−\frac{1}{32π^{3}i}Ψ_{3}(τ; 4, 0, 4, 0; 0, 1, 1, 1; 0, 0, 0, 0; 2, 4, 2, 4) =

= 4Q + 72Q^{5} − 444Q^{9} + 712Q^{13} + · · · .  (3.24)

From (3.7) and (3.20) follows

−\frac{1}{512π^{3}i}Ψ_{4}(τ; 8, 8, 8, 0; 1, 1, 1, 1; 0, 0, 0, 0; 4, 4, 4, 4) =

= 8Q^{3} − 16Q^{7} − 72Q^{11} + · · · .  (3.25)

According to (3.1), (3.9), (3.10), and (3.21)–(3.25) it is easy to verify that all coefficients by Q^{n} (n ≤ 10) in the expansion of the function ψ(τ; f) by powers of Q are zero.

Theorem 2a. Let n = 2^{α}u. Then

r(n; f) = \frac{16}{5} ∑_{d_{1}d_{2}=u} (\frac{−1}{d})d_{2}^{4} − \frac{12}{5} ∑_{\begin{matrix}x^{2}+y^{2}=2u \\ 2†x,2†y,x>0y>0 \end{matrix}}(x^{4} − 3x^{2}y^{2}) +

+ 8 ∑_{\begin{matrix}x^{2}+y^{2}+8(z^{2}+t^{2})=2u \\ 2†x,2†y,x>0,y>0 \end{matrix}}(−1)^{(y−1)/2+z+t}(x^{2} − 8z^{2})y)

for α = 0, u ≡ 1 (mod 4),

= \frac{16}{5} ∑_{d_{1}d_{2}=u} (\frac{−1}{d})d_{2}^{4} + 192 ∑_{\begin{matrix}x^{2}+y^{2}+z^{2}+4t^{2}=u \\ 2†x,2†y,2†z, \\ x>0,y>0,z>0 \end{matrix}}(−1)^{(x+y+z−3)/1+t}xyz

for α = 0, u ≡ 3 (mod 4),

= \frac{224}{5} ∑_{d_{1}d_{2}=u} (\frac{−1}{d})d_{2}^{4} − \frac{168}{5} ∑_{\begin{matrix}x^{2}+y^{2}=2u \\ 2†x,2†y,x>0,y>0 \end{matrix}}(x^{4} − 3x^{2}y^{2})

for α = 0 = 0, u ≡ 1 (mod 4),

= \frac{224}{5} ∑_{d_{1}d_{2}=u} (\frac{−1}{d})d_{2}^{4} for α = 1, u ≡ 3 (mod 4),

= \frac{4612}{5} ∑_{d_{1}d_{2}=u} (\frac{−1}{d})d_{2}^{4} + \frac{1088}{5} ∑_{\begin{matrix}x^{2}+y^{2}=u \\ 2†x,x>0,y=0 \end{matrix}}(x^{4} − 3x^{2}y^{2}) +

+ \frac{2176}{5} ∑_{\begin{matrix}x^{2}+y^{2}=u \\ x≢y (mod 2) \\ x>0,y>0 \end{matrix}}(x^{4} − 3x^{2}y^{2}) for α = 2, u ≡ 1 (mod 8),

= \frac{4612}{5} ∑_{d_{1}d_{2}=u} (\frac{−1}{d})d_{2}^{4} + \frac{2176}{5} ∑_{\begin{matrix}x^{2}+y^{2}=u \\ x≢y (mod 2) \\ x>0,y>0 \end{matrix}}(x^{4} − 3x^{2}y^{2})

for α = 2, u ≡ 5 (mod 8),

= \frac{4}{5}(2^{4α−1} · 9 − 1) ∑_{d_{1}d_{2}=u} (\frac{−1}{d})d_{2}^{4} for α ≥ 2, u ≡ 3 (mod 4),

= \frac{73732}{5} ∑_{d_{1}d_{2}=u} (\frac{−1}{d})d_{2}^{4} + \frac{2176}{5} ∑_{\begin{matrix}x^{2}+y^{2}=2u \\ 2†x,2†y,x>0,y>0 \end{matrix}}(x^{4} − 3x^{2}y^{2})

for α = 3, u ≡ 1 (mod 4),

= \frac{4}{5}(2^{4α−1} · 9+1) ∑_{d_{1}d_{2}=u}(\frac{−1}{d})d_{2}^{4}+ \frac{2176}{5} ∑_{\begin{matrix}x^{2}+y^{2}=2^{α−2}u \\ 2|x,2|y,x>0,y>0 \end{matrix}}(x^{4} − 3x^{2}y^{2})

for 2†α, α > 3, u ≡ 1 (mod 4),

= \frac{4}{5}(2^{4α−1} · 9+1) ∑_{d_{1}d_{2}=u}(\frac{−1}{d})d_{2}^{4}+ \frac{1088}{5} ∑_{\begin{matrix}x^{2}+y^{2}=2^{α−2}u \\ 2|x,x>0,y=0 \end{matrix}}(x^{4} − 3x^{2}y^{2})+

+ \frac{2176}{5} ∑_{\begin{matrix}x^{2}+y^{2}=2^{α−2}u \\ 2|x,2|y,x>0,y>0 \end{matrix}}(x^{4} − 3x^{2}y^{2}) for 2|α, α > 3, u ≡ 1 (mod 8),

= \frac{4}{5}(2^{4α−1} · 9+1) ∑_{d_{1}d_{2}=u} (\frac{−1}{d})d_{2}^{4}+ \frac{2176}{5} ∑_{\begin{matrix}x^{2}+y^{2}=2^{α−2}u \\ 2|x,2|y,x>0,y>0 \end{matrix}}(x^{4} − 3x^{2}y^{2})

for 2|α, α > 3, u ≡ 5 (mod 8).

Proof. Equating the coefficients by Q^{n} in both sides of (3.8), we obtain

r(n; f) = ρ(n; f) − \frac{42}{5}ν_{1}(n) + \frac{544}{5}ν_{2}(n) −
− \frac{3}{5}ν_{3}(n) + 2ν_{4}(n) + 24ν_{5}(n),  (3.26)

where ν_{k}(n) (k = 1, 2, 3, 4, 5) denote respectively the coefficients by Q^{n} in the expansions of the functions \frac{1}{512π^{4}} Ψ_{2}(τ; 8, 8; 0, 0; 0, 0; 4, 4), \frac{1}{8192π^{4}} Ψ_{2}(τ; 0, 0; 0, 0; 0, 0; 4, 4), \frac{1}{128π^{4}} Ψ_{2}(τ; 4, 4; 0, 0; 0, 0; 2, 2), −\frac{1}{32π^{3}i}Ψ_{3}(τ; 4, 0, 4, 0; 0, 1, 1, 1; 0, 0, 0, 0; 2, 4, 2, 4), −\frac{1}{512π^{3}i}Ψ_{4}(τ; 8, 8, 8, 0; 1, 1, 1, 1; 0, 0, 0, 0; 4, 4, 4, 4) by powers of Q.

From (3.4), (3.11), and (3.12) it follows that

\frac{1}{512π^{4}} Ψ_{2}(τ; 8, 8; 0, 0; 0, 0; 4, 4)= ∑_{m_{1},m_{2}=−∞}^{∞}(2m_{1} + 1)^{4}Q^{(2m_{1}+1)^{2}+(2m_{2}+1)^{2}} −

−3 ∑_{m_{1},m_{2}=−∞}^{∞}(2m_{1} + 1)^{2}(2m_{2} + 1)^{2}Q^{(2m_{1}+1)^{2}+(2m_{2}+1)^{2}},

i.e.,

ν_{1}(n) = 4 ∑_{\begin{matrix}x^{2}+y^{2}=n \\ 2†x,2†y,x>0,y>0 \end{matrix}}(x^{4} − 3x^{2}y^{2});  (3.27)

514 G. LOMADZE

it is not difficult to verify that

ν_{1}(n) ≠ 0 for α = 1, u ≡ 1 (mod 4) only.

From (1.16), (3.13), and (3.14) follows

\frac{1}{8192π^{4}} Ψ_{2}(τ; 0, 0; 0, 0; 0, 0; 4, 4) = ∑_{m_{1},m_{2}=−∞}^{∞} m_{1}^{4}Q^{4m_{1}+4m_{2}^{2}} −

−3 ∑_{m_{1},m_{2}=−∞}^{∞} m_{1}^{2}m_{2}^{2}Q^{4m_{1}^{2}+4m_{2}^{2}},

i.e.,

ν_{2}(n) = 4 ∑_{4(x^{2}+y^{2})=n}(x^{4} − 3x^{2}y^{2}) =

= \begin{cases}∑_{x^{2}+y^{2}=2^{α−2}u}(x^{4} − 3x^{2}y^{2}) & for α ≥ 2, \\ 0 & for α = 0, 1; \end{cases}  (3.28)

it is obvious that

∑_{x^{2}+y^{2}=2^{α−2}u}(x^{4}−3x^{2}y^{2})= ∑_{\begin{matrix}x^{2}+y^{2}=u \\ x≢y (mod 2) \end{matrix}}(x^{4}−3x^{2}y^{2}) for α=2, u≡1 (mod 4),

= 0 for α ≥ 2, u ≡ 3 (mod 4),

= ∑_{\begin{matrix}x^{2}+y^{2}=2^{α−2}u \\ x≡y (mod 2) \end{matrix}}(x^{4}−3x^{2}y^{2}) for α≥3, u≡1 (mod 4),

From (3.5), (3.15), and (3.16) it follows that

\frac{1}{128π^{4}} Ψ_{2}(τ; 4, 4; 0, 0; 0, 0; 2, 2)= ∑_{m_{1},m_{2}=−∞}^{∞}(2m_{1}+1)^{4}Q^{(2m_{1}+1)^{2}/2+(2m_{2}+1)^{2}/2}−

−3 ∑_{m_{1},m_{2}=−∞}^{∞}(2m_{1} + 1)^{2}(2m_{2} + 1)^{2}Q^{(2m_{1}+1)^{2}/2+(2m_{2}+1)^{2}/2},

i.e.,

ν_{3}(n) = ∑_{\begin{matrix}x^{2}+y^{2}=2u \\ 2†x,2†y \end{matrix}}(x^{4} − 3x^{2}y^{2}) = 4 ∑_{\begin{matrix}x^{2}+y^{2}=2u \\ 2†x,2†y,x>0,y>0 \end{matrix}}(x^{4} − 3x^{2}y^{2})
for α = 0, u ≡ 1 (mod 4),
= 0 for α = 0, u ≡ 3 (mod 4) and for α > 0.

(3.29)

ON THE NUMBER OF REPRESENTATIONS OF INTEGERS 515

From (3.6) and (3.17)–(3.19) it follows that

−\frac{1}{32π^{3}i}Ψ_{3}(τ; 4, 0, 4, 0; 0, 1, 1, 1; 0, 0, 0, 0; 2, 4, 2, 4) =

= { ∑_{m_{1},m_{3}=−∞}^{∞}(−1)^{m_{3}}(2m_{1} + 1)^{2}Q^{(2m_{1}+1)^{2}/2+4m_{3}^{2}}−

− 8 ∑_{m_{1},m_{3}=−∞}^{∞}(−1)^{m_{3}}2m_{3}^{2}Q^{(2m_{1}+1)^{2}/2+4m_{3}^{2}}} ×

× ∑_{m_{2},m_{4}=−∞}^{∞}(−1)^{m_{2}+m_{4}}(2m_{2} + 1)Q^{(2m_{2}+1)^{2}/2+4m_{4}^{2}} =

= ∑_{m_{1},m_{2},m_{3},m_{4}=−∞}^{∞}(−1)^{m_{2}+m_{3}+m_{4}}((2m_{1}+1)^{2}(2m_{2}+1) − 8(2m_{2}+1)m_{3}^{2})×

×Q^{\frac{1}{2} \{(2m_{1}+1)^{2}+(2m_{2}+1)^{2}+8m_{3}^{2}+8m_{4}^{2}},

i.e.,

ν_{4}(n) = ∑_{\begin{matrix}x^{2}+y^{2}+8(z^{2}+t^{2})=2n \\ 2†x,2†y \end{matrix}}(−1)^{(y−1)/2+z+t}(x^{2} − 8z^{2})y =

= 4 ∑_{\begin{matrix}x^{2}+y^{2}+8(z^{2}+t^{2})=2u \\ 2†x,2†y,x>0,y>0 \end{matrix}}(−1)^{(y−1)/2+z+t}(x^{2} − 8z^{2})y

for α = 0, u ≡ 1 (mod 4),

= 0 for α = 0, u ≡ 3 (mod 4) and for α > 0.

(3.30)

From (3.7) and (3.20) follows

−\frac{1}{512π^{3}i}Ψ_{3}(τ; 8, 8, 8, 0; 1, 1, 1, 1; 0, 0, 0, 0; 4, 4, 4, 4) =

= ∑_{m_{1},m_{2},m_{3},m_{4}=−∞}^{∞}(−1)^{∑_{k=1}^{4} m_{k}} ∏_{k=1}^{3}(2m_{k} + 1)Q^{∑_{k=1}^{3}(2m_{k}+1)^{2}+4m_{4}^{2}},

i.e.,

ν_{5}(n) = 8 ∑_{\begin{matrix}x^{2}+y^{2}+z^{2}+4t^{2}=n \\ 2†x,2†y,2†z \\ x>0,y>0,z>0 \end{matrix}}(−1)^{(x+y+z−3)/2+t}xyz;  (3.31)

it is easy to verify that ν_{5}(n) ≠ 0 only for α = 0, u ≡ 3 (mod 4).

The above theorem follows from (3.26), formulas for ρ(n; f), and (3.27)– (3.31).

References

1. E. Hecke, Analytische Arithmetik der positiven quadratischen Formen. Mathematische Werke, Zweite Auflage, Göttingen, 1970.

2. G. A. Lomadze, On the representation of numbers by some quadratic forms with six variables. (Russian) Trudy Tbilis. Univ., Ser. Mat. Mekh. Astron. 117(1966), 7–43.

3. G. A. Lomadze, On the number of representations of numbers by positive quadratic forms with six variables. (Russian) Trudy Tbilis. Mat. Inst. Razmadze 45(1974), 111–133.

4. G. Lomadze, On some entire modular forms of weights 5 and 6 for the congruence group Γ_{0}(4N). Georgian Math. J. 1(1994), No. 1, 53–75.

5. G. Lomadze, Construction of some modular forms of weights 5 and 6 for the congruence group Γ_{0}(4N). Georgian Math. J. 2(1995), No. 2, 189–199.

6. A. Walfisz, Gitterpunkte in mehrdimensionalen Kugeln. Panstwowe Wydawnictwo Naukowe, Warszawa, 1957.

(Received 01.12.1993)

Author’s address: Faculty of Mechanics and Mathematics I. Javakhishvili Tbilisi State University 2, University St., Tbilisi 380043 Republic of Georgia