Average Distributions and Product of Special Values of L-Series

Amir Akbary^{∗}, University of Lethbridge Chantal David^{†}, Concordia University Robert Juricevic^{‡}, University of Waterloo

1 Introduction

Let E be an elliptic curve defined over the rationals. For any prime p of good reduction, let E_{p} be the elliptic curve over 𝔽_{p} obtained by reducing E mod p. Let a_{p}(E) be the trace of the Frobenius morphism of E_{p}. Then, Hasse proved that #E(𝔽_{p}) = p + 1 − a_{p}(E) with |a_{p}(E)| ≤ 2\sqrt{p}. The case a_{p}(E) = 0 corresponds to supersingular reduction mod p. Let N be a positive integer. For a fixed r ∈ ℤ, and fixed curves E_{1}, …, E_{N}, we define

π_{E_{1},...,E_{N}}^{r} (x) = # {p ≤ x : a_{p}(E_{1}) = … = a_{p}(E_{N}) = r} .

There is a simple heuristic that can be used to predict the asymptotic behavior of π_{E_{1},...,E_{N}}^{r} (x). From Hasse’s bound, the probability that a_{p}(E) = r is

Prob \{a_{p}(E) = r\} ∼ \{ \begin{matrix}\frac{1}{4\sqrt{p}} & if |r| ≤ 2\sqrt{p}; \\ 0 & if |r| > 2\sqrt{p}. \end{matrix}

This suggests the asymptotic behavior

π_{E}^{r}(x) ∼ ∑_{p≤x} Prob \{a_{p}(E) = r\} ∼ C_{E,r} \frac{\sqrt{x}}{log x}

∗Research partially supported by University of Lethbridge Research Fund and NSERC

†Research partially supported by NSERC and FCAR

‡Research partially supported by NSERC and FCAR

where C_{E,r} is a constant depending on E and r. Similarly, assuming that a_{p}(E_{1}) = r and a_{p}(E_{2}) = r are independent events for non-isogenous curves E_{1} and E_{2}, we have for |r| ≤ 2\sqrt{p}

Prob\{a_{p}(E_{1}) = a_{p}(E_{2}) = r\} ∼ \frac{1}{16p}

and more generally

Prob\{a_{p}(E_{1}) = … = a_{p}(E_{N}) = r\} ∼ \frac{1}{4^{N}p^{N/2}} .

Summing the probabilities as above leads to the following conjecture.

Conjecture 1.1 (Lang-Trotter conjecture) Let N be a positive integer, let r ∈ ℤ, and let E_{1}, …, E_{N} be elliptic curves over ℚ, not \overline{ℚ}-isogenous and if r = 0 without complex multiplication. Then,

π_{E_{1},...,E_{N}}^{r} (x) ∼ \{ \begin{matrix}C_{E_{1},r} \frac{\sqrt{x}}{log x} & if N = 1; \\ C_{E_{1},E_{2},r} log log x & if N = 2; \\ is finite & if N > 2. \end{matrix}

For N = 1, there is a more precise conjecture by Lang and Trotter [LT]. Their conjecture is based on a probabilistic model more refined than the simple heuristic above, and they then get a conjectural value for the constant C_{E,r}. In particular, the constant can be 0, and the asymptotic relation is then interpreted to mean that there are only finitely many primes p such that a_{p}(E) = r. This can happen, for example, if E has rational torsion over ℚ. Some other such cases were classified in [DKP].

To this date, very little is known about the Lang–Trotter conjecture. It was shown by Elkies [Elk] that for any elliptic curve E over ℚ, there are infinitely many primes such that a_{p}(E) = 0, but this result is not known for any curve E if r ≠ 0. The best (unconditional) lower bound for this case is π_{E}^{0}(x) ≥ log_{3} x/(log_{4} x)^{1+δ} for any positive δ and x sufficiently large [FM1].

For any r ∈ ℤ, it was shown by Serre [S] that π_{E}^{r}(x) has density 0 in the set of primes, and the best result for this case is π_{E}^{r}(x) ≪ x^{4/5}(log x)^{−1/5} [MMS] under the Generalised Riemann Hypothesis. For r = 0, the unconditional bound π_{E}^{0}(x) ≪ x^{3/4} was obtained by Elkies and Ram Murty.

A classical way to get evidence for hard distribution questions like the Lang-Trotter conjecture is to look at average estimates. For any a, b ∈ ℤ such that 4a^{3} + 27b^{2} ≠ 0, let E(a, b) be the elliptic curve

y^{2} = x^{3} + ax + b.

It was shown by Murty and Fouvry [FM1] that for r = 0, the Lang-Trotter conjecture holds on average, i.e. as x → ∞

\frac{1}{4AB} ∑_{|b|≤B^{|a|≤A}} π_{E(a,b)}^{0}(x) ∼ C_{0} \frac{\sqrt{x}}{log x}

where C_{0} is an explicit non-zero constant. This result was extended to all r ∈ ℤ by David and Pappalardi [DP] who showed that as x → ∞

\frac{1}{4AB} ∑_{|b|≤B^{|a|≤A}} π_{E(a,b)}^{r}(x) ∼ C_{r} \frac{\sqrt{x}}{log x}

where

C_{r} = \frac{2}{π} ∏_{p|r} \frac{p^{2}}{(p^{2} − 1)} ∏_{p∤r} \frac{p(p^{2} − p − 1)}{(p − 1)(p^{2} − 1)}.  (1)

We prove in this paper that the Lang-Trotter conjecture holds on average when N = 2. If r = 0, this was done by Fouvry and Murty [FM2]. We extend it in this paper for all r ∈ ℤ. As for all those average results, the key step is a theorem of Deuring which relates the number of elliptic curves over the finite fields 𝔽_{p} with a_{p}(E) = r to the class number of the quadratic imaginary order of discriminant r^{2} − 4p (see Section 2). Using Dirichlet’s class number formula, the averages to consider are then averages of special values of Dirichlet L-functions (for N = 1), or averages of products of special values of Dirichlet L-functions (for N ≥ 2). In the case r = 0, one can compute those averages by splitting the L-functions

L(1, χ) = ∑_{n≥1} \frac{χ(n)}{n}

into 2 sums, depending if n is a square or not, as only the terms with n a square will contribute to the main term. This is not the case when r ≠ 0, because there is a shifting in the characters χ. Then, all the terms of the Dirichlet L-functions will contribute to the main term, and the computations are more delicate. The average Lang-Trotter conjecture for 2 elliptic curves then follows from this average of products of special values of Dirichlet L-functions.

Theorem 1.2 Let ϵ > 0, and let r be an odd integer. Let A, B be positive integers with A, B ⩾ x^{1+ϵ}. Then as x → ∞,

\frac{1}{16A^{2}B^{2}} ∑_{{|a_{1}|,|a_{2}|⩽A \atop |b_{1}|,|b_{2}|⩽B}} π_{E_{1},E_{2}}^{r}(x) ∼ C_{r} log log x

where

C_{r} = \frac{3}{π^{2}} ∏_{p|r} \frac{p^{2}(p^{2} + 1)}{(p^{2} − 1)^{2}} ∏_{p∤r} \frac{p^{2}(p^{4} − 2p^{2} − 3p − 1)}{(p + 1)^{3}(p − 1)^{3}}.  (2)

We remark that for technical reasons, we restrict to the case r odd in the statement of Theorem 1.2. A similar result (with a different constant) would hold for r even, but is not included here, except for the case r = 0 (done previously by Fouvry and Murty) which is done in section 5. The structure of this paper is as follows: in Section 2, we reduce the statement of Theorem 1.2 to an average of product of special values of L-series; in Section 3, we find a precise asymptotic for the average of product of special values of L-series that is necessary for our application; in Section 4, we find the expression for the constant C_{r} as an Euler product; in section 5, we show that our method implies the Fouvry-Murty result in the case r = 0.

Acknowledgment: We would like to thank Ram Murty for reading the manuscript and for commenting on an earlier version of this work.

2 From elliptic curves to L-series

In all the following, we fix an integer r. For any integers a_{1}, a_{2}, b_{1}, b_{2} such that 4a_{1}^{3} +27b_{1}^{2} ≠ 0 and 4a_{2}^{3} + 27b_{2}^{2} ≠ 0, let

E_{1} : y^{2} = x^{3} + a_{1}x + b_{1}
E_{2} : y^{2} = x^{3} + a_{2}x + b_{2}

be two elliptic curves over ℤ. Then, for such a_{1}, b_{1}, a_{2}, b_{2}, we define

π_{E_{1},E_{2}}^{r}(x) = # {p ≤ x : a_{p}(E_{1}) = a_{p}(E_{2}) = r} .

We consider

∑_{|b_{1}|,|b_{2}|≤B^{|a_{1}|,|a_{2}|≤A}} π_{E_{1},E_{2}}^{r}(x)

where a_{1}, a_{2}, b_{1}, b_{2} are such that (4a_{1}^{3} + 27b_{1}^{2})(4a_{2}^{3} + 27b_{2}^{2}) ≠ 0. Reversing the summations, this is

∑_{B_{r}<p≤x} # {|a_{1}|, |a_{2}| ≤ A, |b_{1}|, |b_{2}| ≤ B : a_{p}(E_{1}) = a_{p}(E_{2}) = r} + O(A^{2}B^{2})  (3)

where B_{r} = max(3, r^{2}/4), and the O(A^{2}B^{2}) comes from the fact that we removed the primes 2 and 3 from the sum.

Let E(a, b) be the elliptic curve y^{2} = x^{3}+ax+b with a, b ∈ ℤ. The reduced curve E(a, b)_{p}/𝔽_{p} is the reduction modulo p of a minimal model at p for E(a, b). Write a = p^{4k}a′ and b = p^{6k}b′ with k ≥ 0 and integers a′, b′ such that v_{p}(a′) < 4 or v_{p}(b′) < 6 (v_{p}(n) is the power of p appearing in n). Then, for p > 3, E(a′, b′) : y^{2} = x^{3} + a′x + b′ is a minimal model for E(a, b) at p. Hence, each elliptic curve E_{p} over the finite field 𝔽_{p} is the reduction of

(\frac{2A}{p} + O(1)) (\frac{2B}{p} + O(1)) + O (\frac{AB}{p^{10}} )

curves E(a, b) with a, b ∈ ℤ and |a| ≤ A, |b| ≤ B, where the second term accounts for non-minimal models. It follows that,

# {|a_{1}|, |a_{2}| ≤ A, |b_{1}|, |b_{2}| ≤ B : a_{p}(E_{1}) = a_{p}(E_{2}) = r}
= (\frac{4AB}{p^{2}} + O (\frac{A}{p} + \frac{B}{p} + \frac{AB}{p^{10}} + 1))^{2} N(p, r)^{2}  (4)

where N(p, r) is the number of curves E over the finite field 𝔽_{p} such that a_{p}(E) = r, or equivalently with p + 1 − r points over that field.

Lemma 2.1 (Deuring’s Theorem) Let p be a prime, and r an integer such that r^{2} −4p < 0. Let H(r^{2} − 4p) be the Kronecker class number

H(r^{2} − 4p) = 2 ∑_{f^{2}|r^{2}−4p} \frac{h(d)}{w(d)}

where the sum runs over all positive integers f such that f^{2}|r^{2} − 4p and d = (r^{2} − 4p)/f^{2} ≡ 0, 1 mod 4 and is not a square, and h(d) and w(d) are the class number and the number of units in the order of discriminant d respectively. Then,

N(p, r) = \frac{(p − 1)}{2} H(r^{2} − 4p).

Proof: See [Deu] or [Cox, Theorem 14.18]. QED.

Using the last lemma and the standard bound H(r^{2} − 4p) ≪ \sqrt{p} log^{2} p, we get

N(p, r)^{2} = \frac{p^{2}H^{2}(r − 4p)}{4} + O (p^{2} log^{4} p)
≪ p^{3} log^{4} p.

Replacing in (4) and (3), this gives

∑_{|b_{1}|,|b_{2}|≤B^{|a_{1}|,|a_{2}|≤A}} π_{E_{1},E_{2}}^{r}(x) = 4A^{2}B^{2} ∑_{B_{r}≤p≤x} \frac{H^{2}(r^{2} − 4p)}{p^{2}}

+O(A^{2}B^{2} + (A^{2}B + AB^{2})x log^{4} x + (A^{2} + AB + B^{2})x^{2} log^{4} x + …

… + (A + B)x^{3} log^{4} x + x^{4} log^{4} x).

We take A, B such that

A, B ≥ x^{1+ϵ}  (5)

for any ϵ > 0. Then, we have

∑_{|b_{1}|,|b_{2}|≤B^{|a_{1}|,|a_{2}|≤A}} π_{E_{1},E_{2}}^{r}(x) = 4A^{2}B^{2} ∑_{B_{r}<p≤x} \frac{H^{2}(r^{2} − 4p)}{p^{2}} + O(A^{2}B^{2}).  (6)

We now analyse the main term. By definition of the Kronecker class number, and using the class number formula, we get

\frac{1}{4} ∑_{B_{r}≤p≤x} \frac{H^{2}(r^{2} − 4p)}{p^{2}} = ∑_{B_{r}<p≤x} \frac{1}{p^{2}} ∑_{{f^{2}|r^{2}−4p \atop f^{2}d_{1}=r^{2}−4p}} \frac{h(d_{1})}{w(d_{1})} ∑_{{g^{2}|r^{2}−4p \atop g^{2}d_{2}=r^{2}−4p}} \frac{h(d_{2})}{w(d_{2})}

= \frac{1}{4π^{2}} ∑_{B_{r}<p⩽x} \frac{1}{p^{2}} ∑_{{f^{2}|r^{2}−4p \atop f^{2}d_{1}=r^{2}−4p}} \frac{\sqrt{4p − r^{2}}}{f} L(1, χ_{d_{1}}) ∑_{{g^{2}|r^{2}−4p \atop g^{2}d_{2}=r^{2}−4p}} \frac{\sqrt{4p − r^{2}}}{g} L(1, χ_{d_{2}})

= \frac{1}{4π^{2}} ∑_{{f⩽2\sqrt{x} \atop g⩽2\sqrt{x}}} \frac{1}{fg} ∑_{p∈S_{f,g}(x)} \frac{4p − r^{2}}{p^{2}} L(1, χ_{d_{1}})L(1, χ_{d_{2}}),

where S_{f,g}(x) is the set of primes

S_{f,g}(x) = {B_{r} < p ≤ x : f^{2}|r^{2} − 4p, g^{2}|r^{2} − 4p,
d_{1} = (r^{2} − 4p)/f^{2} ≡ 0, 1 mod 4, d_{2} = (r^{2} − 4p)/g^{2} ≡ 0, 1 mod 4}.

We rewrite the last sum as

\frac{1}{π^{2}} ∑_{{f⩽2\sqrt{x} \atop g⩽2\sqrt{x}}} \frac{1}{fg} ∑_{p∈S_{f,g}(x)} \frac{L(1, χ_{d_{1}})L(1, χ_{d_{2}})}{p} + O ( ∑_{{f⩽2\sqrt{x} \atop g⩽2\sqrt{x}}} \frac{1}{fg} ∑_{p∈S_{f,g}(x)} \frac{L(1, χ_{d_{1}})L(1, χ_{d_{2}})}{p^{2}} ) . (7)

We will prove in the next section (Theorem 3.1) that for any c > 0

∑_{{f⩽2\sqrt{x} \atop g⩽2\sqrt{x}}} \frac{1}{fg} ∑_{p∈S_{f,g}(x)} L(1, χ_{d_{1}})L(1, χ_{d_{2}}) log p = K_{r}x + O (\frac{x}{log^{c} x}) .

Then, using Theorem 3.1 and partial summation, we find that the first sum of (7) is

\frac{1}{π^{2}x log x} (K_{r}x + O (\frac{x}{log^{c} x})) + \frac{1}{π^{2}} ∫_{2}^{x} (K_{r}t + O (\frac{t}{log^{c} t})) (\frac{1 + log t}{t^{2} log^{2} t}) dt

∼ \frac{K_{r}}{π^{2}} log log x

and similarly that

∑_{{f⩽2\sqrt{x} \atop g⩽2\sqrt{x}}} \frac{1}{fg} ∑_{p∈S_{f,g}(x)} \frac{L(1, χ_{d_{1}})L(1, χ_{d_{2}})}{p^{2}} = O(1).

Then,

\frac{1}{4} ∑_{B_{r}≤p≤x} \frac{H^{2}(r^{2} − 4p)}{p^{2}} ∼ \frac{K_{r}}{π^{2}} log log x

and replacing in (6), we get

\frac{1}{16A^{2}B^{2}} ∑_{|b_{1}|,|b_{2}|≤B^{|a_{1}|,|a_{2}|≤A}} π_{E_{1},E_{2}}^{r}(x) ∼ \frac{K_{r}}{π^{2}} log log x

for A, B ≥ x^{1+ϵ}. Notice, assuming Theorem 3.1, this shows Theorem 1.2. The next section consists of a proof of Theorem 3.1.

3 Average values of product of Dirichlet L-functions

Theorem 3.1 Let r be an odd integer. Then, for any c > 0,

∑_{f⩽2\sqrt{x}} ∑_{g⩽2\sqrt{x}} \frac{1}{fg} ∑_{p∈S_{f,g}(x)} L(1, χ_{d_{1}})L(1, χ_{d_{2}}) log p = K_{r}x + O (\frac{x}{log^{c} x}) ,

where

K_{r} = 3 ∏_{p|r} \frac{p^{2}(p^{2} + 1)}{(p^{2} − 1)^{2}} ∏_{p∤r} \frac{p^{2}(p^{4} − 2p^{2} − 3p − 1)}{(p + 1)^{3}(p − 1)^{3}}.

This section consists of a proof of Theorem 3.1. As r is odd, it follows from the definition of S_{f,g}(x) that f, g are also odd, and that d_{1}, d_{2} are congruent to 1 modulo 4. Also, any common factor between r and f would divide the primes p ∈ S_{f,g}(x), which is impossible because p > B_{r} = max (3, r^{2}/4). Then, the sum is empty unless (2r, fg) = 1, and we can rewrite the sum of Theorem 3.1 as

∑_{{f,g≤2\sqrt{x} \atop (2r,fg)=1}} \frac{1}{fg} ∑_{p∈S_{f,g}(x)} L(1, χ_{d_{1}})L(1, χ_{d_{2}}) log p

where

S_{f,g}(x) = {B_{r} < p ≤ x : f^{2} | r^{2} − 4p, g^{2} | r^{2} − 4p} .

Let

L(s) = L(s, χ_{d_{1}})L(s, χ_{d_{2}}) = ∑_{{m=1 \atop n=1}}^{∞} \frac{χ_{d_{1}}(m)χ_{d_{2}}(n)}{(mn)^{s}} = ∑_{ℓ=1}^{∞} \frac{a_{d_{1},d_{2}}(ℓ)}{ℓ^{s}},

where

a_{d_{1},d_{2}}(ℓ) = ∑_{mn=ℓ} χ_{d_{1}}(m)χ_{d_{2}}(n).  (8)

We then have the trivial bound

a_{d_{1},d_{2}}(ℓ) ≪ d(ℓ) ≪ ℓ^{ϵ}  (9)

for any ϵ > 0, where d(ℓ) is the number of divisors of ℓ. We need an expression for the truncated L-series of L(1).

Lemma 3.2 Let U > 0. Then, for any ϵ > 0,

L(1) = ∑_{ℓ=1}^{∞} \frac{a_{d_{1},d_{2}}(ℓ)}{ℓ}e^{−ℓ/U} + O (\frac{|d_{1}d_{2}|^{3/16+ϵ}}{U^{1/2}} )

where the error term depends on ϵ.

Proof: We have the integral representation

e^{− \frac{1}{U}} = \frac{1}{2πi} ∫_{(1)} Γ(s + 1)U^{s} \frac{ds}{s}

(see [M], p. 353 for a proof). Using this we have

∑_{ℓ=1}^{∞} \frac{a_{d_{1},d_{2}}(ℓ)}{ℓ}e^{−ℓ/U} = \frac{1}{2πi} ∫_{(1)} L(s + 1)Γ(s + 1)U^{s} \frac{ds}{s}.

Now moving the line of integration from (1) to (−\frac{1}{2}) and calculating the residue at s = 0 yields

∑_{ℓ=1}^{∞} \frac{a_{d_{1},d_{2}}(ℓ)}{ℓ}e^{−ℓ/U} = L(1) + \frac{1}{2πi} ∫_{(− \frac{1}{2} )} L(s + 1)Γ(s + 1)U^{s} \frac{ds}{s}.  (10)

Recalling Burgess’s result (see [Bur]), we have for any ϵ > 0,

L(1/2 + it) = L(1/2 + it, χ_{d_{1}})L(1/2 + it, χ_{d_{2}}) ≪_{ϵ} |d_{1}d_{2}|^{3/16+ϵ},

and then

\frac{1}{2πi} ∫_{(− \frac{1}{2} )} L(s + 1)Γ(s + 1)U^{s} \frac{ds}{s} ≪_{ϵ} \frac{|d_{1}d_{2}|^{3/16+ϵ}}{U^{1/2}} .

Replacing this in (10) completes the proof. QED.

Using Lemma 3.2, we write, for any ϵ > 0,

∑_{{f,g≤2\sqrt{x} \atop (2r,fg)=1}} \frac{1}{fg} ∑_{p∈S_{f,g}(x)} L(1) log p

= ∑_{{f,g≤2\sqrt{x} \atop (2r,fg)=1}} \frac{1}{fg} ∑_{p∈S_{f,g}(x)} \{∑_{ℓ=1}^{∞} \frac{a_{d_{1},d_{2}}(ℓ)}{ℓ}e^{−ℓ/U} + O (\frac{|d_{1}d_{2}|^{3/16+ϵ}}{U^{1/2}} )\} log p

= ∑_{{f,g≤2\sqrt{x} \atop (2r,fg)=1}} \frac{1}{fg} ∑_{ℓ=1}^{∞} \frac{e^{−ℓ/U}}{ℓ} ∑_{p∈S_{f,g}(x)} a_{d_{1},d_{2}}(ℓ) log p + O (\frac{1}{U^{1/2}} ∑_{{f,g≤2\sqrt{x} \atop (2r,fg)=1}} \frac{1}{fg} ∑_{p∈S_{f,g}(x)} |d_{1}d_{2}|^{3/16+ϵ} log p) .

Replacing d_{1} and d_{2} by their definition, we can bound the sum in the error term by

≪ \frac{1}{U^{1/2}} ∑_{{f,g≤2\sqrt{x} \atop (2r,fg)=1}} \frac{1}{(fg)^{11/8+2ϵ}} ∑_{p∈S_{f,g}(x)} p^{3/8+2ϵ} log p

≪ \frac{x^{3/8+2ϵ} log x}{U^{1/2}} ∑_{{f,g≤2\sqrt{x} \atop (2r,fg)=1}} \frac{1}{(fg)^{11/8+2ϵ}} ∑_{p∈S_{f,g}(x)} 1 ≪ \frac{x^{11/8+2ϵ}}{U^{1/2}} ,

and we have

∑_{{f,g≤2\sqrt{x} \atop (2r,fg)=1}} \frac{1}{fg} ∑_{p∈S_{f,g}(x)} L(1) log p = ∑_{{f,g≤2\sqrt{x} \atop (2r,fg)=1}} \frac{1}{fg} ∑_{ℓ=1}^{∞} \frac{e^{−ℓ/U}}{ℓ} ∑_{p∈S_{f,g}(x)} a_{d_{1},d_{2}}(ℓ) log p + O (\frac{x^{11/8+2ϵ}}{U^{1/2}} )

(11)

for any ϵ > 0.

Let 1 < V ⩽ 2\sqrt{x} be a parameter to be chosen later. We write the sum in (11) as

∑_{{f,g≤V \atop (2r,fg)=1}} ∑_{ℓ=1}^{∞} \frac{e^{−ℓ/U}}{ℓ} ∑_{p∈S_{f,g}(x)} a_{d_{1},d_{2}}(ℓ) log p + ∑_{{V <f,g⩽2\sqrt{x} \atop (2r,fg)=1}} \frac{1}{fg} ∑_{ℓ=1}^{∞} \frac{e^{−ℓ/U}}{ℓ} ∑_{p∈S_{f,g}(x)} a_{d_{1},d_{2}}(ℓ) log p.

For the sum over large values of f and g, we first notice that for such f and g, we have [f^{2}, g^{2}]|r^{2} − 4p which implies that [f^{2}, g^{2}] ≤ 4x. We also have that 4p ≡ r^{2} mod f^{2} and 4p ≡ r^{2} mod g^{2} ⟺ 4p ≡ r^{2} mod [f^{2}, g^{2}]. Then,

| ∑_{{V <f,g⩽2\sqrt{x} \atop (2r,fg)=1} ; [f^{2},g^{2}]≤4x} \frac{1}{fg} ∑_{ℓ=1}^{∞} \frac{e^{−ℓ/U}}{ℓ} ∑_{p∈S_{f,g}(x)} a_{d_{1},d_{2}}(ℓ) log p|

≤ log x |∑_{ℓ=1}^{∞} \frac{d(ℓ)}{ℓ}e^{−ℓ/U} | ∑_{{V <f,g⩽2\sqrt{x} \atop (2r,fg)=1} ; [f^{2},g^{2}]≤4x} \frac{1}{fg} ∑_{{p⩽x \atop 4p≡r^{2} mod [f^{2},g^{2}]}} 1

≤ x log x |∑_{ℓ=1}^{∞} \frac{d(ℓ)}{ℓ}e^{−ℓ/U} | ∑_{{V <f,g⩽2\sqrt{x} \atop (2r,fg)=1} ; [f^{2},g^{2}]≤4x} \frac{1}{fg[f^{2}, g^{2}]}.  (12)

Lemma 3.3

∑_{ℓ=1}^{∞} \frac{d(ℓ)}{ℓ}e^{−ℓ/U} ≪ log^{2} U

Proof: As in Lemma 3.2, we have the integral representation

∑_{ℓ=1}^{∞} \frac{d(ℓ)}{ℓ}e^{−ℓ/U} = \frac{1}{2πi} ∫_{(1)} ζ^{2}(s + 1)Γ(s + 1)U^{s} \frac{ds}{s}

for the infinite sum that we want to bound, where ζ(s) is the Riemann zeta function. Note that since

ζ(s) = \frac{1}{s − 1} + γ + c_{1}(s − 1) + ...

(see [M], p. 63), the residue of the integrand at s = 0 is

\frac{1}{2} log^{2} U + 2γ log U + c_{0},

where γ is the Euler constant and c_{0} a constant. Now by moving the line of integration from (1) to (−\frac{1}{2}) and calculating the residue at s = 0 we get the desired bound. QED.

Using this lemma, we can bound (12) by

x log x log^{2} U ∑_{{V <f,g⩽2\sqrt{x} \atop (2r,fg)=1}} \frac{(f^{2}, g^{2})}{f^{3}g^{3}} ⩽ x log x log^{2} U ∑_{{V <f,g⩽2\sqrt{x} \atop (2r,fg)=1}} \frac{1}{f^{2}g^{2}} ≪ \frac{x log x log^{2} U}{V^{2}}

to get that

∑_{{f,g≤2\sqrt{x} \atop (2r,fg)=1}} \frac{1}{fg} ∑_{p∈S_{f,g}(x)} L(1) log p = ∑_{{f,g≤V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{ℓ=1}^{∞} \frac{e^{−ℓ/U}}{ℓ} ∑_{p∈S_{f,g}(x)} a_{d_{1},d_{2}}(ℓ) log p

+O (\frac{x^{11/8+2ϵ}}{U^{1/2}} ) + O (\frac{x log x log^{2} U}{V^{2}} ) .  (13)

We now write the sum on the right hand side of (13) as

∑_{{f,g≤V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{ℓ≤U log U} \frac{e^{−ℓ/U}}{ℓ} ∑_{p∈S_{f,g}(x)} a_{d_{1},d_{2}}(ℓ) log p + ∑_{{f,g≤V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{ℓ>U log U} \frac{e^{−ℓ/U}}{ℓ} ∑_{p∈S_{f,g}(x)} a_{d_{1},d_{2}}(ℓ) log p

for some parameter U = U(x) to be chosen later. We first estimate the sum for large values of ℓ. For any ϵ > 0, we have

∑_{ℓ>U log U} \frac{d(ℓ)}{ℓ}e^{−ℓ/U} ≪ ∑_{ℓ>U log U} \frac{e^{−ℓ/U}}{ℓ^{1−ϵ}} ≪ \frac{1}{(U log U)^{1−ϵ}} ∑_{ℓ>U log U} e^{−ℓ/U}
≪ \frac{1}{(U log U)^{1−ϵ}} ∫_{U log U}^{∞} e^{−t/U} dt = \frac{1}{(U log U)^{1−ϵ}}

and then

∑_{{f,g≤V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{ℓ>U log U} \frac{1}{ℓ} e^{−ℓ/U} ∑_{p∈S_{f,g}(x)} a_{d_{1},d_{2}}(ℓ) log p

≪ x log x ∑_{ℓ>U log U} \frac{d(ℓ)}{ℓ}e^{−ℓ/U} ∑_{{f,g≤V \atop (2r,fg)=1}} \frac{1}{fg[f^{2}, g^{2}]} ≪ \frac{x log x log^{2} V}{(U log U)^{1−ϵ}} .

Using this last result and (13), we get that for any ϵ > 0,

∑_{{f,g≤2\sqrt{x} \atop (2r,fg)=1}} \frac{1}{fg} ∑_{p∈S_{f,g}(x)} L(1) log p = ∑_{{f,g≤V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{ℓ≤U log U} \frac{1}{ℓ}e^{−ℓ/U} ∑_{p∈S_{f,g}(x)} a_{d_{1},d_{2}}(ℓ) log p

+O (\frac{x^{11/8+2ϵ}}{U^{1/2}} ) + O (\frac{x log x log^{2} U}{V^{2}} ) + O (\frac{x log x log^{2} V}{(U log U)^{1−ϵ}} ) .  (14)

We now estimate the sum of the right-hand side of (14). By quadratic reciprocity,

χ_{d_{1}}(m) = χ_{d′_{1}}(m) if d_{1} ≡ d′_{1} mod (4m).

We then have

∑_{{f,g≤V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{ℓ≤U log U} \frac{e^{−ℓ/U}}{ℓ} ∑_{p∈S_{f,g}(x)} log p ∑_{mn=ℓ} χ_{d_{1}}(m)χ_{d_{2}}(n)

= ∑_{{f,g≤V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{mn=ℓ^{ℓ≤U log U}} \frac{e^{−ℓ/U}}{ℓ} ∑_{{a mod 4m \atop b mod 4n}} (\frac{a}{m}) (\frac{b}{n}) ∑_{p}^{∗} log p

where ∑_{p}^{∗} runs over primes p such that p ∈ S_{f,g}(x) and d_{1} ≡ a mod (4m), d_{2} ≡ b mod (4n), i.e. the primes p such that B_{r} < p ≤ x and

p ≡ (r^{2} − af^{2})/4 mod mf^{2} and p ≡ (r^{2} − bg^{2})/4 mod ng^{2}.

If (r^{2} − af^{2})/4 ≢ (r^{2} − bg^{2})/4 mod (mf^{2}, ng^{2}), there are no such primes. If the above congruence is satisfied, let θ = θ(a, b, m, n, f, g) be the unique residue modulo [mf^{2}, ng^{2}] which is congruent to (r^{2} − af^{2})/4 modulo mf^{2}, and congruent to (r^{2} − bg^{2})/4 modulo ng^{2}. If (r^{2} − af^{2})/4 ≢ (r^{2} − bg^{2})/4 mod (mf^{2}, ng^{2}), we set θ = 0. Then, we can rewrite the last sum as

∑_{{f,g≤V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{{ℓ⩽U log U \atop mn=ℓ}} \frac{1}{ℓ}e^{−ℓ/U} ∑_{{a mod 4m \atop b mod 4n}} (\frac{a}{m}) (\frac{b}{n}) ∑_{{Br<p⩽x \atop p≡θ mod [mf^{2},ng^{2}]}} log p.

Let a, n be positive integers with (a, n) = 1. Following the standard notation, we write

ψ(x; n, a) = ∑_{{p≤x \atop p≡a mod n}} log p = \frac{x}{ϕ(n)} + E(x; n, a).

With this notation, we rewrite the last sum as

∑_{{f,g≤V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{{ℓ⩽U log U \atop mn=ℓ}} \frac{1}{ℓ}e^{−ℓ/U} ∑_{{a mod 4m \atop b mod 4n}}^{∗} (\frac{a}{m}) (\frac{b}{n}) (\frac{x}{ϕ([mf^{2}, ng^{2}])} + E(x; [mf^{2}, ng^{2}], θ))

where ∑∗_{{a mod 4m \atop b mod 4n}} means that the sum runs over invertible residues a, b modulo m, n respectively such that (r^{2} − af^{2})/4 ≡ (r^{2} − bg^{2})/4 mod (mf^{2}, ng^{2}), and θ is invertible modulo [mf^{2}, ng^{2}], or equivalently (r^{2} − af^{2}, 4m) = 4 and (r^{2} − bg^{2}, 4n) = 4. We then define

c_{f,g}^{r}(m, n) = ∑_{{a (4m)^{∗} \atop (r^{2}−af^{2},4m)=4}} ∑_{{b (4n)^{∗} \atop (r^{2}−bg^{2},4n)=4} ; (r^{2}−af^{2})/4≡(r^{2}−bg^{2})/4 mod (mf^{2},ng^{2})} (\frac{a}{m}) (\frac{b}{n}) .  (15)

Using this notation, we have

∑_{{f,g≤V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{ℓ≤U log U} \frac{e^{−ℓ/U}}{ℓ} ∑_{p∈S_{f,g}(x)} a_{d_{1},d_{2}}(ℓ) log p = x ∑_{{f,g≤V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{{ℓ≤U log U \atop mn=ℓ}} \frac{e^{−ℓ/U}}{ℓ} \frac{c_{f,g}^{r}(m, n)}{ϕ([mf^{2}, ng^{2}])}

+ ∑_{{f,g≤V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{{ℓ≤U log U \atop ℓ=mn}} \frac{e^{−ℓ/U}}{ℓ} ∑_{{a mod 4m \atop b mod 4n}}^{∗} (\frac{a}{m}) (\frac{b}{n}) E(x; [mf^{2}, ng^{2}], θ).  (16)

We first deal with the second sum of (16) which is bounded by

∑_{{f,g≤V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{mn≤U log U} \frac{1}{mn} ∑_{{a mod 4m \atop b mod 4n}}^{∗} |E(x; [mf^{2}, ng^{2}], θ)|.

In the sum ∑∗_{{a mod 4m \atop b mod 4n}}, each pair of residues a, b modulo 4m and 4n respectively yields a different residue θ modulo [mf^{2}, ng^{2}]. We then have

∑_{mn≤U log U} \frac{1}{mn} ∑_{{a mod 4m \atop b mod 4n}}^{∗} |E(x; [mf^{2}, ng^{2}], θ)| ≤ ∑_{mn≤U log U} \frac{1}{mn} ∑_{θ mod [mf^{2},ng^{2}]} |E(x; [mf^{2}, ng^{2}], θ)|

≪ f^{2}g^{2} ∑_{ℓ≤U log Uf^{2}g^{2}} \frac{1}{ℓ} ∑_{θ mod ℓ} c(ℓ) |E(x; ℓ, θ)|

where c(ℓ) is the number of ways that we can write ℓ = [mf^{2}, ng^{2}]. More generally, we have

Lemma 3.4 Let n be a positive integer, and let C(n) be the number of ways to write n = [n_{1}, n_{2}] for any positive integers n_{1} and n_{2}. Then, C(n) ≤ 2^{ν(n)}d(n), where ν(n) is the number of distinct prime factors of n and d(n) is the number of divisors of n.

Proof: Let n = ∏_{i=1}^{r} p_{i}^{α_{i}} with α_{i} ≥ 1 for i = 1, …, r. Then, n = [n_{1}, n_{2}] implies that n_{1} = ∏_{i=1}^{r} p_{i}^{β_{i}} and n_{2} = ∏_{i=1}^{r} p_{i}^{γ_{i}} with 0 ≤ β_{i}, γ_{i} ≤ α_{i} and max (β_{i}, γ_{i}) = α_{i} for i = 1, …, r. As there are 2α_{i} + 1 such pairs (β_{i}, γ_{i}) for each i, we have

C(n) = ∏_{i=1}^{r}(2α_{i} + 1) ≤ ∏_{i=1}^{r} 2(α_{i} + 1) = 2^{ν(n)}d(n).

QED.

Using this result in the last bound, we get

∑_{mn≤U log U} \frac{1}{mn} ∑_{{a mod 4m \atop b mod 4n}}^{∗} |E(x; [mf^{2}, ng^{2}], θ)| ≪ f^{2}g^{2} ∑_{ℓ≤U log Uf^{2}g^{2}} \frac{d^{2}(ℓ)}{ℓ} ∑_{θ mod ℓ} |E(x; ℓ, θ)|

≤ f^{2}g^{2} ( ∑_{{ℓ≤U log Uf^{2}g^{2} \atop θ mod ℓ}} \frac{d^{4}(ℓ)}{ℓ^{2}} )^{1/2} ( ∑_{{ℓ≤U log Uf^{2}g^{2} \atop θ mod ℓ}} E^{2}(x; ℓ, θ))^{1/2}

using the Cauchy-Schwartz inequality. For the first parenthesis, we use the result of Ramanujan [Wil]

∑_{ℓ≤N} d^{r}(ℓ) ∼ A_{r} N log^{2^{r}−1} (N), for r ≥ 2 and A_{r} an absolute constant

with r = 4. Using partial summation, and the fact that f, g ≤ V, this gives

( ∑_{{ℓ≤U log Uf^{2}g^{2} \atop θ mod ℓ}} \frac{d^{4}(ℓ)}{ℓ^{2}} )^{1/2} ≤ ( ∑_{ℓ≤U log Uf^{2}g^{2}} \frac{d^{4}(ℓ)}{ℓ})^{1/2} ≪ log^{8} (V^{4}U log U).

For the second parenthesis, we apply the theorem of Barban-Davenport-Halberstam [Dav, p. 169]. This gives

( ∑_{{ℓ≤V^{4}U log U \atop θ mod ℓ}} E^{2}(x; ℓ, θ))^{1/2} ≪ (V^{4}Ux log U log x)^{1/2}

whenever

\frac{x}{log^{A} x} ⩽ V^{4}U log U ⩽ x,  (17)

for some A > 0. Finally, summing over f, g, this gives

∑_{{f,g≤V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{{ℓ≤U log U \atop ℓ=mn}} \frac{e^{−ℓ/U}}{ℓ} ∑_{{a mod 4m \atop b mod 4n}}^{∗} (\frac{a}{m}) (\frac{b}{n}) E(x; [mf^{2}, ng^{2}], θ)

≪ (V^{4}Ux log U log x)^{1/2} log^{8} (V^{4}U log U) ∑_{{f,g≤V \atop (2r,fg)=1}} fg

≪ V^{6}(Ux log U log x)^{1/2} log^{8} x  (18)

whenever (17) holds.

We now have to evaluate the first sum of (16). We first rewrite the sum as

x ∑_{{f,g=1 \atop (2r,fg)=1}}^{∞} \frac{1}{fg} ∑_{m,n=1}^{∞} \frac{c_{f,g}^{r}(m, n)e^{−mn/U}}{mnϕ([mf^{2}, ng^{2}])} − x ∑_{{f,g⩽V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{{ℓ>U log U \atop mn=ℓ}} \frac{e^{−ℓ/U}}{ℓ} \frac{c_{f,g}^{r}(m, n)}{ϕ([mf^{2}, ng^{2}])}

−x ∑_{{f,g>V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{m,n=1}^{∞} \frac{c_{f,g}^{r}(m, n)e^{−mn/U}}{mnϕ([mf^{2}, ng^{2}])}.  (19)

We first deal with the two error terms of (19). This is done using the bound

c_{f,g}^{r}(m, n) ≪ \frac{mn}{κ(mn)(m, n)}  (20)

which is shown in Lemma 4.8. Using the notation of Section 4, we write k = (f, g) and f = kf′ and g = kg′. If (f′, n) ≠ 1 or (g′, m) ≠ 1, we have c_{f,g}^{r}(m, n) = 0 by Lemma 4.3(i). If (f′, n) = (g′, m) = 1, then (mf^{2}, ng^{2}) = (m, n)(f^{2}, g^{2}). This gives

\frac{c_{f,g}^{r}(m, n)}{ϕ([mf^{2}, ng^{2}])} = \frac{(mf^{2}, ng^{2}) c_{f,g}^{r}(m, n)}{ϕ(mnf^{2}g^{2})} = \frac{(m, n)(f^{2}, g^{2}) c_{f,g}^{r}(m, n)}{ϕ(mnf^{2}g^{2})}
≤ \frac{(m, n)(f^{2}, g^{2}) c_{f,g}^{r}(m, n)}{ϕ(mn)ϕ(f^{2})ϕ(g^{2})} ≪ \frac{mn (f^{2}, g^{2})}{κ(mn)ϕ(mn)ϕ(f^{2})ϕ(g^{2})}

using the bound (20) for c_{f,g}^{r}(m, n). Replacing in the first error term of (19), we get that

x ∑_{{f,g⩽V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{{ℓ>U log U \atop mn=ℓ}} \frac{e^{−ℓ/U}}{ℓ} \frac{c_{f,g}^{r}(m, n)}{ϕ([mf^{2}, ng^{2}])} ≪ x ∑_{{f,g⩽V \atop (2r,fg)=1}} \frac{(f^{2}, g^{2})}{fgϕ(f^{2})ϕ(g^{2})} ∑_{ℓ>U log U} \frac{d(ℓ)}{κ(ℓ)ϕ(ℓ)}.

It is shown in [DP, Lemma 3.4] that

∑_{ℓ=1}^{∞} \frac{ℓ^{3/2}}{κ(ℓ)ϕ(ℓ)}ℓ^{−s}  (21)

converges for Re(s) > 1. Clearly, this implies that ∑_{ℓ=1}^{∞} \frac{d(ℓ)}{κ(ℓ)ϕ(ℓ)} converges. Furthermore,

using the Wiener-Ikehara Tauberian Theorem and partial summation as in the proof of [DP, Lemma 3.4], we can show that for any ϵ > 0,

∑_{ℓ>U log U} \frac{d(ℓ)}{κ(ℓ)ϕ(ℓ)} ≪ (U log U)^{−1/2+ϵ}.  (22)

Also,

∑_{{f,g≤V \atop (2r,fg)=1}} \frac{(f^{2}, g^{2})}{fgϕ(f^{2})ϕ(g^{2})} ≤ 2 ∑_{{f,g≤V \atop f≤g}} \frac{1}{g^{2}ϕ(f)ϕ(g)} ≤ 2 (∑_{f≤V} \frac{1}{fϕ(f)})^{2} = O(1)

and then

x ∑_{{f,g⩽V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{{ℓ>U log U \atop mn=ℓ}} \frac{e^{−ℓ/U}}{ℓ} \frac{c_{f,g}^{r}(m, n)}{ϕ([mf^{2}, ng^{2}])} = O (\frac{x}{(U log U)^{1/2−ϵ}} ) .  (23)

We now look at the second error term of (19). As above, we have

x ∑_{{f,g>V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{m,n=1}^{∞} \frac{c_{f,g}^{r}(m, n)e^{−mn/U}}{mnϕ([mf^{2}, ng^{2}])} ≪ x ∑_{f,g>V} \frac{(f^{2}, g^{2})}{fgϕ(f^{2})ϕ(g^{2})}

≤ x (∑_{f>V} \frac{1}{fϕ(f)})^{2} ≪ \frac{x}{V^{2−2ϵ}}

for any positive ϵ > 0, as ϕ(n) ≫ n^{1−ϵ} for any positive ϵ > 0 [HW, p. 267]. Then, replacing in (19), we get

x ∑_{{f,g≤V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{ℓ⩽U log U} \frac{e^{−ℓ/U}}{ℓ} ∑_{mn=ℓ} \frac{c_{f,g}^{r}(m, n)}{ϕ([mf^{2}, ng^{2}])}

= x ∑_{{f,g=1 \atop (2r,fg)=1}}^{∞} \frac{1}{fg} ∑_{m,n=1}^{∞} \frac{c_{f,g}^{r}(m, n)e^{−mn/U}}{mnϕ([mf^{2}, ng^{2}])} + O (\frac{x}{(U log U)^{1/2−ϵ}} ) + O (\frac{x}{V^{2−2ϵ}} ) . (24)

Finally, we remove the exponential e^{−ℓ/U} from the main term. We have, for any c_{1} > 0,

x ∑_{{f,g,m,n=1 \atop (2r,fg)=1}}^{∞} \frac{c_{f,g}^{r}(m, n)e^{−mn/U}}{mnfgϕ([mf^{2}, ng^{2}])} = \frac{x}{2πi} ∑_{{f,g,m,n=1 \atop (2r,fg)=1}}^{∞} \frac{c_{f,g}^{r}(m, n)}{fgmnϕ([mf^{2}, ng^{2}])} ∫_{(c_{1})} Γ(s) (\frac{U}{mn})^{s} ds

= \frac{x}{2πi} ∫_{(c_{1})} ( ∑_{{f,g,m,n=1 \atop (2r,fg)=1}}^{∞} \frac{c_{f,g}^{r}(m, n)}{fg(mn)^{s+1}ϕ([mf^{2}, ng^{2}])}) Γ(s)U^{s}ds.

Using the bound (20) and working as above, we get

∑_{{f,g,m,n=1 \atop (2r,fg)=1}}^{∞} \frac{c_{f,g}^{r}(m, n)}{(mn)^{s+1}fgϕ([mf^{2}, ng^{2}])} ≪ ∑_{ℓ=1}^{∞} \frac{d(ℓ)}{κ(ℓ)ϕ(ℓ)ℓ^{s}}

and from (21), the sum converges for Re(s) > −1/2 + ϵ, for any ϵ > 0. Then we can move the line of integration to any −1/2 + ϵ < γ < 0, say γ = −1/4. As Γ(s) has a simple pole at s = 0, by using Cauchy’s residue theorem and working as in the proof of Lemma 3.2, we get

x ∑_{{f,g,m,n=1 \atop (2r,fg)=1}}^{∞} \frac{c_{f,g}^{r}(m, n)}{fgmnϕ([mf^{2}, ng^{2}])}e^{−mn/U} = x ∑_{{f,g,m,n=1 \atop (2r,fg)=1}}^{∞} \frac{c_{f,g}^{r}(m, n)}{fgmnϕ([mf^{2}, ng^{2}])} + O (\frac{x}{U^{1/4}} )

and replacing in (24), we have

x ∑_{{f,g≤V \atop (2r,fg)=1}} \frac{1}{fg} ∑_{ℓ⩽U log U} \frac{e^{−ℓ/U}}{ℓ} ∑_{mn=ℓ} \frac{c_{f,g}^{r}(m, n)}{ϕ([mf^{2}, ng^{2}])}

= x ∑_{{f,g,m,n=1 \atop (2r,fg)=1}}^{∞} \frac{c_{f,g}^{r}(m, n)}{fgmnϕ([mf^{2}, ng^{2}])} + O (\frac{x}{(U log U)^{1/2−ϵ}} + \frac{x}{V^{2−2ϵ}} + \frac{x}{U^{1/4}} ) . (25)

This finishes the proof of Theorem 3.1. Indeed, replacing (25) and (18) in (16) and (14), we get that

∑_{{f,g⩽2\sqrt{x} \atop (2r,fg)=1}} \frac{1}{fg} ∑_{p∈S_{f,g}^{r}(x)} L(1, χ_{d_{1}})L(1, χ_{d_{2}}) log p = K_{r}x + O (\frac{x}{(U log U)^{1/2−ϵ}} + \frac{x}{V^{2−2ϵ}} + …

… + \frac{x}{U^{1/4}} + V^{6}(Ux log U log x)^{1/2} log^{8} x + \frac{x^{11/8+2ϵ}}{U^{1/2}} + \frac{x log x log^{2} U}{V^{2}} + \frac{x log x log^{2} V}{(U log U)^{1−ϵ}} )

for all ϵ > 0, with

K_{r} = ∑_{{f,g=1 \atop (2r,fg)=1}}^{∞} \frac{1}{fg} ∑_{mn=1}^{∞} \frac{c_{f,g}^{r}(m, n)}{mnϕ([mf^{2}, ng^{2}])}.  (26)

We choose U = x/log^{α} x and V = log^{β} x for positive integers α, β such that α − 4β − 1 ≥ 1 insuring that the condition (17) is satisfied. Then,

∑_{{f,g⩽2\sqrt{x} \atop (2r,fg)=1}} \frac{1}{fg} ∑_{p∈S_{f,g}^{r}(x)} L(1, χ_{d_{1}})L(1, χ_{d_{2}}) log p = K_{r}x + O (\frac{x}{log^{β} x} + \frac{x}{log^{α/2−6β−9} x})

= K_{r}x + O (\frac{x}{log^{c} x})

for any c > 0 for an appropriate choice of α and β. This proves Theorem 3.1, provided that we get the Euler product expansion for the constant K_{r} of (26). This is done in the next section.

4 The constant

In this section, we express the constant K_{r} as an Euler product of local factors. We first prove that the coefficients c_{f,g}^{r}(m, n) are multiplicative, and we then use this result to prove a bound on the size of c_{f,g}^{r}(m, n) needed to complete the proof of Theorem 3.1 (see Lemma 4.8). Moreover, we also use the multiplicativity of these coefficients to derive the Euler product for the constant K_{r} in Theorem 3.1.

4.1 Multiplicativity of the coefficients c_{f,g}^{r}(m, n)

For all this section, let r be an odd integer, and let f and g be positive odd integers. Let k = (f, g), and let f′, g′ be such that f = f′k and g = g′k. Let m and n be positive integers. For a prime p and an integer n, the valuation v_{p}(n) is the power of p appearing in the integer n.

Definition 4.1 (1) Let

c_{f}^{r}(m) = ∑_{(r^{2}−af^{2},4m)=4^{a(4m)^{∗}}} (\frac{a}{m});

(2) For any invertible residue a modulo 4m, let

c_{f,g}^{r}(n; m, a) = ∑_{(r^{2}−bg^{2})/4≡(r^{2}−af^{2})/4 mod (mf^{2},ng^{2})^{(r^{2}−bg^{2},4n)=4}^{b(4n)^{∗}}} (\frac{b}{n});

(3) Let

c_{f,g}^{r}(m, n) = ∑_{(r^{2}−af^{2},4m)=4^{a(4m)^{∗}}} (\frac{a}{m}) c_{f,g}^{r}(n; m, a).

Of course, this definition agrees with the previous definition of c_{f,g}^{r}(m, n) in (15).

Definition 4.2 A function F (m, n) defined on the set of positive integers m, n is multiplicative if it satisfies

F (m, n) = ∏_{p|mn} F (p^{v_{p}(m)}, p^{v_{p}(n)}).

Lemma 4.3 (i) If (m, g′) ≠ 1 or (n, f′) ≠ 1, then c_{f,g}^{r}(m, n) = 0. (ii) If (n_{1}, n_{2}) = 1, then c_{f,g}^{r}(n_{1}n_{2}; m, a) = c_{f,g}^{r}(n_{1}; m, a) c_{f,g}^{r}(n_{2}; m, a).

Proof: (i) As (r^{2} − bg^{2})/4 ≡ (r^{2} − af^{2})/4 mod (mf^{2}, ng^{2}) ⟺ (af^{′2} − bg^{′2})/4 ≡ 0 mod (mf^{′2}, ng^{′2}), we have

c_{f,g}^{r}(m, n) = ∑_{(r^{2}−af^{2},4m)=4^{a(4m)^{∗}}} (\frac{a}{m}) ∑_{(af^{′2}−bg^{′2})/4≡0 mod (mf^{′2},ng^{′2})^{(r^{2}−bg^{2},4n)=4}^{b(4n)^{∗}}} (\frac{b}{n}) .

Suppose there is a prime p dividing (n, f′). Then, c_{f,g}^{r}(m, n) = 0 because b ≡ 0 mod p, as p divides (mf^{′2}, ng^{′2}) and (g′, p) = 1. The case (m, g′) ≠ 1 is similar.

(ii) From the Generalised Chinese Remainder Theorem, there is a bijection between the set of invertible residues b modulo 4n_{1}n_{2} such that (r^{2} − bg^{2}, 4n_{1}n_{2}) = 4 and the set of pairs (b_{1}, b_{2}) of invertible residues modulo 4n_{1} and 4n_{2} respectively such that (r^{2} − b_{1}g^{2}, 4n_{1}) = 4 and (r^{2} − b_{2}g^{2}, 4n_{2}) = 4. Furthermore,

(af^{2} − bg^{2})/4 ≡ 0 mod (mf^{2}, n_{1}n_{2}g^{2})

if and only if

(af^{2} − b_{1}g^{2})/4 ≡ 0 mod (mf^{2}, n_{1}g^{2}) and (af^{2} − b_{2}g^{2})/4 ≡ 0 mod (mf^{2}, n_{2}g^{2})

as the least common multiple of (mf^{2}, n_{1}g^{2}) and (mf^{2}, n_{2}g^{2}) is (mf^{2}, n_{1}n_{2}g^{2}). This proves the result. QED.

Lemma 4.4 Let m_{1}, m_{2}, n_{1}, n_{2} be positive integers such that (m_{1}, m_{2}) = (n_{1}, n_{2}) = (m_{1}, n_{2}) = (m_{2}, n_{1}) = 1. Then,

c_{f,g}^{r}(m_{1}m_{2}, n_{1}n_{2}) = c_{f,g}^{r}(m_{1}, n_{1}) c_{f,g}^{r}(m_{2}, n_{2}).

Equivalently, the functions c_{f,g}^{r}(m, n) are multiplicative.

Proof: Let n = n_{1}n_{2} and m = m_{1}m_{2}. If (m, g′) ≠ 1, or (n, f′) ≠ 1, then c_{f,g}^{r}(m_{1}m_{2}, n_{1}n_{2}) = 0 by Lemma 4.3(i). But then, one of (m_{1}, g′), (m_{2}, g′), (n_{1}, f′), (n_{2}, f′) is not 1, and either

c_{f,g}^{r}(m_{1}, n_{1}) = 0 or c_{f,g}^{r}(m_{2}, n_{2}) = 0

by Lemma 4.3(i). This proves the lemma in this case, and we now suppose that (m, g′) = (n, f′) = 1. Using Lemma 4.3(ii), we have

c_{f,g}^{r}(m, n_{1}n_{2}) = ∑_{(r^{2}−af^{2},4m)=4^{a (4m)^{∗}}} (\frac{a}{m}) c_{f,g}^{r}(n_{1}; m, a) c_{f,g}^{r}(n_{2}; m, a)

with

c_{f,g}^{r}(n_{1}; m, a) = ∑_{(af^{′2}−b_{1}g^{′2})/4≡0 mod (mf^{′2},n_{1}g^{′2})^{(r^{2}−b_{1}g^{2},4n_{1})=4}^{b_{1}(4n_{1})^{∗}}} (\frac{b_{1}}{n_{1}} ) .

By hypothesis, (mf^{′2}, n_{1}g^{′2}) = (m_{1}f^{′2}, n_{1}g^{′2}), and c_{f,g}^{r}(n_{1}; m, a) = c_{f,g}^{r}(n_{1}; m_{1}, a_{1}) where a_{1} is the reduction of a modulo 4m_{1}. Similarly, we have c_{f,g}^{r}(n_{2}; m, a) = c_{f,g}^{r}(n_{2}; m_{2}, a_{2}) where a_{2} is the reduction of a modulo 4m_{2}. Then, applying the Generalised Chinese Reminder theorem, we have

c_{f,g}^{r}(m, n) = ∑_{(r^{2}−af^{2},4m_{1}m_{2})=4^{a (4m_{1}m_{2})^{∗}}} (\frac{a}{m_{1}m_{2}} ) c_{f,g}^{r}(n_{1}; m_{1}, a_{1}) c_{f,g}^{r}(n_{2}; m_{2}, a_{2})

= ∑_{(r^{2}−a_{1}f^{2},4m_{1})=4^{a_{1} (4m_{1})^{∗}}} (\frac{a_{1}}{m_{1}} ) c_{f,g}^{r}(n_{1}; m_{1}, a_{1}) ∑_{(r^{2}−a_{2}f^{2},4m_{2})=4^{a_{2} (4m_{2})^{∗}}} (\frac{a_{2}}{m_{2}} ) c_{f,g}^{r}(n_{2}; m_{2}, a_{2})

which proves the lemma. QED.

4.2 Bounds for the coefficients c_{f,g}^{r}(m, n)

We prove in this section that the functions c_{f,g}^{r}(m, n) satisfy the bound (20). This is the result needed to complete the proof of Theorem 3.1.

Lemma 4.5 Let p be a prime, and let α, β ≥ 0 be integers. Then,

(i) c_{f,g}^{r}(1, 1) = 1; (ii) If p ∤ fg (i.e. v_{p}(f) = v_{p}(g) = 0), then c_{f,g}^{r}(p^{α}, p^{β}) = c_{1,1}^{r}(p^{α}, p^{β}); (iii) If p | fg and v_{p}(f) = v_{p}(g), then c_{f,g}^{r}(p^{α}, p^{β}) = c_{p,p}^{r}(p^{α}, p^{β}); (iv) Suppose p | fg and v_{p}(f) ≠ v_{p}(g). If α, β ≥ 1, then c_{f,g}^{r}(p^{α}, p^{β}) = 0. If α = 0 and β ≥ 1, then c_{f,g}^{r}(p^{α}, p^{β}) = 0 when v_{p}(g) < v_{p}(f) and c_{f,g}^{r}(p^{α}, p^{β}) = c_{p}^{r}(p^{β}) when v_{p}(g) > v_{p}(f). If α ≥ 1 and β = 0, then c_{f,g}^{r}(p^{α}, p^{β}) = 0 when v_{p}(f) < v_{p}(g) and c_{f,g}^{r}(p^{α}, p^{β}) = c_{p}^{r}(p^{α}) when v_{p}(f) > v_{p}(g).

Proof: (i) By definition. (ii) By definition,

c_{1,1}^{r}(p^{α}, p^{β}) = ∑_{(r^{2}−a,4p^{α})=4^{a(4p^{α})^{∗}}} (\frac{a}{p^{α}} ) ∑_{(a−b)/4≡0 mod (p^{α},p^{β})^{(r^{2}−b,4p^{β})=4}^{b(4p^{β})^{∗}}} (\frac{b}{p^{β}} ) .

As (f, 2p) = (g, 2p) = 1, there is a bijection between the invertible residues modulo 4p^{α} (respectively 4p^{β}) and the set of af^{2} (respectively bg^{2}), where a (respectively b) runs over the set of invertible residues modulo 4p^{α} (respectively 4p^{β}). This gives

c_{1,1}^{r}(p^{α}, p^{β}) = ∑_{(r^{2}−af^{2},4p^{α})=4^{a(4p^{α})^{∗}}} (\frac{af^{2}}{p^{α}} ) ∑_{(af^{2}−bg^{2})/4≡0 mod (p^{α},p^{β})^{(r^{2}−bg^{2},4p^{β})=4}^{b(4p^{β})^{∗}}} (\frac{bg^{2}}{p^{β}} ) .

As

(af^{2} − bg^{2})/4 ≡ 0 mod (p^{α}, p^{β}) ⟺ (af^{2} − bg^{2})/4 ≡ 0 mod (p^{α}f^{2}, p^{β}g^{2})

and

(\frac{af^{2}}{p^{α}} ) = (\frac{a}{p^{α}} ) , (\frac{bg^{2}}{p^{β}} ) = (\frac{b}{p^{β}} )

we get that c_{1,1}^{r}(p^{α}, p^{β}) = c_{f,g}^{r}(p^{α}, p^{β}). (iii) As p | fg, and v_{p}(f) = v_{p}(g), p is odd, and we have

(af^{2} − bg^{2})/4 ≡ 0 mod (p^{α}f^{2}, p^{β}g^{2}) ⟺ af^{′2} ≡ bg^{′2} mod (p^{α}, p^{β}).

Let h = f^{′−2}g^{′2} modulo 4p^{β}. Then, there is a bijection between the set of invertible residues b modulo 4p^{β} and the set of hb, where b runs over the invertible residues b modulo 4p^{β}. Then,

c_{p,p}^{r}(p^{α}, p^{β}) = ∑_{(r^{2}−ap^{2},4p^{α})=4^{a(4p^{α})^{∗}}} (\frac{a}{p^{α}} ) ∑_{a≡b mod (p^{α},p^{β})^{(r^{2}−bp^{2},4p^{β})=4}^{b(4p^{β})^{∗}}} (\frac{b}{p^{β}} )

= ∑_{(r^{2}−ap^{2},4p^{α})=4^{a(4p^{α})^{∗}}} (\frac{a}{p^{α}} ) ∑_{a≡hb mod (p^{α},p^{β})^{(r^{2}−hbp^{2},4p^{β})=4}^{b(4p^{β})^{∗}}} (\frac{bh}{p^{β}}) .

As (r^{2} − ap^{2}, 4p^{α}) = 4 if and only if (r^{2} − af^{2}, 4p^{α}) = 4, (r^{2} − hbp^{2}, 4p^{β}) = 4 if and only if (r^{2} − bg^{2}, 4p^{β}) = 4, and

(\frac{bh}{p^{β}}) = (\frac{b}{p^{β}} )

we get that c_{p,p}^{r}(p^{α}, p^{β}) = c_{f,g}^{r}(p^{α}, p^{β}). (iv) Suppose that p|fg, and v_{p}(f) ≠ v_{p}(g). If α, β ≥ 1, then one of (p^{α}, g′) or (p^{β}, f′) is divisible by p. Then, c_{f,g}^{r}(p^{α}, p^{β}) = 0 by Lemma 4.3(i).

If α = 0, β ≥ 1 and v_{p}(f) > v_{p}(g), then (p^{β}, f′) is divisible by p and c_{f,g}^{r}(p^{α}, p^{β}) = 0 by Lemma 4.3(i). If α = 0, β ≥ 1 and v_{p}(f) < v_{p}(g), then

c_{f,g}^{r}(1, p^{β}) = ∑_{(af^{′2}−bg^{′2})/4≡0 mod (f^{′2},p^{β}g^{′2})^{(r^{2}−bg^{2},4p^{β})=4}^{b(4p^{β})^{∗}}} (\frac{b}{p^{β}} )

is equal to c_{g}(p^{β}) as (f^{′2}, p^{β}g^{′2}) = 1. Finally, from [DP, Lemma 3.3(3)], c_{g}^{r}(p^{β}) = c_{p}^{r}(p^{β}). The proof is similar for α ≥ 1, β = 0 and v_{p}(f) ≠ v_{p}(g). QED.

Lemma 4.6 Let α ≥ 0.

(i) For p odd,

\frac{c_{1}^{r}(p^{α})}{p^{α−1}} = \{ \begin{matrix}− (\frac{r^{2}}{p} ) & when α is odd; \\ p − 1 − (\frac{r^{2}}{p} ) & when α is even. \end{matrix}

(ii) For p odd,

\frac{c_{p}^{r}(p^{α})}{p^{α−1}} = \{ \begin{matrix}0 & when p | r; \\ p − 1 & when α is even and p ∤ r; \\ 0 & when α is odd and p ∤ r. \end{matrix}

(iii)

\frac{c_{1}^{r}(2^{α})}{2^{α−1}} = (−1)^{α}.

Proof: This is [DP, Lemma 3.3]. QED.

Lemma 4.7 Let α, β ≥ 0, not both 0.

(i) For p odd,

\frac{c_{1,1}^{r}(p^{α}, p^{β})}{p^{max (α,β)−1}} = \{ \begin{matrix}− (\frac{r^{2}}{p} ) & when α + β is odd; \\ p − 1 − (\frac{r^{2}}{p} ) & when α + β is even. \end{matrix}

(ii) For p odd,

\frac{c_{p,p}^{r}(p^{α}, p^{β})}{p^{max (α,β)−1}} = \{ \begin{matrix}0 & when p | r; \\ p − 1 & when α + β is even and p ∤ r; \\ 0 & when α + β is odd and p ∤ r. \end{matrix}

(iii)

\frac{c_{1,1}^{r}(2^{α}, 2^{β})}{2^{max (α,β)−1}} = (−1)^{α+β}.

Proof: (i) If α = 0, then c_{1,1}^{r}(1, p^{β}) = c_{1}^{r}(p^{β}), and the result follows from Lemma 4.6(i). Similarly for β = 0. We then suppose that α, β ≥ 1, and without loss of generality that α ≤ β. As p is odd, we have

c_{1,1}^{r}(p^{α}, p^{β}) = ∑_{(r^{2}−a,p)=1^{a (p^{α})^{∗}}} (\frac{a}{p})^{α} ∑_{b≡a mod p^{α}^{(r^{2}−b,p)=1}^{b(p^{β})^{∗}}} (\frac{b}{p})^{β}

= p^{β−α} ∑_{(r^{2}−a,p)=1^{a (p^{α})^{∗}}} (\frac{a}{p})^{α+β} = p^{β−1} ∑_{a≢r^{2} mod p^{a (p)^{∗}}} (\frac{a}{p})^{α+β} .

This proves (i). (ii) As in (i), we can suppose that 1 ≤ α ≤ β. As p is odd, we have

c_{p,p}^{r}(p^{α}, p^{β}) = ∑_{(r^{2}−ap^{2},p)=1^{a (p^{α})^{∗}}} (\frac{a}{p})^{α} ∑_{bp^{2}≡ap^{2} mod p^{α+2}^{(r^{2}−bp^{2},p)=1}^{b(p^{β})^{∗}}} (\frac{b}{p})^{β} .

If p | r, then p | (r^{2} − ap^{2}, p), and c_{p,p}^{r}(p^{α}, p^{β}) = 0. If p ∤ r, then (r^{2} − ap^{2}, p) = 1, and

c_{p,p}^{r}(p^{α}, p^{β}) = ∑_{a (p^{α})^{∗}} (\frac{a}{p})^{α} ∑_{b≡a mod p^{α}^{b(p^{β})^{∗}}} (\frac{b}{p})^{β}

= p^{β−α} ∑_{a (p^{α})^{∗}} (\frac{a}{p})^{α+β} = p^{β−1} ∑_{a (p)^{∗}} (\frac{a}{p})^{α+β} .

This proves (ii). (iii) As above, we can suppose that 1 ≤ α ≤ β. We have

c_{1,1}^{r}(2^{α}, 2^{β}) = ∑_{(r^{2}−a,2^{α+2})=4^{a(2^{α+2})^{∗}}} (\frac{a}{2})^{α} ∑_{a≡b mod 2^{α+2}^{(r^{2}−b,2^{β+2})=4}^{b(2^{β+2})^{∗}}} (\frac{b}{2})^{β}

= 2^{β−α} ∑_{(r^{2}−a,2^{α+2})=4^{a(2^{α+2})^{∗}}} (\frac{a}{2})^{α+β}.

As the value of the character depends only on the value of a modulo 8, and as r^{2} ≡ 1 mod 8, we have

c_{1,1}^{r}(2^{α}, 2^{β}) = 2^{β−α}2^{α−1} ∑_{a≡5 mod 8^{a(8)^{∗}}} (\frac{a}{2})^{α+β} = 2^{β−1}(−1)^{α+β}.

This proves (iii). QED. Lemma 4.8 For any integers m, n ≥ 1, we have

c_{f,g}^{r}(m, n) = O (\frac{mn}{κ(mn) (m, n)}) .

Here κ(n) is the multiplicative arithmetic function generated by the identity

κ(p^{α}) = \{ \begin{matrix}p & α odd \\ 1 & α even \end{matrix}

for any prime p and any positive integer α. Proof: From Lemma 4.4, c_{f,g}^{r}(m, n) is multiplicative, i.e.

c_{f,g}^{r}(m, n) = ∏_{p|mn} c_{f,g}^{r}(p^{α(p)}, p^{β(p)}).

Let p be any prime. It then follows from Lemmas 4.5, 4.6 and 4.7 that for integers α, β ≥ 0, we have

c_{f,g}^{r}(p^{α}, p^{β}) ≪ (\frac{p^{α+β}}{κ(p^{α+β}) (p^{α}, p^{β})})

with an absolute constant. We then have

c_{f,g}^{r}(m, n) ≪ ∏_{p|mn} \frac{p^{α(p)+β(p)}}{κ(p^{α(p)+β(p)}) (p^{α(p)}, p^{β(p)})} = \frac{mn}{κ(mn) (m, n)}.

QED.

4.3 Euler Product

We compute in this section the Euler product for the constant C_{r}. We recall from Section 2 that

C_{r} = \frac{K_{r}}{π^{2}}

and from (26)

K_{r} = ∑_{{f,g=1 \atop (2r,fg)=1}}^{∞} \frac{1}{fg} ∑_{m,n=1}^{∞} \frac{c_{f,g}^{r}(m, n)}{mnϕ([mf^{2}, ng^{2}])}.

¿From Lemma 4.3(i), c_{f,g}^{r}(m, n) = 0 when (f′, n) ≠ 1 or (g′, m) ≠ 1. Now, if (f′, n) = (g′, m) = 1,

[mf^{2}, ng^{2}] = k^{2} [mf^{′2}, ng^{′2}] = k^{2} [m, n] f^{′2}g^{′2} = [m, n] [f^{2}, g^{2}] .

Using that and the identity

ϕ(ab) = \frac{ϕ(a)ϕ(b)(a, b)}{ϕ(a, b)}

we get that

K_{r} = ∑_{{f,g⩾1 \atop (2r,fg)=1}} \frac{1}{fgϕ([f^{2}, g^{2}])} ∑_{m,n=1}^{∞} \frac{c_{f,g}^{r}(m, n)ϕ([f^{2}, g^{2}], [m, n])}{mnϕ([m, n])([f^{2}, g^{2}], [m, n])}.  (27)

One can check that the function in the inside sum is a multiplicative function of m and n. For such functions, we have the following.

Lemma 4.9 (Euler product) Let F (m, n) be a multiplicative function. Then,

∑_{m,n≥1} F (m, n) = ∏_{p} ∑_{α,β≥0} F (p^{α}, p^{β}).

We then write the inside sum of (27) as

∑_{m,n=1}^{∞} \frac{c_{f,g}^{r}(m, n)ϕ([f^{2}, g^{2}], [m, n])}{mnϕ([m, n])([f^{2}, g^{2}], [m, n])} = ∏_{p} ∑_{α,β≥0} \frac{c_{f,g}^{r}(p^{α}, p^{β})ϕ([f^{2}, g^{2}], [p^{α}, p^{β}])}{p^{α}p^{β}ϕ([p^{α}, p^{β}])([f^{2}, g^{2}], [p^{α}, p^{β}])}.

We now break the product in 3 parts, depending on the p-adic valuations of f and g. We first notice that for any prime p

\frac{ϕ([f^{2}, g^{2}], [p^{α}, p^{β}])}{([f^{2}, g^{2}], [p^{α}, p^{β}])} = \{ \begin{matrix}1 & if α = β = 0; \\ 1 & if p ∤ fg; \\ \frac{p − 1}{p} & p | fg, α, β not both 0. \end{matrix}

Then, using Lemma 4.5, we can rewrite the last product as

∏_{p∤fg} ∑_{α,β≥0} \frac{c_{1,1}^{r}(p^{α}, p^{β})}{p^{α}p^{β}ϕ([p^{α}, p^{β}])} ∏_{v_{p}(f)=v_{p}(g)^{p|fg}} (1 + \frac{p − 1}{p} ∑_{(α,β)≠(0,0)^{α,β≥0}} \frac{c_{p,p}^{r}(p^{α}, p^{β})}{p^{α}p^{β}ϕ([p^{α}, p^{β}])})

× ∏_{v_{p}(f)<v_{p}(g)^{p|fg}} (1 + \frac{p − 1}{p} ∑_{β>0} \frac{c_{p}^{r}(p^{β})}{p^{β}ϕ(p^{β})}) ∏_{v_{p}(f)>v_{p}(g)^{p|fg}} (1 + \frac{p − 1}{p} ∑_{α>0} \frac{c_{p}^{r}(p^{α})}{p^{α}ϕ(p^{α})})

= ∏_{p} E_{1}(p) ∏_{v_{p}(f)=v_{p}(g)^{p|fg}} \frac{E_{2}(p)}{E_{1}(p)} ∏_{v_{p}(f)<v_{p}(g)^{p|fg}} \frac{E_{3}(p)}{E_{1}(p)} ∏_{v_{p}(f)>v_{p}(g)^{p|fg}} \frac{E_{3}(p)}{E_{1}(p)}

where

E_{1}(p) = ∑_{α,β≥0} \frac{c_{1,1}^{r}(p^{α}, p^{β})}{p^{α}p^{β}ϕ([p^{α}, p^{β}])}

E_{2}(p) = 1 + \frac{p − 1}{p} ∑_{(α,β)≠(0,0)^{α,β≥0}} \frac{c_{p,p}^{r}(p^{α}, p^{β})}{p^{α}p^{β}ϕ([p^{α}, p^{β}])}

E_{3}(p) = 1 + \frac{p − 1}{p} ∑_{β>0} \frac{c_{p}^{r}(p^{β})}{p^{β}ϕ(p^{β})} .

Replacing the last equation in (27), we get

K_{r} = ∏_{p} E_{1}(p) ∑_{{f,g⩾1 \atop (2r,fg)=1}} \frac{1}{fgϕ([f^{2}, g^{2}])} ∏_{v_{p}(f)=v_{p}(g)^{p|fg}} \frac{E_{2}(p)}{E_{1}(p)} ∏_{v_{p}(f)<v_{p}(g)^{p|fg}} \frac{E_{3}(p)}{E_{1}(p)} ∏_{v_{p}(f)>v_{p}(g)^{p|fg}} \frac{E_{3}(p)}{E_{1}(p)}.

One can check that the function

F (f, g) = \frac{1}{fgϕ([f^{2}, g^{2}])} ∏_{v_{p}(f)=v_{p}(g)^{p|fg}} \frac{E_{2}(p)}{E_{1}(p)} ∏_{v_{p}(f)<v_{p}(g)^{p|fg}} \frac{E_{3}(p)}{E_{1}(p)} ∏_{v_{p}(f)>v_{p}(g)^{p|fg}} \frac{E_{3}(p)}{E_{1}(p)}

is a multiplicative function of f and g. We compute F (1, 1) = 1, and for γ, δ ≥ 0 not both 0

F (p^{γ}, p^{δ}) = \{ \begin{matrix}\frac{1}{p^{γ}p^{δ}ϕ([p^{2γ}, p^{2δ}])} \frac{E_{2}(p)}{E_{1}(p)} & if γ = δ; \\ \frac{1}{p^{γ}p^{δ}ϕ([p^{2γ}, p^{2δ}])} \frac{E_{3}(p)}{E_{1}(p)} & if γ < δ; \\ \frac{1}{p^{γ}p^{δ}ϕ([p^{2γ}, p^{2δ}])} \frac{E_{3}(p)}{E_{1}(p)} & if γ > δ. \end{matrix}

Using Lemma 4.9, this gives

K_{r} = ∏_{p} E_{1}(p) ∏_{p∤2r} ∑_{γ,δ≥0} F (p^{γ}, p^{δ})

= ∏_{p|2r} E_{1}(p) ∏_{p∤2r} (E_{1}(p) + E_{2}(p) ∑_{γ≥1} \frac{1}{p^{2γ}ϕ(p^{2γ})} + 2E_{3}(p) ∑_{γ<δ^{γ,δ≥0}} \frac{1}{p^{γ}p^{δ}ϕ([p^{2γ}, p^{2δ}])}) .

One computes

E_{1}(2) = \frac{4}{9}
E_{1}(p) = \frac{p^{2}(p^{2} + 1)}{(p^{2} − 1)^{2}} for p | r
E_{1}(p) = \frac{p^{5} − p^{4} − p^{3} − 4p^{2} + 1}{(p − 1)^{3}(p + 1)^{2}} for p ∤ 2r
E_{2}(p) = \frac{p^{4} + p^{3} + 2p^{2} − p − 1}{p(p − 1)(p + 1)^{2}} for p ∤ 2r
E_{3}(p) = 1 + \frac{1}{p(p + 1)} = \frac{p^{2} + p + 1}{p(p + 1)}
∑_{γ≥1} \frac{1}{p^{2γ}ϕ(p^{2γ})} = \frac{p}{(p − 1)(p^{4} − 1)}
∑_{γ<δ^{γ,δ≥0}} \frac{1}{p^{γ}p^{δ}ϕ([p^{2γ}, p^{2δ}])} = (\frac{p}{p − 1})^{2} (\frac{1}{p^{3} − 1} − \frac{1}{p^{4} − 1}) = \frac{p^{5}}{(p^{4} − 1)(p^{3} − 1)(p − 1)}.

Replacing in the last expression for K_{r}, this gives

K_{r} = \frac{4}{9} ∏_{p|r} \frac{p^{2}(p^{2} + 1)}{(p^{2} − 1)^{2}} ∏_{p∤2r} \frac{p^{2}(p^{4} − 2p^{2} − 3p − 1)}{(p + 1)^{3}(p − 1)^{3}}
= 3 ∏_{p|r} \frac{p^{2}(p^{2} + 1)}{(p^{2} − 1)^{2}} ∏_{p∤r} \frac{p^{2}(p^{4} − 2p^{2} − 3p − 1)}{(p + 1)^{3}(p − 1)^{3}}

and finally

C_{r} = \frac{3}{π^{2}} ∏_{p|r} \frac{p^{2}(p^{2} + 1)}{(p^{2} − 1)^{2}} ∏_{p∤r} \frac{p^{2}(p^{4} − 2p^{2} − 3p − 1)}{(p + 1)^{3}(p − 1)^{3}}.

5 The supersingular case

The case r = 0 was considered by Fouvry and Murty in [FM2], and we verify here that our method gives the same asymptotic. We start by considering Equation (3.1) of [FM2]

T (x) = ∑_{p≤x} \frac{h^{2}(−p)}{p^{2}} + 2 ∑_{p≤x} \frac{h(−p)h(−4p)}{p^{2}} + ∑_{p≤x} \frac{h^{2}(−4p)}{p^{2}}
= T_{1,1}(x) + 2T_{1,4}(x) + T_{4,4}(x).

Proceeding as in Section 2, we write

T_{1,1}(x) = ∑_{p∈S_{2,2}(x)} \frac{L(1, χ_{−p})L(1, χ_{−p})}{p}

T_{1,4}(x) = 2 ∑_{p∈S_{2,1}(x)} \frac{L(1, χ_{−p})L(1, χ_{−4p})}{p}

T_{4,4}(x) = 4 ∑_{p∈S_{1,1}(x)} \frac{L(1, χ_{−4p})L(1, χ_{−4p})}{p}.

We replace 1/p by log p in the above sums, and we call the corresponding new sums T̂_{i,j}(x). One can easily get the asymptotic for T from T̂ by partial summation as in Section 2. We now calculate each of the sums T̂_{i,j}(x). Proceeding as in Section 3, we get

T̂_{1,1}(x) ∼ ( ∑_{m,n=1}^{∞} \frac{c_{2,2}^{0}(m, n)}{mnϕ([4m, 4n])}) x

where

c_{2,2}^{0}(m, n) = ∑_{{a(4m)^{∗} \atop a≡1 mod 4}} (\frac{a}{m}) ∑_{b(4n)^{∗} ; {b≡1 mod 4 \atop (a−b)/4≡0 mod (m,n)}} (\frac{b}{n}) .

When m and n are odd, we have

c_{2,2}^{0}(m, n) = ∑_{a(m)^{∗}} (\frac{a}{m}) ∑_{{b(n)^{∗} \atop a≡b mod (m,n)}} (\frac{b}{n})

and for 1 ≤ α ≤ β, we have

c_{2,2}^{0}(2^{α}, 2^{β}) = 2^{β−1}(1 + (−1)^{α+β}).

Using these and following the arguments of Section 4, we get the Euler product

∑_{m,n=1}^{∞} \frac{c_{2,2}^{0}(m, n)}{mnϕ([4m, 4n])} = \frac{1}{2} ∏_{p} \frac{1 + 1/p^{2}}{(1 − 1/p^{2})^{2}} = \frac{5π^{2}}{24} .

Proceeding similarly, we get

T̂_{1,4}(x) ∼ (2 ∑_{{m,n=1 \atop n odd}}^{∞} \frac{c_{2,1}^{0}(m, n)}{mnϕ([4m, n])}) x

where

c_{2,1}^{0}(m, n) = ∑_{{a(4m)^{∗} \atop a≡1 mod 4}} (\frac{a}{m}) ∑_{{b(n)^{∗} \atop a≡b mod (4m,n)}} (\frac{b}{n}) .

When m is odd, we have

c_{2,1}^{0}(m, n) = ∑_{a(m)^{∗}} (\frac{a}{m}) ∑_{{b(n)^{∗} \atop a≡b mod (m,n)}} (\frac{b}{n})

and for α ≥ 1, we have

c_{2,1}^{0}(2^{α}, 1) = 2^{α−1}(−1)^{α}.

Using these and following the arguments of Section 4, we get the Euler product

2 ∑_{{m,n=1 \atop n odd}}^{∞} \frac{c_{2,1}^{0}(m, n)}{mnϕ([4m, n])} = 2 (\frac{1}{2}) (\frac{1}{1 − 1/2^{2}} ) ∏_{p≥3} \frac{1 + 1/p^{2}}{(1 − 1/p^{2})^{2}} = \frac{π^{2}}{4} .

Proceeding in the same way, we get

T̂_{4,4}(x) ∼ (4 ∑_{{m,n=1 \atop m,n odd}}^{∞} \frac{c_{1,1}^{0}(m, n)}{mnϕ([m, n])}) x

where

c_{1,1}^{0}(m, n) = ∑_{a(m)^{∗}} (\frac{a}{m}) ∑_{{b(n)^{∗} \atop a≡b mod (m,n)}} (\frac{b}{n}) .

Here m and n are odd, and we have

4 ∑_{{m,n=1 \atop m,n odd}}^{∞} \frac{c_{1,1}^{0}(m, n)}{mnϕ([m, n])} = 4 ∏_{p≥3} \frac{1 + 1/p^{2}}{(1 − 1/p^{2})^{2}} = \frac{3π^{2}}{4} .

Finally, putting the last 3 estimates together, we get

T (x) ∼ (\frac{5}{24} + \frac{1}{2} + \frac{3}{4}) \frac{x}{log x} = \frac{35}{24} \frac{x}{log x}

and then Theorem 1.2 also holds for r = 0 with C_{0} = 35/96. This is the result obtained by Fouvry and Murty in [FM2].

References

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Amir Akbary, Department of Mathematics and Computer Science, University of Lethbridge, 4401 University Drive West, Lethbridge, Alberta, T1K 3M4, CANADA E–mail: akbary@cs.uleth.ca

Chantal David, Department of Mathematics and Statistics, Concordia University, 1455 de Maisonneuve Blvd. West, Montréal, Quebec, H3G 1M8, CANADA E–mail: cdavid@mathstat.concordia.ca

Robert Juricevic, Department of Pure Mathematics, University of Waterloo, Waterloo, Ontario, N2L 3G1, CANADA E–mail: rjuricevic@math.uwaterloo.ca