2 Background and two CM-forms

Let k and N be positive integers and $\chi $ a Dirichlet character modulo N. We denote by $M_k(\Gamma _0(N),\chi )$ the space of holomorphic modular forms of weight k on $\Gamma _0(N)$ with Nebentypus $\chi $ , and by $S_k(\Gamma _0(N),\chi )$ the subspace of cusp forms (we suppress $\chi $ when it is the trivial character). Denote by $\chi _{23}$ the character defined by $\chi _{23}( \bullet ) = \big(\frac{23}{\bullet}\big)$ and write $q := e^{2 \pi i z}$ for complex z with $\mathrm {Im}(z)>0$ .

We begin by stating a slight refinement of Theorem 2.1 in [Reference Penniston5] which implies that the values of $b_{23}(n)$ with $11 \nmid n$ can be realised modulo $11$ as the Fourier coefficients of a modular form of integer weight. (Since this result can be established in the same way as Proposition 2.1 in [Reference Penniston6], we do not include the proof.)

Proposition 2.1. There exists a cusp form

$$ \begin{align*} H(z) \in S_{1330}(\Gamma_0(144 \cdot 11^3 \cdot 23),\chi_{23}) \cap {\mathbb Z}[[q]] \end{align*} $$

such that

$$ \begin{align*} H(z) \equiv \sum_{n=0}^{\infty} \bigg( \frac{n}{11} \bigg) b_{23}(n) q^{12n+11} \pmod{11}. \end{align*} $$

Following closely the exposition in [Reference Serre7], we now construct two normalised Hecke eigenforms with complex multiplication that turn out to be related to $H(z)$ modulo $11$ . Let $K = {\mathbb Q}( \sqrt {-69} )$ , a number field of discriminant $d=-276$ , and define the ideal ${\mathcal F}$ of the ring of integers ${\mathcal O}_K = {\mathbb Z}[\sqrt {-69}]$ by

$$ \begin{align*} {\mathcal F} = ( 6, 2 \sqrt{-69} ). \end{align*} $$

We note that $N({\mathcal F})=12$ , and from now on, we write $\theta := \sqrt {-69}$ . One can verify that

$$ \begin{align*} \{ 1,5,2+\theta,4+\theta \} \end{align*} $$

is a set of representatives of $( {\mathcal O}_K / {\mathcal F})^*$ , and since $( {\mathcal O}_K / {\mathcal F})^* \cong {\mathbb Z}_2 \times {\mathbb Z}_2$ , it follows that the function $\rho : ( {\mathcal O}_K / {\mathcal F})^* \to {\mathbb C}^*$ defined by

$$ \begin{align*} \rho(1)=1 , \quad \rho(5)=-1, \quad \rho(2+\theta)=1, \quad \rho(4+\theta)=-1 \end{align*} $$

is a group homomorphism.

Next, we define Hecke characters $c_1$ and $c_2$ on the fractional ideals of K coprime to ${\mathcal F}$ . We begin by defining $c_1$ on principal ideals of ${\mathcal O}_K$ via

(2.1) $$ \begin{align} c_1((a+b \theta)) = \rho(a+b \theta) (a+b \theta)^{49} \end{align} $$

for $a,b \in {\mathbb Z}$ . Since ${\mathcal O}_K$ has class number eight, we must extend $c_1$ . One can check that the class group is isomorphic to ${\mathbb Z}_2 \oplus {\mathbb Z}_4$ and is generated by the nonprincipal ideals $(5,1+\theta )$ and $(23,\theta )$ which satisfy $(5,1+\theta )^4 = (2-3\theta )$ and $(23,\theta )^2=(23)$ . Letting

(2.2) $$ \begin{align} c_1((5,1+\theta)) = \bigg( \frac{1}{2}\bigg(\sqrt{10+3\sqrt{6}} - i \sqrt{10-3\sqrt{6}}\bigg) \bigg)^{49} \end{align} $$

and

(2.3) $$ \begin{align} c_1((23,\theta)) = ( \sqrt{-23})^{49} = 23^{24} \sqrt{-23} \end{align} $$

allows us to complete $c_1$ to a Hecke character of K with exponent $49$ and conductor ${\mathcal F}$ , which in turn yields a Dirichlet character $\omega _1$ defined by

$$ \begin{align*} \omega_1(n) = c_1( n {\mathcal O}_K ) / n^{49} \end{align*} $$

for $n \in {\mathbb Z}$ coprime to $6$ . We define companion characters $c_2$ and $\omega _2$ similarly, with $c_2$ extending to nonprincipal ideals via

$$ \begin{align*} c_2((5,1+\theta)) = i c_1((5,1+\theta)) \end{align*} $$

and

$$ \begin{align*} c_2((23,\theta)) = - c_1((23,\theta)). \end{align*} $$

For $v \in \{1,2\}$ , let

$$ \begin{align*} g_{v}(z) = \sum c_{v}({\mathcal I}) q^{N({\mathcal I})} : = \sum_{n=1}^{\infty} s_{v}(n) q^n, \end{align*} $$

where the first sum is over all ideals ${\mathcal I}$ of ${\mathcal O}_K$ coprime to ${\mathcal F}$ . Then $g_v$ is a normalised Hecke eigenform of weight $50$ , level $|d| \cdot N({\mathcal F}) = 144 \cdot 23$ and character $\epsilon _K \omega _v$ , where $\epsilon _K(p) = \big(\frac{-69}{p}\big)$ for any prime $p \not \in \{2,3,23\}$ . Note that for such a prime, we have $\omega _v(p) = \rho (p) = \big(\frac{p}{3}\big)$ . Then since $\big(\frac{-69}{p}\big) \big(\frac{p}{3}\big) = \big(\frac{23}{p}\big)$ by quadratic reciprocity, it follows that

$$ \begin{align*} g_v(z) \in S_{50}( \Gamma_0(144 \cdot 23),\chi_{23}). \end{align*} $$

The fact that $g_v(z)$ is an eigenform implies that

(2.4) $$ \begin{align} s_v(mn)=s_v(m)s_v(n) \quad\mbox{when } (m,n)=1 \end{align} $$

and

(2.5) $$ \begin{align} s_v(p^{\, j+1}) = s_v(p) s_v(p^{\,j}) - \chi_{23}(p) p^{49} s_v(p^{\,j-1}) \end{align} $$

for all primes $p \geq 5$ and $j \geq 1$ .

3 Proofs of the main results

We begin by finding explicit expressions for the Fourier coefficients $s_v(p)$ of the CM-forms $g_v(z)$ which can be leveraged to evaluate the coefficients of two auxiliary forms (see $h_v(z)$ below) modulo $11$ . An odd prime p satisfying $\big(\frac{-69}{p}\big)=-1$ is inert in K, which implies that $s_v(p)=0$ . Since $23$ ramifies in K, we have $s_v(23)=c_v((23,\theta ))$ . Now suppose p is an odd prime with $\big(\frac{-69}{p}\big)=1$ and that $x,y \in {\mathbb Z}$ are as in (1.1). Define $\delta _p$ by

(3.1) $$ \begin{align} \delta_p = \begin{cases} \rho(x+y \theta) & \mathrm{if}\ (a,b,c)=(1,0,69), \\ \rho((9x+16y)-(x-y) \theta) & \mathrm{if}\ (a,b,c)=(6,6,13), \\ \rho((5x+y)-y \theta) & \mathrm{if}\ (a,b,c)=(5,2,14), \\ \rho((7x+y)+y \theta) & \mathrm{if}\ (a,b,c)=(7,2,10), \\ \rho(23y+x\theta) & \mathrm{if}\ (a,b,c)=(3,0,23), \\ \rho((5x-32y)+(x+3y)\theta) & \mathrm{if}\ (a,b,c)=(2,2,35), \end{cases} \end{align} $$

and for $n \geq 1$ , let

$$ \begin{align*} t_1(n) = \begin{cases} s_1(n) & \mathrm{if}\ n \equiv 1 \pmod{12}, \\ s_1(n) / i\sqrt{10 - 3\sqrt{6}} & \mathrm{if}\ n \equiv 5 \pmod{12}, \\ s_1(n) / \sqrt{14-5\sqrt{6}} & \mathrm{if}\ n \equiv 7 \pmod{12}, \\ s_1(n) / \sqrt{-23} & \mathrm{if}\ n \equiv 11 \pmod{12}. \end{cases} \end{align*} $$

We define $t_2(n)$ in the same way as $t_1(n)$ except that we divide $s_2(n)$ by $i\sqrt {10 + 3\sqrt {6}}$ when $n \equiv 5 \pmod {12}$ , and by $\sqrt {14 + 5\sqrt {6}}$ when $n \equiv 7 \pmod {12}$ . Our next result (in the proof of which we will see that $t_1(n) \in {\mathbb Z}[\sqrt {6}]$ ) gives expressions for $t_1(p)$ modulo $11$ in terms of the six reduced binary quadratic forms of discriminant $-276$ mentioned in the introduction. Recall that $\theta = \sqrt {-69}$ .

Theorem 3.1. Suppose p is an odd prime with $\big(\frac{-69}{p}\big)=1$ . Let $x,y \in {\mathbb Z}$ be defined by (1.1), replacing x and y by $-x$ and $-y$ if necessary so that $-x-2y+ \sqrt {6}y>0$ if $(a,b,c)=(5,2,14)$ and $x-2y - \sqrt {6}y>0$ if $(a,b,c)=(7,2,10)$ . Then, modulo the ideal of ${\mathbb Z}[\sqrt {6}]$ generated by $11$ ,

$$ \begin{align*} \delta_p t_1(p) \equiv \begin{cases} 2xp^4 & \mathrm{if}\ (a,b,c)=(1,0,69), \\ (2x+y)p^4 \cdot \sqrt{6} & \mathrm{if}\ (a,b,c)=(6,6,13), \\ (x+2y+\sqrt{6}y) p^4 & \mathrm{if}\ (a,b,c)=(5,2,14), \\ (x-2y-\sqrt{6}y) p^4 & \mathrm{if}\ (a,b,c)=(7,2,10), \\ 9yp^4 & \mathrm{if}\ (a,b,c)=(3,0,23), \\ yp^4 \cdot \sqrt{6} & \mathrm{if}\ (a,b,c)=(2,2,35). \end{cases} \end{align*} $$

Proof. For the sake of brevity, we provide details for only two of the cases, as the others can be handled similarly.

For $(a,b,c)=(2,2,35)$ , we begin with the ideal calculation

$$ \begin{align*} (5,1+\theta)^2 (23,\theta)(47,5+\theta) = (46-19\theta). \end{align*} $$

Combining this with (2.1), (2.2), (2.3) and the multiplicativity of $c_1$ yields

$$ \begin{align*} c_1((47,5+\theta)) = \bigg( \frac{5+\theta}{\sqrt{2}} \bigg)^{49}. \end{align*} $$

Now suppose that $p>47$ is a prime with $p=2x^2+2xy+35y^2$ for some $x,y \in {\mathbb Z}$ , and denote by ${\mathfrak P}$ and ${\mathfrak P}'$ the prime ideals of ${\mathcal O}_K$ lying above p. It is easy to verify that if $a+b \theta \in (47,5+\theta )$ for some $a,b \in {\mathbb Z}$ , then $a \equiv 5b \pmod {47}$ . Using this and the equalities

$$ \begin{align*} 47p = (5x-32y)^2+69(x+3y)^2 = (5x+37y)^2+69(x-2y)^2, \end{align*} $$

we find that the elements

$$ \begin{align*} \mu = (5x-32y)+(x+3y)\theta \quad \mathrm{and} \quad \nu = (5x+37y)+(x-2y)\theta \end{align*} $$

generate the principal ideals $(47,5+\theta ) {\mathfrak P}$ and $(47,5+\theta ){\mathfrak P}'$ . Since

$$ \begin{align*} \mu \nu = -44x^2-44xy-770y^2+(10x^2+10xy+175y^2)\theta , \end{align*} $$

it follows that

$$ \begin{align*} \rho(\,\mu)\rho(\nu) = \rho(\,\mu \nu) = \rho(4+\theta)=-1, \end{align*} $$

and hence $s_1(p) = c_1({\mathfrak P})+c_1({\mathfrak P}')$ is equal to

$$ \begin{align*} \rho(\,\mu) \bigg[ \bigg( \sqrt{2}x + \frac{\sqrt{2}}{2} y + \frac{y}{2} \sqrt{-138} \bigg)^{49} - \bigg( \sqrt{2}x + \frac{\sqrt{2}}{2} y - \frac{y}{2} \sqrt{-138} \bigg)^{49} \bigg]. \end{align*} $$

A calculation now shows that $s_1(p)$ has the form $\rho (\,\mu ) yC(x,y) \sqrt {-138}$ , where $C(x,y) \in {\mathbb Z}[x,y]$ is a homogeneous polynomial of degree $48$ . Using Fermat’s Little Theorem we find that $yC(x,y)$ is congruent modulo $11$ to $y(2x^2+2xy+35y^2)^4$ , and since $p \equiv 11 \pmod {12}$ , our result follows.

Now let $(a,b,c)=(7,2,10)$ . We begin by using the equality

$$ \begin{align*} (5,1+\theta)(23,\theta)(7,6+\theta) = (23-2\theta) \end{align*} $$

as above to deduce that

$$ \begin{align*} c_1((7,6+\theta)) = \bigg( \frac{1}{2}\bigg(\sqrt{14-5\sqrt{6}} + i \sqrt{14+5\sqrt{6}}\bigg) \bigg)^{49}. \end{align*} $$

Combining this with

$$ \begin{align*} c_1((7,6+\theta))c_1((7,1+\theta)) = c_1((7)) = \rho(7) \cdot 7^{49} = 7^{49} \end{align*} $$

then gives

$$ \begin{align*} c_1((7,1+\theta)) = \bigg( \frac{1}{2}\bigg(\sqrt{14-5\sqrt{6}} - i \sqrt{14+5\sqrt{6}}\bigg) \bigg)^{49}. \end{align*} $$

Next suppose that $p>7$ is a prime with $p=7x^2+2xy+10y^2$ . Then $7p = (7x+y)^2 + 69y^2$ , and by replacing x by $-x$ and y by $-y$ if necessary we may ensure that $x-2y-\sqrt {6}y>0$ . Let z be an integer such that

$$ \begin{align*} z(7x+y) \equiv -69y \pmod{p}. \end{align*} $$

One can verify that $z^2 \equiv -69 \pmod {p}$ (which implies that $(p, \pm z+\theta )$ are the prime ideals of ${\mathcal O}_K$ above p) and

$$ \begin{align*} (7,1+\theta)(p,z+\theta) = ((7x+y)+y\theta). \end{align*} $$

It follows that

$$ \begin{align*} c_1((p,& \,z+\theta)) = \rho((7x+y)+y\theta)) \\ & \cdot\bigg( \frac{1}{2}\bigg(\sqrt{2p+(-5x^2-8xy+6y^2)\sqrt{6}} + s \cdot \sqrt{-2p+(-5x^2-8xy+6y^2)\sqrt{6}}\bigg) \bigg)^{49}, \end{align*} $$

where $s = \mathrm {sign}(x^2-4xy-2y^2)$ .

Since $p \equiv 7 \pmod {12}$ ,

$$ \begin{align*} c_1((p,z+\theta))c_1((p,-z+\theta)) = c_1((p)) = \rho(p) p^{49} = p^{49}. \end{align*} $$

Hence, $c_1((p,-z+\theta )) = \overline {c_1((p,z+\theta ))}$ , and a calculation reveals that

(3.2) $$ \begin{align} s_1(p) = \sqrt{14-5 \sqrt{6}} \cdot \rho((7x+y)+y \theta)(x-2y-\sqrt{6}y) D(x,y) \end{align} $$

with $D(x,y) \in {\mathbb Z}[\sqrt {6}][x,y]$ . Our result now follows as in the previous case.

Remark 3.2. For our claim that $t_1(n) \in {\mathbb Z}[\sqrt {6}]$ , note that if $p=5x^2+2xy+14y^2$ , then $p \equiv 5 \pmod {12}$ , and an argument similar to the one given in our proof of (3.2) for primes represented by $F_{7,2,10}$ shows that

(3.3) $$ \begin{align} s_1(p) = i \sqrt{10 - 3\sqrt{6}} \cdot \rho((5x+y)-y \theta) (x+2y+\sqrt{6}y) E(x,y) \end{align} $$

for some $E(x,y) \in {\mathbb Z}[\sqrt {6}][x,y]$ . Moreover, observe that

$$ \begin{align*} i \sqrt{10 - 3\sqrt{6}} \cdot \sqrt{14-5\sqrt{6}} = (-2+\sqrt{6}) \sqrt{-23}. \end{align*} $$

Remark 3.3. The parallel result to Theorem 3.1 for $t_2(p)$ states that for each odd prime p with $\big(\frac{-69}{p}\big)=1$ ,

$$ \begin{align*} \delta_p t_2(p) \equiv \begin{cases} 2xp^4 & \mathrm{if}\ (a,b,c)=(1,0,69), \\ -(2x+y)p^4 \cdot \sqrt{6} & \mathrm{if}\ (a,b,c)=(6,6,13), \\ (-x-2y+\sqrt{6}y) p^4 & \mathrm{if}\ (a,b,c)=(5,2,14), \\ (-x+2y-\sqrt{6}y) p^4 & \mathrm{if}\ (a,b,c)=(7,2,10), \\ -9yp^4 & \mathrm{if}\ (a,b,c)=(3,0,23), \\ yp^4 \cdot \sqrt{6} & \mathrm{if}\ (a,b,c)=(2,2,35). \end{cases} \end{align*} $$

We now show that the behaviour of $b_{23}(n)$ modulo $11$ for $11 \nmid n$ is controlled by the Fourier coefficients of the CM-forms $g_v(z)$ defined in Section 2. We do this by establishing a relation between the coefficients of the form $H(z)$ from Proposition 2.1 and those of two auxiliary forms built from the $g_v(z)$ which we now define.

For $n \geq 1$ , write

$$ \begin{align*} t_1(n) = \alpha_n + \beta_n \sqrt{6} \end{align*} $$

with $\alpha _n,\beta _n \in {\mathbb Z}$ . Using Theorem 3.1 and Remark 3.3 along with (2.4) and (2.5), it is straightforward to show that $t_2(n) = - \alpha _n + \beta _n \sqrt {6}$ when $n \equiv 11 \pmod {12}$ . For each $v \in \{1,2\}$ , we define the modular form

$$ \begin{align*} h_v(z) = \sum_{n \equiv 11 \!\!\pmod{12}} t_v(n) q^n \in M_{50}(\Gamma_0(144 \cdot 23),\chi_{23}). \end{align*} $$

Next, let

$$ \begin{align*} F(z) := \frac{h_1(z)-h_2(z)}{2} = \sum_{n \equiv 11 \!\!\pmod{12}} \alpha_n q^n \end{align*} $$

and

$$ \begin{align*} G(z) := \frac{h_1(z)+h_2(z)}{2\sqrt{6}} = \sum_{n \equiv 11 \!\!\pmod{12}} \beta_n q^n. \end{align*} $$

Theorem 3.4. Suppose n is a positive integer with $11 \nmid n$ . Then

$$ \begin{align*} b_{23}(n) \equiv \bigg( \frac{n}{11} \bigg) (\alpha_{12n+11} + 9 \beta_{12n+11}) \pmod{11}. \end{align*} $$

Proof. Note first that if $H(z)$ is the modular form in Proposition 2.1, then our result is equivalent to the integral power series congruence

$$ \begin{align*} H(z) \equiv F(z) + 9G(z) \pmod{11}. \end{align*} $$

Recall the weight ten normalised Eisenstein series

$$ \begin{align*} E_{10}(z) = 1 - 264 \sum_{n=1}^{\infty} \sigma_9(n)q^n \in M_{10}(\Gamma_0(1)), \end{align*} $$

where $\sigma _9(n) = \sum _{d \mid n} d^9$ . Since $F(z)+9G(z) \in M_{50}( \Gamma _0(144 \cdot 23),\chi _{23})$ ,

$$ \begin{align*} (F(z)+9G(z)) E_{10}(z)^{128} \in M_{1330}( \Gamma_0(144 \cdot 23),\chi_{23}), \end{align*} $$

and hence both $(F(z)+9G(z)) E_{10}(z)^{128}$ and $H(z)$ lie in the space

$$ \begin{align*} M_{1330}( \Gamma_0(144 \cdot 11^3 \cdot 23),\chi_{23}). \end{align*} $$

Since $E_{10}(z) \equiv 1 \pmod {11}$ , on checking that the coefficients of the Fourier expansions of $H(z)$ and $F(z)+9G(z)$ agree modulo $11$ as far as the Sturm bound [Reference Sturm, Chudnovsky, Chudnovsky, Cohn and Nathanson8], our proof is complete.

Remark 3.5. Theorems 3.1 and 3.4, when used in conjunction with (2.4), (2.5), the value of $s_1(23)$ and the fact that $s_1(p)=0$ for odd primes p with $\big(\frac{-69}{p}\big)=-1$ , allow one to evaluate $b_{23}(n)$ modulo $11$ for any n not divisible by $11$ .

To illustrate, first let $n=197$ . Since $12n+11=5^3 \cdot 19$ , we begin by noting that

$$ \begin{align*} 5=F_{5,2,14}(-1,0) \quad \mathrm{and} \quad 19=F_{7,2,10}(-1,-1). \end{align*} $$

By (3.3) and (3.2),

$$ \begin{align*} s_1(5) = i \sqrt{10-3 \sqrt{6}} \cdot \rho(-5)(-1)E(-1,0) = - i \sqrt{10-3 \sqrt{6}} \cdot E(-1,0) \end{align*} $$

and

$$ \begin{align*} s_1(19) = \sqrt{14-5 \sqrt{6}} \cdot \rho(-8 - \theta) (1 + \sqrt{6}) D(-1,-1) = \sqrt{14-5 \sqrt{6}} \cdot (-1 - \sqrt{6}) D(-1,-1). \end{align*} $$

Then (2.5) yields

$$ \begin{align*} s_1(5^2) = (-10+3 \sqrt{6}) E(-1,0)^2 + 5^{49} \end{align*} $$

and

$$ \begin{align*} s_1(5^3) = -i \sqrt{10-3 \sqrt{6}} \cdot [ (-10+3 \sqrt{6}) E(-1,0)^3 + 2 \cdot 5^{49}E(-1,0)], \end{align*} $$

and so (2.4) allows us to conclude that $s_1(5^3 \cdot 19)$ is equal to

$$ \begin{align*} ( 4 - \sqrt{6} ) D(-1,-1) [(-10+3 \sqrt{6} ) E(-1,0)^3 + 2 \cdot 5^{49}E(-1,0) ] \sqrt{-23}. \end{align*} $$

Since $D(-1,-1) \equiv 19^4 \equiv 4 \pmod {11}$ , $E(-1,0) \equiv 5^4 \equiv 9 \pmod {11}$ and $5^{49} \equiv 9 \pmod {11}$ , it follows that

$$ \begin{align*} t_1( 5^3 \cdot 19 ) \equiv 4 + \sqrt{6} \pmod{11}. \end{align*} $$

Theorem 3.4 then yields

$$ \begin{align*} b_{23}(197) \equiv \bigg( \frac{197}{11} \bigg) (4 + 9 \cdot 1) \equiv 9 \pmod{11}, \end{align*} $$

which one may verify using the exact value $b_{23}(197)=2626664703430$ .

Now let $n=102$ . Since $12n+11=5 \cdot 13 \cdot 19$ , we proceed by noting that $13=F_{6,6,13}(0,1)$ , which gives

$$ \begin{align*} s_1(13) = t_1(13) \equiv \rho (16 + \theta) (1) (13^4) \sqrt{6} \equiv 6 \sqrt{6} \pmod{11}. \end{align*} $$

Combining this with our values of $s_1(5)$ and $s_1(19)$ , we can calculate

$$ \begin{align*} t_1(5 \cdot 13 \cdot 19) \equiv 2 + 6 \sqrt{6} \pmod{11}, \end{align*} $$

and hence Theorem 3.4 implies that

$$ \begin{align*} b_{23}(102) \equiv \bigg( \frac{102}{11} \bigg) (2 + 9 \cdot 6) \equiv 1 \pmod{11}, \end{align*} $$

which can be checked against $b_{23}(102)= 226889906$ .