A remark on algebraic curves derived from convolution sums

Abstract

Hahn (Rocky Mt. J. Math. 37:1593-1622, 2007) established three differential equations according to $\mathcal{P}(q)$, $\mathcal{E}(q)$, and $\mathcal{Q}(q)$, which allows us to obtain the values of the formulas for

$$\sum _{l+m+n=N}\tilde{\sigma }(l)\tilde{\sigma }(m)\tilde{\sigma }(n),\sum _{l+m+n=N}\hat{\sigma }(l)\hat{\sigma }(m)\tilde{\sigma }(n),$$

etc. Finally, by using the above equations, we derive the algebraic curves.

MSC:11A67.

1 Introduction

Let ℕ denote the sets of positive integers. In number theory, divisor functions are defined as

$$\begin{array}{c}{\sigma }_s(N)=\sum _{d|N}{d}^s,\sigma (N):={\sigma }_1(N)=\sum _{d|N}d,\hfill \\ {\tilde{\sigma }}_s(n)=\sum _{d|n}{(-1)}^{d-1}{d}^s,{\hat{\sigma }}_s(n)=\sum _{d|n}{(-1)}^{(n/d)-1}{d}^s\hfill \end{array}$$

for $N,s,d\in \mathbb{N}$. The convolution sum is special one of divisor functions begun by Ramanujan’s effort and expanded by many authors; e.g., see [1]. For example,

$$\sum _{m=1}^{n-1}\sigma (m)\sigma (n-m)=\frac{1}{12}(5{\sigma }_3(n)+(1-6n)\sigma (n))$$

(1)

is the result of Ramanujan and also Huard, Ou, Spearman, and Williams [[1], (3.10)]. We can see

$$\sum _{m=1}^{n-1}m\sigma (m)\sigma (n-m)=\frac{n}{24}(5{\sigma }_3(n)+(1-6n)\sigma (n)),$$

(2)

$$\sum _{m=1}^{n-1}\sigma (m){\sigma }_3(n-m)=\frac{1}{240}(21{\sigma }_5(n)+(10-30n){\sigma }_3(n)-\sigma (n))$$

(3)

in [[1], (3.11), (3.12)], respectively. Moreover, there are

$$\sum _{m<n/2}\sigma (m)\sigma (n-2m)=\frac{1}{24}(2{\sigma }_3(n)+(1-3n)\sigma (n)+8{\sigma }_3\left(\frac{n}{2}\right)+(1-6n)\sigma \left(\frac{n}{2}\right))$$

(4)

in [[1], (4.4)] and

$$\begin{array}{r}\sum _{m<n/2}{\sigma }_3(m)\sigma (n-2m)=\frac{1}{240}({\sigma }_5(n)-\sigma (n)+20{\sigma }_5\left(\frac{n}{2}\right)+(10-30n){\sigma }_3\left(\frac{n}{2}\right)),\\ \sum _{m<n/2}\sigma (m){\sigma }_3(n-2m)=\frac{1}{240}(5{\sigma }_5(n)+(10-15n){\sigma }_3(n)+16{\sigma }_5\left(\frac{n}{2}\right)-\sigma \left(\frac{n}{2}\right))\end{array}$$

(5)

in [[1], Theorem 6]. In addition, Lahiri [2] gave the value of the sum

$$\sum _{m_1+\cdots +m_r=n}{m}_1^{a_1}\cdots m_r^{a_r}{\sigma }_{b_1}(m_1)\cdots {\sigma }_{b_r}(m_r)(r\ge 3),$$

where the sum is over all positive integers $m_1,\dots ,m_r$ satisfying $m_1+\cdots +m_r=n$, $a_i\in \mathbb{N}\cup \{0\}$, and $b_i\in \mathbb{N}$. In this paper we substitute ${\hat{\sigma }}_{b_i}(m_i)$ for ${\sigma }_{b_i}(m_i)$ and obtain the formulas as Lahiri’s evaluation. So, we refer to Hahn’s paper [3]. Officially, we look into Hanh’s definition of three functions for $|q|<1$,

$$\mathcal{P}(q):=1+8\sum _{n=1}^{\mathrm{\infty }}\tilde{\sigma }(n)q^n,$$

(6)

$$\mathcal{E}(q):=1+24\sum _{n=1}^{\mathrm{\infty }}\hat{\sigma }(n)q^n,$$

(7)

$$\mathcal{Q}(q):=1-16\sum _{n=1}^{\mathrm{\infty }}{\tilde{\sigma }}_3(n)q^n.$$

(8)

Three functions (6), (7), and (8) satisfy the following differential equations:

$$q\frac{d\mathcal{P}(q)}{dq}=\frac{{\mathcal{P}}^2(q)-\mathcal{Q}(q)}{4},$$

(9)

$$q\frac{d\mathcal{E}(q)}{dq}=\frac{\mathcal{E}(q)\mathcal{P}(q)-\mathcal{Q}(q)}{2},$$

(10)

$$q\frac{d\mathcal{Q}(q)}{dq}=\mathcal{P}(q)\mathcal{Q}(q)-\mathcal{E}(q)\mathcal{Q}(q).$$

(11)

The paper is organized as follows. In Section 2, we obtain the values of the formulas for

$$\begin{array}{c}\sum _{l+m+n=N}\tilde{\sigma }(l)\tilde{\sigma }(m)\tilde{\sigma }(n),\hfill \\ \sum _{l+m+n=N}\hat{\sigma }(l)\hat{\sigma }(m)\tilde{\sigma }(n),\hfill \end{array}$$

etc. and insist on the following. Let

$$\begin{array}{rl}f(N)\in & \{\sum _{l+m+n=N}\tilde{\sigma }(l)\tilde{\sigma }(m)\tilde{\sigma }(n),\sum _{m=1}^{N-1}\hat{\sigma }(m)\hat{\sigma }(N-m),\\ \sum _{l+m+n=N}\hat{\sigma }(l)\hat{\sigma }(m)\hat{\sigma }(n),\sum _{m=1}^{N-1}\hat{\sigma }(m){\tilde{\sigma }}_3(N-m),\\ \sum _{l+m+n=N}\hat{\sigma }(l)\hat{\sigma }(m)\tilde{\sigma }(n),\sum _{m=1}^{N-1}m\hat{\sigma }(m)\hat{\sigma }(N-m)\}.\end{array}$$

If N is an odd integer and $n\in \mathbb{N}\cup \{0\}$, then there exist $u,a,b,c,d,e,g\in \mathbb{Z}$ satisfying

$$f(N)=\frac{1}{u}[a{\sigma }_5(N)+(bN+c){\sigma }_3(N)+(d{N}^2+eN+g)\sigma (N)]$$

with $a+b+c+d+e+g=0$ (see Theorem 2.9).

In Section 2, we derive the convolution sums of the restricted divisor functions. In Section 3, we obtain the algebraic curves by using the convolution sums in Section 2.

2 Convolution sum ${\sum }_{l+m+n=N}\tilde{\sigma }(l)\tilde{\sigma }(m)\tilde{\sigma }(n)$

Theorem 2.1 Let N (≥3) be any positive integer. Then we have

$$\sum _{l+m+n=N}\tilde{\sigma }(l)\tilde{\sigma }(m)\tilde{\sigma }(n)=\frac{1}{64}\{{\tilde{\sigma }}_5(N)+6(1-N){\tilde{\sigma }}_3(N)+(8N^2-12N+3)\tilde{\sigma }(N)\}.$$

Proof Multiplying $\mathcal{P}(q)$ on both sides in (9), we obtain

$${\mathcal{P}}^3(q)=4\mathcal{P}(q)\cdot q\frac{d\mathcal{P}(q)}{dq}+\mathcal{P}(q)\mathcal{Q}(q).$$

(12)

Employing the definition of $\mathcal{P}(q)$ and $\mathcal{Q}(q)$, we can rewrite (12) as

$$\begin{array}{rl}{(1+8\sum _{n=1}^{\mathrm{\infty }}\tilde{\sigma }(n)q^n)}^3=& 4(1+8\sum _{n=1}^{\mathrm{\infty }}\tilde{\sigma }(n)q^n)q(8\sum _{m=1}^{\mathrm{\infty }}m\tilde{\sigma }(m)q^{m-1})\\ +(1+8\sum _{n=1}^{\mathrm{\infty }}\tilde{\sigma }(n)q^n)(1-16\sum _{m=1}^{\mathrm{\infty }}{\tilde{\sigma }}_3(m)q^m).\end{array}$$

Thus, we have

$$\begin{array}{c}512\sum _{N=3}^{\mathrm{\infty }}\sum _{L=2}^{N-1}\sum _{l=1}^{L-1}\tilde{\sigma }(N-L)\tilde{\sigma }(L-l)\tilde{\sigma }(l)q^N\hfill \\ =-16\sum _{N=1}^{\mathrm{\infty }}\{\tilde{\sigma }(N)+{\tilde{\sigma }}_3(N)-2N\tilde{\sigma }(N)\}{q}^N-192\sum _{N=2}^{\mathrm{\infty }}\sum _{l=1}^{N-1}\tilde{\sigma }(l)\tilde{\sigma }(N-l)q^N\hfill \\ -128\sum _{N=2}^{\mathrm{\infty }}\sum _{l=1}^{N-1}\tilde{\sigma }(l){\tilde{\sigma }}_3(N-l)q^N+256\sum _{N=2}^{\mathrm{\infty }}\sum _{l=1}^{N-1}l\tilde{\sigma }(l)\tilde{\sigma }(N-l)q^N.\hfill \end{array}$$

Then we refer to

$$4\sum _{m<n}\tilde{\sigma }(m)\tilde{\sigma }(n-m)=-{\tilde{\sigma }}_3(n)+(2n-1)\tilde{\sigma }(n)$$

in [[3], (4.4)],

$$16\sum _{m<n}\tilde{\sigma }(m){\tilde{\sigma }}_3(n-m)=-{\tilde{\sigma }}_5(n)+2(n-1){\tilde{\sigma }}_3(n)+\tilde{\sigma }(n)$$

in [[3], (4.9)] and by direct calculation we get

$$\sum _{l=1}^{N-1}l\tilde{\sigma }(l)\tilde{\sigma }(N-l)=\frac{N}{2}\sum _{l=1}^{N-1}\tilde{\sigma }(l)\tilde{\sigma }(N-l).$$

This completes the proof of the theorem. □

Corollary 2.2 We obtain

$$ln|\mathcal{Q}(q)|=8\sum _{n=1}^{\mathrm{\infty }}\frac{\tilde{\sigma }(n)-3\hat{\sigma }(n)}{n}{q}^n,$$

where $\mathcal{Q}(q)=1-16{\sum }_{n=1}^{\mathrm{\infty }}{\tilde{\sigma }}_3(n)q^n$ as (8).

Proof By separating the variables in (11), we can have

$$-{\int }_0^q\frac{d\mathcal{Q}(x)}{\mathcal{Q}(x)}={\int }_0^q\frac{\mathcal{E}(x)-\mathcal{P}(x)}{x}dx.$$

Thus,

$$-ln|\mathcal{Q}(q)|={\int }_0^{q}8\sum _{n=1}^{\mathrm{\infty }}(3\hat{\sigma }(n)-\tilde{\sigma }(n))x^{n-1}dx.$$

We note that $3\hat{\sigma }(n)-\tilde{\sigma }(n)$ is positive because

$$3\hat{\sigma }(n)-\tilde{\sigma }(n)=2(\sigma (n)-\sigma (n/2))$$

according to

$${\tilde{\sigma }}_s(n)={\sigma }_s(n)-2^{s+1}{\sigma }_s(n/2)\text{and}{\hat{\sigma }}_s(n)={\sigma }_s(n)-2{\sigma }_s(n/2)$$

in [[3], (1.12), (1.13)]. Therefore, we can exchange the summation and the integral. This completes the proof of the corollary. □

We introduce the following Lemma 2.3 to deduce Theorem 2.6.

Lemma 2.3 Let N (≥2) be any positive integer. Then we have

• (a) ${\sum }_{m=1}^{N-1}\hat{\sigma }(m)\hat{\sigma }(N-m)=\frac{1}{12}({\sigma }_3(N)+4{\sigma }_3(\frac{N}{2})-\hat{\sigma }(N))$.

• (b) ${\sum }_{m=1}^{N-1}\hat{\sigma }(m){\tilde{\sigma }}_3(N-m)=-\frac{1}{48}({\tilde{\sigma }}_5(N)+2{\tilde{\sigma }}_3(N)-3\hat{\sigma }(N))$.

• (c) ${\sum }_{m=1}^{N-1}m\hat{\sigma }(m)\hat{\sigma }(N-m)=\frac{N}{24}({\sigma }_3(N)+4{\sigma }_3(\frac{N}{2})-\hat{\sigma }(N))$.

Proof (a) The proof is the same as Remark in [[3], p.13]. Also, Hahn showed that

$$36\sum _{m<n}\hat{\sigma }(m)\hat{\sigma }(n-m)=\{\begin{array}{cc}-3\hat{\sigma }(n)+3{\tilde{\sigma }}_3(n),\hfill & \text{if }n\text{ is odd},\hfill \\ -3\hat{\sigma }(n)-5{\tilde{\sigma }}_3(n)+4{\tilde{\sigma }}_3(n/2),\hfill & \text{if }n\text{ is even}\hfill \end{array}$$

in [[3], Theorem 4.2] derived by the identity ${\mathcal{E}}^2(q)=z^4{(1+x)}^2$, which is the same result (a).

• (b) The convolution sum can be written as

  $$\begin{array}{rl}\sum _{m=1}^{N-1}\hat{\sigma }(m){\tilde{\sigma }}_3(N-m)=& \sum _{m=1}^{N-1}\{\sigma (m)-2\sigma \left(\frac{m}{2}\right)\}\{{\sigma }_3(N-m)-16{\sigma }_3\left(\frac{N-m}{2}\right)\}\\ =& \sum _{m=1}^{N-1}\sigma (m){\sigma }_3(N-m)-16\sum _{m=1}^{N-1}\sigma (m){\sigma }_3\left(\frac{N-m}{2}\right)\\ -2\sum _{m=1}^{N-1}\sigma \left(\frac{m}{2}\right){\sigma }_3(N-m)+32\sum _{m=1}^{N-1}\sigma \left(\frac{m}{2}\right){\sigma }_3\left(\frac{N-m}{2}\right).\end{array}$$

Then we obtain

$$\begin{array}{r}\sum _{m=1}^{N-1}\sigma (m){\sigma }_3\left(\frac{N-m}{2}\right)=\sum _{t<N/2}\sigma (N-2t){\sigma }_3(t),\\ \sum _{m=1}^{N-1}\sigma \left(\frac{m}{2}\right){\sigma }_3(N-m)=\sum _{t<N/2}\sigma (t){\sigma }_3(N-2t),\\ \sum _{m=1}^{N-1}\sigma \left(\frac{m}{2}\right){\sigma }_3\left(\frac{N-m}{2}\right)=\sum _{t<N/2}\sigma (t){\sigma }_3(\frac{N}{2}-t).\end{array}$$

Therefore, we have proved (b) by using (3) and (5).

• (c) We obtain the proof by direct calculation of the index m. □

Corollary 2.4 Let $N=2q+1$ be an odd prime in Lemma  2.3 and let $T_i(n)={\sum }_{k=1}^n{k}^i$ ($i\ge 1$). Then we have

• (a) ${\sum }_{m=1}^{2q}\hat{\sigma }(m)\hat{\sigma }(2q+1-m)=2T_2(q)$.

• (b) ${\sum }_{m=1}^{2q}\hat{\sigma }(m){\tilde{\sigma }}_3(2q+1-m)=-2(q^2+q+1)T_2(q)$.

• (c) ${\sum }_{m=1}^{2q}m\hat{\sigma }(m)\hat{\sigma }(2q+1-m)=N{T}_2(q)$.

Proof

From Lemma 2.3(a), (b), and (c), we obtain

$$\begin{array}{r}\sum _{m=1}^{2q}\hat{\sigma }(m)\hat{\sigma }(2q+1-m)=\frac{1}{12}\{{(2q+1)}^3-(2q+1)\},\\ \sum _{m=1}^{2q}\hat{\sigma }(m){\tilde{\sigma }}_3(2q+1-m)=-\frac{1}{48}\{{(2q+1)}^5+2{(2q+1)}^3-3(2q+1)\},\\ \sum _{m=1}^{2q}m\hat{\sigma }(m)\hat{\sigma }(2q+1-m)=\frac{2q+1}{24}\{{(2q+1)}^3-(2q+1)\}.\end{array}$$

This completes the proof of the corollary. □

Example 2.5 The first eleven values of ${\sum }_{m=1}^{2q}\hat{\sigma }(m)\hat{\sigma }(2q+1-m)$ and $2T_2(q)$ are listed in Table 1.

Table 1 Examples for ${\mathbf{\sum }}_{\mathit{m}\mathbf{=}\mathbf{1}}^{\mathbf{2}\mathit{q}}\stackrel{\mathbf{ˆ}}{\mathit{\sigma }}\mathbf{(}\mathit{m}\mathbf{)}\stackrel{\mathbf{ˆ}}{\mathit{\sigma }}\mathbf{(}\mathbf{2}\mathit{q}\mathbf{+}\mathbf{1}\mathbf{-}\mathit{m}\mathbf{)}$ and $\mathbf{2}{\mathit{T}}_{\mathbf{2}}\mathbf{(}\mathit{q}\mathbf{)}$ ( $\mathbf{1}\mathbf{\le }\mathit{q}\mathbf{\le }\mathbf{11}$ )

Theorem 2.6 Let N (≥3) be any positive integer. Then we have

$$\begin{array}{rl}\sum _{l+m+n=N}\hat{\sigma }(l)\hat{\sigma }(m)\tilde{\sigma }(n)=& \frac{1}{576}\{{\tilde{\sigma }}_5(N)+4{\tilde{\sigma }}_3(N)+\tilde{\sigma }(N)-6(2N-1)\hat{\sigma }(N)\\ +6(N-1)({\sigma }_3(N)+4{\sigma }_3\left(\frac{N}{2}\right))\}.\end{array}$$

Proof Multiplying $\mathcal{E}(q)$ in (10), we obtain

$${\mathcal{E}}^2(q)\mathcal{P}(q)=\mathcal{E}(q)\mathcal{Q}(q)+2\mathcal{E}(q)\cdot q\frac{d\mathcal{E}(q)}{dq}.$$

So, we have

$$\begin{array}{c}{(1+24\sum _{n=1}^{\mathrm{\infty }}\hat{\sigma }(n)q^n)}^2(1+8\sum _{m=1}^{\mathrm{\infty }}\tilde{\sigma }(m)q^m)\hfill \\ =(1+24\sum _{n=1}^{\mathrm{\infty }}\hat{\sigma }(n)q^n)(1-16\sum _{m=1}^{\mathrm{\infty }}{\tilde{\sigma }}_3(m)q^m)\hfill \\ +2(1+24\sum _{n=1}^{\mathrm{\infty }}\hat{\sigma }(n)q^n)q(24\sum _{m=1}^{\mathrm{\infty }}m\hat{\sigma }(m)q^{m-1}).\hfill \end{array}$$

Therefore

$$\begin{array}{c}4\text{,}608\sum _{N=3}^{\mathrm{\infty }}\sum _{L=2}^{N-1}\sum _{l=1}^{L-1}\hat{\sigma }(N-L)\hat{\sigma }(L-l)\tilde{\sigma }(l)q^N\hfill \\ =-8\sum _{N=1}^{\mathrm{\infty }}\{\tilde{\sigma }(N)+2{\tilde{\sigma }}_3(N)+3\hat{\sigma }(N)-6N\hat{\sigma }(N)\}{q}^N\hfill \\ -384\sum _{N=2}^{\mathrm{\infty }}\sum _{l=1}^{N-1}\hat{\sigma }(l)\tilde{\sigma }(N-l)q^N-384\sum _{N=2}^{\mathrm{\infty }}\sum _{l=1}^{N-1}\hat{\sigma }(l){\tilde{\sigma }}_3(N-l)q^N\hfill \\ -576\sum _{N=2}^{\mathrm{\infty }}\sum _{l=1}^{N-1}\hat{\sigma }(l)\hat{\sigma }(N-l)q^N+1\text{,}152\sum _{N=2}^{\mathrm{\infty }}\sum _{l=1}^{N-1}l\hat{\sigma }(l)\hat{\sigma }(N-l)q^N.\hfill \end{array}$$

Lastly, we use Lemma 2.3. □

Theorem 2.7 Let N (≥3) be any positive integer. Then we have

$$\sum _{l+m+n=N}\hat{\sigma }(l)\hat{\sigma }(m)\hat{\sigma }(n)=\frac{1}{192}\{{\sigma }_5(N)-8{\sigma }_5\left(\frac{N}{2}\right)-2{\sigma }_3(N)-8{\sigma }_3\left(\frac{N}{2}\right)+\hat{\sigma }(N)\}.$$

Proof

Let us consider

$$\begin{array}{r}\sum _{l+m+n=N}\hat{\sigma }(l)\hat{\sigma }(m)\hat{\sigma }(n)\\ =\sum _{l=1}^{N-2}\hat{\sigma }(l)\sum _{m+n=N-l}\hat{\sigma }(m)\hat{\sigma }(n)\\ =\sum _{l=1}^{N-2}\hat{\sigma }(l)\frac{1}{12}\{{\sigma }_3(N-l)+4{\sigma }_3\left(\frac{N-l}{2}\right)-\hat{\sigma }(N-l)\}\\ =\sum _{l=1}^{N-1}\hat{\sigma }(l)\frac{1}{12}\{{\sigma }_3(N-l)+4{\sigma }_3\left(\frac{N-l}{2}\right)-\hat{\sigma }(N-l)\}\end{array}$$

(13)

by Lemma 2.3(a). Then (13) becomes

$$\begin{array}{rl}\sum _{l+m+n=N}\hat{\sigma }(l)\hat{\sigma }(m)\hat{\sigma }(n)=& \frac{1}{12}\{\sum _{l=1}^{N-1}\sigma (l){\sigma }_3(N-l)-2\sum _{l<N/2}\sigma (l){\sigma }_3(N-2l)\\ +4\sum _{t<N/2}{\sigma }_3(t)\sigma (N-2t)-8\sum _{t<N/2}\sigma (t){\sigma }_3(\frac{N}{2}-t)\\ -\sum _{l=1}^{N-1}\hat{\sigma }(l)\hat{\sigma }(N-l)\}\end{array}$$

by $\hat{\sigma }(n)=\sigma (n)-2\sigma (\frac{n}{2})$. □

Remark 2.8 Let $N=2q+1$ be an odd prime in Theorem 2.7. Then we have

$$\sum _{l+m+n=2q+1}\hat{\sigma }(l)\hat{\sigma }(m)\hat{\sigma }(n)=T_1(q)\cdot T_2(q).$$

Proof It is obvious. □

Importantly, we propose the following theorem.

Theorem 2.9 Let

$$\begin{array}{rl}f\in & \{\sum _{l+m+n=N}\tilde{\sigma }(l)\tilde{\sigma }(m)\tilde{\sigma }(n),\sum _{m=1}^{N-1}\hat{\sigma }(m)\hat{\sigma }(N-m),\\ \sum _{l+m+n=N}\hat{\sigma }(l)\hat{\sigma }(m)\hat{\sigma }(n),\sum _{m=1}^{N-1}\hat{\sigma }(m){\tilde{\sigma }}_3(N-m),\\ \sum _{l+m+n=N}\hat{\sigma }(l)\hat{\sigma }(m)\tilde{\sigma }(n),\sum _{m=1}^{N-1}m\hat{\sigma }(m)\hat{\sigma }(N-m)\}.\end{array}$$

If N is an odd integer and $n\in \mathbb{N}\cup \{0\}$, then there exist $u,a,b,c,d,e,g\in \mathbb{Z}$ satisfying

$$f(N)=\frac{1}{u}[a{\sigma }_5(N)+(bN+c){\sigma }_3(N)+(d{N}^2+eN+g)\sigma (N)]$$

with $a+b+c+d+e+g=0$.

Proof It is satisfied by Theorem 2.1, Lemma 2.3, Theorem 2.6, and Theorem 2.7. □

The expressions are shown in Table 2.

Table 2 Formulas for $\mathit{f}\mathbf{(}\mathit{N}\mathbf{)}$ with odd N

3 Algebraic curves derived from convolution sums

To obtain the result of this section, we need a general theory and it is this that we describe. Suppose that the two polynomials

$$\begin{array}{c}{f}_1(x)=c_0{x}^n+\cdots +c_{n-1}x+c_n,\hfill \\ f_2(x)=d_0{x}^m+\cdots +d_{m-1}x+d_m\hfill \end{array}$$

have common zero, say $x_0$. Then each of the equations

$$f_1(x)=x{f}_1(x)=\cdots =x^{m-1}{f}_1(x)=0=f_2(x)=x{f}_2(x)=\cdots =x^{n-1}{f}_2(x)$$

is of the form $p(x)=0$, where p is a polynomial of degree at most $m+n-1$, and as each of these equations is satisfied when $x=x_0$, the determinant of the coefficients must vanish. This $(m+n)\times (m+n)$ determinant is the resultant $R(f_1,f_2)$ of $f_1$ and $g_1$ and

$$R(f_1,f_2)=\left|\begin{array}{cccc}{c}_0& \dots & \dots & c_n\\ \ddots & \ddots \\ c_0& \dots & \dots & c_n\\ d_0& \dots & \dots & d_m\\ \ddots & \ddots \\ d_0& \dots & \dots & d_m\end{array}\right|,$$

where the omitted elements are zero, and the diagonal of $R(f_1,f_2)$ contains m occurrences of $c_0$ and n of $d_m$ [[4], p.206]. We can obtain Table 3 from the results of Section 2.

Table 3 Formulas for ${\mathit{a}}_{\mathbf{1}}\mathbf{\sim }{\mathit{a}}_{\mathbf{12}}$

Corollary 3.1 Let

$$\begin{array}{c}{a}_1(n):=\sum _{m=1}^{n-1}\sigma (m)\sigma (n-m),a_2(n):=\sum _{m=1}^{n-1}m\sigma (m)\sigma (n-m),\hfill \\ a_3(n):=\sum _{m=1}^{n-1}\sigma (m){\sigma }_3(n-m),a_4(n):=\sum _{m<n/2}\sigma (m)\sigma (n-2m),\hfill \\ a_5(n):=\sum _{m<n/2}{\sigma }_3(m)\sigma (n-2m),a_6(n):=\sum _{m<n/2}\sigma (m){\sigma }_3(n-2m),\hfill \\ a_7(n):=\sum _{l+m+s=n}\tilde{\sigma }(l)\tilde{\sigma }(m)\tilde{\sigma }(s),a_8(n):=\sum _{m=1}^{n-1}\hat{\sigma }(m)\hat{\sigma }(n-m),\hfill \\ a_9(n):=\sum _{m=1}^{n-1}\hat{\sigma }(m){\tilde{\sigma }}_3(n-m),a_{10}(n):=\sum _{m=1}^{n-1}m\hat{\sigma }(m)\hat{\sigma }(n-m),\hfill \\ a_{11}(n):=\sum _{l+m+s=n}\hat{\sigma }(l)\hat{\sigma }(m)\tilde{\sigma }(s),a_{12}(n):=\sum _{l+m+s=n}\hat{\sigma }(l)\hat{\sigma }(m)\hat{\sigma }(s).\hfill \end{array}$$

There exists a polynomial $T(x,y)\in \mathbb{Z}[x,y]$ such that $T(a_i(p),a_j(p))=0$ with $i,j\in \{1,2,\dots ,12\}$ where p is an odd prime. We abbreviate $a_1(n)$ to $a_1$ and it is also applied to the other values. Then we get Table 4.

Table 4 Formulas of $\mathit{T}\mathbf{(}\mathit{x}\mathbf{(}\mathit{p}\mathbf{)}\mathbf{,}\mathit{y}\mathbf{(}\mathit{p}\mathbf{)}\mathbf{)}\mathbf{=}\mathbf{0}$

Proof We illustrate the proof for the first $T(x(p),y(p))=0$ in Table 4. In Table 3 we consider ${\sum }_{m=1}^{x-1}\sigma (m)\sigma (x-m)$ and ${\sum }_{m=1}^{x-1}m\sigma (m)\sigma (x-m)$ with an odd prime x put by $2q+1$. As

$$\begin{array}{c}\sum _{m=1}^{2q}\sigma (m)\sigma (2q+1-m)=\frac{1}{3}(10q^3+9q^2-q),\hfill \\ \sum _{m=1}^{2q}m\sigma (m)\sigma (2q+1-m)=\frac{1}{6}(20q^4+28q^3+7q^2-q),\hfill \end{array}$$

the polynomials

$$\begin{array}{c}{f}_1(X)=10X^3+9X^2-X-3a_1(2q+1),\hfill \\ f_2(X)=20X^4+28X^3+7X^2-X-6a_2(2q+1)\hfill \end{array}$$

have common zero, namely, $X=q$. We deduce that for each odd prime p ($=2q+1$),

$$\left|\begin{array}{ccccccc}10& 9& -1& -3a_1& 0& 0& 0\\ 0& 10& 9& -1& -3a_1& 0& 0\\ 0& 0& 10& 9& -1& -3a_1& 0\\ 0& 0& 0& 10& 9& -1& -3a_1\\ 20& 28& 7& -1& -6a_2& 0& 0\\ 0& 20& 28& 7& -1& -6a_2& 0\\ 0& 0& 20& 28& 7& -1& -6a_2\end{array}\right|=0$$

and this simplifies to give $R(f_1,f_2)=108\text{,}000(-3a_1^3+6a_1^4+5a_1^2{a}_2+12a_1{a}_2^2-20a_2^3)=0$, so we can find the irreducible polynomial $T(a_1,a_2)=-3a_1^3+6a_1^4+5a_1^2{a}_2+12a_1{a}_2^2-20a_2^3=0$. Other results in Table 4 can also be obtained by using the resultant. □

Remark 3.2 The plane curves $T(x,y)=0$ in Corollary 3.1 all have zero-genus since x, y are polynomials of p, which leads to a morphism from the projective line ${\mathbb{P}}^1$ to plane curves $T(x,y)=0$. Then it follows easily from the Riemann-Hurwitz theorem (e.g., see Corollary 1 on page 91 of [5]).

Example 3.3 We suggest Figure 1, Figure 2 and Figure 3 for results of $Z_{i,j}=T(a_i,a_j)$ and $T(a_i,a_j)=0$.

Figure 1

${\mathit{Z}}_{\mathbf{1}\mathbf{,}\mathbf{2}}$ and $\mathit{T}\mathbf{(}{\mathit{a}}_{\mathbf{1}}\mathbf{,}{\mathit{a}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\mathbf{0}$ .

Figure 2

${\mathit{Z}}_{\mathbf{1}\mathbf{,}\mathbf{3}}$ and $\mathit{T}\mathbf{(}{\mathit{a}}_{\mathbf{1}}\mathbf{,}{\mathit{a}}_{\mathbf{3}}\mathbf{)}\mathbf{=}\mathbf{0}$ .

Figure 3

${\mathit{Z}}_{\mathbf{1}\mathbf{,}\mathbf{4}}$ and $\mathit{T}\mathbf{(}{\mathit{a}}_{\mathbf{1}}\mathbf{,}{\mathit{a}}_{\mathbf{4}}\mathbf{)}\mathbf{=}\mathbf{0}$ .