Universal sums of generalized octagonal numbers

TL;DR

This paper proves the universality of certain sums of generalized octagonal numbers, confirming conjectures by Sun, and provides a criterion for the universality of arbitrary such sums, extending the 15-theorem.

Contribution

It establishes the universality of specific quaternary sums of generalized octagonal numbers and generalizes the 15-theorem to these sums.

Findings

Proved universality for five specific sums of generalized octagonal numbers.

Provided an effective criterion for the universality of arbitrary sums of generalized octagonal numbers.

Confirmed conjectures posed by Sun regarding these sums.

Abstract

An integer of the form $P_8(x)=3x^2-2x$ for some integer $x$ is called a generalized octagonal number. A quaternary sum $\Phi_{a,b,c,d}(x,y,z,t)=aP_8(x)+bP_8(y)+cP_8(z)+dP_8(t)$ of generalized octagonal numbers is called {\it universal} if $\Phi_{a,b,c,d}(x,y,z,t)=n$ has an integer solution $x,y,z,t$ for any positive integer $n$. In this article, we show that if $a=1$ and $(b,c,d)=(1,3,3), (1,3,6), (2,3,6), (2,3,7)$ or $(2,3,9)$, then $\Phi_{a,b,c,d}(x,y,z,t)$ is universal. These were conjectured by Sun in \cite {sun}. We also give an effective criterion on the universality of an arbitrary sum $a_1 P_8(x_1)+a_2P_8(x_2)+\cdots+a_kP_8(x_k)$ of generalized octagonal numbers, which is a generalization of "$15$-theorem" of Conway and Schneeberger.

Equations

$$\begin{array}[]{ll}1,\ 2,\ 3,\ 5,\ 6,\ 7,\ 10,\ 13,\ 14,\ 15,\ 17,\ 19,\ 21,\ 22,\ 23,\ 26,\ 29,\\ 30,\ 31,\ 34,\ 35,\ 37,\ 42,\ 58,\ 93,\ 110,\ 145,\ 203,\ \ \text{and}\ \ 290.\end{array}$$

$$P_m(x)=\frac{(m-2)x^2-(m-4)x}{2}$$

$$P_4(x)+P_4(y)+P_4(z)+P_4(t)=n$$

$${\Phi }_{a,b,c,d}(x,y,z,t)=a{P}_8(x)+b{P}_8(y)+c{P}_8(z)+d{P}_8(t)=n$$

$${\begin{array}[]{lllllllll}&(1,1,1),&(1,1,2),&(1,1,3),&(1,1,4),&(1,2,2),&(1,2,3),&(1,2,4),&(1,2,5),\\ &(1,2,6),&(1,2,7),&(1,2,8),&(1,2,9),&(1,2,10),&(1,2,11),&(1,2,12),&(1,2,13),\\ &(1,3,3),&(1,3,5),&(1,3,6),&(2,2,2),&(2,2,3),&(2,2,4),&(2,2,5),&(2,2,6),\\ &(2,3,4),&(2,3,5),&(2,3,6),&(2,3,7),&(2,3,8),&(2,3,9),&(2,4,4),&(2,4,5),\\ &(2,4,6),&(2,4,7),&(2,4,8),&(2,4,9),&(2,4,10),&(2,4,11),&(2,4,12),&(2,4,13)\end{array}}$$

$$(1,3,3),(1,3,6),(2,3,6),(2,3,7),\text{and}(2,3,9).$$

$$1,2,3,4,6,7,9,12,13,14,18,\text{and}60.$$

$$P_8(x)+b{P}_8(y)+c{P}_8(z)+d{P}_8(t)=n$$

$$(3x-1{)}^2+b(3y-1{)}^2+c(3z-1{)}^2+d(3t-1{)}^2=3n+1+b+c+d$$

$$x^2+b{y}^2+c{z}^2+d{t}^2=3n+1+b+c+d$$

$$3(3m)+8=(3y+1{)}^2+(3y-1{)}^2+3z^2+3t^2,\text{where}zt\not\equiv 0(\text{mod 3}).$$

$$3(3m+2)+8=(3y+2{)}^2+(3y-2{)}^2+3z^2+3t^2,\text{where}zt\not\equiv 0(\text{mod 3}).$$

$$3(3m+1)+8=(3c-5{)}^2+(3c+5{)}^2+3a^2+3b^2,\text{where}ab\not\equiv 0(\text{mod 3}).$$

$$3n+11-18\cdot 2^2\to ⟨1,1,9⟩\text{ if 3n+11≡1 (mod 2) and 3n+11≡7 (mod 8)},$$

$$3n+11-18\to ⟨1,1,9⟩\text{ if 3n+11≡7 (mod 8)},$$

$$3n+11-18\cdot 2^2\to ⟨1,1,9⟩\text{ if 3n+11≡2 (mod 4)},$$

$$3n+11-18\to ⟨1,1,9⟩\text{ if 3n+11≡0 (mod 4)}.$$

$$a^2+b^2+3(c-2d{)}^2+6(c+d{)}^2=a^2+b^2+9c^2+18d^2=3n+11,$$

$$a^2+2b^2+3(c-2d{)}^2+6(c+d{)}^2=a^2+2b^2+9c^2+18d^2=3n+12,$$

$$a^2+2b^2\ne 0\text{and}(c-2d{)}^2+2(c+d{)}^2\ne 0,$$

$$d\not\equiv 0(\text{mod 3}),d\equiv 3n+13(\text{mod 2})\text{and}3n+13-7d^2\ne 0.$$

$$3a^2+3b^2+6c^2=3n+13-7\cdot d^2\equiv 0(\text{mod 2})\text{and}3n+13-7d^2>0.$$

$$3a^2+6c^2=(a+2c{)}^2+2(a-c{)}^2\ne 0.$$

$$e^2+3b^2+2f^2={\left(\frac{e+3b}{2}\right)}^2+3\cdot {\left(\frac{e-b}{2}\right)}^2+2f^2=3n+13-7d^2,$$

$${\Phi }_{a_1,a_2,\dots ,a_k}(x_1,x_2,\dots ,x_k)=a_1{P}_8(x_1)+a_2{P}_8(x_2)+\cdots +a_k{P}_8(x_k)=n$$

$$1,2,3,4,6,7,9,12,13,14,18,\text{and}60.$$

$$x(3x-2)+y(3y-2)+2z(3z-2)+14t(3t-2)=n$$

$$\alpha \in \{3n+18-14\cdot 1^2,3n+18-14\cdot 2^2,3n+18-14\cdot 4^2\}$$

$$\beta \in \{3n+18-14\cdot 5^2,3n+18-14\cdot 7^2,3n+18-14\cdot 8^2\}$$

$$x^2+y^2+4t^2=3n+9-3c^2.$$

$$(2t{)}^2+(12t{)}^2+4(-5t{)}^2=(14t{)}^2+(6t{)}^2+4(2t{)}^2=248t^2,$$

$$a^2+b^2+3c^2=a^2+{\left(\frac{b+3c}{2}\right)}^2+3\cdot {\left(\frac{b-c}{2}\right)}^2=3n+19-7d^2-7e^2,$$

Full text

This work of the first author was supported by BK21 PLUS SNU Mathematical Sciences Division.

This work of the second author was supported by the National Research Foundation of Korea (NRF-2017R1A2B4003758).

1. introduction

The famous Lagrange’s four square theorem states that every positive integer can be written as a sum of at most four integral squares. Motivated by Lagrange’s four square theorem, Ramanujan provided a list of 55 candidates of diagonal quaternary quadratic forms that represent all positive integers. In [4], Dickson pointed out that the quaternary form $x^{2}+2y^{2}+5z^{2}+5t^{2}$ that is included in Ramanujan’s list represents all positive integers except $15$ and confirmed that Ramanujan’s assertion for all the other $54$ forms is true. A positive definite integral quadratic form is called universal if it represents all non-negative integers. The problem on determining all universal quaternary forms was completed by Conway and Schneeberger. They proved that there are exactly $204$ universal quaternary quadratic forms. Furthermore, they proved the so called “15-theorem”, which states that every positive definite integral quadratic form that represents $1,2,3,5,6,7,10,14$ and $15$ is, in fact, universal, irrespective of its rank (see [1]). Recently, Bhargava and Hanke [2] proved, so called, “290-theorem”, which states that every positive definite integer-valued quadratic form is universal if it represents

[TABLE]

In general, a polygonal number of order $m$ (or an $m$-gonal number) for $m\geq 3$ is defined by

[TABLE]

for some non-negative integer $x$. If we admit that $x$ is a negative integer, then $P_{m}(x)$ is called a generalized polygonal number of order $m$ (or a generalized $m$-gonal number). Then, Lagrange’s four square theorem implies that the diophantine equation

[TABLE]

has an integer solution $x,y,z$ and $t$, for any non-negative integer $n$.

Recently, Sun in [11] proved that every positive integer can be written as a sum of four generalized octagonal numbers, which is also considered as a generalization of Lagrange’s four square theorem. He also defined that for positive integers $a\leq b\leq c\leq d$, a quaternary sum $aP_{8}(x)+bP_{8}(y)+cP_{8}(z)+dP_{8}(t)$ (simply, $\Phi_{a,b,c,d}(x,y,z,t)$) of generalized octagonal numbers is universal over $\mathbb{Z}$ if the diophantine equation

[TABLE]

has an integer solution for any non-negative integer $n$. Then, he showed that if $\Phi_{a,b,c,d}$ is universal over $\mathbb{Z}$, then $a=1$, and $(b,c,d)$ is one of the following 40 triples:

[TABLE]

and proved the universalities over $\mathbb{Z}$ except the following 5 triples:

[TABLE]

He conjectured that for each of these 5 triples, the sum $\Phi_{1,b,c,d}$ is also universal over $\mathbb{Z}$.

The aim of this article is to prove this conjecture. In Section 3, we further prove that the sum $a_{1}P_{8}(x_{1})+a_{2}P_{8}(x_{2})+\cdots+a_{k}P_{8}(x_{k})$ of generalized octagonal numbers is universal over $\mathbb{Z}$ if and only if it represents

[TABLE]

This might be considered as a generalization of “$15$-theorem” of Conway and Schneeberger. For universal sums of triangular numbers, see [3].

For a quadratic form $f(x_{1},x_{2},\dots,x_{n})=\sum_{1\leq i,j\leq n}a_{ij}x_{i}x_{j}\ (a_{ij}=a_{ji})$, the corresponding symmetric matrix of $f$ is defined by $M_{f}=(a_{ij})$. In particular, for a diagonal quadratic form $f(x_{1},x_{2},\dots,x_{n})=a_{1}x_{1}^{2}+a_{2}x_{2}^{2}+\cdots+a_{n}x_{n}^{2}$, we simply write $M_{f}=\langle a_{1},a_{2},\dots,a_{n}\rangle$. For an integer $k$, if the diophantine equation $f(x_{1},x_{2},\dots,x_{n})=k$ has an integer solution, then we say $k$ is represented by $f$, and we write $k{\,\rightarrow\,}f$. The genus of $f$, denoted by $\text{gen}(f)$, is the set of all quadratic forms that are locally isometric to $f$. The number of isometry classes in $\text{gen}(f)$ is called the class number of $f$.

Any unexplained notations and terminologies can be found in [7] or [10].

2. Quaternary universal sums of generalized octagonal numbers

In this section, we prove that if $(b,c,d)$ is one of the triples in (1.1), then the quaternary sum $\Phi_{1,b,c,d}$ of generalized octagonal numbers is universal over $\mathbb{Z}$.

Let $0<b\leq c\leq d$ be positive integers. Recall that a quaternary sum $\Phi_{1,b,c,d}$ of generalized octagonal numbers is said to be universal over $\mathbb{Z}$ if the diophantine equation

[TABLE]

has an integer solution for any non-negative integer $n$. Note that $\Phi_{1,b,c,d}$ is universal over $\mathbb{Z}$ if and only if the equation

[TABLE]

has an integer solution for any non-negative integer $n$. This is equivalent to the existence of an integer solution $x,y,z,t$ of the diophantine equation

[TABLE]

such that $xyzt\not\equiv 0\ (\text{mod 3})$.

Theorem 2.1**.**

The quaternary sum $\Phi_{1,1,3,3}$ of generalized octagonal numbers is universal over $\mathbb{Z}$.

Proof.

It is enough to show that $x^{2}+y^{2}+3z^{2}+3t^{2}=3n+8$ has an integer solution $(x,y,z,t)\in{\mathbb{Z}}^{4}$ such that $zt\not\equiv 0\ (\text{mod 3})$. First, assume that $n=3m$. Note that the class number of $\langle 1,1,6\rangle$ is $1$ and $z^{2}+t^{2}+6y^{2}=3m+2$ has an integer solution such that $zt\not\equiv 0\ (\text{mod 3})$. Hence

[TABLE]

Furthermore, we also have

[TABLE]

Now assume that $n=3m+1$. Note that for any $m\leq 4$, the diophatine equation $x^{2}+y^{2}+3z^{2}+3t^{2}=3(3m+1)+8$ has an integer solution $(x,y,z,t)=(a,b,c,d)$ such that $zt\not\equiv 0\ (\text{mod 3})$. Therefore we assume that $m\geq 5$. Similarly as above, $z^{2}+t^{2}+6y^{2}=3m-13$ has an integer solution $(z,t,y)=(a,b,c)\in{\mathbb{Z}}^{3}$ such that $ab\not\equiv 0\ (\text{mod 3})$. Hence

[TABLE]

Therefore, the diophantine equation $(3x-1)^{2}+(3y-1)^{2}+3(3z-1)^{2}+3(3t-1)^{2}=3n+8$ has always an integer solution. ∎

Theorem 2.2**.**

The quaternary sum $\Phi_{1,1,3,6}$ of generalized octagonal numbers is universal over $\mathbb{Z}$.

Proof.

First, we show that $x^{2}+y^{2}+9z^{2}+18t^{2}=3n+11$ has an integer solution such that $9z^{2}+18t^{2}\neq 0$. If $n\leq 20$, then one may directly check that such an integer solution exists. Note that the class number of $\langle 1,1,9\rangle$ is $1$ and it represents all integers $n$ such that $n\equiv 2\ (\text{mod 3})$ and $n\neq 4^{s}(8t+7)$ for any non-negative integers $s,t$. Note that for any $n\geq 21$,

[TABLE]

Hence $3n+11=x^{2}+y^{2}+9z^{2}+18t^{2}$ has an integer solution $(x,y,z,t)=(a,b,c,d)\in{\mathbb{Z}}^{4}$ such that $ab\not\equiv 0\ (\text{mod 3})$ and $d=1$ or $2$. Since

[TABLE]

$x^{2}+y^{2}+3z^{2}+6t^{2}=3n+11$ has an integer solution $(x,y,z,t)=(a,b,c-2d,c+d)$. Now, by Theorem 9 of [6] (see also [5], and for more generalization see [9]), there are integers $e,f$ such that $e^{2}+2f^{2}=(c-2d)^{2}+2(c+d)^{2}$ and $ef\not\equiv 0\ (\text{mod 3})$. ∎

Theorem 2.3**.**

The quaternary sum $\Phi_{1,2,3,6}$ of generalized octagonal numbers is universal over $\mathbb{Z}$.

Proof.

First, we show that the diophantine equation $x^{2}+2y^{2}+9z^{2}+18t^{2}=3n+12$ has an integer solution $(x,y,z,t)=(a,b,c,d)\in{\mathbb{Z}}^{4}$ such that $a^{2}+2b^{2}\neq 0$ and $c^{2}+2d^{2}\neq 0$. If $n\leq 44$, then one may directly check that such an integer solution exists. From now on, we assume that $n\geq 45$.

Assume that an integer $3m$, for some positive integer $m$ is represented by the genus of $\langle 1,2,9\rangle$. Since the class number of $\langle 1,1,2\rangle$ is one, $3m$ is represented by $\langle 1,1,2\rangle$, that is, $x^{2}+y^{2}+2z^{2}=3m$ has an integer solution $(x,y,z)=(a,b,c)\in{\mathbb{Z}}^{3}$. Note that $a$ or $b$ is divisible by $3$. Therefore $3m$ is represented by $\langle 1,2,9\rangle$. Similarly, every integer of the form $3m$, for some positive integer $m$ that is represented by the genus of $\langle 1,2,18\rangle$ is represented by itself.

If $3n+12\equiv 1\ (\text{mod 2})$ and $n\geq 21$, then both $3n+12-18\cdot 1^{2}$ and $3n+12-18\cdot 2^{2}$ are represented by $\langle 1,2,9\rangle$ by the above observation. Furthermore, at least one of them is not of the form $9\cdot N^{2}$ for any integer $N$. If $3n+12\equiv 2\ (\text{mod 4})$ and $n\geq 45$, then both $3n+12-9\cdot 2^{2}$ and $3n+12-9\cdot 4^{2}$ are represented by $\langle 1,2,18\rangle$. Furthermore, at least one of them is not of the form $18\cdot M^{2}$ for any integer $M$. Finally, assume that $3n+12=2^{\delta}4^{m}a$, where $a$ is odd, $\delta=0$ or $1$, and $m\geq 1$. Then the assertion follows from the fact that the assertion holds if $2^{\delta}a\geq 12$ or $2^{\delta}4^{m}a=3\cdot 4,6\cdot 4$ and $9\cdot 4$.

Now, since

[TABLE]

and

[TABLE]

the diophantine equation $(3x-1)^{2}+2(3y-1)^{2}+3(3z-1)^{2}+6(3t-1)^{2}=3n+12$ has always an integer solution by Theorem 9 of [6]. ∎

Theorem 2.4**.**

The quaternary sums $\Phi_{1,2,3,7}$ and $\Phi_{1,2,3,9}$ of generalized octagonal numbers are all universal over $\mathbb{Z}$.

Proof.

Since the proofs are quite similar to each other, we only provide the proof of the former case.

It suffices to show that the diophantine equation $x^{2}+2y^{2}+3z^{2}+7t^{2}=3n+13$ has an integer solution $(x,y,z,t)=(a,b,c,d)\in\mathbb{Z}^{4}$ such that $abcd\not\equiv 0\ (\text{mod 3})$ for any non-negative integer $n$. If $n\leq 110$, then one may directly check that such an integer solution exists. Therefore we assume that $n\geq 111$. First, we show that the diophantine equation $3x^{2}+3y^{2}+6z^{2}+7t^{2}=3n+13$ has an integer solution $(x,y,z,t)=(a,b,c,d)\in\mathbb{Z}^{4}$ such that

[TABLE]

Note that an integer $n$ is represented by $\langle 3,3,6\rangle$ if and only if $n$ is divisible by $3$ and is not of the form $6\cdot 4^{s}(8t+7)$ for any non-negative integers $s,t$. Then one may easily show that there is an integer $d\in\{1,5,7\}$ such that $3n+13-7\cdot d^{2}$ is represented by $\langle 3,3,6\rangle$. Furthermore, one may also show that there is an integer $d\in\{2,4\}$ such that $3n+13-7\cdot d^{2}$ is represented by $\langle 3,3,6\rangle$. Hence By taking a suitable integer $d\in\{1,5,7,2,4\}$, we assume that there are integers $a,b$ and $c$ such that

[TABLE]

Since $3n+13-7d^{2}>0$, we may assume that

[TABLE]

Hence there are integers $e,f$ such that $e^{2}+2f^{2}=(a+2c)^{2}+2(a-c)^{2}$ and $ef\not\equiv 0\ (\text{mod 3})$ by Theorem 9 of [6]. Therefore, if $b\not\equiv 0\ (\text{mod 3})$, then we are done. Assume that $b\equiv 0\ (\text{mod 3})$. Since $e\equiv b\ (\text{mod 2})$, we have

[TABLE]

where $\left(\frac{e+3b}{2}\right)\left(\frac{e-b}{2}\right)\not\equiv 0\ (\text{mod 3})$. Therefore, the diophantine equation $(3x-1)^{2}+2(3y-1)^{2}+3(3z-1)^{2}+7(3t-1)^{2}=3n+13$ has an integer solution for any non-negative integer $n$. ∎

3. The octagonal theorem of sixty

Let $a_{1},a_{2},\dots,a_{k}$ be any positive integers. For an integer $n$, if the diophantine equation

[TABLE]

has an integer solution, then we say the sum $\Phi_{a_{1},a_{2},\dots,a_{k}}$ of generalized octagonal numbers represents $n$ and we write $n{\,\rightarrow\,}\Phi_{a_{1},a_{2},\dots,a_{k}}$. The sum $\Phi_{a_{1},a_{2},\dots,a_{k}}$ of generalized octagonal numbers is called universal over $\mathbb{Z}$ if it represents all non-negative integers. In this section, we prove:

Theorem 3.1**.**

The sum $\Phi_{a_{1},a_{2},\dots,a_{k}}$ of generalized octagonal numbers is universal over $\mathbb{Z}$ if and only if it represents the integers

[TABLE]

To prove this, we need the following six lemmas:

Lemma 3.2**.**

The quaternary sum $\Phi_{1,1,2,14}$ of generalized octagonal numbers represents all positive integers except $60$.

Proof.

It suffices to show that the diophantine equation $x^{2}+y^{2}+2z^{2}+14t^{2}=3n+18$ has an integer solution $(x,y,z,t)=(a,b,c,d)\in\mathbb{Z}^{4}$ such that $abcd\not\equiv 0\ (\text{mod 3})$ for any non-negative integer $n\neq 60$. Note that an integer $n$ is represented by $\langle 1,1,2\rangle$ if and only if it is not of the form $2^{2s+1}(8t+7)$ for any non-negative integers $s,t$. If $n\leq 292$ and $n\not\eq 60$, then one may directly check that the equation

[TABLE]

has an integer solution. Hence we may assume that $n\geq 293$. One may easily show that there is an integer

[TABLE]

such that $\alpha{\,\rightarrow\,}\langle 1,1,2\rangle$. Furthermore, one may also show that there is an integer

[TABLE]

such that $\beta{\,\rightarrow\,}\langle 1,1,2\rangle$. For all possible cases when both $\alpha$ and $\beta$ are squares of integers, one may easily check that the equation $x^{2}+y^{2}+2z^{2}=\alpha$ has an integer solution $(x,y,z)=(a,b,c)$ such that $abc\not\equiv 0\ (\text{mod 3})$. Therefore, we may assume that the equation $x^{2}+y^{2}+2z^{2}+14t^{2}=3n+18$ has an integer solution $(x,y,z,t)=(a,b,c,d)\in\mathbb{Z}^{4}$ such that $d\not\equiv 0\ (\text{mod 3})$ and $3n+18-14\cdot d^{2}$ is not a square of an integer. If $abc\not\equiv 0\ (\text{mod 3})$, then we are done. If $abc\equiv 0\ (\text{mod 3})$, then without loss of generality, we may assume that $b\equiv c\equiv 0\ (\text{mod 3})$ and $a\not\equiv 0\ (\text{mod 3})$. Since $b^{2}+2c^{2}\not\eq 0$, there are integers $e,f$ such that $e^{2}+2f^{2}=b^{2}+2c^{2}$ and $ef\not\equiv 0\ (\text{mod 3})$ by Theorem 9 of [6]. The lemma follows from this. ∎

Lemma 3.3**.**

The quaternary sum $\Phi_{1,1,3,4}$ of generalized octagonal numbers represents all positive integers except $18$.

Proof.

It suffices to show that the diophantine equation $x^{2}+y^{2}+3z^{2}+4t^{2}=3n+9$ has an integer solution $(x,y,z,t)=(a,b,c,d)\in\mathbb{Z}^{4}$ such that $abcd\not\equiv 0\ (\text{mod 3})$ for any non-negative integer $n\neq 18$. If $n\leq 46$ and $n\neq 18$, then one may directly check that such an integer solution exists. From now on, we assume that $n\geq 47$. One may easily show that there is an integer $c\in\{2,5,7\}$ such that $3n+9-3\cdot c^{2}$ is represented by $\langle 1,1,4\rangle$. Therefore, the equation $x^{2}+y^{2}+3z^{2}+4t^{2}=3n+9$ has an integer solution $(x,y,z,t)=(a,b,c,d)\in\mathbb{Z}^{4}$ such that $a^{2}+b^{2}+4d^{2}\neq 0$ and $c\not\equiv 0\ (\text{mod 3})$. If $abd\not\equiv 0\ (\text{mod 3})$, then we are done. Suppose that $a\equiv b\equiv d\equiv 0\ (\text{mod 3})$. If $\tau=\frac{1}{3}\begin{pmatrix}1&2&4\\ -2&-1&4\\ -1&1&-1\end{pmatrix}$, then one may easily check that $\tau(a,b,d)^{t}=(a_{1},b_{1},d_{1})^{t}$ is also an integer solution of

[TABLE]

Therefore, there is a positive integer $m$ such that $\tau^{m}(a,b,d)^{t}=(a_{m},b_{m},d_{m})^{t}$ is an integer solution of (3.1) such that $a_{m}b_{m}d_{m}\not\equiv 0\ (\text{mod 3})$ or for any positive integer $m$, $\tau^{m}(a,b,d)^{t}=(a_{m},b_{m},d_{m})^{t}$ is an intger solution of (3.1) each of whose component is divisible by $3$. Since there are only finitely many integer solution of (3.1) and $\tau$ has an infinite order, the latter is impossible unless $(a,b,d)$ is an eigenvector of $\tau$. Note that $(a,b,d)=(2t,12t,-5t)$ is an eigenvector of $\tau$ and

[TABLE]

where $(14t,6t,2t)$ is not an eigenvector of $\tau$. Therefore the equation $x^{2}+y^{2}+3z^{2}+4t^{2}=3n+9$ has always an integer solution $(x,y,z,t)=(a,b,c,d)\in\mathbb{Z}^{4}$ such that $abcd\not\equiv 0\ (\text{mod 3})$. This completes the proof. ∎

Lemma 3.4**.**

The quaternary sum $\Phi_{1,2,3,3}$ of generalized octagonal numbers represents all positive integers except $12$.

Proof.

Since the proof is quite similar to that of Theorem 2.4, it is left to the reader. ∎

Lemma 3.5**.**

The quinary sum $\Phi_{1,1,3,7,\alpha}$ of generalized octagonal numbers is universal over $\mathbb{Z}$ for any integer $\alpha\in\{7,9,10,11,13,14\}$.

Proof.

Since the proofs are quite similar to each other, we only provide the proof of the sum $\Phi_{1,1,3,7,7}$.

It suffices to show that the diophantine equation $x^{2}+y^{2}+3z^{2}+7t^{2}+7s^{2}=3n+19$ has an integer solution $(x,y,z,t,s)=(a,b,c,d,e)$ such that $abcde\not\equiv 0\ (\text{mod 3})$ for any non-negative integer $n$. If $n\leq 12$, then one may directly check that such an integer solution exists. Therefore we assume that $n\geq 13$. Note that there are integers $d,e\in\{1,2\}$ such that $3n+19-7d^{2}-7e^{2}\equiv 0,1\ (\text{mod 4})$. Since the class number of $\langle 1,1,3\rangle$ is one and $3n+19-7d^{2}-7e^{2}\equiv 2\ (\text{mod 3})$, $3n+19-7d^{2}-7e^{2}$ is represented by $\langle 1,1,3\rangle$. Therefore the diophantine equation $x^{2}+y^{2}+3z^{2}+7t^{2}+7s^{2}=3n+19$ has an integer solution $(x,y,z,t,s)=(a,b,c,d,e)$ such that $d,e\in\{1,2\}$ and $3n+19-7d^{2}-7e^{2}\equiv 0,1\ (\text{mod 4})$. If $c\not\equiv 0\ (\text{mod 3})$, then we are done. Assume that $c\equiv 0\ (\text{mod 3})$. Without loss of generalitiy, we may assume that $b\equiv c\ (\text{mod 2})$, for $3n+19-7d^{2}-7e^{2}\equiv 0,1\ (\text{mod 4})$. Then we have

[TABLE]

where $\left(\frac{b+3c}{2}\right)\left(\frac{b-c}{2}\right)\not\equiv 0\ (\text{mod 3})$. Therefore, the equation $(3x-1)^{2}+(3y-1)^{2}+3(3z-1)^{2}+7(3t-1)^{2}+7(3s-1)^{2}=3n+19$ has an integer solution for any non-negative integer $n$. ∎

Lemma 3.6**.**

The quinary sum $\Phi_{1,1,3,7,8}$ of generalized octagonal numbers is universal over $\mathbb{Z}$.

Proof.

It suffices to show that the diophantine equation $x^{2}+y^{2}+3z^{2}+7t^{2}+8s^{2}=3n+20$ has an integer solution $(x,y,z,t,s)=(a,b,c,d,e)$ such that $abcde\not\equiv 0\ (\text{mod 3})$ for any non-negative integer $n$. If $n\leq 117$, then one may directly check that such an integer solution exists. Therefore we assume that $n\geq 118$. First, we show that the the diophantine equation $x^{2}+y^{2}+3z^{2}+7t^{2}+8s^{2}=3n+20$ has an integer solution $(x,y,z,t,s)=(a,b,c,d,e)$ such that $cd\not\equiv 0\ (\text{mod 3})$ and $3n+20-3c^{2}-7d^{2}$ is not a square of an integer. Note that the class number of $\langle 1,1,8\rangle$ is one. Then one may easily show that for any integer $n\geq 118$,

[TABLE]

Furthermore, in each case, at least one of two integers of the form $3n+20-3c^{2}-7d^{2}$ is not a square of an integer, for we are assuming that $n\geq 118$. Therefore the equation $x^{2}+y^{2}+3z^{2}+7t^{2}+8s^{2}=3n+20$ has an integer solution $(x,y,z,t,s)=(a,b,c,d,e)$ such that $cd\not\equiv 0\ (\text{mod 3})$. If $abe\not\equiv 0\ (\text{mod 3})$, then we are done. Assume that $a\not\equiv 0\ (\text{mod 3})$ and $be\equiv 0\ (\text{mod 3})$. Since $3n+20-3c^{2}-7d^{2}$ is not a square of an integers, $b^{2}+8s^{2}\neq 0$. Therefore the equation $(3x-1)^{2}+(3y-1)^{2}+3(3z-1)^{2}+7(3t-1)^{2}+8(3s-1)^{2}=3n+20$ has always an integer solution by Theorem 4.1 of [9]. ∎

Lemma 3.7**.**

The quinary sum $\Phi_{1,1,3,7,12}$ of generalized octagonal numbers is universal over $\mathbb{Z}$.

Proof.

It suffices to show that the equation $x^{2}+y^{2}+3z^{2}+7t^{2}+12s^{2}=3n+24$ has an integer solution $(x,y,z,t,s)$ such that $xyzts\not\equiv 0\ (\text{mod 3})$. If $n\leq 142$, then one may directly check that such an integer solution exists. Therefore we assume that $n\geq 143$. First, assume that $n=3m$. Note that the class number of $\langle 1,4,6\rangle$ is one and every odd positive integer that is not divisible by $3$ is represented by $\langle 1,4,6\rangle$. Therefore $9m+22-7\cdot 1^{2}$ or $9m+22-7\cdot 8^{2}$ is represented by $\langle 3,12,18\rangle$. Hence the equation

[TABLE]

has an integer solution such that $zts\not\equiv 0\ (\text{mod 3})$. Assume that $n=3m+1$. Since $9m+25-7\cdot 2^{2}$ or $9m+25-7\cdot 7^{2}$ is represented by $\langle 3,12,18\rangle$, the equation

[TABLE]

has an integer solution such that $zts\not\equiv 0\ (\text{mod 3})$. Assume that $n=3m+2$. Since $9m+28-7\cdot 4^{2}$ or $9m+28-7\cdot 5^{2}$ is represented by $\langle 3,12,18\rangle$, the equation

[TABLE]

has an integer solution such that $zts\not\equiv 0\ (\text{mod 3})$. Therefore the diophantine equation $(3x-1)^{2}+(3y-1)^{2}+3(3z-1)^{2}+7(3t-1)^{2}+12(3s-1)^{2}=3n+24$ has always an integer solution. ∎

Proof of Theorem 3.1. Without loss of generality, we may assume that $0<a_{1}\leq a_{2}\leq\cdots\leq a_{k}$. Since $1{\,\rightarrow\,}\Phi_{a_{1},a_{2},\dots,a_{k}}$, we have $a_{1}=1$. Since $2{\,\nrightarrow\,}\Phi_{1}$, we have $a_{2}=1$ or $2$. Assume that $(a_{1},a_{2})=(1,1)$. Since $3{\,\nrightarrow\,}\Phi_{1,1}$ and $3{\,\rightarrow\,}\Phi_{a_{1},a_{2},\dots,a_{k}}$, we have $1\leq a_{3}\leq 3$. Assume that $(a_{1},a_{2})=(1,2)$. Since $4{\,\nrightarrow\,}\Phi_{1,2}$ and $4{\,\rightarrow\,}\Phi_{a_{1},a_{2},\dots,a_{k}}$, we have $2\leq a_{3}\leq 4$. Now, note that $a{\,\nrightarrow\,}\Phi_{a_{1},a_{2},a_{3}}$, where

[TABLE]

Therefore, $a_{3}\leq a_{4}\leq a$ for each possible case, where $a$ is the integer given above.

If $(a_{1},a_{2},a_{3},a_{4})$ is one of possible quadruples given above except the quadruples $(1,1,2,14)$, $(1,1,3,4)$, $(1,1,3,7)$ and $(1,2,3,3)$, then the quaternary sum $\Phi_{a_{1},a_{2},a_{3},a_{4}}$ of generalized octagonal numbers is universal over $\mathbb{Z}$ by [11] and Section 2. Now, note that $b{\,\nrightarrow\,}\Phi_{a_{1},a_{2},a_{3},a_{4}}$ and $b{\,\rightarrow\,}\Phi_{a_{1},a_{2},a_{3},a_{4},c}$ for any integer $c$ satisfying $a_{4}\leq c\leq b$, where

[TABLE]

Furthermore, for each case, $\Phi_{a_{1},a_{2},a_{3},a_{4}}$ represents all positive integers except $b$ by Lemmas 3.2, 3.3 and 3.4. Therefore, the sum $\Phi_{a_{1},a_{2},a_{3},a_{4},c}$ of generalized octagonal numbers is universal over $\mathbb{Z}$ for any $c$ such that $a_{4}\leq c\leq b$. Note that $14\not{\,\rightarrow\,}\Phi_{1,1,3,7}$. Therefore $7\leq a_{5}\leq 14$ if $(a_{1},a_{2},a_{3},a_{4})=(1,1,3,7)$. The sum $\Phi_{1,1,3,7,d}$ of generalized octagonal numbers is universal over $\mathbb{Z}$ for any $d$ such that $7\leq d\leq 14$ by Lemmas 3.5, 3.6 and 3.7. This completes the proof. ∎