On sums of four pentagonal numbers with coefficients

1.   Introduction

For each $m = 3,4,5,\ldots$, the polygonal numbers of order $m$ are given by

In particular, those $p_5(n)$ with $n\in{\Bbb N}$ are called pentagonal numbers. A famous claim of Fermat states that each $n\in{\Bbb N}$ can be written as a sum of $m$ polygonal numbers of order $m$. This was proved by Lagrange for $m = 4$ in 1770, by Gauss for $m = 3$ in 1796, and by Cauchy for $m{\geq} 5$ in 1813. For Cauchy's polygonal number theorem, one may consult Nathanson [6] and [7,Chapter 1,pp. 3-34] for details. In 1830 Legendre refined Cauchy's polygonal number theorem by showing that for any $m = 5,6,\ldots$ every sufficiently large integer is a sum of five polygonal numbers of order $m$ one of which is $0$ or $1$ (cf. [7,p. 33]).

In 2016 Sun [9,Conjecture 5.2(ⅱ)] conjectured that each $n\in{\Bbb N}$ can be written as

provided that $(b,c,d)$ is among the following 15 triples:

In 2017, Meng and Sun [5] confirmed this for $(b,c,d) = (1,2,2),(1,2,4)$. In this paper we prove the conjecture for

Theorem 1.1. Each $n\in{\Bbb N}$ can be written as a sum of five pentagonal numbers two of which are equal, that is, there are $x,y,z,w\in{\Bbb N}$ such that

Remark 1.1. Clearly, Theorem 1.1 is stronger than the classical result that any nonnegative integer is a sum of five pentagonal numbers. In Feb. 2019 the second author even conjectured that any integer $n>33066$ is a sum of three pentagonal numbers.

Theorem 1.2. Any $n\in{\Bbb N}$ can be written as $p_5(w)+2p_5(x)+3p_5(y)+4p_5(z)$ with $w,x,y,z\in{\Bbb N}$.

Theorem 1.3. Let ${\delta}\in\{1,2\}$. Then any $n\in{\Bbb N}$ can be written as $p_5(w)+p_5(x)+2p_5(y)+3{\delta} p_5(z)$ with $w,x,y,z\in{\Bbb N}$.

We will prove Theorems 1.1-1.3 in Sections 2-4 respectively. Our proofs use some known results on ternary quadratic forms.

Those $p_5(x) = x(3x-1)/2$ with $x\in{\Bbb Z}$ are called generalized pentagonal numbers. Clearly,

Recently, Ju [3] showed that for any positive integers $a_1,\ldots,a_k$ the set

contains all nonnegative integers whenever it contains the twelve numbers

The generalized octagonal numbers are those $p_8(x) = x(3x-2)$ with $x\in{\Bbb Z}$. In 2016, Sun [9] proved that any positive integer can be written as a sum of four generalized octagonal numbers one of which is odd. See also Sun [11] and [10] for representations of nonnegative integers in the form $x(ax+b)/2+y(cy+d)/2+z(ez+f)/2$ with $x,y,z$ integers or nonnegative integers, where $a,c,e$ are positive integers and $b,d,f$ are integers with $a+b,c+d,e+f$ all even.

2.   Proof of Theorem 1.1

Lemma 2.1. Any positive even number $n$ not in the set $\{{{5}^{2k+1}}m:\ k,m\in \mathbb{N}\ and\ m\equiv \pm 2\ (\text{mod}\ 5)\}$ can be written as $x^2+y^2+z^2+(x+y+z)^2/2$ with $x,y,z\in{\Bbb Z}$.

Proof. By Dickson [2,pp. 112-113],

Thus $8n = s^2+2t^2+10z^2$ for some $s,t,z\in{\Bbb Z}$. Clearly, $2\mid s$ and $t{\equiv} z\ ({\rm{mod}}\ 2)$. Without loss of generality, we may assume that $t\not{\equiv} z\ ({\rm{mod}}\ 4)$ if $2\nmid z$. (If $t{\equiv} z\ ({\rm{mod}}\ 4)$ with $z$ odd, then $-t\not{\equiv} z\ ({\rm{mod}}\ 4)$.) Write $s = 2r$ and $t = 2w+z$ with $r,w\in{\Bbb Z}$. Then $2\nmid w$ if $2\nmid z$. Since

both $r-z$ and $w(w+z)$ are even. If $2\mid z$ then $2\mid w$. Recall that $2\nmid w$ if $2\nmid z$. So $w{\equiv} z{\equiv} r\ ({\rm{mod}}\ 2)$. Now, both $x = (r+w)/2$ and $y = (w-r)/2$ are integers. Observe that

and hence $n = x^2+y^2+z^2+(x+y+z)^2/2$. This ends the proof.

Lemma 2.2. Let $n\in{\Bbb N}$. Suppose that there are $B\in{\Bbb N}$ and $x,y,z\in{\Bbb Z}$ such that $3\mid n+B$ and

Then $n = p_5(x_0)+p_5(y_0)+p_5(z_0)+2p_5(w_0)$ for some $x_0,y_0,z_0,w_0\in{\Bbb N}$.

Proof. Clearly, $w = -(x+y+z)/2\in{\Bbb Z}$. As $|x|,|y|,|z|,|w|{\leq} B$, all the numbers

are nonnegative integers. Observe that

This concludes the proof.

Proof of Theorem 1.1. We can easily verify the desired result for $n = 0,\ldots,8891$. Below we assume that $n{\geq} 8892$. If

then

and

Case 1. $5\nmid n$.

As

we have

and hence there is an integer $B$ satisfying (2.1) with $B{\equiv}-n\ ({\rm{mod}}\ 3)$. By the above,

As the even number $\frac23(n+B)+B-5B^2$ is not divisible by $5$, in light of Lemma 2.1 there are $x,y,z\in{\Bbb Z}$ such that

Now, by applying Lemma 2.2 we find that $n = p_5(x_0)+p_5(y_0)+p_5(z_0)+2p_5(w_0)$ for some $x_0,y_,z_0,w_0\in{\Bbb N}$.

Case 2. $n = 5q$ for some $q\in{\Bbb N}$.

In this case, we can easily verify the desired result when $8892{\leq} n{\leq} 222288$. Below we assume that

Choose ${\delta}\in\{0,\pm1\}$ such that $1-q-{\delta}\not{\equiv}0,\pm2\ ({\rm{mod}}\ 5)$. As

there is an integer $B$ satisfying (2.1) such that $B{\equiv}-n\ ({\rm{mod}}\ 3)$ and $(B-1)^2{\equiv}{\delta}\ ({\rm{mod}}\ 5)$. Note that

and

Thus, by applying Lemmas 2.1 and 2.2 we get that $n = p_5(x_0)+p_5(y_0)+p_5(z_0)+2p_5(w_0)$ for some $x_0,y_,z_0,w_0\in{\Bbb N}$.

In view of the above, we have completed the proof of Theorem 1.1.

3.   Proof of Theorem 1.2

Lemma 3.1. Let $q\in{\Bbb N}$ with $q$ odd and not squarefree, or $2\mid q$ and $q\not\in\{4^k(16l+6):\ k,l\in{\Bbb N}\}$. Then there are $x,y,z\in{\Bbb Z}$ such that

Proof. By K. Ono and K. Soundararajan [8], and Dickson [1], the Ramanujan form $x^2+y^2+10z^2$ represents $q$. Write $q = a^2+b^2+10c^2$ with $a,b,c\in{\Bbb Z}$. Then, for

we have

This concludes the proof.

Lemma 3.2. Let $n\in{\Bbb N}$. Suppose that there are $B\in{\Bbb N}$ and $x,y,z\in{\Bbb Z}$ such that

Then $n = p_5(w_0)+2p_5(x_0)+3p_5(y_0)+4p_5(z_0)$ for some $w_0,x_0,y_0,z_0\in{\Bbb N}$.

Proof. Set $w = -(2x+3y+4z)$. As $|w|,|x|,|y|,|z|{\leq} B$, all the numbers

are nonnegative integers. Observe that

This ends the proof.

Proof of Theorem 1.2. We can verify the result for $n = 0,\ldots,45325137$ directly via a computer. Below we assume that

Since

there is an integer $B$ with

such that

if $n$ is odd, and $B{\equiv} 3n-1\ ({\rm{mod}}\ 8)$ and $B{\equiv}3n^2-2n\ ({\rm{mod}}\ 9)$ if $n$ is even. Note that

and

Let $q = (n+5B-15B^2)/9$. When $n$ is odd, we can easily see that $q$ is an odd integer divisible by $9$. When $n$ is even, $q$ is an even integer with $q{\equiv}4\ ({\rm{mod}}\ 8)$, and hence $q\not = 4^k(16l+6)$ for any $k,l\in{\Bbb N}$. By Lemma 3.1, we can write $6q = (2n+10B)/3-10B^2$ as $2x^2+3y^2+4z^2+(2x+3y+4z)^2$ with $x,y,z\in{\Bbb Z}$. Applying Lemma 3.2, we see that $n = p_5(w_0)+2p_5(x_0)+3p_5(y_0)+4p_5(z_0)$ for some $w_0,x_0,y_0,z_0\in{\Bbb N}$.

The proof of Theorem 1.2 is now complete.

4.   Proof of Theorem 1.3

Lemma 4.1. Let $q\in{\Bbb N}$ be a multiple of $9$ with $7\nmid q$ or

Then there are $x,y,z\in{\Bbb Z}$ such that

Proof. Since $9\mid q$ and

by [4,Theorem 2] we can write $q$ as $a^2+b^2+7c^2$ with $a,b,c\in{\Bbb Z}$. For

we have

This concludes the proof.

Lemma 4.2. Let $n\in{\Bbb N}$ and ${\delta}\in\{1,2\}$. Suppose that there are $B\in{\Bbb N}$ and $x,y,z\in{\Bbb Z}$ such that

Then $n = p_5(w_0)+p_5(x_0)+2p_5(y_0)+3{\delta} p_5(z_0)$ for some $w_0,x_0,y_0,z_0\in{\Bbb N}$.

Proof. Set $w = -(x+2y+3{\delta} z)$. As $|w|,|x|,|y|,|z|{\leq} B$, all the numbers

are nonnegative integers. Observe that

This ends the proof.

Proof of Theorem 1.3 with ${\delta} = 1$. We can verify the desired result for $n = 0,1,\ldots,808834880$ directly via a computer. Below we assume that

Since

there is an integer $B$ with

such that $B{\equiv} 18n^3+3n^2-35n\ ({\rm{mod}}\ {81})$, and $3n/7+1-(B+1)^2{\equiv}3,5,6\ ({\rm{mod}}\ 7)$ if $7\mid n$. Such an integer $B$ exists in view of the Chinese Remainder Theorem and the simple observations

Note that

and

It is easy to see that

is an integer divisible by $9$. If $n = 7n_0$ for some $n_0\in{\Bbb N}$, then

By Lemma 4.1, we can write $6q = (2n+7B)/3-7B^2$ as $x^2+2y^2+3z^2+(x+2y+3z)^2$ with $x,y,z\in{\Bbb Z}$. Applying Lemma 4.2 with ${\delta} = 1$, we see that $n = p_5(w_0)+p_5(x_0)+2p_5(y_0)+3p_5(z_0)$ for some $w_0,x_0,y_0,z_0\in{\Bbb N}$. This completes the proof.

Lemma 4.3. Let $q\in{\Bbb N}$ with $q\not{\equiv}7\ ({\rm{mod}}\ 8)$ or

Then there are $x,y,z\in{\Bbb Z}$ such that

Proof. By Dickson [2,pp. 112-113], we can write $q$ as $a^2+2b^2+10c^2$ with $a,b,c\in{\Bbb Z}$. For

we have

This ends the proof.

Proof of Theorem 1.3 with ${\delta} = 2$. We can verify the desired result for $n = 0,1,\ldots,897099188$ directly via a computer. Below we assume that

Since

there is an integer $B$ with

such that $B{\equiv} 3n^2-2n\ ({\rm{mod}}\ 9)$ and $B{\equiv} n^2-n-1\ ({\rm{mod}}\ 8)$, and $(B-1)^2\not{\equiv}2n_0\pm1,2n_0-2\ ({\rm{mod}}\ 5)$ if $n = 5n_0$ with $n_0\in{\Bbb N}$. Then

and $q\not{\equiv}7\ ({\rm{mod}}\ 8)$. If $n = 5n_0$ for some $n_0\in{\Bbb N}$, then

As in the proof of Theorem 1.2, we also have

Now applying Lemma 4.3 and Lemma 4.2 with ${\delta} = 2$, we obtain that $n = p_5(w_0)+p_5(x_0)+2p_5(y_0)+6p_5(z_0)$ for some $w_0,x_0,y_0,z_0\in{\Bbb N}$.

The proof of Theorem 1.3 with ${\delta} = 2$ is now complete.