4
$\begingroup$

Let $\mathbb N=\{0,1,2,\ldots\}$. Several years ago I proved that $$\{aw^3+bx^3+cy^3+dz^3:\ w,x,y,z\in\mathbb N\}\not=\mathbb N$$ for any positive integers $a,b,c,d$ (cf. http://maths.nju.edu.cn/~zwsun/179b.pdf ).

I'm curious whether there are positive integers $m$ and $n$ such that $$\left\{\left\lfloor\frac{a^3+b^3}m+\frac{c^3+d^3}n\right\rfloor:\ a,b,c,d\in\mathbb N\right\}=\mathbb N.$$ My computation suggests that $(m,n)=(2,6)$ might meet my purpose. Moreover, I have formulated the following conjecture.

Conjecture. Each $n\in\mathbb N$ can be written as the integral part of $(a^3+b^3)/2+(c^3+d^3)/6$ with $a,b,c,d\in\mathbb N$, $a\ge\max\{b,1\}$ and $c\ge\max\{d,1\}$.

I have verified this for all $n=0,\ldots,60000$. For example, $219$ has a unique required representation: $$219=\left\lfloor \frac{4^3+0^3}2 +\frac{10^3+5^3}6\right\rfloor.$$ For the number of ways to write $n\in\mathbb N$ in the given form, one may consult http://oeis.org/A343326.

QUESTION. Is the above conjecture true? How to prove it?

Your comments are welcome!

Improve this question
$\endgroup$
5
  • $\begingroup$ I also conjecture that each natural number can be written as $$\left\lfloor \frac{a^3+b^3}3\right\rfloor+\left\lfloor \frac{c^3+d^3}5\right\rfloor$$ with $a,b,c,d\in\mathbb N$. The pair $(3,5)$ may be replaced by some other pairs like $(4,5)$. $\endgroup$
    – Zhi-Wei Sun
    Commented Apr 12, 2021 at 3:07
  • 1
    $\begingroup$ There are four cubes. The transformation can be reduced to an equation with squares. And there will also be four squares. And as you know, four squares can make up any number. $\endgroup$
    – individ
    Commented Apr 12, 2021 at 5:11
  • 1
    $\begingroup$ @individ: I don't completely follow. I see how you can reduce to two squares, but then from those cannot further reduce to one linear (due to $2/6$, or $3/5$ or $4/5$, not being square). Could you kindly clarify your comment? $\endgroup$
    – Yaakov Baruch
    Commented Apr 12, 2021 at 7:25
  • $\begingroup$ I don't think individ's vague arguments are reasonable. Below $n$ one only has about $\root 3\of{n}$ cubes while one has about $\sqrt{n}$ squares. $\endgroup$
    – Zhi-Wei Sun
    Commented Apr 12, 2021 at 11:36
  • $\begingroup$ I have extended the verification of the conjecture in the posting to $10^5$. $\endgroup$
    – Zhi-Wei Sun
    Commented Apr 12, 2021 at 11:39

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .